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https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/19.E%3A_Transition_Metals_and_Coordination_Chemistry_(Exercises) |
Write the electron configurations for each of the following elements: The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and sub-shells. The electron configuration of each element is unique to its position on the periodic table where the energy level is determined by the period and the number of electrons is given by the atomic number of the element. There are four different types of orbitals (s, p, d, and f) which have different shapes and each orbital can hold a maximum of 2 electrons, but the p, d and f orbitals have different sub-levels, meaning that they are able to hold more electrons. The periodic table is broken up into groups which we can use to determine orbitals and thus, write electron configurations: Group 1 & 2: S orbital Group 13 - 18: P orbital Group 3 - 12: D orbital Lanthanide & Actinides: F orbital Each orbital (s, p, d, f) has a maximum number of electrons it can hold. An easy way to remember the electron maximum of each is to look at the periodic table and count the number of periods in each collection of groups. Group 1 & 2: 2 (2 electrons total = 1 orbital x max of 2 electrons = 2 electrons) Group 13 - 18: 6 (6 electrons total = 3 orbitals x 2 electrons max = 6 electrons) Group 3 - 12: 10 (10 electrons total = 5 orbitals x 2 electrons max = 10 electrons) Lanthanide & Actinides: 14 (14 electrons total = 7 orbitals x 2 electrons max = 14 electrons) Electron fills the orbitals in a specific pattern that affects the order in which the long-hand versions are written: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f An easier and faster way to write electron configurations is to use noble gas configurations as short-cuts. We are able to do this because the electron configurations of the noble gases always have all filled orbitals. He: 1s 2s Ne: 1s 2s 2p Ar: 1s 2s 2p 3s 3p Kr: 1s 2s 2p 3s 3p 4s 3d 4p Xe: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p Rn: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p The most common noble gas configuration used is Ar. When you want to use the noble gas configuration short-cut, you place the noble gas's symbol inside of brackets: [Ar] and then write it preceding the rest of the configuration, which is solely the orbitals the proceed after that of the noble gas. a. Sc Let's start off by identifying where Scandium sits on the periodic table: row 4, group 3. This identification is the critical basis we need to write its electron configuration. By looking at Scandium's atomic number, 21, it gives us both the number of protons and the number of electrons. At the end of writing its electron configuration, the electrons should add up to 21. At row 4, group 3 Sc, is a transition metal; meaning that its electron configuration will include the D orbital. Now, we can begin to assign the 21 electrons of Sc to orbitals. As you assign electrons to their orbitals, you move right across the periodic table. Its first 2 electrons are in the 1s orbital which is denoted as 1s where the "1" preceding the s denotes the fact that it is of row one, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 21-2=19 more electrons to assign. Its next 2 electrons are in the 2s orbital which is denoted as 2s where the "2" preceding the s indicates that it is of row two, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 19-2=17 more electrons to assign. Its next 6 electrons are in the 2p orbital which is denoted as 2p where the "2" preceding the p indicates that it is of row two, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 17-6=11 more electrons to assign. Its next 2 electrons are in the 3s orbital which is denoted as 3s where the "3" preceding the s indicates that it is of row three, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 11-2=9 more electrons to assign. Its next 6 electrons are in the 3p orbital which is denoted as 3p where the "3" preceding the p indicates that it is of row three, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 9-6=3 more electrons to assign. Its next 2 electrons are in the 4s orbital which is denoted as 4s where the "4" preceding the s indicates that it is of row four, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 3-2=1 more electron to assign. Its last electron would be alone in the 3 d orbital which is denoted as 3d where the "3" preceding the d indicates that, even though it is technically of row 4, by disregarding the first row of H and He, this is the third row and it has an exponent of 1 because there is only 1 electron to be placed in the d orbital. Now we have assigned all of the electrons to the appropriate orbitals and sub-orbitals, so that the final, entire electron configuration is written as: 1s 2s 2p 3s 3p 4s 3d This is the long-hand version of its electron configuration. So for Sc, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d b. Ti Start off by identifying where Titanium sits on the periodic table: row 4, group 4, meaning it has 22 electrons total. Titanium is one element to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 2 electrons remaining, so therefore the final orbital will be denoted as: 3d If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Titanium will be: 1s 2s 2p 3s 3p 4s 3d So for Ti, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d c. Cr Start off by identifying where Chromium sits on the periodic table: row 4, group 6, that means it has a total of 24 electrons. But first, Cr, along with Mo, Nb, Ru, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 24 electrons that its configuration would look like: 1s 2s 2p 3s 3p 4s 3d which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that there are 4 electrons in its 3 d orbital. The most stable configurations are half-filled (d ) and full orbitals (d ), so the elements with electrons resulting in ending with the d or d are so unstable that we write its stable form instead, where an electron from the preceding s orbital will be moved to fill the d orbital, resulting in a stable orbital. If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Chromium will be: 1s 2s 2p 3s 3p So for Cr, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d d. Fe Start off by identifying where Iron sits on the periodic table: row 4, group 8, meaning it has 26 electrons total. This is 5 elements to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 6 electrons remaining, so therefore the final orbital will be denoted as: 3d If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Iron will be: 1s 2s 2p 3s 3p 4s 3d So for Fe, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d e. Ru Start off by identifying where Ruthenium sits on the periodic table: row 5, group 8, that means it has a total of 44 electrons. But first, as stated earlier, Ru, along with Cr, Mo, Nb, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 44 electrons that its configuration would look like: 1s 2s 2p 3s 3p 4s 3d 4p which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that, even though there are 4 paired electrons, there are also 4 electrons unpaired. This results in a very unstable configuration, so to restore stability, we have to use a configuration that has the most paired electrons, which would be to take an electron from the s orbital and place it in the d orbital to create: 5s 4d If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Ru will be: 1s 2s 2p 3s 3p 4s 3d 4p So for Cr, its short-hand version of its electron configuration would therefore be: [Kr] 5s 4d Write the electron configurations for each of the following elements and its ions: Electrons are distributed into molecular orbitals, the \(s, p, d, and f\) blocks. An orbital will have a number in front of it and a letter that corresponds to the block. The s block holds two electrons, the p block holds six, the d block holds ten, and the f block holds fourteen. So, based on the number of electrons an atom has, the molecular orbitals are filled up in a certain way. The order of the orbitals is \(1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p\). An exponent will be put after the letter for each orbital to signify how many electrons are in that orbital. Noble gas notation can also be used by putting the noble gas prior to the element you are writing the configuration for, and then proceed by writing the orbitals filled after the noble gas. Metal ions of the d-block will have the two electrons removed from the s block prior to any electrons being removed from the proceeding d-block. Solutions: 1. \(Ti\) Titanium has an atomic number of 22, meaning it has 22 electrons. The noble gas prior to Titanium is Argon. Looking at row 4 of the periodic table, Titanium still has 4 electrons to be placed in orbitals since Argon has 18 electrons that are already placed. The remaining electrons will fill the \(4s\) orbital and the remaining two electrons will go into the \(3d\) orbital. [Ar]4s 3d 2. \(Ti^{+2}\) This is an ion with a plus 2 charge, meaning 2 electrons have been removed. The electrons will be removed from the \(4s\) orbital and the 2 remaining electrons will be placed in the \(3d\) orbital. Like number 1, the prior noble gas is Argon. [Ar]3d 3. \(Ti^{+3}\) This is an ion with a plus 3 charge, meaning 3 electrons have been removed. The first 2 electrons will be removed from the \(4s\) orbital, and the third will be taken from the \(3d\) orbital, and the 1 remaining electron will be placed in the 3d orbital. Like number 1, the prior noble gas is Argon. [Ar]3d 4. \(Ti^{+4}\) This is an ion with a plus 4 charge, meaning 4 electrons have been removed. The first 2 electrons will be removed from the \(4s\) orbital and the second 2 will be removed from the \(3d\) orbital. This results in the ion having the same electron configuration as Argon. [Ar] Answers: Write the electron configurations for each of the following elements and its 3+ ions: In order to write the electron configuration, we begin by finding the element on the periodic table. Since La, Sm, and Lu are all a period below the noble gas Xenon, we can abbreviate \({1s^2}{2s^2}{2p^6}{3s^2}{3p^6}{3d^{10}}{4s^2}{4p^6}{4d^{10}}{5s^2}{5p^6}\) as [Xe] when writing the orbital configurations. We then find the remaining of the orbital configurations using the Aufbau Principle. For other elements not just those in period 6, the shorthand notation using noble gases would be the noble gas in the period above the given element. 1. La has three additional electrons. Two of them fill the 6s shell and the other single electron is placed on the 5d shell. \(La:\) [Xe] \({6{s}^2} {5{d}^1}\) 2. Sm has eight more electrons. The 6s orbital is filled as previously and the 4f orbital receives 6 electrons because pairing electrons requires lower energy on the 4f shell than on the 5d shell. \(Sm:\) [Xe] \({6{s}^2} {4{f}^6}\) 3. Lu has seventeen more electrons. Two electrons fill the 6s orbital, 14 electrons fill the 4f orbital, and extra single electron goes to the 5d orbital . \(Lu:\) [Xe] \({6s^2}{4f^{14}}{5d^1}\) To find the 3+ ion electron configuration, we remove 3 electrons from the neutral configuration, starting with the 6s orbital. 1. The ionization of La removes the three extra electrons. So it reverts back to the stable Xenon configuration. \({La^{3+}:}\) [Xe] 2. The ionization of Sm removes two electrons from the 6s shell and one from the outermost (4f) shell \({Sm^{3+}}:\) [Xe] \({4f^5}\) 3. The ionization of Lu removes its two 6s shell and one from the outermost (5d) shell, leaving only a full 4f shell \(Lu^{3+}:\) [Xe] \(4f^{14}\) La: [Xe]6 5 , La : [Xe]; Sm: [Xe]6 4 , Sm : [Xe]4 ; Lu: [Xe]6 4 5 , Lu : [Xe]4 Why are the lanthanoid elements not found in nature in their elemental forms? Lanthanides are rarely found in their elemental forms because they readily give their electrons to other more electronegative elements, forming compounds instead of staying in a pure elemental form. They have very similar chemical properties with one another, are often found deep within the earth, and difficult to extract. They are the inner transition elements and have partially filled d orbitals that can donate electrons. Because of this, they are very reactive and electropositive. Which of the following elements is most likely to be used to prepare La by the reduction of La O : Al, C, or Fe? Why? An is a list of elements in decreasing order of their reactivity. Elements on the top of the list are good because they easily give up an electron, and elements on the bottom of the series are good because they are highly electronegative would really want to accept an electron. Compare Aluminum, Carbon, and Iron on an activity series. Many activity series include carbon and hydrogen as references. An activity series can be found here The activity series goes in the order (from top to bottom): Aluminum, Carbon, and Iron. Identify which element is the best reducing agent. Elements on the top of the list are the best reducing agents, because they give up electrons the best. Aluminum is the best reducing agent of the options available. Therefore aluminum will be the best reducing agent to prepare La by the reduction of La O because it is the most reactive in the series amongst the three elements. Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La(III) into La. Which of the following is the strongest oxidizing agent: \(\ce{VO4^{3-}}\), \(\ce{CrO4^2-}\), or \(\ce{MnO4-}\)? Oxidizing agents oxidize other substances. In other words, they gain electrons or become reduced. These agents should be in their highest oxidation state. In order to determine, the strength of the compounds above as oxidizing agents, determine the oxidation numbers of each constituent elements. \(\\\mathrm{VO_4^{3-}}\) We know that \(\mathrm{O}\) has a -2 oxidation state and the overall charge of the ion is -3. We just need to determine Vanadate's oxidation number in this compound. \(\\\mathrm{V} + \mathrm{-2(4)} = \mathrm{-3}\) \(\\\mathrm{V} = \mathrm{+5}\) Vanadate has an oxidation number of +5, which is its highest possible oxidation state. \(\\\mathrm{CrO_4^{2-}}\) Like in the previous calculation, \(\mathrm{O}\) has a -2 oxidation state. The overall charge is -2. So calculate for chromium. \(\\\mathrm{Cr} + \mathrm{-2(4)} = \mathrm{-2}\) \(\\\mathrm{Cr} = \mathrm{+6}\) Chromium is in its highest possible oxidation state of +6 in this compound. \(\\\mathrm{MnO_4^-}\) \(\mathrm{O}\) has a -2 oxidation state and the overall charge is -1. \(\\\mathrm{Mn} + \mathrm{-2(4)} = \mathrm{-1}\) \(\\\mathrm{Mn} = \mathrm{+7}\) Manganese is also in its highest oxidation state, +7. An oxidizing agent has to be able to gain electrons which, in turn, reduces its oxidation state. Here manganese has the greatest oxidation state which allows it to experience a greater decrease in its oxidation state if needed, meaning it can gain the most electrons. So among the three compounds, \(\mathrm{MnO_4^-}\) is the strongest oxidizing agent. This method assumes the metals have similar electronegativities. Alternatively, check a redox table. \(MnO_4^-\) Which of the following elements is most likely to form an oxide with the formula MO : Zr, Nb, or Mo? Mo because Zr has an oxidation state of +4 and Nb has a oxidation state of +5 and those would not balance out the charge of 3 oxygens in the state of -2 which creates a charge of -6. Mo however has multiple oxidation states, the most common being +6 which balances out the -6 charge created by 3 oxygen ions. This is why its most likely to form an oxide with the formula MO or \(\ce{MoO3}\). Mo The following reactions all occur in a blast furnace. Which of these are redox reactions? o identify redox reaction, we have to determine if have to see if the equation is an oxidation-reduction reaction-meaning that the species are changing oxidation states during the reaction, which involves the transfer of electrons between two species. If a species is losing electrons, then that species is being oxidized. If a species is gaining electrons, then that species is being reduced. A way to remember this is using the acronyms OIL RIG. xidation s oss, and eduction s ain, referring to electrons. Both of these must occur for an equation to be a redox reaction. Let's see if these equations are redox reactions or not: a. In the reactants side \(\ce{Fe2O3}\), Fe is has an oxidation number of +3. In the product \(\ce{Fe3O4}\), Fe has an oxidation number of +2.67. Since Fe changed from +3 to +2.67, we can say that Fe had gained electrons and therefore reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . b. In the reactant, \(\ce{Fe3O4}\), Fe has an oxidation number of +2.67. In the product, FeO, Fe has an oxidation number of +2. Since the oxidation of Fe has changed from +2.67 to +2, electrons have been added therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . c. In the reactant side, in FeO, Fe has an oxidation number of +2 and in the products side Fe has an oxidation number of 0. Since the oxidation number of Fe changed from +2 to 0, electrons have been gained and therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . d. In the reactants C has an oxidation number of 0, and in the products side in \(\ce{CO2}\), C has an oxidation number of +4. Since the oxidation number of C has changed from 0 to +4, we can say that C has been oxidized. In the reactants, in \(\ce{O2}\) oxygen has an oxidation number of 0, and in the products CO2, oxygen has an oxidation number of -2. Since the oxidation number of oxygen has changed from 0 to -2, oxygen has been reduced. Since there is oxidation and reduction of species- we can conclude that this . e. In the reactants \(\ce{CO2 }\) has an oxidation number of +4, and in the products side in CO, C has an oxidation number of +2. Since carbon went from +4 to +2, carbon has been reduced. In the reactants, in \(\ce{CO2 }\) oxygen has an oxidation number of -4 and in the products CO carbon has an oxidation number of -2. Since oxygen went from -4 to -2, it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . f. In the reactants, \(\ce{CaCO3}\) Ca has an oxidation number of +2, and in products CaO Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation g. In the products CaO Ca has an oxidation number of +2, and in the products \(\ce{CaSiO3}\) Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation a, b, c, d, e Why is the formation of slag useful during the smelting of iron? Slag is a substance formed as a byproduct of iron ore or iron pellets melting together in a blast furnace. Slag is also the byproduct that is formed when a desired metal has been separated from its raw ore. It is important to note that slag from steel mills is created in a manner that reduces the loss of the desired iron ore. The \(\ce{CaSiO3}\)slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to \(\ce{O2}\), which would oxidize the \(\ce{Fe}\) back to \(\ce{Fe2O3}\). Since Fe has a low reduction potential of -0.440 this means it has a high oxidation potential so it would easily oxidize in the presence of O2. Creating a barrier between iron and oxygen allows the maximum product of iron to be obtained in the end of smelting. The CaSiO slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to O , which would oxidize the Fe back to Fe O . Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer. Manganese(VII) oxide, can be written as Mn O In relation to the Lewis acid-base theory, a Lewis acid accepts lone pair electrons, and is also known as the electron pair acceptor. Based on this theory, acidity can be measured by the element's ability to accept electron pairs. By doing the math, we find that Manganese has an oxidation state of +7 (Oxygen has an oxidation state of -2, and 2x-7=-14 , in order to balance the ion's charge, Mn must be +7). Therefore Mn has high capability of accepting electrons due to its high positive charge. For most metals, as the oxidation number increases, so does its acidity, because of its increased ability to accept electrons. In relation to the Lewis acid-base theory, the Lewis acid accepts lone pair electrons; thus, it is also known as the electron pair acceptor. This may be any chemical species. Acids are substances that must be lower than 7. Therefore, oxides of manganese is most likely going to become more acidic in (aq) solutions if the oxidation number increases. Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 Na Cr O is required in the titration. What percentage of the ore sample was iron? To answer this question, we must first identify the net ionic equation from the given half-reactions. We can write the oxidation and reduction half-reactions: \[ \text{ oxidation:} \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+} \] \[ \text{ reduction: }\ce{Cr2O7^2-} \rightarrow \text{ Cr}^\text{ 3+} \] We can quickly balance the oxidation half-reaction by adding the appropriate number of electrons to get \[ \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}+\text{ e}^- \] The first step in balancing the reduction half-reaction is to balance elements in the equation other than O and H. In doing so, we get \[\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+} \] The second step would be to add enough water molecules to balance the oxygen. \[ \ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \] Next, we add the correct amount of H to balance the hydrogen atoms. \[ \ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \] Finally, we add enough electrons to balance charge. \[ \ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \] The electrons involved in both half-reactions must be equal in order for us to combine the two to get the net ionic equation. This can be done by multiplying each equation by the appropriate coefficient. Scaling the oxidation half-reaction by 6, we get \[ 6 \text{ Fe}^\text{ 2+} \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^- \] Now we can combine both half-reactions to get \[ 6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}\] The electrons cancel out, so you get: \[ 6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 6 \text{ Fe}^\text{ 3+}+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}\] From this we can see that the mole ratio of Cr O to Fe is 1:6. Given that 19.17 mL (or 0.01917 L) of 0.01 M Na Cr O was needed for titration we know that \[ 0.01917\text{ L} \times 0.01 \text{ M} = 1.917 \times 10^{-4} \text{ mol} \] of Na Cr O reacted. Also, since any number of moles of Na Cr O produces the same number of moles of Cr O in solution \[ 1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Na2Cr2O7} =1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\] We can use the mole ratio of Cr O to Fe to determine how many moles of iron (ii) was in the solution. The number of moles of iron (ii) is the same as the number of moles of pure iron in the sample since all of the iron was converted into iron (ii). \[ 1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\times \frac{6\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}}{1\text{ mol}\text{ of }\ce{Cr2O7^2-}} = 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} \] \[ 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} = 0.0011502\text{ mol}\text{ of } \text{ Fe} \] Now we can find the number of grams of iron that were present in the 2.5 g iron ore sample. \[ 0.0011502\text{ mol}\text{ of } \text{ Fe}\times\frac{55.847\text { g}}{1\text{ mol}} = 0.0642352194\text{ g}\text{ of }\text{ Fe} \] Finally, we can answer the question and find what percentage of the ore sample was iron. \[ \frac{0.0642352194\text{ g}}{2.5\text{ g}} \times 100 \approx 2.57\text{%} \] 2.57% How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe O to convert that Fe O into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume. This question uses a series of unit conversions and the \(PV=nRT\) equation. The first step is to write out the balanced chemical equation for the conversion of Fe O to pure iron. \[2\;Fe_2O_3(s)\rightarrow 4\;Fe(s)+3\;O_2(g)\] Next, we need to analyze the original question to determine the value that we need to solve for. Because the question asks for a value of cubic feet, we know we need to solve for volume. We can manipulate \(PV=nRT\) to solve for volume. \[V={nRT}/P\] Now determine the known variables and convert into units that will be easy to deal with. \[n = 2000\:lbs\; Fe_2O_3\frac{453.592\: grams\: Fe_2O_3}{1\: lb \:Fe_2O_3}\frac{1\: mole \:Fe_2O_3}{159.69\: grams \:Fe_2O_3}\frac{3 \;moles\: O_2}{2\; moles \:Fe_2O_3}\] \[n=8521\: moles\: of \:O_2\] Convert to atm for easier calculations \[R=\frac{.0821\:L\:atm}{mol\:K}\] \[T=0^{\circ}C=273\:K\] \[P=760 \:torr= 1 \:atm\] Now plug the numbers into the manipulated gas law to get to an answer for V. \[V=190991.8\: liters \:of\: O_2\] From here we convert liters to cubic feet. use the conversion \[1\;L=.0353 ft^3\] thus we have 6744.811 ft of O We then refer back to the initial question and remember that this value is only 19% of the volume of the total air. So use a simple equation to determine the total volume of air in cubic feet. \[6744.811ft^3=.19x\] x=35499 ft of air 35499 ft of air Find the potentials of the following electrochemical cell: Cd | Cd ( = 0.10) ‖ Ni ( = 0.50) | Ni Write out your two half reactions and identify which is oxidation and which is reduction using the acronym OIL RIG to remember that oxidation is loss of electrons and reduction is gain of electrons Cd(s)⟶Cd (aq)+2e (oxidation) Ni (aq)+2e ⟶Ni(s) (reduction) Write out complete balanced equation Cd(s)+ Ni (aq)⟶Cd (aq)+Ni(s) Find E E = E -E oxidation: Cd(s)⟶Cd (aq)+2e E =-0.40V reduction: Ni (aq)+2e ⟶Ni(s) E =-0.26V * E values come from standard reduction potentials table given above. Also, remember anode is where oxidation happens, and cathode is where reduction happens. E cell=-0.26-(-.40) E =0.14V Find Q Q=[products]/[reactants] (look at complete balanced equation) (remember that [x] means the concentration of x typically given in molarity and that we ignore solids or liquids) Q=[Cd ]/[Ni ] Q=0.10/0.50 Q=0.2 Calculate E using E= E -(.0592/n)logQ (n is number of moles of electrons transferred and in our case the balanced reaction transfers 2 electrons) E= 0.14-(.0592/2)log(0.2) E= 0.14-(-.207) 0.16 V A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt? Assuming that metal chloride is XCl The balance equation for the reaction would be: \[XCl(aq)+AgNO_{3}(aq)\rightarrow XNO_{3}(aq) +AgCl(s)\] The mass of AgCl = 3.03707g To find the moles of AgCl present: Next, we can determine the moles of AgCl present in the reaction since 1) the mass of the precipitate is given to us and 2) this value can help us determine the moles of alkali metal chloride compound present. Given the mass of AgCl is 3.03707g in the problem and the molecular mass of AgCl per mole is 143.32g, we can solve for how many moles of AgCl is in the reaction: \[moles of Agcl=\tfrac{3.03707g}{143.32g/mol}=0.0211 mol\] Since the molar ratio of the compounds are 1:1 so the number of moles of XCl used = 0.0211 mol We can calculate the weight of Cl with the equation: \[0.0211 mol \times 35.5g/mol = 0.7490g\] the amount of metal present in the original compound is the weight of the compound subtracted by weight of the Cl ion: \[(2.5624- 0.7490)g= 1.8134g\] And the percentage can be calculate by \[\frac{0.7490}{2.5624}\times 100= 29.23 \%\] the molar ratio of XCl is 1:1 so then Atomic mass of metal = \(=\frac{1.8134\;g\; metal}{0.0211\; mol\; RbCl}=85.943g/mol\) So the atomic mass is 85.943 g/mol which is of Rb hence the identity of the salt is RbCl The standard reduction potential for the reaction \(\ce{[Co(H2O)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(H2O)6]^2+}(aq)\) is about 1.8 V. The reduction potential for the reaction \(\ce{[Co(NH3)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(NH3)6]^2+}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H O) ] and/or [Co(NH ) ] , can be oxidized to the corresponding cobalt(III) complex by oxygen. To calculate the cell potential, we need to know the potentials for each half reaction. After doing so, we need to determine which one is being oxidized and which one is being reduced. The one that is oxidized is the anode and the one that is reduced is the cathode. To find the cell potential, you use this formula and the reduction potential values found in a reduction potential table. If E° is positive, \(\Delta\)G is negative and the reaction is spontaneous. E° = E°cathode - E°anode Because it states that \([Co(H_{2}O)_{6}]^{3+}\) will be oxidized, this means it is the anode. \(O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} \rightarrow 2 H_{2}O\) +1.229 V \(O_{2}\) is being reduced, so it is the cathode. 1.229V - 1.8V= -.571 V, or -0.6 V using significant figures. This cannot happen spontaneously because E° is negative. For \([Co(NH_{3})_{6}]^{3+}\), it is again being oxidized, meaning it’s the anode. 1.229-.1= 1.129 V or 1.1 V using significant figures. This reaction is spontaneous because E° is positive. ° = −0.6 V, ° is negative so this reduction is not spontaneous. ° = +1.1 V, ° is positive so this reduction is spontaneous. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) There is a myriad of reactions that can occur, which include: single replacement, double replacement, combustion, acid-base/neutralization, decomposition or synthesis. The first step to determine the products of a reaction is to identify the type of reaction. From then on, the next steps you take to predict the products will vary based on the reaction type. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) Predict the products of each of the following reactions. Whenever a metal reacts with an acid, the products are salt and hydrogen. Because Fe is lower on the activity series, we know that when it reacts with an acid it will result in the formation of Hydrogen gas. To simplify the equation is: \[Metal + Acid ⟶ Salt + Hydrogen\] The salt produced will depend on the metal and in this case, the metal is iron (Fe) so the resulting equation would be: \[\ce{Fe}(s)+\ce{H2SO4}(aq)⟶ \ce{FeSO4}(aq) + \ce{H2}(g)\] This equation works out as the H2 is removed from H SO , resulting in a SO ion where Fe will take on an oxidation state of Fe to form FeSO which will be the salt in this example. But since FeSO and H2SO are aqueous, the reactants and products can also be written as its ions where the overall equation can be: \[\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO2^{−4}}(aq)⟶\ce{Fe2+}(aq)+\ce{SO2^{−4}}(aq)+\ce{H2}(g)+\ce{2H2O}(l)\] In this case, adding a metal hydroxide (NaOH) to a solution with a transition metal ion (Fe) will form a transition metal hydroxide (XOH). As iron is bonded to three chlorine atoms in the reactants side, it has the oxidation state of +3 where three hydroxide ions (OH ) are needed to balance out the charges when they are bonded in the products. The remaining ions are Na and Cl where they bond together in a 1:1 ratio where there are 3 molecules of NaCl once the reaction is balanced. The overall reaction will be: \[\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶ \ce{Fe(OH)3}(s) + \ce{3NaCl}(aq)\] NOTE: Fe(OH) (s) is a solid as it is a rule that all all transition metal hydroxides are insoluble and a precipitate is formed. Since NaOH(aq) and NaCl(aq) are aqueous, we can write them out in their ion forms: \[\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH^{−}}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)\] This is an example of a metal hydroxide reacting with an acid where a metal salt and water will always be formed: \[Metal Hydroxide + Acid ⟶ Metal Salt + Water\] When this rule is applied to this equation, we will get the following: \[\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶ \ce{MnBr2}(aq)+\ce{2H2O}(l)\] But to follow through with this question, the aqueous solutions such as HBr(aq) and MnBr (aq) can be re-written as: \[\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)\] This is the general reaction of a metal reacting with oxygen which will always result in a metal oxide. However, the metal oxide is determined by the oxidation state of the metal so there may be several outcomes of this reaction such as: \[\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)\] \[\ce{Cr}(s)+\ce{O2}(g)⟶\ce{2CrO}(s)\] \[\ce{Cr}(s)+\ce{O2}(g)⟶\ce{CrO2}(s)\] \[\ce{2Cr}(s)+\ce{3O2}(g)⟶\ce{CrO3}(s)\] However, Cr O is the main oxide of chromium so it can be assumed that this is the general product of this reaction. This follows the general reaction of a metal oxide and an acid will always result in a salt and water \[Metal Oxide + Acid ⟶ Salt + Water\] Using this general reaction, similar to the general reactions above, the reaction will result in: \[\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)\] However, since HCl is an aqueous solution, the overall equation can also be re-written as: \[\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)\] Titanium is able to react with the halogens where there are two oxidation state that titanium can be: +3 and +4. The following reactions follow each oxidation state accordingly: \[\ce{2Ti}(s)+\ce{3F2}(g)⟶ \ce{2TiF3}(s)\] \[\ce{Ti}(s)+\ce{2F2}(g)⟶\ce{TiF4}(s)\] However, since there is the symbol "xs", this indicates that F is added in excess so the second reaction is favored more as it drives the reaction to completion. OVERALL: \[\ce{Ti}(s)+\ce{ Describe the electrolytic process for refining copper. By electrolysis, copper can be refined and purely made. The reason why copper needs to remove the impurities is because it helps increase the electrical conductivity in electrical wire. You can refine copper and remove the impurities through electrolysis. Pure copper is important in making electrical wire, because it creates better electrical conductivity when transferring electricity. In order for better electrical conductivity, the impurities needs to be removed and this can be done by firing the impure copper to remove the impurities, such as sulfur, oxygen, etc. and shaping them into electrical anodes that can be used in electrolysis. Then the copper electrodes are placed into an electrical cell (into separate beakers) where electrical current can pass through the beakers and onto the electrodes. Through this process, the copper is stripped off of the anode and deposited onto the cathode. This process helps remove the impurities and refine copper because all the copper has been deposited onto the cathode all in one electrode. This process increases the weight of the cathode due to copper being deposited onto the cathode. This is a prime example of how to tell if an electrode is a cathode or an anode, as stated in Q17.2.9 above. Predict the products of the following reactions and balance the equations. What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid? Formula for iron(II) sulfide: \[FeS\] Definition of non-oxidizing acid: A non-oxidizing acid is an acid that doesn't act the oxidizing agent. Its anion is a weaker oxidizing agent than H+, thus it can't be reduced. Examples of non-oxidizing acids: \[HCl, HI, HBr, H_3PO_4, H_2SO_4\] Step 2: Choose one of the non-oxidizing acid, in this case HCl, and write the chemical reaction: \[FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)\] The gas produced when iron (II) sulfide treated with a non-oxidizing acid, HCl, is H S (dihydrogen sulfide) gas. Predict the products of each of the following reactions and then balance the chemical equations. Steam is water (\(\ce{H_{2}O}\)) We can write out the reaction as: \(\ce{Fe}\) + \(\ce{H_{2}O}\) → ? This is a single replacement reaction, so \(\ce{Fe}\) replaces \(\ce{H_{2}}\). So, one of the products is \(\ce{Fe_{3}O_{4}}\) since it is a combination of iron(II) oxide, \(\ce{FeO}\), and iron(III) oxide, \(\ce{Fe_{2}O_{3}}\). The \(\ce{Fe}\) is heated in an atmosphere of steam. \(\ce{H_{2}}\) becomes neutrally charged and becomes another product. After balancing the coefficients, the final reaction is: \(\ce{3Fe}(s)\) + \(\ce{4H_{2}O}(g)\) → \(\ce{Fe_{3}O_{4}}(s)\) + \(\ce{4H_{2}}(g)\) \(\ce{NaOH}\) added to a solution of \(\ce{Fe(NO_{3})_{3}}\) is a double replacement and precipitation reaction. We can write out the reaction as: \(\ce{NaOH}\) + \(\ce{Fe(NO_{3})_{3}}\) → ? The \(\ce{Na}\) and \(\ce{Fe}\) switch to form \(\ce{Fe(OH)_{3}}(s)\) and \(\ce{NaNO_{3}}(aq)\). \(\ce{Fe(OH)_{3}}\) is solid because it is insoluble according to solubility rules. After balancing the coefficients in the reaction, the final reaction is: \(\ce{Fe(NO_{3})_{3}}(aq)\) + \(\ce{3NaOH}(aq)\) → \(\ce{Fe(OH)_{3}}(s)\) + \(\ce{NaNO_{3}}(aq)\) For instance, the acid used to make the acidic solution is \(\ce{H_{2}SO_{4}}\), then the reaction is: \(\ce{FeSO_{4}}\) + \(\ce{KMnO_{4}}\) + \(\ce{H_{2}SO_{4}}\) → \(\ce{Fe_{2}(SO_{4})_{3}}\) + \(\ce{MnSO_{4}}\) + \(\ce{H_{2}O}\) + \(\ce{K_{2}SO_{4}}\) Next, the net ionic reaction has to be written to get rid of the spectator ions in the reaction, this is written as: \(\ce{Fe^{2+}}\) + \(\ce{MnO_{4}^{-}}\) + \(\ce{H^{+}}\) → \(\ce{Fe^{3+}}\) + \(\ce{Mn^{2+}}\) + \(\ce{H_{2}O}\) As seen in the net ionic equation above, \(\ce{Fe^{2+}}\) is oxidized to \(\ce{Fe^{3+}}\) and \(\ce{MnO_{4}^{-}}\) is reduced to \(\ce{Mn^{2+}}\). These can be written as two half reactions: \(\ce{Fe^{2+}}\) → \(\ce{Fe^{3+}}\) \(\ce{MnO_{4}^{-}}\) → \(\ce{Mn^{2+}}\) To balance the oxidation half reaction, one electron as to be added to the \(\ce{Fe^{3+}}\), this is shown as: \(\ce{Fe^{2+}}\) → \(\ce{Fe^{3+}}\) + \(\ce{e^{-}}\) The reduction half reaction also has to be balanced, but with \(\ce{H^{+}}\) ions and \(\ce{H_{2}O}\), this is shown as: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) After the charge of the \(\ce{Mn}\) atoms are balanced, the overall charge has to be balanced on both sides because on the reactants side, the charge is \(\ce{7+}\), and the charge on the products side is \(\ce{2+}\). The overall charge can be balanced by adding electrons, this is shown as: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5e^{-}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) Now since both half reactions are balanced, the electrons in both half reactions have to be equal, and then the half reactions are added together. After this is done, the reaction looks like this: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5Fe^{2+}}\) + \(\ce{5e^{-}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) + \(\ce{5Fe^{3+}}\) + \(\ce{5e^{-}}\) The \(\ce{5e^{-}}\) on both sides cancel out and the final balanced reaction is: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5Fe^{2+}}\) →\(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) + \(\ce{5Fe^{3+}}\) \(\ce{Fe}\) added to a dilute solution of \(\ce{H_{2}SO_{4}}\) is a single replacement reaction. The \(\ce{Fe}\) is added to a dilute solution so the \(\ce{H_{2}SO_{4}}\) is written as separate ions. We can write out the reaction as: \(\ce{Fe}(s)\) + \(\ce{2H^+}(aq)\) + \(\ce{(SO_{4})^{2-}}(aq)\) → ? The Fe replaces the \(\ce{H^+}\) ion, and becomes an \(\ce{Fe^{2+}}\) ion. \(\ce{H_{2}O}\) is also a product because the solution is dilute. Furthermore, the \(\ce{FeSO_{4}}\) also has to be separated into ions as a result of the \(\ce{Fe}\) being added to a dilute solution. After balancing all of the coefficients, the final reaction is: \(\ce{Fe}(s)\) + \(\ce{(2H_{3}O)^+}(aq)\) + \(\ce{(SO_{4})^{2-}}(aq)\) → \(\ce{Fe^{2+}}(aq)\) + \(\ce{SO_{4}^{2-}}(aq)\) + \(\ce{H_{2}}(g)\) + \(\ce{2H_{2}O}(l)\) \(\ce{H^+}\) can also be written as the the hydronium ion, \(\ce{(H_{3}O)^{+}}\). We initially can initially write out: \(\ce{4Fe(NO_{3})_{2}}\) + \(\ce{4HNO_{3}}\) + \(\ce{O_{2}}\) → ? We write the oxygen term in the reactants because it is stated that the solution is allowed to stand in air. We just have to analyze the possible products that can be formed and we can see that the hydrogen from nitric acid can combine with oxygen gas to form water and then combining everything together, we get the final reaction to be: \(\ce{4Fe(NO_{3})_{2}}(aq)\) + \(\ce{4HNO_{3}}(aq)\) + \(\ce{O_{2}}(g)\) → \(\ce{2H_{2}O}(l)\) + \(\ce{4Fe(NO_{3})_{3}}(aq)\) When \(\ce{FeCO_{3}}\) is added to \(\ce{HClO_{4}}\), a double replacement reaction occurs. The \(\ce{Fe^{2+}}\) ion switches spots with the \(\ce{H^+}\) ion to form \(\ce{Fe(ClO_{4})_{2}}\) as a product. When the \(\ce{H^+}\) ion is added to the \(\ce{(CO_{3})^{2-}}\) ion, \(\ce{H_{2}CO_{3}}\) is formed. After balancing the coefficients, the final reaction is: \(\ce{FeCO_{3}}(s)\) + \(\ce{HClO_{4}}(aq)\) → \(\ce{Fe(ClO_{4})_{2}}(aq)\) + \(\ce{H_{2}O}(l)\) + \(\ce{CO_{2}}(g)\) Air is composed of oxygen gas, which is a diatomic molecule, so it is \(\ce{O_{2}}\). Adding \(\ce{Fe}\) to \(\ce{O_{2}}\) will cause a synthesis reaction to occur forming \(\ce{Fe_{2}O_{3}}\). After balancing coefficients, the final reaction is: \(\ce{3Fe}(s)\) + \(\ce{2O_{2}}(g)\) → \(\ce{Fe_{2}O_{3}}(s)\) Balance the following equations by oxidation-reduction methods; note that elements change oxidation state. \[\ce{Co(NO3)2}(s)⟶\ce{Co2O3}(s)+\ce{NO2}(g)+\ce{O2}(g)\] Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state. \[Co(NO_3){_2(s)}⟶Co_2O{_3(s)}+NO{_2(g)}+O{_2(g)}\] In this reaction, N changes oxidation states from +5 to +4 (reduced), Co changes oxidation states from +2 to +3 (oxidized), and O changes oxidation states from -2 to 0 (also oxidized). First, split this reaction into an oxidation and reduction half reaction set, and balance all of the elements that are not hydrogen or oxygen (we will deal with these later): \[Reduction: 2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2\] Now, for the oxidation reaction, we are only dealing with O on the products side. In order to balance this, we will need to add water and hydrogen to both sides: \[Oxidation: 2H_2O\rightarrow O_2+4H^+ \] Balance the amount of oxygens on each side by adding the correct number of water molecules (H O), and balance the amount of hydrogen by adding the correct number of H atoms: \[Reduction: 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O\] \[Oxidation: 2H_2O\rightarrow O_2+4H^+\] Finally, balance the charges by adding electrons to each side of the equation. For the reduction reaction, we will add 2 electrons to balance out the 2H , and to the oxidation reaction, we will add 4 electrons to balance out the 4H . Remember, the goal of this step is to make sure that the charges are balanced, so we can cancel them out in the end. \[Reduction: 2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O\] \[Oxidation: 2H_2O\rightarrow O_2+4H^+ +4e^-\] Multiply the reduction reaction by two, in order to balance the charges so there are 4 electrons on each side of the reaction. \[Reduction: 2(2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O)\] and combine both reactions which comes out to: \[2H_2O + 4Co(NO_3)_2 + 4H^+ \rightarrow 2CO_2O_3 + 8NO_2 + 2H_2O + O_2 + 4H^+\] Cancel out like terms: \[4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}\] Both sides have overall charges of 0 and can be checked to see if they are balanced. \[4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}\] Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility. Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility. A: Step 1: look at the question and begin to write out a general product to reactant formula for this reaction. Step 2: try to reason out why a precipitate will form but only for a finite period of time before reforming in an aqueous substance. Step 3: With step 2 you should have noticed that the reaction is a multiple step reaction and using the rough formula that you derived in step 1, you should try and see what the series of steps are that lead to the overall product of liquid AgCN In this reaction we see how NaCN is added to AgNO .A precipitate forms but then disappears with the addition of even more NaCN, this must mean that its an intermediate reaction which will not appear as the final product. The silver and the cyanide temporarily bond, but the bond is too weak to hold them together so they are pulled apart again when NaCN is added because a new, more stronger and stable compound is formed: [Ag(CN) ] (aq). The actual reaction equation when it is first taking place is \[AgCl(aq)+NaCN(aq)\rightarrow AgCN(s)+NaCl(aq)\] This can be written out in the following way: as CN is added, the silver and the cyanide combine : Ag (aq)+CN (aq)→AgCN (s) As more CN is added the silver and two cyanide combine to create a more stable compound: Ag (aq)+2CN (aq)→[Ag(CN) ] (aq) AgCN(s) + CN (aq) → [Ag(CN) ] (aq) As CN is added, \[\ce{Ag+}(aq)+\ce{CN-}(aq)⟶\ce{AgCN}(s)\] As more CN is added, \[\ce{Ag+}(aq)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)\] \[\ce{AgCN}(s)+\ce{CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)\] Predict which will be more stable, [CrO ] or [WO ] , and explain. According to the rules associated with Crystal Field Stabilizing Energies, stable molecules contain more electrons in the lower-energy molecular orbitals than in the high-energy molecular orbitals. In this case, both complexes have O as ligands, and both have a -2 charge. Therefore, you determine stability by comparing the metals. Chromium is in the 3d orbital, according to the periodic table. Tungsten (W) is in the 5d orbital. 3d is a lower energy level than 5d.Higher-level orbitals are more easily ionized, and make their base elemental form more stable. If the elemental form is more stable the oxidized form is less stable. Therefore, [CrO ] is more stable than [WO ] . [CrO ] is more stable because Chromium is in the 3d orbital while Tungsten is in the 4d orbital, which has a higher energy level and makes it less stable. Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M O are examples of in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M O , to permit estimation of the metal’s two oxidation states.) The first step to solving this problem is looking at the rules of Oxidizing states for various elements: chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidation_State The main rules that will be used in these problems will be the oxidation state rule 6 which states that oxidation state for Oxygen is (-2) and rule 2 which is that the total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species. Solving these problems requires simple algebra. The oxidation states of both elements in the compound is equal to zero, so set the unknown oxidation of the element that is not oxygen to a variable \({x}\), and the oxidation state of Oxygen equal to \({-2}\). Then multiply both oxygen states by the number of atoms of the element present. Add the values together, set the equation equal to zero and solve for \({x}\). \(\ce{FeO}={-2}+{x}={0}⟶{x}={Fe}={+2}\) \(Fe^{2+}\) \(\ce{Fe2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Fe}={+3}\) \(Fe^{3=}\) (One Fe Atom has an oxidation state of +2 and the other 2 Fe atoms have an oxidation state of +3) 7. \(\ce{Co3O4}=\ce{CoO}·\ce{Co2O3}=\) \(\ce{CoO}={-2}+{x}={0}⟶{x}={Co}={+2}\) \(Co^{2+}\) \(\ce{Co2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Co}={+3}\) \(Co^{3+}\) (One Co Atom has an oxidation state of +2 and the other 2 Co atoms have an oxidation state of +3) 8. \(\ce{NiO}={-2}+{x}={0}⟶{x}={Ni}={+2}\) \(Ni^{2+}\) 9. \(\ce{Cu2O}={-2}+{2{x}}={0}⟶{-2}+{2x}={0}⟶{x}={Cu}={+1}\) \(Cu^{1+}\) Sc ; Ti ; V ; Cr ; Mn ; Fe and Fe ; Co and Co ; Ni ; Cu Indicate the coordination number for the central metal atom in each of the following coordination compounds: First we must identify whether or not the ligand has more than one bonded atom (bidentate/polydentate). Using the table below we are able to do this. Now that we have identified the number of bonded atoms from each ligand, we can find the total number of atoms bonded to the central metal ion, giving us the coordination number. Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate: To determine coordination numbers we must count the total number of ligands bonded to the central metal and distinguish monodentate and polydentate ligands. To determine the formulas, we use the nomenclature rules and work backwards. Give the coordination number for each metal ion in the following compounds: You can determine a compound's coordination number based on how many ligands are bound to the central atom. 1) In this compound, Cobalt is the central atom, and it has 3 CO molecules attached to it. However, CO is a bidentate ligand, which means it binds to the central atom in two places rather than one. This means that the coordination number of [Co(CO ) ] is 6. A coordination number of 6 means that the structure is most likely octahedral. 2) In this compound, Copper is the central atom. 4 ammonia molecules are attached to it. This means the coordination number is 4, and the structure is likely tetrahedral. 3) For this compound, we can ignore the (SO ) because it is not bound to the central atom. The central atom is cobalt, and it has 4 ammonia molecules and 2 bromine molecules bound to it. The coordination number is 6. 4) There are two compounds here, indicated by the brackets. The central atom for both is platinum. One of them has 4 ammonia molecules attached, and the other has 4 chlorine atoms attached. Both complexes have a coordination number of 4. 5) We can ignore (NO ) for this compound. The central atom is Chromium. There are 3 ethylenediamine molecules attached to the chromium. Ethylenediamine is a bidentate ligand, so the coordination number is 6. 6) Palladium is the central atom. 2 ammonia molecules and 2 bromine atoms are bound to the palladium atom. The coordination number is 4. 7) We can ignore the K structure. Copper is the central atom, and there are 5 chlorine molecules attached to it. The coordination number is 5, so the structure is either trigonal bipyramidal or square pyramidal. 8) In this compound, zinc is the central atom. There are 2 ammonia molecules and 2 chlorine atoms attached. This means that the coordination number is 4. Sketch the structures of the following complexes. Indicate any , , and optical isomers. Cis and trans are a type of geometric isomer, meaning there is a difference in the orientation in which the ligands are attached to the central metal. In cis, two of the same ligands are adjacent to one another and in trans, two of the same ligands are directly across from one another. Optical isomers → have the ability to rotate light, optical isomers are also chiral. Only chiral complexes have optical isomers Chiral → asymmetric, structure of its mirror image is not superimposable Enantiomers: chiral optical isomers (compound can have multiple enantiomers) Tetrahedral complex with 4 distinct ligands → always chiral Solutions: a. \([Pt(H_2O)_2Br_2]\) (square planar) This complex has 2 kinds of ligands. The matching ligands can either be adjacent to each other and be cis, or they can be across from each other and be trans. b. \([Pt(NH_3)(py)(Cl)(Br)]\) (square planar, py = pyridine, \(C_5H_5N\)) This complex has 4 different ligands. There is no plane of symmetry in any of the enantiomers, making the structures chiral and therefore has optical isomers. c. \([Zn(NH_3)_3Cl]^+\) (tetrahedral) There is a plane of symmetry from \(NH_3\) through \(Zn\) to the other \(NH_3\), therefore it is not chiral. d. \([Pt(NH_3)_3Cl]^+\) (square planar) There is a plane of symmetry from \(NH_3\) through \(Pt\) to the other \(NH_3\), therefore it is not chiral. e. \([Ni(H_2O)_2Cl_2]\) The \(Cl\) ligands can either be right next to each other, or directly across from one another allowing for both cis and trans geometries. f. \([Co(C_2O_4)_2Cl_2]^-3\) (note that \(C_2O_4^-2\) is the bidentate oxalate ion, \(^−O_2CCO_2^-\) There is a plane of symmetry from \(Cl\) through Co to the other \(Cl\) in a "trans" chlorine configuration, therefore it is not chiral in a chlorine "trans" configuration. However, there is no symmetry in the chlorine "cis" configuration, indicating multiple "cis" isomers. a. [Pt(H O) Br ]: b. [Pt(NH )(py)(Cl)(Br)]: c. [Zn(NH ) Cl] : d. [Pt(NH ) Cl] : e. [Ni(H O) Cl ]: f. [Co(C O ) Cl ] : Draw diagrams for any , , and optical isomers that could exist for the following (en is ethylenediamine): We are instructed to draw all geometric isomers and optical isomers for the specified compound. Optical isomers exist when an isomer configuration is not superimposable on its mirror image. This means there are two distinct molecular shapes. Often a left and right hand are cited as an example; if you were to take your right hand and place it upon your left, you cannot make the major parts of your hand align on top of one another. The basic idea when deciding whether something is optically active is to look for a plane of symmetry--if you are able to bisect a compound in a manner that establishes symmetry, then the compound does not have an optical isomer. Cis isomers exist when there are 2 ligands of the same species placed at 90 degree angles from each other. Trans isomers exist when there are 2 ligands of the same species placed at 180 degree angles from each other. Problem 1 This compound is an octahedral molecule, so the six ligands (atoms in the complex that are not the central transition metal) are placed around the central atom at 90 degree angles. Two optical isomers exist for [Co(en) (NO )Cl] . The second isomer is drawn by taking the mirror image of the first. Problem 2 This compound is also an octahedral molecule. Two cis (optical) isomers and one trans isomer exist for [Co(en) Cl ] . The trans isomer can be drawn by placing the chlorine ligands in positions where they form a 180 degree angle with the central atom. The first cis isomer can be drawn by placing the chlorine ligands in positions where they form a 90 degree angle with the central atom. The second cis isomer can be found by mirroring the first cis isomer, like we did in problem 1. Problem 3 This compound is also an octahedral molecule. One trans isomer and one cis isomer of [Pt(NH ) Cl ] exist. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle with the central atom. Problem 4 This compound is also an octahedral molecule. Two optical isomers for [Cr(en) ] exist. The second optical isomer can be drawn by taking the mirror image of the first optical isomer. Problem 5 This compound is a square planar complex, so the ligands are placed around the central atom in a plane, at 90 angles. A trans isomer and a cis isomer exist for the complex [Pt(NH ) Cl ]. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle in the plane with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle in the plane with the central atom. Name each of the compounds or ions given in Exercise Q19.2.3, including the oxidation state of the metal. Rules to follow for coordination complexes 1. Cations are always named before the anions. 2. Ligands are named before the metal atom or ion. 3. Ligand names are modified with an ‐o added to the root name of an anion. For neutral ligands the name of the molecule is used, with the exception of OH2, NH3, CO and NO. 4. The prefixes mono‐, di‐, tri‐, tetra‐, penta‐, and hexa‐ are used to denote the number of simple ligands. 5. The prefixes bis‐, tris‐, tetrakis‐, etc., are used for more complicated ligands or ones that already contain di‐, tri‐, etc. 6. The oxidation state of the central metal ion is designated by a Roman numeral in parentheses. 7. When more than one type of ligand is present, they are named alphabetically. Prefixes do not affect the order. 8. If the complex ion has a negative charge, the suffix –ate is added to the name of the metal. 9. In the case of complex‐ion isomerism the names cis, trans, fac, or mer may precede the formula of the complex‐ion name to indicate the spatial arrangement of the ligands. Cis means the ligands occupy adjacent coordination positions, and trans means opposite positions just as they do for organic compounds. The complexity of octahedral complexes allows for two additional geometric isomers that are peculiar to coordination complexes. Fac means facial, or that the three like ligands occupy the vertices of one face of the octahedron. Mer means meridional, or that the three like ligands occupy the vertices of a triangle one side of which includes the central metal atom or ion. Name each of the compounds or ions given in Exercise Q19.2.5. S19.2.7 Names of the above compounds. 1. [Co(en) (NO )Cl] Attain the names of the ligands and metal cation. Names can be found here. Co: Cobalt en: Ethylenediamine NO : Nitro Cl: Chloro Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (en) : bis(Ethylenediamine) Find the charges of the ligands. Charges can be found here. en: 0 NO : -1 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Co + 2(en) + (NO ) + Cl = 1 Co +2(0) + (-1) + (-1) = 1 Co = 3 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 2. [Co(en) Cl ] Attain the names of the ligands and metal cation. Names can be found here. Co: Cobalt en: Ethylenediamine Cl: Chloro Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (en) : bis(Ethylenediamine) Cl : dichloro Find the charges of the ligands. Charges can be found here. en: 0 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Co + 2(en) +2(Cl) = 1 Co + 2(0) + 2(-1) = 1 Co = 3 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 3. [Pt(NH ) Cl ] Attain the names of the ligands and metal cation. Names can be found here. Pt: Platinum NH : Ammine Cl: Chloro Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (NH ) : diammine Cl : tetrachloro Find the charges of the ligands. Charges can be found here. NH : 0 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Pt + 2(NH ) + 4(Cl) = 0 Pt + 2(0) + 4(-1) = 0 Pt = 4 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 4. [Cr(en) ] Attain the names of the ligands and metal cation. Names can be found here. Cr: Cromium en: ethylenediamine Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (en) : tris(ethylenediamine) Find the charges of the ligands. Charges can be found here. en: 0 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Cr + 3(en) = 3 Cr + 3(0) = 3 Cr = 3 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 5. [Pt(NH ) Cl ] Attain the names of the ligands and metal cation. Names can be found here. NH : Ammine Cl: Chloro Pt: Platinum Add the appropriate pre-fixes to each ligand depending on the number. (NH ) : diammine Cl : dichloro Find the charges of the ligands. Charges can be found here. NH : 0 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Pt + 2(NH ) + 2(Cl) = 0 Pt + 2(0) + 2(-1) = 0 Pt = 2 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. Specify whether the following complexes have isomers. Isomers are compounds that have the same number of atoms, but have different structures. Structural isomers (linkage, ionization, coordination) and stereoisomers (geometric and optical) can occur with several compounds. 1. tetrahedral \(\mathrm{[Ni(CO)_2(Cl)_2]}\) (Fig 1.) Immediately, we can cancel out the possibility of linkage, ionization, and coordination isomers. There are no other coordination complex for coordination isomerism, there is no ligand that can bond to the atom in more than one way for it to exhibit linkage isomerism, and there are no ions outside the coordination sphere for ionization isomerism. This is a tetrahedral structure which immediately rules out any geometric isomers since they require 90° and/or 180° bond angles. Tetrahedral structures have 109.5° angles. To confirm that the structure has no optical isomer, we must determine if there is a plane of symmetry. Structures that have no plane of symmetry are considered chiral and would have optical isomers. (Fig 2.) Since there is a plane of symmetry, we can conclude that there are no optical isomers. Overall, there are no isomers that exist for this compound. 2. trigonal byprimidal \(\mathrm{[Mn(CO)_4(NO)]}\) (Fig 3.) There are no ions, other coordination complex, and ambidentate ligands. Therefore, no structural isomers exist for this structure. Geometric isomers do not exist for this compound because there is only one nitrosyl ligand. (Fig 4.) In the image above, after the structure has been rotated, we can see that there is a plane of symmetry. Thus, there are no optical isomers. No isomers (the ones mentioned above) exist for this compound. 3. \(\mathrm{[Pt(en)_{2}Cl_2]Cl_2}\) (Fig 5.) Coordination isomerism cannot exist for this complex because there are no other complexes. There are no linkage isomers because there are no ambidentate ligands. Ionization isomers cannot exist in this complex either, even though there is a neutral molecule outside the coordination sphere. If we exchange \(\mathrm{Cl_2}\) with one ethyldiamine molecule, There would be 5 ligands in the coordination sphere instead of 4. This difference in the ratio of metal atom to ligands means that an ionization isomer cannot exist. (Fig 6.) The image above, shows the chloro and ethyldiamine ligands at a 90° angle with its other identical ligand. This is the isomer, while Fig. 5 shows the isomer. Fig 5. shows that there is a plane of symmetry in the isomer. Therefore, that structure does not have an optical isomer. On the other hand, the isomer does not have a plane of symmetry and therefore has an optical isomer. none; none; The two Cl ligands can be or . When they are , there will also be an optical isomer. Predict whether the carbonate ligand \(\ce{CO3^2-}\) will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand. \(\ce{CO3^2−}\) can be either monodentate or bidentate, since two of its oxygen atoms have lone pairs as shown above and can form covalent bonds with a transition metal ion. In most cases carbonate is monodentate because of its trigonal planar geometry (there is 120 degrees between the oxygens so it's hard for both to bind to the same metal). However, in some cases it will bind to two different metals, making it bidentate. CO will coordinate to a metal center as a monodentate ligand. Draw the geometric, linkage, and ionization isomers for [CoCl CN,CN]. Isomers are compounds with same formula but different atom arrangement. There are two subcategories: , which are isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another, and , isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space. There are three subcategories under structural isomers: , which are isomers that are identical except for a ligand has exchanging places with an anion or neutral molecule that was originally outside the coordination complex; , isomers that have an interchange of some ligands from the cationic part to the anionic part; and , in two or more coordination compounds in which the donor atom of at least one of the ligands is different. There are also two main kinds of stereoisomers: , metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal, and , which occurs when the mirror image of an object is non-superimposable on the original object.
Some of the isomers look almost identical, but that is because the CN ligand can be attached by both (but not at the same time) the C or N. Determine the number of unpaired electrons expected for [Fe(NO ) ] and for [FeF ] in terms of crystal field theory. Step 1: Determine the oxidation state of the Fe For \([Fe(NO_{2}){_{6}}]^{3-}\) and \([FeF_{6}]^{3-}\), both \(NO_{2}\) and \(F_{6}\) have a charge of -1. Since there is 6 of them then that means the charge is -6 and in order for there to be an overall charge of -3, Fe has to have a +3 charge. Step 2: Determine type of ligand Based on the spectrochemical series we can see that \(NO_{2}^{-}\) is a stronger field ligand than F\(^{-}\), and therefore is a low spin complex because it has a high \(\Delta_{\circ}\) unlike F\(^{-}\) which is a high spin. Step 3: Draw the crystal field \([Fe(NO_{2}){_{6}}]^{3-}\) \([FeF_{6}]^{3-}\) There is 1 unpaired electron for \([Fe(NO_{2}){_{6}}]^{3-}\), and 5 for \([FeF_{6}]^{3-}\) based on the crystal field theory. [Fe(NO ) ] :1 electron [FeF ] :5 electrons Draw the crystal field diagrams for [Fe(NO ) ] and [FeF ] . State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δ to P for each complex. a) \[[Fe(NO_2)_6]^{_{-4}}\] b) \[[FeF_6]^{_{-3}}\] Give the oxidation state of the metal, number of electrons, and the number of unpaired electrons predicted for [Co(NH ) ]Cl . The oxidation state of the metal can be found by identifying the charge of one of each molecule in the coordinate compound, multiplying each molecule's charge by the respective number of molecules present, and adding the products. This final sum represents the charge of the overall coordination compound. You can then solve for the oxidation state of the metal algebraically. In this case, one chloride anion Cl has a charge of -1. So three chloride anions have a total charge of -3. One ammine ligand NH has no charge so six ammine ligands have a total charge of zero. Finally, we are trying to solve for the oxidation state of a cobalt ion. Now we can write the equation that adds the total charges of each molecule or ion and is equal to the total charge of the overall coordinate compound. \[ (\text{oxidation}\text{ state}\text{ of}\text{ Co})+(-3)+0=0 \] \[ \text{oxidation}\text{ state}\text{ of}\text{ Co}=+3 \] . Now we need to identify the number of d-electrons in the Co ion. The electron configuration for cobalt that has no charge is \[ [\text{Ar}]4\text{s}^23\text{d}^7 \] However, a Co ion has 3 less electrons than its neutral counterpart and has an electron configuration of \[ [\text{Ar}]3\text{d}^6 \] For transition metals, the \( \text{s} \) electrons are lost first. So cobalt loses its two \( 4\text{s} \) electrons first and then loses a single \( 3\text{d} \) electron meaning To predict the number of unpaired electrons, we must first determine if the complex is high spin or low spin. Whether the complex is high spin or low spin is determined by the ligand in the coordinate complex. Specifically, the ligand must be identified as either a weak-field ligand or a strong-field ligand based on the spectrochemical series. Weak-field ligands induce high spin while strong-field ligands induce low spin. We can then construct the energy diagram or crystal field diagram of the designated spin that has the proper electron placings. The geometric shape of the compound must also be identified to construct the correct diagram. Finally, from this crystal field diagram we can determine the number of unpaired electrons. The number of electrons in the diagram is equal to the number of \( \text{d} \) electrons of the metal. The ligand in this case is NH , which is a strong field ligand according to the spectrochemical series. This means that the complex is low spin. Additionally, six monodentate ligands means the ligand field is octahedral. The number of electrons that will be in the diagram is 6 since the metal ion Co has 6 \( \text{d} \) electrons. Now the proper crystal field diagram can be constructed. From the crystal field diagram, we can tell that . a) 3+ b) 6 d electrons c) No unpaired electrons The solid anhydrous solid CoCl is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have. From our knowledge of ligands and coordination compounds (or if you need a refresher Coordination Compounds), we can assume the product of CoCl in water. H O is a common weak field ligand that forms six ligand bonds around the central Cobalt atom while the Chloride stays on the outer sphere. We can use this to determine the complex: \([Co(H_2O)_6]Cl_2\) From this formation, we can use the Crystal Field Theory (CFT)(Crystal Field Theory) to determine the number of unpaired electrons. This coordination compound has six ligand bonds attached to the central atom which means the CFT model will follow the octahedral splitting. Keep in mind that we know H O is a weak field ligand and will produce a high spin. High spin is when the electrons pairing energy (P) is greater than the octahedral splitting energy. Thus, the electrons spread out and maximize spin. In order to fill out our crystal field diagram, we need to determine the charge of cobalt. Because the H O ligand is neutral, and there are two chlorine ions, we can deduce the charge of cobalt is plus two in order to make the coordination complex neutral. From here, we can use the electron configuration of Co [Ar]4s 3d . The electrons that are taken away from the cobalt atom in order to form the plus two charge will from the 4s orbital and leave the 3d orbital untouched. Thus, there will be 7 electrons in the crystal field diagram and appear as: We can see here that there are 3 unpaired electrons. [Co(H O) ]Cl with three unpaired electrons. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain. It is not possible for a metal in the transition series to have six unpaired electrons. This is because transition metals have a general electron configuration of (n-1)d ns where n is the quantum number. The last electron will go into the d orbital which has 5 orbitals that can each contain 2 electrons, yielding 10 electrons total. According to Hund's Rule, electrons prefer to fill each orbital singly before they pair up. This is more energetically favorable. Since there are only 5 orbitals and due to Hund's Rule, the maximum number of unpaired electrons a transition metal can have is 5. Therefore, there cannot be a complex of a transition metal that has 6 unpaired electrons. For example, lets look at iron's electron configuration. Iron has an electron configuration of 1s 2s 2p 3s 3p 4s 3d . Now the most important orbital to look at is the d orbital which has 6 electrons in it, but there are only 4 unpaired electrons as you can see by this diagram: 3d: [↿⇂] [↿] [↿,↿,↿] Each [ ] represents an orbital within the d orbital. This diagram follows Hund's rule and shows why no transition metal can have 6 unpaired electrons. How many unpaired electrons are present in each of the following? 1. For [CoF ] , we first found the oxidation state of Co, which is 3+ since F has a 1- charge and since there is 6 F, Co's charge has to be 3+ for the overall charge to be 3-. \[\textrm{charge of Co} + \textrm{-6} = \textrm{-6}\] \[\textrm{charge of Co} = \textrm{+3}\] After finding the oxidation state, I then go to the periodic table to find its electron configuration: [Ar]3d We distribute the 6 d-orbital electrons along the complex and since it is high spin, the electrons is distributed once in each energy level before it is paired. There is only one pair and the other 4 electrons are unpaired, making the answer 4. 2. The same process is repeated. We find the charge of Mn, which is 3+, making the electron configuration: [Ar]3d \[\textrm{charge of Mn} + \textrm{-6} = \textrm{-3}\] \[\textrm{charge of Mn} = \textrm{+3}\] There is a difference between this and number 1. This is low spin so instead of distributing one electron in each level before pairing it, I must distribute one electron on the bottom and then pair them all up before I'm able to move to the top portion. So since there is 4, there is only a pair at d and the other two electrons are unpaired. Making the number of unpaired electrons 2. 3.The same process as number 2 is applied. The only difference is that the charge of Mn is now 2+ so the electron configuration: [Ar]3d . \[\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}\] \[\textrm{charge of Mn} = \textrm{+2}\] Since is it low spin like number 2, I only need to add an extra electron to the next level, making that 2 pairs of electron and only 1 electron unpaired. 4. Since Cl has a -1 charge like CN, Mn's charge is also 2+ with the same electron configuration as number 3, which is 5. \[\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}\] \[\textrm{charge of Mn} = \textrm{+2}\] With 5 electrons, this is high spin instead of low. So as stated in number 1, we pair distribute the electrons on all levels first. Since there are 5 electrons and 5 levels and they are al distributed, there are zero pairs, making that 5 unpaired electrons. 5. Using the same process as the problems above, Rh's charge is 3+, with the electron configuration: [Kr]4d . \[\textrm{charge of Rh} + \textrm{-6} = \textrm{-3}\] \[\textrm{charge of Mn} = \textrm{+3}\] With a low spin and 6 electrons, all electrons are paired up, making it 0 electrons that are unpaired. 4; 2; 1; 5; 0 Explain how the diphosphate ion, [O P−O−PO ] , can function as a water softener that prevents the precipitation of Fe as an insoluble iron salt. The diphosphate ion, [O P−O−PO ] can function as a water softener keeping the iron in a water soluble form because of its more negative electrochemical potential than water's. This is similar to the way plating prevents metals from reacting with oxygen to corrode. Mineral deposits are formed by ionic reactions. The Fe will form an insoluble iron salt of iron(III) oxide-hydroxide when a salt of ferric iron hydrolyzes water. However, with the addition of [O P−O−PO ] , the Fe cations are more attracted to the PO group, forming a Fe(PO ) complex. The excess minerals in this type of water is considered hard thus its name hard water. For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the orbitals increases. Which complex in each of the following pairs of complexes is more stable? The Spectrochemical Series is as follows \[I^{-}<Br^{-}<SCN^{-}\approx Cl^{-}<F^{-}<OH^{-}<ONO^{-}<ox<H_{2}O<SCN^{-}<EDTA<NH_{3}<en<NO_{2}^{-}<CN\] The strong field ligands (on the right) are low spin which fills in more electrons in the orbitals. The weak field ligands (on the left) are high spin so it can fill electrons in the orbitals and orbitals. In conclusion, more electrons are filled up from the strong field ligands because the electrons don't move up to the orbitals. a. \([Fe(CN)_{6}]^{4-}\) \(CN\) is a stronger ligand than \(H_{2}O\) so it is low spin, which fills up the orbitals. b. \([Co(NH_{3})_{6}]^{3+}\) \(NH_{3}\) is a stronger ligand than \(F\). c. \([Mn(CN)_{6}]^{4-}\) \(CN\) is a stronger ligand than \(Cl^{-}\). For more information regarding the shape of the complex and d-electron configuration, libretext provides more information on how to classify high and low spin complexes. [Fe(CN) ] ; [Co(NH ) ] ; [Mn(CN) ] Trimethylphosphine, P(CH ) , can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH ) can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound? a) b) Tetrahedral Would you expect the complex [Co(en) ]Cl to have any unpaired electrons? Any isomers? Assign oxidation states to each element. Cl- has a -1 oxidation state. En is neutral, so 0. The entire complex is also neutral, so in order to balance the charges out, Co must be +3 because there are 3 chlorides, which gives a -3 charge. Write the electron configuration for \(Co^{3+}\). \([Ar]3d^6\). There are 6 electrons. Check where en lies on the spectrochemical series. Does it have a strong field strength? It does, so these electrons will exist at the d-level with high splitting energy because the magnitude of the pairing energy is less than the crystal field splitting energy in the octahedral field. You will the notice that there aren't any unpaired electrons when you draw the Crystal Field Theory (CFT) diagram. This complex does not have any geometric isomers because cis-trans structures cannot be formed. The mirror image is nonsuperimpoasable, which means the enantiomers are chiral molecules; if the mirror image is placed on top on the original molecule, then they will never be perfectly aligned to give the same molecule. The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer. Would you expect the Mg [Cr(CN) ] to be diamagnetic or paramagnetic? Explain your reasoning. The first step to determine the magnetism of the complex is to calculate the oxidation state of the transition metal. In this case, the transition metal is Cr. Before doing so, we need to find charge of the of the complex ion [Cr(CN) ] given that the oxidation state of Mg [Cr(CN) ] [Cr(CN) ] Mg [Cr(CN) ] A19.3.11 a) Paramagnetic Would you expect salts of the gold ion, Au , to be colored? Explain. No. . A partially filled d orbital, for example, can yield various colors. After completing the noble gas configuration, we see that Au has a configuration of [Xe] 4f . Since Au has a , we are certain that any salts of the gold ion, Au will be *An example of a colored ion would be copper(II). Cu has an electron configuration of [Ar]3d . It has one unpaired electron. Copper(II) appears blue. No. Au has a complete 5 sublevel. [CuCl ] is green. [Cu(H O) ] is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting? Although a color might appear a certain way, it actual absorbs a different color, opposite of it on the color wheel. In this case; [CuCl ] appears green but is opposite of red on the color wheel which is absorbed and is characterized by wavelengths 620-800 nanometers. [Cu(H O) ] appears blue but is opposite of orange on the color wheel which is absorbed and is characterized by wavelengths 580-620 nanometers. When determining which absorbs the higher energy photons, one must look at the complex itself. A higher energy indicates a high energy photon absorbed and a lower energy indicates a lower energy photon absorbed. How can we determine this? By looking at the complex and more specifically the ligand attached and its location in the spectrochemical series. The ligands attached are Water and Chlorine and since Water is a stronger ligand than Chlorine according to the series, it also has larger energy, indicating a higher energy. This means that the complex [Cu(H O) ] absorbs a higher energy photon because of its a stronger ligand than chlorine. Part 2 of this question also asks which complex is predicted to have a larger crystal field splitting. To determine this you also use the spectrochemical series and see which ligand is stronger. Since H O is stronger than Cl on the spectrochemical series, we can say [Cu(H O) ] has a higher crystal field splitting. a) [Cu(H O) ] [Cu(H O) ] has a higher crystal field splitting | 84,938 | 2,916 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.04%3A_Properties_of_Solutions |
Solutions are likely to have properties similar to those of their major component—usually the solvent. However, some solution properties differ significantly from those of the solvent. Here, we will focus on liquid solutions that have a solid solute, but many of the effects we will discuss in this section are applicable to all solutions. Solutes affect some properties of solutions that depend only on the concentration of the dissolved particles. These properties are called . Four important colligative properties that we will examine here are vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure. compounds separate into individual molecules when they are dissolved, so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast, compounds separate into their constituent ions when they dissolve, so 1 mol of an ionic compound will produce more than 1 mol of dissolved particles. For example, that dissolves yields and , for a of particles in solution. Thus, the effect on a solution’s properties by dissolving NaCl may be twice as large as the effect of dissolving the same amount of moles of glucose (C H O ). All liquids evaporate. In fact, given enough volume, a liquid will turn completely into a vapor. If enough volume is not present, a liquid will evaporate only to the point where the rate of evaporation equals the rate of vapor condensing back into a liquid. The pressure of the vapor at this point is called the of the liquid. The presence of a dissolved solid lowers the characteristic vapor pressure of a liquid so that it evaporates more slowly. (The exceptions to this statement are if the solute itself is a liquid or a gas, in which case the solute will also contribute something to the evaporation process. We will not discuss such solutions here.) This property is called vapor pressure depression and is depicted in Figure \(\Page {1}\). A related property of solutions is that their boiling points are higher than the boiling point of the pure solvent. Because the presence of solute particles decreases the vapor pressure of the liquid solvent, a higher temperature is needed to reach the boiling point. This phenomenon is called boiling point elevation. The addition of one mole of sucrose (molecular compound) in one liter of water will raise the boiling point from 100 C to 100.5 C but the addition of one mole of NaCl in one liter of water will raise the boiling point by 2 x 0.5 C = 1 C. Furthermore, the addition of one mole of \(\ce{CaCl2}\) in one liter of water will raise the boiling point by 3 x 0.5 C = 1.5 C. Some people argue that putting a pinch or two of salt in water used to cook spaghetti or other pasta makes a solution that has a higher boiling point, so the pasta cooks faster. In actuality, the amount of solute is so small that the boiling point of the water is practically unchanged. The presence of solute particles has the opposite effect on the freezing point of a solution. When a solution freezes, only the solvent particles come together to form a solid phase, and the presence of solute particles interferes with that process. Therefore, for the liquid solvent to freeze, more energy must be removed from the solution, which lowers the temperature. Thus, solutions have lower freezing points than pure solvents do. This phenomenon is called freezing point depression. Both boiling point elevation and freezing point depression have practical uses. For example, solutions of water and ethylene glycol (C H O ) are used as coolants in automobile engines because the boiling point of such a solution is greater than 100°C, the normal boiling point of water. In winter, salts like NaCl and \(\ce{CaCl2}\) are sprinkled on the ground to melt ice or keep ice from forming on roads and sidewalks (Figure \(\Page {2}\)). This is because the solution made by dissolving sodium chloride or calcium chloride in water has a lower freezing point than pure water, so the formation of ice is inhibited. Which solution’s freezing point deviates more from that of pure water—a 1 M solution of NaCl or a 1 M solution of \(\ce{CaCl2}\)? Colligative properties depend on the number of dissolved particles, so the solution with the greater number of particles in solution will show the greatest deviation. When NaCl dissolves, it separates into two ions, Na and Cl . But when \(\ce{CaCl2}\) dissolves, it separates into three ions—one Ca ion and two Cl ions. Thus, mole for mole, \(\ce{CaCl2}\) will have 50% more impact on freezing point depression than NaCl. Which solution’s boiling point deviates more from that of pure water—a 1 M solution of \(\ce{CaCl2}\) or a 1 M solution of MgSO ? \(\ce{CaCl2}\) Estimate the boiling point of 0.2 M \(\ce{CaCl2}\) solution. The boiling point increases 0.5 C for every mole of solute per liter of water. For this estimation, let's assume that 1 liter of solution is roughly the same volume as 1 liter of water. A 0.2 M \(\ce{CaCl2}\) solution contains 0.2 moles of \(\ce{CaCl2}\) solution formula units per liter of solution. Each \(\ce{CaCl2}\) unit separates into three ions. \[\mathrm{0.2\: mol\: CaCl_2\times\dfrac{3\: mol\: ions}{1\: mol\: CaCl_2}\times\dfrac{0.5\: deg\: C}{1\: mol\: ion}=0.3\: deg\: C} \nonumber \] The normal boiling point of water is 100 C, so the boiling point of the solution is raised to 100.3 C. Estimate the freezing point of 0.3 M \(\ce{CaCl2}\) solution. minus 1.7 C The last colligative property of solutions we will consider is a very important one for biological systems. It involves , the process by which solvent molecules can pass through certain membranes but solute particles cannot. When two solutions of different concentration are present on either side of these membranes (called ), there is a tendency for solvent molecules to move from the more dilute solution to the more concentrated solution until the concentrations of the two solutions are equal. This tendency is called . External pressure can be exerted on a solution to counter the flow of solvent; the pressure required to halt the osmosis of a solvent is equal to the osmotic pressure of the solution. Osmolarity (osmol) is a way of reporting the total number of particles in a solution to determine osmotic pressure. It is defined as the molarity of a solute times the number of particles a formula unit of the solute makes when it dissolves (represented by \(i\)): \[osmol = M \times i\label{Eq1} \] If more than one solute is present in a solution, the individual osmolarities are additive to get the total osmolarity of the solution. Solutions that have the same osmolarity have the same osmotic pressure. If solutions of differing osmolarities are present on opposite sides of a semipermeable membrane, solvent will transfer from the lower-osmolarity solution to the higher-osmolarity solution. Counterpressure exerted on the high-osmolarity solution will reduce or halt the solvent transfer. An even higher pressure can be exerted to force solvent from the high-osmolarity solution to the low-osmolarity solution, a process called . Reverse osmosis is used to make potable water from saltwater where sources of fresh water are scarce. A 0.50 M NaCl aqueous solution and a 0.30 M Ca(NO ) aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow. The solvent will flow into the solution of higher osmolarity. The NaCl solute separates into two ions—Na and Cl —when it dissolves, so its osmolarity is as follows: The Ca(NO ) solute separates into three ions—one Ca and two NO —when it dissolves, so its osmolarity is as follows: The osmolarity of the Ca(NO ) solution is lower than that of the NaCl solution, so water will transfer through the membrane from the Ca(NO ) solution to the NaCl solution. A 1.5 M C H O aqueous solution and a 0.40 M Al(NO ) aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow. osmol C H O = 1.5; osmol Al(NO ) = 1.6 The solvent flows from C H O solution (lower osmolarity) to \(\ce{Al(NO3)3}\) solution (higher osmolarity). The main function of the kidneys is to filter the blood to remove wastes and extra water, which are then expelled from the body as urine. Some diseases rob the kidneys of their ability to perform this function, causing a buildup of waste materials in the bloodstream. If a kidney transplant is not available or desirable, a procedure called dialysis can be used to remove waste materials and excess water from the blood. In one form of dialysis, called , a patient’s blood is passed though a length of tubing that travels through an (also called a ). A section of tubing composed of a semipermeable membrane is immersed in a solution of sterile water, glucose, amino acids, and certain electrolytes. The osmotic pressure of the blood forces waste molecules and excess water through the membrane into the sterile solution. Red and white blood cells are too large to pass through the membrane, so they remain in the blood. After being cleansed in this way, the blood is returned to the body. Dialysis is a continuous process, as the osmosis of waste materials and excess water takes time. Typically, 5–10 lb of waste-containing fluid is removed in each dialysis session, which can last 2–8 hours and must be performed several times a week. Although some patients have been on dialysis for 30 or more years, dialysis is always a temporary solution because waste materials are constantly building up in the bloodstream. A more permanent solution is a kidney transplant. Cell walls are semipermeable membranes, so the osmotic pressures of the body’s fluids have important biological consequences. If solutions of different osmolarity exist on either side of the cells, solvent (water) may pass into or out of the cells, sometimes with disastrous results. Consider what happens if red blood cells are placed in a solution, meaning a solution of lower osmolarity than the liquid inside the cells. The cells swell up as water enters them, disrupting cellular activity and eventually causing the cells to burst. This process is called . If red blood cells are placed in a solution, meaning one having a higher osmolarity than exists inside the cells, water leaves the cells to dilute the external solution, and the red blood cells shrivel and die. This process is called . Only if red blood cells are placed in solutions that have the same osmolarity as exists inside the cells are they unaffected by negative effects of osmotic pressure. The concentration of an isotonic sodium chloride (NaCl) solution is only half that of an isotonic glucose (C H O ) solution because NaCl produces two ions when a formula unit dissolves, while molecular C H O produces only one particle when a formula unit dissolves. The osmolarities are therefore the same even though the concentrations of the two solutions are different. Osmotic pressure explains why you should not drink seawater if you are abandoned in a life raft in the middle of the ocean. Its osmolarity is about three times higher than most bodily fluids. You would actually become thirstier as water from your cells was drawn out to dilute the salty ocean water you ingested. Our bodies do a better job coping with hypotonic solutions than with hypertonic ones. The excess water is collected by our kidneys and excreted. Osmotic pressure effects are used in the food industry to make pickles from cucumbers and other vegetables and in brining meat to make corned beef. It is also a factor in the mechanism of getting water from the roots to the tops of trees! A perfusionist is a medical technician trained to assist during any medical procedure in which a patient’s circulatory or breathing functions require support. The use of perfusionists has grown rapidly since the advent of open-heart surgery in 1953. Most perfusionists work in operating rooms, where their main responsibility is to operate heart-lung machines. During many heart surgeries, the heart itself must be stopped. In these situations, a heart-lung machine keeps the patient alive by aerating the blood with oxygen and removing carbon dioxide. The perfusionist monitors both the machine and the status of the blood, notifying the surgeon and the anesthetist of any concerns and taking corrective action if the status of the blood becomes abnormal. Despite the narrow parameters of their specialty, perfusionists must be highly trained. Certified perfusion education programs require a student to learn anatomy, physiology, pathology, chemistry, pharmacology, math, and physics. A college degree is usually required. Some perfusionists work with other external artificial organs, such as hemodialysis machines and artificial livers. | 12,918 | 2,917 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_from_Dehydrohalogenation_of_Haloalkanes |
Alkenes can be obtained from haloalkanes (alkyl halides). These haloalkanes are usually bromo and iodo and less commonly, chloro derivatives. Haloalkanes on heating with alcoholic \(KOH\) loses one molecule of hydrogen halide to give alkene. If two alkenes may be formed due to dehydrohalogenation of a haloalkane, the one which is most substituted is the main product. For example, dehydrohalogenation of 2-bromobutane gives, The order of reactivity of haloalkanes in dehydrohalogenation is, Tertiary > Secondary > Primary. Note: Reactions in which a small molecule like \(H_2O\) or \(HX\) is lost are known as elimination reactions. Vicinal dihaloalkanes are those dihalogen derivatives of alkanes in which two halogen atoms are on the adjacent carbon atoms. Alkenes can be obtained from vicinal dihaloalkanes by dehalogenation. When such a dihaloalkane is heated with zinc in methanol, an alkene is formed. 1,2-dibromoethane gives 1,2-dibromoethane ethene | 978 | 2,918 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/07%3A_Mixtures_and_Solutions/7.02%3A_Partial_Molar_Volume |
The of compound A in a mixture of A and B can be defined as \[ V_A = \left (\dfrac{\partial V}{\partial n_A} \right)_{p,T,n_B} \nonumber \] Using this definition, a change in volume for the mixture can be described using the total differential of \(V\): \[ dV = \left( \dfrac{\partial V}{\partial n_A}\right)_{p,T,n_B} dn_A + \left( \dfrac{\partial V}{\partial n_B}\right)_{p,T,n_A} dn_B \nonumber \] or \[ dV = V_a \, dn_A + V_b\,dn_B \nonumber \] and integration yields \[V = \int _0^{n_A} V_a \, dn_A + \int _0^{n_B} V_b\,dn_B \nonumber \] \[ V = V_a \, n_A + V_b\,n_B \nonumber \] This result is important as it demonstrates an important quality of partial molar quantities. Specifically, if \(\xi_i\) represents the partial molar property \(X\) for component i of a mixture, The total property \(X \) for the mixture is given by \[X = \sum_{i} \xi_in_i \nonumber \] It should be noted that while the volume of a substance is never negative, the partial molar volume can be. An example of this appears in the dissolution of a strong electrolyte in water. Because the water molecules in the solvation sphere of the ions are physically closer together than they are in bulk pure water, there is a volume decrease when the electrolyte dissolves. This is easily observable at high concentrations where a larger fraction of the water in the sample is tied up in solvation of the ions. | 1,397 | 2,919 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.11%3A_Oscillating_Reactions |
In most cases, the conversion of reactants into products is a fairly smooth process, in that the concentrations of the reactants decrease in a regular manner, and those of the products increase in a similar regular manner. However, some reactions can show irregular behavior in this regard. One particularly peculiar (but interesting!) phenomenon is that of , in which reactant concentrations can rise and fall as the reaction progresses. One way this can happen is when the products of the reaction (or one of the steps) catalyzes the reaction (or one of the steps. This process is called . An example of an autocatalyzed mechanism is the . This is a three-step mechanism defined as follows: \[ A +X \xrightarrow{k_1} X + X \nonumber \] \[ X +Y \xrightarrow{k_2} Y + Y \nonumber \] \[ Y \xrightarrow{k_3} B \nonumber \] In this reaction, the concentration of reactant \(A\) is held constant by continually adding it to the reaction mixture. The first step is , so as it proceeds, it speeds up. However, an increase in the production of \(X\) by the first reaction increases the rate of the second reaction as well, which is also autocatalyzed. Finally, the removal of \(Y\) through the third reaction brings things to a halt, until the first reaction can again produce a build up of \(X\) to start the cycle over. A plot of the concentration of \(X\) and \(Y\) as a function of time looks as follows: This mechanism follows kinetics predicted by what is called the relationship. In this case, \(X\) represents the “prey” and \(Y\) represents the “predator”. The population of the predator cannot build up unless there is a significant population of prey on which the predators can feed. Likewise, the population of predators decreases when the population of the prey falls. And finally, there is a lag, as the rise and decline of the prey population controls the rise and fall of the predator population. The equations have been studied extensively and have applications not just in chemical kinetics, but in biology, economics, and elsewhere. One wonders if the equations can be applied to help to understand politics! | 2,138 | 2,920 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Structure_of_Organic_Molecules/Cis_and_Trans_Isomers_of_Alkanes |
Several organic compounds may have identical compositions but will have widely different physical and chemical properties because the arrangement of the atoms is different. Isomers and identical compounds both have the same number of each kind of element in a formula. A simple count will establish this fact. As a consequence of the double bond, some alkene compounds exhibit a unique type of isomerism. Rotation around a single bond occurs readily, while rotation around a double bond is restricted. The pi bond prevents rotation because of the electron overlap both above and below the plane of the atoms. A single bond is analogous to two boards nailed together with one nail. A double bond is analogous to two boards nailed together with two nails. In the first case you can twist the boards, while in the second case you cannot twist them. are compounds with different spatial arrangements of groups attached to the carbons of a double bond. In alkenes, the carbon-carbon double bond is rigidly fixed. Even though the attachment of atoms is the same, the geometry (the way the atoms "see" each other) is different. When looking for geometric isomers, a guiding principle is that there . A "group" can be hydrogen, alkyls, halogens, etc. may appear to have different arrangements as written, but closer examination by rotation or turning will result in the molecules being superimposed. If they are super impossible or if they have identical names, then the two compounds are in fact identical. of compounds have a different arrangement of the atoms. Isomer compounds will differ from identical compounds by the arrangement of the atoms. See Both compounds have the same number of atoms, C H . They are isomers because in the left molecule the root is 4 carbons with one branch. In the right molecule, the root is 3 carbons with 2 branches. They are because they have the same number of atoms but different arrangements of those atoms. If the number of each element is different, the two compounds are merely completely different. A simple count of the atoms will reveal them as different. In the example on the left, the chlorine atoms can be opposite or across from each other in which case it is called the . If the the chlorine atoms are next to or adjacent each other, the isomer is called " ". If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH etc. there cannot be any geometric isomers. Consider the longest chain containing the double bond: If two groups (attached to the carbons of the double bond) are on the same side of the double bond, the isomer is a alkene. If the two groups lie on opposite sides of the double bond, the isomer is a alkene. One or more of the "groups" may or may not be part of the longest chain. In the case on the left, the "group" is a methyl - but is actually part of the longest chain. A common mistake is to name this compound as 1,2-dimethylethene. Look at all carbons for the longest continuous chain - the root is 4 carbons - butene. For the structures below:
a. Draw both cis/trans isomers, if any, of the structure based upon the name. b. Look at the graphic and state whether the compound is cis, trans, or not cis/trans isomers. | 3,252 | 2,921 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.09%3A_Temperature_Dependence |
In general, increases in temperature increase the rates of chemical reactions. It is easy to see why, since most chemical reactions depend on molecular collisions. And as we discussed in Chapter 2, the frequency with which molecules collide increases with increased temperature. But also, the kinetic energy of the molecules increases, which should increase the probability that a collision event will lead to a reaction. An empirical model was proposed by Arrhenius to account for this phenomenon. The (Arrhenius, 1889) can be expressed as \[ k = A e^{-E_a/RT} \nonumber \] Although the model is empirical, some of the parameters can be interpreted in terms of the energy profile of the reaction. \(E_a\), for example, is the , which represents the energy barrier that must be overcome in a collision to lead to a reaction. If the rate constant for a reaction is measure at two temperatures, the activation energy can be determined by taking the ratio. This leads to the following expression for the Arrhenius model: \[ \ln \left( \dfrac{k_1}{k_2} \right) = - \dfrac{E_a}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{Arrhenius} \] For a given reaction, the rate constant doubles when the temperature is increased form 25 °C to 35 °C. What is the Arrhenius activation energy for this reaction? The energy of activation can be calculated from the Arrhenius Equation (Equation \ref{Arrhenius}). \[ \ln \left( \dfrac{2k_1}{k_1} \right) = - \dfrac{E_a}{8.314 \, \dfrac{J}{mol\, K}} \left( \dfrac{1}{308\,K} - \dfrac{1}{298\,K} \right) \nonumber \] From this reaction: \[E_a = 52.9\, kJ/mol \nonumber \] Preferably, however, the rate constant is measured at several temperatures, and then the activation energy can be determined using all of the measurements, by fitting them to the expression \[ \ln (k) = - \dfrac{E_a}{RT} + \ln (A) \nonumber \] This can be done graphically by plotting the natural logarithm of the rate constant as a function of \(1/T\) (with the temperature measured in K). The result should be a straight line (for a well-behaved reaction!) with a slope of \(–E_a/R\). There are some theoretical models (such as collision theory and transition state theory) which suggest the form of the Arrhenius model, but the model itself is purely empirical. A general feature, however, of the theoretical approaches is to interpret the activation energy as an energy barrier which a reaction must overcome in order to lead to a chemical reaction. | 2,480 | 2,922 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/01%3A_The_Basics/1.04%3A_The_Zeroth_Law_of_Thermodynamics |
is an important property when it comes to measuring energy flow through a system. But how does one use or measure temperature? Fortunately, there is a simple and intuitive relationship which can be used to design a thermometer – a device to be used to measure temperature and temperature changes. The zeroth law of thermodynamics can be stated as follows: This basic principle has been used to define standard temperature scales by the International Committee on Weights and Measures (BIPM) to guide the adoption of the International Practical Temperature Scale of 1990 (Mangum & Furukawa, 1990). | 610 | 2,924 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/RR._Radical_Reactions/RR2._Initiation%3A_Bond_Homolysis |
Sometimes, radicals form because a covalent bond simply splits in half. Two atoms that used to be bonded to each other go their separate ways. Each atom takes with it one electron from the former bond. This process is called homolysis, meaning the bond is breaking evenly. In contrast, heterolysis is the term for a bond that breaks via ionization, with one atom getting both electrons from the bond. In pictures, we show this process using curved arrows, but the arrows we use are slightly different from the ones you may be used to seeing in polar reaction chemistry. Instead of a regular arrowhead, we use a half arrowhead. This kind of arrow looks a little more like a fish hook. It is easy to remember the roles of the two kinds of arrows, because a full arrowhead describes the movement of an electron pair, whereas a half arrowhead describes the movement of only one electron. Why would a covalent bond simply break apart? There are really a number of factors and a number of events that may result in this situation. The simple part of the story is that the bond must have been weak in the first place. There was enough energy available in the form of heat transferred from the surroundings (or sometimes in the form of light) to overcome the stabilization energy of the bond. What makes a bond weak or strong? That is a complicated question. Many factors influence bond strength. However, two of the main factors responsible for covalent bond strength are the degree of electron sharing because of "overlap" and the degree of bond polarity resulting from "exchange". Most strong covalent bonds rely on a mixture of these two factors. One fairly common feature in homolysis is a bond between two atoms of the same kind. For example, elemental halogens often undergo homolysis pretty easily. The ease with which these bonds can be split in half is illustrated by their low bond dissociation energies. Not much energy needs to be added in order to overcome the bonds between these atoms. This propensity for radical formation can be understood in terms of the lack of a polar component in these bonds. These atoms rely solely on atomic overlap to share electrons with each other. There is a notable exception to the rule that homoatomic bonds are inherently weak, and that is a carbon-carbon bond. Its bond dissociation energy is listed in the table for comparison with the halogens. The relative strength of carbon-carbon bonds gives rise to a multitude of carbon-based "organic" compounds in nature. The formation of bonds between like atoms is called "catenation"; carbon is the world champion. Draw structures for the following reagents and show curved arrows to illustrate the initiation of radicals in each case. a) Br b) H O c) (CH ) CO H d) (CH CH ) S Silicon (BDE = 53 kcal/mol) and sulfur (BDE = 54 kcal/mol) are also capable of catenation, but the bonds that these atoms form between themselves are much weaker than C-C bonds. It seems to be generally true that larger atoms form weaker bonds, at least in the main group of the periodic table. After all, I-I bonds are weaker than Br-Br bonds, which are weaker than Cl-Cl bonds. It is sometimes argued that this trend is a result of poor spatial overlap between the more diffuse p orbitals nearer the bottom of the periodic table. Draw structures of the following reagents and show curved arrows to illustrate the initiation of radicals in each case. a) (CH CH ) Pb b) (CH ) SbSb(CH ) c) (CH CH CH CH ) SnH More information on bond strengths is available at . , | 3,547 | 2,925 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Nuclear_Magnetic_Resonance_Spectroscopy/Supplemental_NMR_Topics |
Nuclear spin may be related to the nucleon composition of a nucleus in the following manner: Spin 1/2 nuclei have a spherical charge distribution, and their NMR behavior is the easiest to understand. Other spin nuclei have nonspherical charge distributions and may be analyzed as prolate or oblate spinning bodies. All nuclei with non-zero spins have magnetic moments ( ), but the nonspherical nuclei also have an electric ( ). Some characteristic properties of selected nuclei are given in the following table. Isotope Natural %
Abundance Spin (I) Magnetic
Moment (μ)* Magnetogyric
Ratio (γ) H H B C O F Si P * in units of nuclear magnetons = 5.05078•10 JT
γ in units of 10 rad T sec The model of a spinning nuclear magnet aligned with or against an external magnetic field (for I = 1/2 nuclei) must be refined for effective interpretation of NMR phenomena. Just as a spinning mass will precess in a gravitational field (a gyroscope), the magnetic moment associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The frequency of precession is proportional to the strength of the magnetic field, as noted by the equation: . The frequency is called the and has units of radians per second. The proportionality constant is known as the and is proportional to the magnetic moment ( ). Some characteristic γ's were listed in the preceding table of nuclear properties. A Spinning Gyroscope in a Gravity Field A Spinning Charge in a Magnetic Field If rf energy having a frequency matching the Larmor frequency is introduced at a right angle to the external field (e.g. along the x-axis), the precessing nucleus will absorb energy and the magnetic moment will flip to its I = 1/2 state. This excitation is shown in the following diagram. Note that frequencies in radians per second may be converted to Hz (cps) by dividing by 2 . The energy difference between nuclear spin states is small compared with the average kinetic energy of room temperature samples, and the +1/2 and 1/2 states are nearly equally populated. Indeed, in a field of 2.34 T the excess population of the lower energy state is only six nuclei per million. Although this is a very small difference , when we consider the number of atoms in a practical sample (remember the size of Avogadro's number), the numerical excess in the lower energy state is sufficient for selective and sensitive spectroscopic measurements. The diagram on the left below illustrates the macroscopic magnetization of a sample containing large numbers of spin 1/2 nuclei at equilibrium in a strong external magnetic field (B ). A slight excess of +1/2 spin states precess randomly in alignment with the external field and a smaller population of 1/2 spin states precess randomly in an opposite alignment. An overall net magnetization therefore lies along the z-axis. The diagram and animation on the right show the changes in net macroscopic magnetization that occur as energy is introduced by rf irradiation at right angles to the external field. It is convenient to show the rf transmitter on the x-axis and the receiver-detector coil on the y-axis. On clicking the " " button the animation will begin, and will repeat five times. An inherent problem of the NMR experiment must be pointed out here. We have noted that the population difference between the spin states is proportionally very small. A fundamental requirement for absorption spectroscopy is a population imbalance between a lower energy ground state and a higher energy excited state. This can be expressed by the following equation, where is a proportionality constant. If the mole fractions of the spin states are equal (η = η ) then the population difference is zero and no absorption will occur. If the rf energy used in an NMR experiment is too high this of the higher spin state will result and useful signals will disappear. For NMR spectroscopy to be practical, an efficient mechanism for nuclei in the higher energy 1/2 spin state to return to the lower energy +1/2 state must exist. In other words, the spin population imbalance existing at equilibrium must be restored if spectroscopic observations are to continue. Now an isolated spinning nucleus will not spontaneouly change its spin state in the absence of external perturbation. Indeed, hydrogen gas (H ) exists as two stable spin isomers: ortho (parallel proton spins) and para (antiparallel spins). NMR spectroscopy is normally carried out in a liquid phase (solution or neat) so that there is close contact of sample molecules with a rapidly shifting crowd of other molecules (Brownian motion). This thermal motion of atoms and molecules generates local fluctuating electromagnetic fields, having components that match the Larmor frequency of the nucleus being studied. These local fields stimulate emission/absorption events that establish spin equilibrium, the excess spin energy being detected as it is released. This relaxation mechanism is called (or Longitudinal Relaxation). The efficiency of spin-lattice relaxation depends on factors that influence molecular movement in the lattice, such as viscosity and temperature. The relaxation process is kinetically first order, and the reciprocal of the rate constant is a characteristic variable designated , the spin-lattice relaxation time. In non-viscous liquids at room temperature T ranges from 0.1 to 20 sec. A larger T indicates a slower or more inefficient spin relaxation. Another relaxation mechanism called spin-spin relaxation (or transverse relaxation) is characterized by a relaxation time T . This process, which is actually a spin exchange, will not be discussed here. In a given strong external magnetic field, each structurally distinct set of hydrogens in a molecule has a characteristic resonance frequency, just as each tubular chime in percussion instrument has a characteristic frequency. To discover the frequency of a chime we can strike it with a mallet and measure the sound emitted. This procedure can be repeated for each chime in the group so that all the characteristic frequencies are identified. An alternative means of aquiring the same information is to strike all the chimes simultaneously, and to subject the complex collection of frequencies produced to mathematical analysis. In the following diagram the four frequencies assigned to our set of chimes are added together to give a complex summation wave. This is a straightforward conversion; and the reverse transformation, while not as simple, is readily accomplished, provided the combination signal is adequately examined and characterized. A CW NMR spectrometer functions by irradiating each set of distinct nuclei in turn, a process analagous to striking each chime independently. For a high resolution spectrum this must be done slowly, and a 12 ppm sweep of the proton region takes from 5 to 10 minutes. It has proven much more efficient to excite all the proton nuclei in a molecule at the same time, followed by mathematical analysis of the complex rf resonance frequencies emitted as they relax back to the equilibrium state. This is the principle on which a pulse Fourier transform spectrometer operates. By exposing the sample to a very short (10 to 100 μsec), relatively strong (about 10,000 times that used for a CW spectrometer) burst of rf energy along the x-axis, as described above, all of the protons in the sample are excited simultaneously. The macroscopic magnetization model remains useful if we recognize it is a combination of megnetization vectors for all the nuclei that have been excited. The overlapping resonance signals generated as the excited protons relax are collected by a computer and subjected to a Fourier transform mathematical analysis. As shown in the diagram on the left, the Fourier transform analysis, abbreviated FT, converts the complex time domain signal emitted by the sample into the frequency (or field) domain spectrum we are accustomed to seeing. In this fashion a complete spectrum can be acquired in a few seconds. Because the is a first order process, the rf signal emitted by the sample decays exponentially. This is called a signal, abbreviated . Free Induction Decay Signal Since, the FID signal collected after one pulse, may be stored and averaged with the FID's from many other identical pulses prior to the Fourier transform, the NMR signal strength from a small sample may be enhanced to provide a useable spectrum. This has been essential to acquiring spectra from low abundance isotopes, such as C. In practice, the pulse FT experiment has proven so versatile that many variations of the technique, suited to special purposes, have been devised and used effectively. The compound on the left has a chain of ten methylene groups linking para carbons of a benzene ring. Such bridged benzenes are called . The meta analogs are also known. The structural constraints of the bridging chain require the middle two methylene groups to lie over the face of the benzene ring, which is a NMR shielding region. The four hydrogen atoms that are part of these groups display resonance signals that are more than two ppm higher field than the two methylene groups bonded to the edge of the ring (a deshielding region). The 14 π-electron bridged annulene on the right is an aromatic (4n + 2) system, and has the same anisotropy as benzene. Nuclei located over the face of the ring are shielded, and those on the periphery are deshielded. The ring hydrogens give resonance signals in the range 8.0 to 8.7 δ, as expected from their deshielded location (note that there are three structurally different hydrogens on the ring). The two propyl groups are structurally equivalent ( ), and are free to rotate over the faces of the ring system (one above and one below). On average all the propyl hydrogens are shielded, with the innermost methylene being the most affected. The negative chemical shifts noted here indicate that the resonances occurs at a higher field than the TMS reference signal. A remarkable characteristic of annulenes is that antiaromatic 4n π-electron systems are anisotropic in the opposite sense as their aromatic counterparts. A dramatic illustration of this fact is provided by the dianion derivative of the above bridged annulene. This dianion, formed by the addition of two electrons, is a 16 π-electron (4n) system. In the NMR spectrum of the dianion, the ring hydrogens resonate at high field (they are shielded), and the hydrogens of the propyl group are all shifted downfield (deshielded). The innermost methylene protons (magenta) give an NMR signal at +22.2 ppm, and the signals from the adjacent methylene and methyl hydrogens also have unexpectedly large chemical shifts. Compounds in which two or more benzene rings are fused together include examples such as naphthalene, anthracene and phenanthrene, shown in the following diagram, present interesting insights into aromaticity and reactivity. The resonance stabilization of these compounds, calculated from heats of hydrogenation or combustion, is given beneath each structure. Unlike benzene, the structures of these compounds show measurable double bond localization, which is reflected in their increased reactivity both in substitution and addition reactions. However, the HNMR spectra of these aromatic hydrocarbons do not provide much insight into the distribution of their pi-electrons. As expected, naphthalene displays two equally intense signals at δ 7.46 & 7.83 ppm. Likewise, anthracene shows three signals, two equal intensity multiplets at δ 7.44 & 7.98 ppm and a signal half as intense at δ 8.4 ppm. Thus, the influence of double bond localization or competition between benzene and higher annulene stabilization cannot be discerned. The much larger C H fused benzene ring cycle, named "kekulene" by Heinz Staab and sometimes called "superbenzene" by others, serves to probe the relative importance of benzenoid versus annulenoid aromaticity. A generic structure of this remarkable compound is drawn on the left below, together with two representative Kekule contributing structures on its right. There are some 200 Kekule structures that can be drawn for kekulene, but these two canonical forms represent extremes in aromaticity. The central formula has two [4n+2] annulenes, an inner [18]annulene and an outer [30]annulene (colored pink and blue respectively). The formula on the right has six benzene rings (colored green) joined in a ring by meta bonds, and held in a planar configuration by six cis-double bond bridges. The coupled annulene contributor in the center has an energetically equivalent canonical form in which the single and double bonds making up the annulenes are exchanged. If these contributors dominate the aromatic character of kekulene, the 6 inside hydrogens should be shielded by the ring currents, and the 18 hydrogens on the periphery should be deshielded. Furthermore, the C:C bonds composing each annulene ring should have roughly equal lengths. If the benzene contributor on the right (and its equivalent Kekule form) dominate the aromaticity of kekulene, all the benzene hydrogens will be deshielded, and the six double bond links on the periphery will have bond lengths characteristic of fixed single and double bonds The extreme insolubility of kekulene made it difficult to grow suitable crystals for X-ray analysis or obtain solution NMR spectra. These problems were eventually solved by using high boiling solvents, the HNMR spectrum being taken at 150 to 200° C in deuterated tetrachlorobenzene solution. The experimental evidence demonstrates clearly that the hexa-benzene ring structure on the right most accurately represents kekulene. This evidence is shown below. The extremely low field resonance of the inside hydrogens is assigned from similar downfield shifts in model compounds. Hydrogen bonding of hydroxyl and amino groups not only causes large variations in the chemical shift of the proton of the hydrogen bond, but also influences its coupling with adjacent C-H groups. As shown on the right, the 60 MHz proton NMR spectrum of pure (neat) methanol exhibits two signals, as expected. At 30° C these signals are sharp singlets located at δ 3.35 and 4.80 ppm, the higher-field methyl signal (magenta) being three times as strong as the OH signal (orange) at lower field. When cooled to -45 ° C, the larger higher-field signal changes to a doublet (J = 5.2 Hz) having the same chemical shift. The smaller signal moves downfield to δ 5.5 ppm and splits into a quartet (J = 5.2 Hz). The relative intensities of the two groups of signals remains unchanged. This interesting change in the NMR spectrum, which is shown in the two spectra below, is due to increased stability of hydrogen bonded species at lower temperature. Since hydrogen bonding not only causes a resonance shift to lower field, but also decreases the rate of intermolecular proton exchange, the hydroxyl proton remains bonded to the alkoxy group for a sufficient time to exert its spin coupling influence.
Under routine conditions, rapid intermolecular exchange of the OH protons of alcohols often prevents their coupling with adjacent hydrogens from being observed. Intermediate rates of proton exchange lead to a broadening of the OH and coupled hydrogen signals, a characteristic that is useful in identifying these functions. Since traces of acid or base catalyze this hydrogen exchange, pure compounds and clean sample tubes must be used for experiments of the kind described here. Another way of increasing the concentration of hydrogen bonded methanol species is to change the solvent from chloroform-d to a solvent that is a stronger hydrogen bond acceptor. Examples of such solvents are given in the following table. In contrast to the neat methanol experiment described above, very dilute solutions are used for this study. Since chloroform is a poor hydrogen bond acceptor and the dilute solution reduces the concentration of methanol clusters, the hydroxyl proton of methanol generates a resonance signal at a much higher field than that observed for the pure alcohol. Indeed, the OH resonance signal from simple alcohols in dilute chloroform solution is normally found near δ 1.0 ppm. The exceptionally strong hydrogen bond acceptor quality of DMSO is demonstrated here by the large downfield shift of the methanol hydroxyl proton, compared with a slight upfield shift of the methyl signal. The expected spin coupling patterns shown above are also observed in this solvent. Although acetone and acetonitrile are better hydrogen-bond acceptors than chloroform, they are not as effective as DMSO.
The solvent effect shown above suggests a useful diagnostic procedure for characterizing the OH resonance signals from alcohol samples. For example, a solution of ethanol in chloroform-d displays the spectrum shown on the left below, especially if traces of HCl are present (otherwise broadening of the OH and CH signals occurs). Note that the chemical shift of the OH signal (red) is less than that of the methylene group (blue), and no coupling of the OH proton is apparent. The vicinal coupling (J = 7 Hz) of the methyl and methylene hydrogens is typical of ethyl groups. In DMSO-d solution small changes of chemical shift are seen for the methyl and methylene group hydrogens, but a dramatic downfield shift of the hydroxyl signal takes place because of hydrogen bonding. Coupling of the OH proton to the adjacent methylene group is evident, and both the coupling constants can be measured. Because the coupling constants are different, the methylene signal pattern is an overlapping doublet of quartets (eight distinct lines) rather than a quintet. Note that residual hydrogens in the give a small broad signal near δ 2.5 ppm. For many alcohols in dilute chloroform-d solution, the hydroxyl resonance signal is often broad and obscured by other signals in the δ 1.5 to 3.0 region. The simple technique of using DMSO-d as a solvent, not only shifts this signal to a lower field, but permits 1°-, 2 °- & 3 °-alcohols to be distinguished. Thus, the hydroxyl proton of 2-propanol generates a doublet at δ 4.35 ppm, and the corresponding signal from 2-methyl-2-propanol is a singlet at δ 4.2 ppm. The more acidic OH protons of phenols are similarly shifted – from δ 4 to 7 in chloroform-d to δ 8.5 to 9.5 in DMSO-d . In the spectrum of 1,1-dichloroethane shown on the right, it is clear that the three methyl hydrogens (red) are coupled with the single methyne hydrogen (orange) in a manner that causes the former to appear as a doublet and the latter as a quartet. The light gray arrow points to the unperturbed chemical shift location for each proton set. Full Spectrum of 1,1-dichloroethane 2.06 ppm Signal Explained 5.89 ppm Signal Explained The statistical distribution of spins within each set explains both the n+1 rule and the relative intensities of the lines within a splitting pattern. The action of a single neighboring proton is easily deduced from the fact that it must have one of two possible spins. Interaction of these two spin states with the nuclei under observation leads to a doublet located at the expected chemical shift. The corresponding action of the three protons of the methyl group requires a more detailed analysis. In the display of this interaction four possible arrays of their spins are shown. The mixed spin states are three times as possible as the all +1/2 or all 1/2 collection. Consequently, we expect four signals, two above the chemical shift and two below it. This spin analysis also suggests that the intensity ratio of these signals will be 1:3:3:1. The line separations in splitting patterns are measured in Hz, and are characteristic of the efficiency of the spin interaction; they are referred to as coupling constants (symbol J). In the above example, the common coupling constant is 6.0 Hz. A simple way of estimating the relative intensities of the lines in a first-order coupling pattern is shown on the right. This array of numbers is known as Pascal's triangle, and is easily extended to predict higher multiplicities. The number appearing at any given site is the sum of the numbers linked to it from above by the light blue lines. Thus, the central number of the five quintet values is 3 + 3 = 6. Of course, a complete analysis of the spin distributions, as shown for the case of 1,1-dichloroethane above, leads to the same relative intensities. , and reflect the unique spin interaction characteristics of coupled sets of nuclei in a specific structure. As noted earlier, coupling constants may vary from a fraction of a Hz to nearly 20 Hz, important factors being the nature and spatial orientation of the bonds joining the coupled nuclei. In simple, freely rotating alkane units such as CH CH X or YCH CH X the coupling constant reflects an average of all significant conformers, and usually lies in a range of 6 to 8 Hz. This conformational mobility may be restricted by incorporating the carbon atoms in a rigid ring, and in this way the influence of the dihedral orientation of the coupled hydrogens may be studied. The structures of cis and trans-4-tert-butyl-1-chlorocyclohexane, shown above, illustrate how the coupling constant changes with the dihedral angle (φ) between coupled hydrogens. The inductive effect of chlorine shifts the resonance frequency of the red colored hydrogen to a lower field (δ ca. 4.0), allowing it to be studied apart from the other hydrogens in the molecule. The preferred equatorial orientation of the large tert-butyl group holds the six-membered ring in the chair conformation depicted in the drawing. In the trans isomer this fixes the red hydrogen in an axial orientation; whereas for the cis isomer it is equatorial. The listed values for the dihedral angles and the corresponding coupling constants suggest a relationship, which has been confirmed and clarified by numerous experiments. This relationship is expressed by the equation shown below. Geminal couplings are most commonly observed in cyclic structures, but are also evident when methylene groups have diastereomeric hydrogens. We have noted that rapidly exchanging hydroxyl hydrogens are not spin-coupled to adjacent C-H groups. The reason for this should be clear. As each exchange occurs, there will be an equal chance of the new proton having a +1/2 or a 1/2 spin (remember that the overall populations of the two spin states are nearly identical). Over time, therefore, the hydroxyl hydrogen behaves as though it is rapidly changing its spin, and the adjacent nuclei see only a zero spin average from it. If we could cause other protons in a molecule to undergo a similar spin averaging, their spin-coupling influence on adjacent nuclei would cease. Such NMR experiments are possible, and are called .
When a given set of nuclei is irradiated with strong rf energy at its characteristic Larmor frequency, spin saturation and rapid interconversion of the spin states occurs. Neighboring nuclei with different Larmor frequencies are no longer influenced by specific long-lived spins, so spin-spin signal splitting of the neighbors vanishes. The following spectrum of 1-nitropropane may be used to illustrate this technique. The three distinct sets of hydrogens in this molecule generate three resonance signals (two triplets and a broad sextet). A carefully tuned decoupling signal may be broadcast into the sample while the remaining spectrum is scanned. The region of the decoupling signal is obscured, but resonance signals more than 60 Hz away may still be seen. . Now methyl hydrogens have a smaller chemical shift than methylene hydrogens, so methyl groups (colored black here) can usually be distinguished. However, the chemical shifts of the different methylene groups (blue, red & green) are so similar that many NMR spectrometers will not resolve them. Consequently, a 90 MHz proton spectrum of octane shows a distorted triplet at δ 0.9 ppm, produced by the six methyl protons, and a strong broad singlet at δ 1.2 ppm coming from all twelve methylene protons. A similar failure to resolve structurally different hydrogen atoms occurs in the case of alkyl substituted benzene rings. The chemical shift difference between ortho, meta and para hydrogens in such compounds is often so small that they are seen as a single resonance signal in an NMR spectrum. The 90 MHz spectrum of benzyl alcohol in chloroform-d solution provides an instructive example, shown below. A broad strong signal at δ 7.24 ppm is characteristic of the aromatic protons on alkylbenzenes. Since the chemical shifts of these hydrogens are nearly identical, no spin coupling is observed. If the magnetic field strength is increased to 400 Mz (lower spectrum) the aromatic protons are more dispersed (orange, magenta and green signals), and the spin coupling of adjacent hydrogens (J = 7.6 Hz) causes overlap of the signals (gray shaded enlargement). Anisole, an isomer of benzyl alcohol, has a more dispersed set of aromatic signals, thanks to the electron donating influence of the methoxy substituent. The 90 MHz spectrum of anisole shows this greater dispersion, but the spin coupling of adjacent hydrogens still results in signal overlap. The 400 Mz spectrum at the bottom illustrates the greater dispersion of the chemical shifts, and since the coupling constants remain unchanged, the splitting patterns no longer overlap. In all these examples a very small meta-hydrogen coupling has been ignored. Not all simple compounds have simple proton NMR spectra. The following example not only illustrates this point, but also demonstrates how a careful structural analysis can rationalize an initially complex spectrum. The 100 MHz H NMR spectrum of a C H ClO compound is initially displayed. This spectrum is obviously complex and not easily interpreted, except for concluding that no olefinic C-H protons are present. With a higher field spectrum it is clear that each of the five hydrogen atoms in the molecule is structurally unique, and is producing a separate signal. Also, it is clear there is considerable spin coupling of all the hydrogens. To see the coupling patterns more clearly it is necessary to expand and enhance the spectrum in these regions. For purposes of our demonstration, this can be done by clicking on any one of the signal multiplets. ing in an open area should return the original 500 MHz display. In some of the expanded displays two adjacent groups of signals are shown. Once an enlarged pattern is displayed, the line separations in Hz can be measured (remember that for a 500 MHz spectrum 1 ppm is 500 Hz). The middle signal at 3.2 ppm is the most complex, and overlap of some multiplet lines has occurred. This spectrum has several interesting features. First, hydrogens A, B & C are clearly different, and are spin-coupled to each other. Hydrogens B & C are , whereas A is oriented to B & C in a vicinal manner. Since J and J are similar, the H signal is a broad triplet. Although hydrogens D & E might seem identical at first glance, they are , and should therefore have different chemical shifts. The DE geminal coupling constant is 11.7 Hz, so each of these hydrogens appears as a doublet of doublets.
The splitting of the H signal is complex and not immediately obvious. The diagram shows the consequences of the four operating coupling constants. Broad band decoupling of the hydrogen atoms in a molecule was an essential operation for obtaining simple (single line) carbon NMR spectra. The chemical shifts of the carbon signals provide useful information, but it would also be very helpful to know how many hydrogen atoms are bonded to each carbon. Aside from the fact that carbons having no bonded hydrogens generally give weak resonance signals, this information is not present in a completely decoupled spectrum. Clever methods of retaining the hydrogen information while still enjoying the benefits of proton decoupling have been devised. The techniques involved are beyond the scope of this discussion, but the overall results can still be appreciated. The C NMR spectrum of camphor shown below will serve as an illustration. It will be helpful to view an expanded section of this spectrum from δ 0.0 to 50.0 ppm, and this will be presented in the spectrum. The two lowest field signals are missing in the expanded display. Even though the expanded display now shows the distinct carbon signals clearly, the origin of each is ambiguous. An early method of regaining coupling information was by . In this approach a weaker and more focused proton decoupling frequency is applied as the carbon spectrum is acquired. Vestiges of the C-H coupling remain in the carbon signals, but the apparent coupling constants are greatly reduced. View the spectrum. The results of such an experiment will be displayed. Notice that all the methyl groups are quartets (three coupled hydrogens), the methylene groups are triplets and methine carbons are doublets. Overlap of two quartets near δ 19 ppm and the doublet and triplet near δ 43 ppm are complicating factors. A better way for classifying the carbon signals is by a technique called (insensitive nuclear enhancement by polarization transfer). This method takes advantage of the influence of hydrogen on C relaxation times, and can be applied in several modes. One of the most common applications of INEPT separates the signals of methyl and methine carbons from those of methylene carbons by their sign. Carbons having no hydrogen substituents have a zero signal. | 29,879 | 2,926 |
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The crystal lattices we have described represent an idealized, simplified system that can be used to understand many of the important principles governing the behavior of solids. In contrast, real crystals contain large numbers of defects (typically more than 10 per milligram), ranging from variable amounts of impurities to missing or misplaced atoms or ions. These defects occur for three main reasons: In this section, we discuss how defects determine some of the properties of solids. We begin with solids that consist of neutral atoms, specifically metals, and then turn to ionic compounds. Metals can have various types of defects. A point defect is any defect that involves only a single particle (a lattice point) or sometimes a very small set of points. A line defect is restricted to a row of lattice points, and a plane defect involves an entire plane of lattice points in a crystal. A vacancy occurs where an atom is missing from the normal crystalline array; it constitutes a tiny void in the middle of a solid (Figure \(\Page {1}\)). We focus primarily on point and plane defects in our discussion because they are encountered most frequently. Impurities can be classified as interstitial or substitutional. An interstitial impurity is usually a smaller atom (typically about 45% smaller than the host) that can fit into the octahedral or tetrahedral holes in the metal lattice (Figure \(\Page {1}\)). Steels consist of iron with carbon atoms added as interstitial impurities (Table \(\Page {1}\)). The inclusion of one or more transition metals or semimetals can improve the corrosion resistance of steel. In contrast, a substitutional impurity is a different atom of about the same size that simply replaces one of the atoms that compose the host lattice (Figure \(\Page {1}\)). Substitutional impurities are usually chemically similar to the substance that constitutes the bulk of the sample, and they generally have atomic radii that are within about 15% of the radius of the host. For example, strontium and calcium are chemically similar and have similar radii, and as a result, strontium is a common impurity in crystalline calcium, with the Sr atoms randomly occupying sites normally occupied by Ca. Interstitial impurities are smaller atoms than the host atom, whereas substitutional impurities are usually chemically similar and are similar in size to the host atom. Inserting an extra plane of atoms into a crystal lattice produces an edge dislocation. A familiar example of an edge dislocation occurs when an ear of corn contains an extra row of kernels between the other rows (Figure \(\Page {2}\)). An edge dislocation in a crystal causes the planes of atoms in the lattice to deform where the extra plane of atoms begins (Figure \(\Page {2}\)). The edge dislocation frequently determines whether the entire solid will deform and fail under stress. Deformation occurs when a dislocation moves through a crystal. To illustrate the process, suppose you have a heavy rug that is lying a few inches off-center on a nonskid pad. To move the rug to its proper place, you could pick up one end and pull it. Because of the large area of contact between the rug and the pad, however, they will probably move as a unit. Alternatively, you could pick up the rug and try to set it back down exactly where you want it, but that requires a great deal of effort (and probably at least one extra person). An easier solution is to create a small wrinkle at one end of the rug (an edge dislocation) and gradually push the wrinkle across, resulting in a net movement of the rug as a whole (part (a) in Figure \(\Page {3}\)). Moving the wrinkle requires only a small amount of energy because only a small part of the rug is actually moving at any one time. Similarly, in a solid, the contacts between layers are broken in only one place at a time, which facilitates the deformation process. If the rug we have just described has a second wrinkle at a different angle, however, it is very difficult to move the first one where the two wrinkles intersect (part (b) in Figure \(\Page {3}\)); this process is called pinning. Similarly, intersecting dislocations in a solid prevent them from moving, thereby increasing the mechanical strength of the material. In fact, one of the major goals of materials science is to find ways to pin dislocations to strengthen or harden a material. Pinning can also be achieved by introducing selected impurities in appropriate amounts. Substitutional impurities that are a mismatch in size to the host prevent dislocations from migrating smoothly along a plane. Generally, the higher the concentration of impurities, the more effectively they block migration, and the stronger the material. For example, bronze, which contains about 20% tin and 80% copper by mass, produces a much harder and sharper weapon than does either pure tin or pure copper. Similarly, pure gold is too soft to make durable jewelry, so most gold jewelry contains 75% (18 carat) or 58% (14 carat) gold by mass, with the remainder consisting of copper, silver, or both. If an interstitial impurity forms polar covalent bonds to the host atoms, the layers are prevented from sliding past one another, even when only a small amount of the impurity is present. For example, because iron forms polar covalent bonds to carbon, the strongest steels need to contain only about 1% carbon by mass to substantially increase their strength (Table \(\Page {1}\)). Most materials are polycrystalline, which means they consist of many microscopic individual crystals called grains that are randomly oriented with respect to one another. The place where two grains intersect is called a grain boundary. The movement of a deformation through a solid tends to stop at a grain boundary. Consequently, controlling the grain size in solids is critical for obtaining desirable mechanical properties; fine-grained materials are usually much stronger than coarse-grained ones. Work hardening is the introduction of a dense network of dislocations throughout a solid, which makes it very tough and hard. If all the defects in a single 1 cm3 sample of a work-hardened material were laid end to end, their total length could be 106 km! The legendary blades of the Japanese and Moorish swordsmiths owed much of their strength to repeated work hardening of the steel. As the density of defects increases, however, the metal becomes more brittle (less malleable). For example, bending a paper clip back and forth several times increases its brittleness from work hardening and causes the wire to break. The compound NiTi, popularly known as “memory metal” or nitinol (nickel–titanium Naval Ordinance Laboratory, after the site where it was first prepared), illustrates the importance of deformations. If a straight piece of NiTi wire is wound into a spiral, it will remain in the spiral shape indefinitely, unless it is warmed to 50°C–60°C, at which point it will spontaneously straighten out again. The chemistry behind the temperature-induced change in shape is moderately complex, but for our purposes it is sufficient to know that NiTi can exist in two different solid phases. The high-temperature phase has the cubic cesium chloride structure, in which a Ti atom is embedded in the center of a cube of Ni atoms (or vice versa). The low-temperature phase has a related but kinked structure, in which one of the angles of the unit cell is no longer 90°. Bending an object made of the low-temperature (kinked) phase creates defects that change the pattern of kinks within the structure. If the object is heated to a temperature greater than about 50°C, the material undergoes a transition to the cubic high-temperature phase, causing the object to return to its original shape. The shape of the object above 50°C is controlled by a complex set of defects and dislocations that can be relaxed or changed only by the thermal motion of the atoms. Memory metal has many other practical applications, such as its use in temperature-sensitive springs that open and close valves in the automatic transmissions of cars. Because NiTi can also undergo pressure- or tension-induced phase transitions, it is used to make wires for straightening teeth in orthodontic braces and in surgical staples that change shape at body temperature to hold broken bones together. Another flexible, fatigue-resistant alloy composed of titanium and nickel is Flexon. Originally discovered by metallurgists who were creating titanium-based alloys for use in missile heat shields, Flexon is now used as a durable, corrosion-resistant frame for glasses, among other uses. Because steels with at least 4% chromium are much more corrosion resistant than iron, they are collectively sold as “stainless steel.” Referring to the composition of stainless steel in Table \(\Page {1}\) and, if needed, the atomic radii in Figure 7.7 , predict which type of impurity is represented by each element in stainless steel, excluding iron, that are present in at least 0.05% by mass. : composition of stainless steel and atomic radii : type of impurity : Using the data in Table \(\Page {1}\) and the atomic radii in Figure 7.7, determine whether the impurities listed are similar in size to an iron atom. Then determine whether each impurity is chemically similar to Fe. If similar in both size and chemistry, the impurity is likely to be a substitutional impurity. If not, it is likely to be an interstitial impurity. : According to Table \(\Page {1}\), stainless steel typically contains about 1% carbon, 1%–5% manganese, 0.05% phosphorus, 1%–3% silicon, 5%–10% nickel, and 15%–20% chromium. The three transition elements (Mn, Ni, and Cr) lie near Fe in the periodic table, so they should be similar to Fe in chemical properties and atomic size (atomic radius = 125 pm). Hence they almost certainly will substitute for iron in the Fe lattice. Carbon is a second-period element that is nonmetallic and much smaller (atomic radius = 77 pm) than iron. Carbon will therefore tend to occupy interstitial sites in the iron lattice. Phosphorus and silicon are chemically quite different from iron (phosphorus is a nonmetal, and silicon is a semimetal), even though they are similar in size (atomic radii of 106 and 111 pm, respectively). Thus they are unlikely to be substitutional impurities in the iron lattice or fit into interstitial sites, but they could aggregate into layers that would constitute plane defects. Consider nitrogen, vanadium, zirconium, and uranium impurities in a sample of titanium metal. Which is most likely to form an interstitial impurity? a substitutional impurity? : nitrogen; vanadium All the defects and impurities described for metals are seen in ionic and molecular compounds as well. Because ionic compounds contain both cations and anions rather than only neutral atoms, however, they exhibit additional types of defects that are not possible in metals. The most straightforward variant is a substitutional impurity in which a cation or an anion is replaced by another of similar charge and size. For example, Br can substitute for Cl , so tiny amounts of Br are usually present in a chloride salt such as CaCl or BaCl . If the substitutional impurity and the host have different charges, however, the situation becomes more complicated. Suppose, for example, that Sr (ionic radius = 118 pm) substitutes for K (ionic radius = 138 pm) in KCl. Because the ions are approximately the same size, Sr should fit nicely into the face-centered cubic (fcc) lattice of KCl. The difference in charge, however, must somehow be compensated for so that electrical neutrality is preserved. The simplest way is for a second K ion to be lost elsewhere in the crystal, producing a vacancy. Thus substitution of K+ by Sr in KCl results in the introduction of two defects: a site in which an Sr ion occupies a K position and a vacant cation site. Substitutional impurities whose charges do not match the host’s are often introduced intentionally to produce compounds with specific properties. Virtually all the colored gems used in jewelry are due to substitutional impurities in simple oxide structures. For example, α-Al O , a hard white solid called corundum that is used as an abrasive in fine sandpaper, is the primary component, or matrix, of a wide variety of gems. Because many trivalent transition metal ions have ionic radii only a little larger than the radius of Al (ionic radius = 53.5 pm), they can replace Al in the octahedral holes of the oxide lattice. Substituting small amounts of Cr3+ ions (ionic radius = 75 pm) for Al gives the deep red color of ruby, and a mixture of impurities (Fe , Fe , and Ti ) gives the deep blue of sapphire. True amethyst contains small amounts of Fe in an SiO (quartz) matrix. The same metal ion substituted into different mineral lattices can produce very different colors. For example, Fe ions are responsible for the yellow color of topaz and the violet color of amethyst. The distinct environments cause differences in d orbital energies, enabling the Fe ions to absorb light of different frequencies, a topic discussed later. Substitutional impurities are also observed in molecular crystals if the structure of the impurity is similar to the host, and they can have major effects on the properties of the crystal. Pure anthracene, for example, is an electrical conductor, but the transfer of electrons through the molecule is much slower if the anthracene crystal contains even very small amounts of tetracene despite their strong structural similarities. If a cation or an anion is simply missing, leaving a vacant site in an ionic crystal, then for the crystal to be electrically neutral, there must be a corresponding vacancy of the ion with the opposite charge somewhere in the crystal. In compounds such as KCl, the charges are equal but opposite, so one anion vacancy is sufficient to compensate for each cation vacancy. In compounds such as CaCl , however, two Cl anion sites must be vacant to compensate for each missing Ca cation. These pairs (or sets) of vacancies are called Schottky defects and are particularly common in simple alkali metal halides such as KCl (part (a) in Figure \(\Page {4}\)). Many microwave diodes, which are devices that allow a current to flow in a single direction, are composed of materials with Schottky defects. Occasionally one of the ions in an ionic lattice is simply in the wrong position. An example of this phenomenon, called a Frenkel defect, is a cation that occupies a tetrahedral hole rather than an octahedral hole in the anion lattice (part (b) in Figure \(\Page {4}\)). Frenkel defects are most common in salts that have a large anion and a relatively small cation. To preserve electrical neutrality, one of the normal cation sites, usually octahedral, must be vacant. Frenkel defects are particularly common in the silver halides AgCl, AgBr, and AgI, which combine a rather small cation (Ag , ionic radius = 115 pm) with large, polarizable anions. Certain more complex salts with a second cation in addition to Ag and Br or I have so many Ag+ ions in tetrahedral holes that they are good electrical conductors in the solid state; hence they are called solid electrolytes. Since most ionic compounds do not conduct electricity in the solid state, however they do conduct electricity when molten or dissolved in a solvent that separates the ions, allowing them to migrate in response to an applied electric field.) In response to an applied voltage, the cations in solid electrolytes can diffuse rapidly through the lattice via octahedral holes, creating Frenkel defects as the cations migrate. Sodium–sulfur batteries use a solid Al O electrolyte with small amounts of solid Na O. Because the electrolyte cannot leak, it cannot cause corrosion, which gives a battery that uses a solid electrolyte a significant advantage over one with a liquid electrolyte. In a sample of NaCl, one of every 10,000 sites normally occupied by Na is occupied instead by Ca . Assuming that all of the Cl sites are fully occupied, what is the stoichiometry of the sample? : ionic solid and number and type of defect : stoichiometry : : Pure NaCl has a 1:1 ratio of Na and Cl ions arranged in an fcc lattice (the sodium chloride structure). If all the anion sites are occupied by Cl , the negative charge is −1.00 per formula unit. If 0.01% of the Na sites are occupied by Ca ions, the cation stoichiometry is Na0.99Ca0.01. This results in a positive charge of (0.99)(+1) + (0.01)(+2) = +1.01 per formula unit, for a net charge in the crystal of +1.01 + (−1.00) = +0.01 per formula unit. Because the overall charge is greater than 0, this stoichiometry must be incorrect. The most plausible way for the solid to adjust its composition to become electrically neutral is for some of the Na sites to be vacant. If one Na site is vacant for each site that has a Ca cation, then the cation stoichiometry is Na0.98Ca0.01. This results in a positive charge of (0.98)(+1) + (0.01)(+2) = +1.00 per formula unit, which exactly neutralizes the negative charge. The stoichiometry of the solid is thus Na Ca Cl . In a sample of MgO that has the sodium chloride structure, 0.02% of the Mg ions are replaced by Na ions. Assuming that all of the cation sites are fully occupied, what is the stoichiometry of the sample? : If the formula of the compound is Mg Na O , then x must equal 0.01 to preserve electrical neutrality. The formula is thus Mg Na O . The , states that chemical compounds contain fixed integral ratios of atoms. In fact, nonstoichiometric compounds contain large numbers of defects, usually vacancies, which give rise to stoichiometries that can depart significantly from simple integral ratios without affecting the fundamental structure of the crystal. Nonstoichiometric compounds frequently consist of transition metals, lanthanides, and actinides, with polarizable anions such as oxide (O ) and sulfide (S ). Some common examples are listed in Table \(\Page {2}\), along with their basic structure type. These compounds are nonstoichiometric because their constituent metals can exist in multiple oxidation states in the solid, which in combination preserve electrical neutrality. One example is iron(II) oxide (ferrous oxide), which produces the black color in clays and is used as an abrasive. Its stoichiometry is not FeO because it always contains less than 1.00 Fe per O (typically 0.90–0.95). This is possible because Fe can exist in both the +2 and +3 oxidation states. Thus the difference in charge created by a vacant Fe site can be balanced by two Fe sites that have Fe ions instead [+2 vacancy = (3 − 2) + (3 − 2)]. The crystal lattice is able to accommodate this relatively high fraction of substitutions and vacancies with no significant change in structure. Because a crystal must be electrically neutral, any defect that affects the number or charge of the cations must be compensated by a corresponding defect in the number or charge of the anions. Real crystals contain large numbers of defects. Defects may affect only a single point in the lattice (a point defect), a row of lattice points (a line defect), or a plane of atoms (a plane defect). A point defect can be an atom missing from a site in the crystal (a vacancy) or an impurity atom that occupies either a normal lattice site (a substitutional impurity) or a hole in the lattice between atoms (an interstitial impurity). In an edge dislocation, an extra plane of atoms is inserted into part of the crystal lattice. Multiple defects can be introduced into materials so that the presence of one defect prevents the motion of another, in a process called pinning. Because defect motion tends to stop at grain boundaries, controlling the size of the grains in a material controls its mechanical properties. In addition, a process called work hardening introduces defects to toughen metals. Schottky defects are a coupled pair of vacancies—one cation and one anion—that maintains electrical neutrality. A Frenkel defect is an ion that occupies an incorrect site in the lattice. Cations in such compounds are often able to move rapidly from one site in the crystal to another, resulting in high electrical conductivity in the solid material. Such compounds are called solid electrolytes. Nonstoichiometric compounds have variable stoichiometries over a given range with no dramatic change in crystal structure. This behavior is due to a large number of vacancies or substitutions of one ion by another ion with a different charge. 1. How are defects and impurities in a solid related? Can a pure, crystalline compound be free of defects? How can a substitutional impurity produce a vacancy? 2. Why does applying a mechanical stress to a covalent solid cause it to fracture? Use an atomic level description to explain why a metal is ductile under conditions that cause a covalent solid to fracture. 3. How does work hardening increase the strength of a metal? How does work hardening affect the physical properties of a metal? 4. Work-hardened metals and covalent solids such as diamonds are both susceptible to cracking when stressed. Explain how such different materials can both exhibit this property. 5. Suppose you want to produce a ductile material with improved properties. Would impurity atoms of similar or dissimilar atomic size be better at maintaining the ductility of a metal? Why? How would introducing an impurity that forms polar covalent bonds with the metal atoms affect the ductility of the metal? Explain your reasoning. 6. Substitutional impurities are often used to tune the properties of material. Why are substitutional impurities generally more effective at high concentrations, whereas interstitial impurities are usually effective at low concentrations? 7. If an O ion (ionic radius = 132 pm) is substituted for an F ion (ionic radius = 133 pm) in an ionic crystal, what structural changes in the ionic lattice will result? 8. How will the introduction of a metal ion with a different charge as an impurity induce the formation of oxygen vacancies in an ionic metal-oxide crystal? 9. Many nonstoichiometric compounds are transition metal compounds. How can such compounds exist, given that their nonintegral cation:anion ratios apparently contradict one of the basic tenets of Dalton’s atomic theory of matter? 10. If you wanted to induce the formation of oxygen vacancies in an ionic crystal, which would you introduce as substitutional impurities—cations with a higher positive charge or a lower positive charge than the cations in the parent structure? Explain your reasoning. 5. Impurity atoms of similar size and with similar chemical properties would be most likely to maintain the ductility of the metal, because they are unlikely to have a large effect on the ease with which one layer of atoms can move past another under mechanical stress. Larger impurity atoms are likely to form “bumps” or kinks that will make it harder for layers of atoms to move across one another. Interstitial atoms that form polar covalent bonds with the metal atoms tend to occupy spaces between the layers; they act as a “glue” that holds layers of metal atoms together, which greatly decreases the ductility. 7. Since O and F are both very similar in size, substitution is possible without disruption of the ionic packing. The difference in charge, however, requires the formation of a vacancy on another F site to maintain charge neutrality. 9. Most transition metals form at least two cations that differ by only one electron. Consequently, nonstoichiometric compounds containing transition metals can maintain electrical neutrality by gaining electrons to compensate for the absence of anions or the presence of additional metal ions. Conversely, such compounds can lose electrons to compensate for the presence of additional anions or the absence of metal ions. In both cases, the positive charge on the transition metal is adjusted to maintain electrical neutrality. 1. The ionic radius of K is 133 pm, whereas that of Na is 98 pm. Do you expect K to be a common substitutional impurity in compounds containing Na ? Why or why not? 2. Given Cs (262 pm), Tl (171 pm), and B (88 pm) with their noted atomic radii, which atom is most likely to act as an interstitial impurity in an Sn lattice (Sn atomic radius = 141 pm)? Why? 3. After aluminum, iron is the second most abundant metal in Earth’s crust. The silvery-white, ductile metal has a body-centered cubic (bcc) unit cell with an edge length of 286.65 pm. a. Use this information to calculate the density of iron. b. What would the density of iron be if 0.15% of the iron sites were vacant? c. How does the mass of 1.00 cm of iron without defects compare with the mass of 1.00 cm of iron with 0.15% vacancies? 4. Certain ceramic materials are good electrical conductors due to high mobility of oxide ions resulting from the presence of oxygen vacancies. Zirconia (ZrO ) can be doped with yttrium by adding Y O . If 0.35 g of Y O can be incorporated into 25.0 g of ZrO while maintaining the zirconia structure, what is the percentage of oxygen vacancies in the structure? 5. Which of the following ions is most effective at inducing an O vacancy in crystal of CaO? The ionic radii are O , 132 pm; Ca , 100 pm; Sr , 127 pm; F , 133 pm; La , 104 pm; and K , 133 pm. Explain your reasoning. 1. No. The potassium is much larger than the sodium ion. 3. a. 7.8744 g/cm b. 7.86 g/cm c. Without defects, the mass is 0.15% greater. 5. The lower charge of K makes it the best candidate for inducing an oxide vacancy, even though its ionic radius is substantially larger than that of Ca . Substituting two K ions for two Ca ions will decrease the total positive charge by two, and an oxide vacancy will maintain electrical neutrality. For example, if 10% of the Ca ions are replaced by K , we can represent the change as going from Ca O to K Ca O , which has a net charge of +2. Loss of one oxide ion would give a composition of K Ca O , which is electrically neutral. | 26,122 | 2,927 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO15._The_Anomeric_Center |
When a sugar cyclizes via donation of a hydroxy lone pair to the carbonyl, it forms a " ". We have already seen that hemiacetals are unstable with respect to further substitution. Pi donation from an oxygen in the hemiacetal can displace the other oxygen. A second nucleophile can then donate to the pseudo-carbonyl that results. The hemiacetal position in the sugar is called the anomeric center. The anomeric center is special for two reasons. First, as you have already seen, the anomeric center is a chiral center. This new center can form with either of two configurations. The sugar, which is already chiral, can become either of two diastereomers when it cyclizes. Second, the anomeric center is a site of enhanced reactivity in the sugar, in terms of substitution of the carbonyl. Anomeric reactivity involves pi donation from one oxygen to push off the other oxygen. This mode of reaction should be familiar. The C=O+ unit that forms resembles a carbonyl. Furthermore, the positive charge on the oxygen brings to mind an activated carbonyl. This position is especially attractive for nucleophiles. It isn't an accident that in many sugar-containing biomolecules, substituents are found at the anomeric center. For example, nucleosides sub-units found in DNA and RNA are all substituted at this position. A number of other biological agents contain this motif as well. Substitution at the anomeric position can be accelerated if a proton source is available. Show why. Show the stereochemical results of substitution at the anomeric center of glucose with methanol. Nucleotides, which form DNA and RNA chains, are just like nucleosides, but they all have a phosphate at a specific position. Explain what is special about this position that could make it form a phosphate more easily than the other hydroxyl sites. Show the products of the following substitution reactions. Although sugars contain a number of chiral centers, characterizing them by polarimetry is complicated. Optical rotation measurements are done in solution, in a polarimetry cell. Polarimetry is always a little bit complicated, because the optical rotation varies with the concentration of the solution and the length of the polarimetry cell. When a pure enantiomer of a sugar such as alpha-D-glucose is dissolved, usually in water, its optical rotation also varies with time. In other words, the reading keeps changing, eventually settling out far from the initial value. That means care must be taken in measuring this information, and in interpreting the data. Show why alpha-D-glucose would exhibit a changing optical rotation value after being dissolved. Suppose a one gram sample of alpha-D-glucopyranose is dissolved in 1 mL of water and its optical rotation is measured in a a 1 dm cell. Initially, a value of 100 degrees is recorded. After several hours, the value has stopped changing, and is 48 degrees. The experiment is repeated with beta-D-glucopyranose. In water, alpha-D-glucopyranose predominates over the beta form in solution. Explain why. In less polar solvents (compared to water) such as dichloromethane, beta-D-glucopyranose predominates over the alpha form. This phenomenon is thought to result from the influence of lone pair-lone pair repulsion. , | 3,265 | 2,928 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/06%3A_Properties_of_Gases/6.06%3A_Real_Gases_and_Critical_Phenomena |
Make sure you thoroughly understand the following essential concepts that have been presented above. The "ideal gas laws" as we know them do a remarkably good job of describing the behavior of a huge number chemically diverse substances as they exist in the gaseous state under ordinary environmental conditions, roughly around 1 atm pressure and a temperature of 300 K. But when the temperature is reduced, or the pressure is raised, the relation PV = constant that defines the ideal gas begins to break down, and its properties become unpredictable; eventually the gas condenses into a liquid. Why is this important? It is of obvious interest to a chemical engineer who needs to predict the properties of the gases involved in a chemical reaction carried out at several hundred atmospheres pressure. This is especially so when we consider that some of the basic tenets of the ideal gas model have to be abandoned in order to explain such properties as Even so, many of the common laws such as Boyle's and Charles' continue to describe these gases quite well even under conditions where these phenomena are evident. Under ordinary environmental conditions (moderate pressures and above 0°C), the isotherms of substances we normally think of as gases don't appear to differ very greatly from the hyperbolic form \[\dfrac{PV}{RT} = \text{constant} \label{6.6.1}\] However, over a wider range of conditions, things begin to get more complicated. Thus isopentane (Figure \(\Page {1}\)) behaves in a reasonably ideal manner above 210 K, but below this temperature the isotherms become somewhat distorted, and at 185 K and below they cease to be continuous, showing peculiar horizontal segments in which reducing the volume does not change the pressure. Within this region, any attempt to compress the gas simply "squeezes" some of it into the liquid state whose greater density exactly compensates for the smaller volume, thus maintaining the pressure at a constant value. It turns out that real gases eventually begin to follow their own unique equations of state, and ultimately even cease to be gases. In this unit we will see why this occurs, what the consequences are, and how we might modify the ideal gas equation of state to extend its usefulness over a greater range of temperatures and pressures. According to , the product is a constant at any given temperature, so a plot of as a function of the pressure of an ideal gas yields a horizontal straight line. This implies that any increase in the pressure of the gas is exactly counteracted by a decrease in the volume as the molecules are crowded closer together. But we know that the molecules themselves are finite objects having volumes of their own, and this must place a lower limit on the volume into which they can be squeezed. So we must reformulate the ideal gas equation of state as a relation that is true only in the limiting case of zero pressure: \[\lim_{P \rightarrow 0} PV=nRT \label{6.6.2}\] So what happens when a real gas is subjected to a very high pressure? The outcome varies with both the molar mass of the gas and its temperature, but in general we can see the the effects of both repulsive and attractive intermolecular forces: The universal attractive force described above is known as the , or force. There may also be additional (and usually stronger) attractive forces related to charge imbalance in the molecule or to hydrogen bonding. These various attractive forces are often referred to collectively as . A plot of as a function of pressure is a very sensitive indicator of deviations from ideal behavior, since such a plot is just a horizontal line for an ideal gas. Figures \(\Page {2}\) and \(\Page {3}\) demonstrate how these plots vary with the nature of the gas and with temperature, respectively. , which generally increase with molecular weight, cause the product to decrease as higher pressures bring the molecules closer together and thus within the range of these attractive forces; the effect is to cause the volume to decrease more rapidly than it otherwise would. The repulsive forces always eventually win out. However, as the molecules begin to intrude on each others' territory, the stronger repulsive forces cause the curve to bend upward. The makes a big difference! At higher temperatures, increased thermal motions overcome the effects of intermolecular attractions which normally dominate at lower pressures (Figure \(\Page {3}\)). So all gases behave more ideally at higher temperatures. For any gas, there is a special temperature (the ) at which attractive and repulsive forces exactly balance each other at zero pressure. As you can see in this plot for methane, some of this balance does remain as the pressure is increased. How might we modify the ideal gas equation of state to take into account the effects of intermolecular interactions? The first and most well known answer to this question was offered by the Dutch scientist J.D. van der Waals (1837-1923) in 1873. The ideal gas model assumes that the gas molecules are merely points that occupy no volume; the " " term in the equation is the volume of the container and is independent of the nature of the gas. van der Waals recognized that the molecules themselves take up space that subtracts from the volume of the container (Figure \(\Page {4}\)), so that the “volume of the gas” in the ideal gas equation should be replaced by the term (\(V–b\)), in which \(b\) relates to the , typically of the order of 20-100 cm mol . The excluded volume surrounding any molecule defines the closest possible approach of any two molecules during collision. Note that the excluded volume is greater then the volume of the molecule, its radius being half again as great as that of a spherical molecule. The other effect that van der Waals needed to correct for are the intermolecular attractive forces. These are ignored in the ideal gas model, but in real gases they exert a small cohesive force between the molecules, thus helping to hold the gas together and reducing the pressure it exerts on the walls of the container. Because this pressure depends on both the frequency the intensity of collisions with the walls, the reduction in pressure is proportional to the of the number of molecules per volume of space, and thus for a fixed number of molecules such as one mole, the reduction in pressure is inversely proportional to the square of the volume of the gas. The smaller the volume, the closer are the molecules and the greater will be the effect. The van der Walls equation replaces the \(P\) term in the ideal gas equation with \(P + (a / V^2)\) in which the magnitude of the constant a increases with the strength of the intermolecular attractive forces. The complete van der Waals equation of state can be written as Although most students are not required to memorize this equation, you are expected to understand it and to explain the significance of the terms it contains. You should also understand that the van der Waals constants \(a\) and \(b\) must be determined empirically for every gas. This can be done by plotting the behavior of the gas and adjusting the values of \(a\) and \(b\) until the van der Waals equation results in an identical plot. The constant a is related in a simple way to the molecular radius; thus the determination of \(a\) constitutes an indirect measurement of an important microscopic quantity. The van der Waals equation is only one of many equations of state for real gases. More elaborate equations are required to describe the behavior of gases over wider pressure ranges. These generally take account of higher-order nonlinear attractive forces, and require the use of more empirical constants. Although we will make no use of them in this course, they are widely employed in chemical engineering work in which the behavior of gases at high pressures must be accurately predicted. The most striking feature of real gases is that they cease to remain gases as the temperature is lowered and the pressure is increased. Figure \(\Page {6}\) illustrates this behavior; as the volume is decreased, the lower-temperature isotherms suddenly change into straight lines. Under these conditions, the pressure remains constant as the volume is reduced. This can only mean that the gas is “disappearing" as we squeeze the system down to a smaller volume. In its place, we obtain a new state of matter, the liquid. In the green-shaded region, phases, liquid, and gas, are simultaneously present. Finally, at very small volume all the gas has disappeared and only the liquid phase remains. At this point the isotherms bend strongly upward, reflecting our common experience that a liquid is practically incompressible. To better understand this plot, look at the isotherm labeled . As the gas is compressed from to , the pressure rises in much the same way as Boyle's law predicts. Compression beyond , however, does not cause any rise in the pressure. What happens instead is that some of the gas condenses to a liquid. At , the substance is entirely in its liquid state. The very steep rise to corresponds to our ordinary experience that liquids have very low compressibilities. The range of volumes possible for the liquid diminishes as the critical temperature is approached. Liquid and gas can coexist only within the regions indicated by the green-shaded area in the diagram above. As the temperature and pressure rise, this region becomes more narrow, finally reaching zero width at the . The values of , and at this juncture are known as the , and . The isotherm that passes through the critical point is called the . Beyond this isotherm, the gas and liquids become indistinguishable; there is only a single fluid phase, sometimes referred to as a (Figure \(\Page {7}\)). At temperatures below 31°C (the ), CO acts somewhat like an ideal gas even at a rather high pressure ( ). Below 31°, an attempt to compress the gas to a smaller volume eventually causes condensation to begin. Thus at 21°C, at a pressure of about 62 atm ( ), the volume can be reduced from 200 cm to about 55 cm without any further rise in the pressure. Instead of the gas being compressed, it is replaced with the far more compact liquid as the gas is essentially being "squeezed" into its liquid phase. After all of the gas has disappeared ( ), the pressure rises very rapidly because now all that remains is an almost incompressible liquid. Above this isotherm ( ), CO exists only as a . What happens if you have some liquid carbon dioxide in a transparent cylinder at just under its of 62 atm, and you then compress it slightly? Nothing very dramatic until you notice that the meniscus has disappeared. By successively reducing and increasing the pressure, you can "turn the meniscus on and off". One intriguing consequence of the very limited bounds of the liquid state is that you could start with a gas at large volume and low temperature, raise the temperature, reduce the volume, and then reduce the temperature so as to arrive at the liquid region at the lower left, without ever passing through the two-phase region, and thus The state of matter, as the fluid above the critical point is often called, possesses the flow properties of a gas and the solvent properties of a liquid. The density of a supercritical fluid can be changed over a wide range by adjusting the pressure; this, in turn, changes its solubility, which can thus be optimized for a particular application. The picture at the right shows a commercial laboratory device used for carrying out chemical reactions under supercritical conditions. Supercritical carbon dioxide is widely used to dissolve the caffeine out of coffee beans and as a dry-cleaning solvent. Supercritical water has recently attracted interest as a medium for chemically decomposing dangerous environmental pollutants such as PCBs. Supercritical fluids are being increasingly employed as as substitutes for organic solvents (so-called "green chemistry") in a range of industrial and laboratory processes. Applications that involve supercritical fluids include extractions, nano particle and nano structured film formation, supercritical drying, carbon capture and storage, as well as enhanced oil recovery studies. | 12,309 | 2,929 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Chemical_Reactions_Examples/Electrolytes |
One of the most important properties of water is its ability to dissolve a wide variety of substances. Solutions in which water is the dissolving medium are called . For electrolytes, is the most important solvent. Ethanol, ammonia, and acetic acid are some of the non-aqueous solvents that are able to dissolve electrolytes. Substances that give ions when dissolved in water are called . They can be divided into acids, bases, and salts, because they all give ions when dissolved in water. These solutions conduct electricity due to the mobility of the positive and negative ions, which are called and respectively. completely ionize when dissolved, and no neutral molecules are formed in solution. For example, \(\ce{NaCl}\), \(\ce{HNO3}\), \(\ce{HClO3}\), \(\ce{CaCl2}\) etc. are strong electrolytes. An ionization can be represented by \(\mathrm{NaCl_{\large{(s)}} \rightarrow Na^+_{\large{(aq)}} + Cl^-_{\large{(aq)}}}\) Since \(\ce{NaCl}\) is an (s), which consists of cations \(\ce{Na+}\) and anions \(\ce{Cl-}\), no molecules of \(\ce{NaCl}\) are present in \(\ce{NaCl}\) solid or \(\ce{NaCl}\) solution. The ionization is said to be complete. The solute is one hundred percent (100%) ionized. Some other ionic solids are \(\ce{CaCl2}\), \(\ce{NH4Cl}\), \(\ce{KBr}\), \(\ce{CuSO4}\), \(\ce{NaCH3COO}\) (sodium acetate), \(\ce{CaCO3}\), and \(\ce{NaHCO3}\) (baking soda). Small fractions of molecules ionize when dissolve in water. Some neutral molecules are present in their solutions. For example, \(\ce{NH4OH}\) (ammonia), \(\ce{H2CO3}\) (carbonic acid), \(\ce{CH3COOH}\) (acetic acid), and most organic acids and bases are weak electrolytes. The following ionization is not complete, \(\mathrm{H_2CO_{3\large{(aq)}} \rightleftharpoons H^+_{\large{(aq)}} + HCO^-_{3\large{(aq)}}}\) In a solution, \(\ce{H2CO3}\) molecules are present. The fraction (often expressed as a %) that undergos ionization depends on the concentration of the solution. On the other hand, ionization can be viewed as an equilibrium established for the above reaction, for which the is defined as \(\mathrm{\mathit K = \dfrac{[H^+] [HCO_3^-]}{[H_2CO_3]}}\) where we use [ ] to mean the concentration of the species in the [ ]. For carbonic acid, = 4.2x10 . You can generalize the definition of here to give the equilibrium constant expression for any weak electrolyte. Pure water is a very weak electrolyte. The ionization or of pure water can be represented by the ionization equation \(\mathrm{H_2O \rightleftharpoons H^+ + OH^-}\) and the equilibrium constant is \(\mathrm{\mathit K = \dfrac{[H^+] [OH^-]}{[H_2O]}}\) For pure water, \(\ce{[H2O]}\) is a constant (1000/18 = 55.6 M), and we often use the , , for water, \(\mathrm{\mathit K_w = \mathit K [H_2O] [H^+] [OH^-]}\) The constant depends on temperature. At 298 K, = 1x10 . If there is no solute in water, the solution has equal concentrations of \(\ce{[H+]}\) and \(\ce{[OH-]}\). \(\mathrm{[H^+] = [OH^-] = 1\times10^{-7}}\), and \(\mathrm{pH = -\log [H^+] = 7}\). Note that only at 298 K is the pH of water = 7. At higher temperatures, the pH is slightly less than 7, and at lower temperatures, the pH is greater than 7. Our body fluids are solutions of electrolytes and many other things. The combination of blood and the circulatory system is the , because it coordinates all the life functions. When the heart stops pumping in a heart attack, the life ends quickly. Getting the heart restarted as soon as one can is crucial in order to maintain life. The primary electrolytes required in the body fluid are cations (of calcium, potassium, sodium, and magnesium) and anions (of chloride, carbonates, aminoacetates, phosphates, and iodide). These are nutritionally called . Electrolyte balance is crucial to many body functions. Here's some extreme examples of what can happen with an imbalance of electrolytes: elevated potassium levels may result in cardiac arrhythmias; decreased extracellular potassium produces paralysis; excessive extracellular sodium causes fluid retention; and decreased plasma calcium and magnesium can produce muscle spasms of the extremities. When a patient is dehydrated, a carefully prepared (commercially available) electrolyte solution is required to maintain health and well being. In terms of child health, oral electrolyte is need when a child is dehydrated due to diarrhea. The use of oral electrolyte maintenance solutions, which is responsible for saving millions of lives worldwide over the last 25 years, is one of the most important medical advances in protecting the health of children in the century, explains Juilus G.K. Goepp, MD, assistant director of the Pediatric Emergency Department of the Children's Center at Johns Hopkins Hospital. If a parent provides an oral electrolyte maintenance solution at the very start of the illness, dehydration can be prevented. The functionality of electrolyte solutions is related to their properties, and interest in electrolyte solutions goes far beyond chemistry. Solutions of electrolytes are always required in batteries, even in . The simplest battery consists of two electrodes. The figure here illustrates a copper-zinc battery. The left hand is a zinc electrode. The zinc atoms have a tendency to become ions, leaving the electrons behind. \(\mathrm{Zn_{\large{(s)}} \rightarrow Zn^{2+}_{\large{(aq)}} + 2 e^-}\). As the zinc ions going into the solution, anions move from the copper cell to the zinc cell to compensate for the charge, and at the same time, electrons go from the \(\ce{Zn}\) electrode to the \(\ce{Cu}\) electrode to neutralize the copper ions. \(\mathrm{Cu^{2+}_{\large{(aq)}} + 2 e^- \rightarrow Cu_{\large{(s)}}}\) In dry cells, the solution is replaced by a paste so that the solution will not leak out of the package. In this cell, the \(\ce{Zn}\) and \(\ce{Cu}\) electrode has a voltage of 1.10 V, if the concentrations of the ions are as indicated. When solutions of electrolytes are combined, the cations and anions will meet each other. When the ions are indifferent of each other, there is no reaction. However, some cations and anions may form a molecule or solid, and thus the cations and anions change partners. These are called metathesis reactons, which include: Redox reactions are also possible between the various ions. In fact, the battery operations involve redox reactions. \(\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}\) \(\mathrm{Zn | Zn^{2+} || Cu^{2+}| Cu}\). \(\ce{CuSO4}\) \(\ce{Cu^2+}\) \(\mathrm{Zn | Zn^{2+}}\) Hint: e. any anion
Explain ion movement in a solution of electrolytes. Hint: yes The two types of ions move in opposite directions.
Explain ion movement in a solution of electrolytes. Hint: b. any copper salt
Any salt can be used for the \(\ce{Zn}\)-electrode. But for the \(\ce{Cu}\) electrode, \(\ce{CuSO4}\) or \(\ce{CuCl2}\) is commonly used.
Apply chemical knowledge to battery setups. Hint: b. \(\ce{NaCl}\) solution A salt solution is usually used, but solutions of acids and bases will be all right. \(\ce{NaCl}\) solution is economical and easy to handle. Hint: e. table salt
Distinguish strong and weak electrolytes. Hint: c. water at low temperature See pH.
Define and estimate pH. See Redox.
Explain metathesis reactions. | 7,293 | 2,930 |
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Theoretical models attempting to describe the nature of gases date back to the earliest scientific inquiries into the nature of matter and even earlier! In about 50 BC, Lucretius, a Roman philosopher, proposed that macroscopic bodies were composed of atoms that continually collide with one another and are in constant motion, despite the observable reality that the body itself is as rest. However, Lucretius’ ideas went largely ignored as they deviated from those of Aristotle, whose views were more widely accepted at the time. In 1738, Daniel Bernoulli (Bernoulli, 1738) published a model that contains the basic framework for the modern Kinetic Molecular theory. Rudolf Clausius furthered the model in 1857 by (among other things) introducing the concept of mean free path (Clausius, 1857). These ideas were further developed by Maxwell (Maxwell, Molecules, 1873). But, because atomic theory was not fully embraced in the early 20 century, it was not until Albert Einstein published one of his seminal works describing Brownian motion (Einstein, 1905) in which he modeled matter using a kinetic theory of molecules that the idea of an atomic (or molecular) picture really took hold in the scientific community. In its modern form, the Kinetic Molecular Theory of gasses is based on five basic postulates. Qualitatively, this model predicts the form of the ideal gas law. Putting all of these together yields \[ p \propto \dfrac{nT}{V} =k \dfrac{nT}{V} \nonumber \] which is exactly the form of the ideal gas law! The remainder of the job is to derive a value for the constant of proportionality (\(k\)) that is consistent with experimental observation. For simplicity, imagine a collection of gas particles in a fixed-volume container with all of the particles traveling at the same velocity. What implications would the kinetic molecular theory have on such a sample? One approach to answering this question is to derive an expression for the pressure of the gas. The pressure is going to be determined by considering the collisions of gas molecules with the wall of the container. Each collision will impart some force. So the greater the number of collisions, the greater the pressure will be. Also, the larger force imparted per collision, the greater the pressure will be. And finally, the larger the area over which collisions are spread, the smaller the pressure will be. \[ p \propto \dfrac{ (\text{number of collisions}) \times (\text{force imparted per collision})}{area} \nonumber \] First off, the pressure that the gas exerts on the walls of the container would be due entirely to the force imparted each time a molecule collides with the interior surface of the container. This force will be scaled by the number of molecules that hit the area of the wall in a given time. For this reason, it is convenient to define a “collision volume”. \[ V_{col}= (v_x \cdot \Delta t) \cdot A \nonumber \] where \(v_x\) is the speed the molecules are traveling in the x direction, \(\ t\) is the time interval (the product of \(v_x·\Delta T\) gives the length to the collision volume box) and A is the area of the wall with which the molecules will collide. Half of the molecules within this volume will collide with the wall since half will be traveling toward it and half will be traveling away from it. The number of molecules in this collision volume will be given by the total number of molecules in the sample and the fraction of the total volume that is the collision volume. And thus, the number of molecules that will collide with the wall is given by \[ N_{col} =\dfrac{1}{2} N_{tot} \dfrac{V_{col}}{V} \nonumber \] And thus the number of molecules colliding with the wall will be \[ N_{col} =\dfrac{1}{2} N_{tot} \dfrac{(v_x \cdot \Delta t) \cdot A}{V} \nonumber \] The magnitude of that force imparted per collision will be determined by the time-rate of change in momentum of each particle as it hits the surface. It can be calculated by determining the total momentum change and dividing by the total time required for the event. Since each colliding molecule will change its velocity from v to –v , the magnitude of the momentum change is 2(mv ). Thus the force imparted per collision is given by \[ F = \dfrac{2(mv_x)}{\Delta t} \nonumber \] and the total force imparted is \[\begin{align} F_{tot} &= N_{col} \dfrac{2 (mv_x)}{\Delta t} \\[4pt] &= \dfrac{1}{2} N_{tot} \left[ \dfrac{(v_x\Delta t)A}{V} \right] \left[ \dfrac{2(m v_x)}{\Delta t} \right] \\[4pt] &= N_{tot} \left(\dfrac{mv_x^2}{V} \right) A \end{align} \nonumber \] Since the pressure is given as the total force exerted per unit area, the pressure is given by \[p = \dfrac{F_{tot}}{A} = N_{tot} \left( \dfrac{mv_x^2}{V} \right) = \dfrac{N_{tot}m}{V} v_x^2 \nonumber \] The question then becomes how to deal with the velocity term. Initially, it was assumed that all of the molecules had the same velocity, and so the magnitude of the velocity in the x-direction was merely a function of the trajectory. However, real samples of gases comprise molecules with an entire distribution of molecular speeds and trajectories. To deal with this distribution of values, we replace (\(v_x^2\)) with the squared average of velocity in the x direction \(\langle v_x \rangle ^2\). \[p = \dfrac{N_{tot}m}{V} \langle v_x \rangle ^2 \label{press} \] The distribution function for velocities in the x direction, known as the , is given by: \[f(v_x) = \underbrace{\sqrt{ \dfrac{m}{2\pi k_BT} }}_{\text{normalization term}} \underbrace{\exp \left(\dfrac{-mv_x^2}{2k_BT} \right)}_{\text{exponential term}} \nonumber \] This function has two parts: a and an exponential term. The normalization constant is derived by noting that \[\int _{-\infty}^{\infty} f(v_x) dv_x =1 \label{prob} \] The Maxwell-Boltzmann distribution has to be normalized because it is a . As such, the sum of the probabilities for all possible values of v be unity. And since \(v_x\) can take any value between -∞ and ∞, then Equation \ref{prob} must be true. So if the form of \(f(v_x)\) is assumed to be \[f(v_x) = N \exp - \left(\dfrac{mv_x^2}{2k_BT} \right) \nonumber \] The normalization constant \(N\) can be found from \[ \int_{-\infty}^{\infty} f(v_x) dv_x = \int_{-\infty}^{\infty} N \exp \left(\dfrac{-mv_x^2}{2k_BT} \right) dv_x =1 \nonumber \] The expression can be simplified by letting \(\alpha = m/2k_BT\). It is then more simply written \[ N \int_{-\infty}^{\infty} \exp \left(\dfrac{-mv_x^2}{2k_BT} \right) dv_x =1 \nonumber \] A table of definite integrals says that \[ \int_{-\infty}^{\infty} e^{- a x^2} dx = \sqrt{\dfrac{\pi}{a}} \nonumber \] So \[ N \sqrt{\dfrac{\pi}{\alpha}} = \left( \dfrac{m}{2\pi k_BT} \right) ^{1/2} \nonumber \] And thus the normalized distribution function is given by \[ f(v_x) = \left( \dfrac{m}{2\pi k_BT} \right) ^{1/2} \text{exp} \left( \dfrac{m v_x^2}{2 k_BT} \right) \label{MB} \] Calculating an average for a finite set of data is fairly easy. The average is calculated by \[ \bar{x} = \dfrac{1}{N} \sum_{i=1}^N x_i \nonumber \] But how does one proceed when the set of data is infinite? Or how does one proceed when all one knows are the probabilities for each possible measured outcome? It turns out that that is fairly simple too! \[ \bar{x} = \sum_{i=1}^N x_i P_i \nonumber \] where \(P_i\) is the probability of measuring the value \(x_i\). This can also be extended to problems where the measurable properties are not discrete (like the numbers that result from rolling a pair of dice) but rather come from a continuous parent population. In this case, if the probability is of measuring a specific outcome, the average value can then be determined by \[\bar{x} = \int x P(x) dx \nonumber \] where \(P(x)\) is the function describing the probability distribution, and with the integration taking place across all possible values that x can take. A value that is useful (and will be used in further developments) is the average velocity in the x direction. This can be derived using the probability distribution, as shown in the mathematical development box above. The average value of \(v_x\) is given by \[ \langle v_x \rangle = \int _{-\infty}^{\infty} v_x (f(v_x) dx \nonumber \] This integral will, by necessity, be zero. This must be the case as the distribution is symmetric, so that half of the molecules are traveling in the +x direction, and half in the –x direction. These motions will have to cancel. So, a more satisfying result will be given by considering the magnitude of \(v_x\), which gives the speed in the x direction. Since this cannot be negative, and given the symmetry of the distribution, the problem becomes \[ \langle |v_x |\rangle = 2 \int _{0}^{\infty} v_x (f(v_x) dx \nonumber \] In other words, we will consider only half of the distribution, and then double the result to account for the half we ignored. For simplicity, we will write the distribution function as \[ f(v_x) = N \exp(-\alpha v_x^2) \nonumber \] where \[ N= \left( \dfrac{m}{2\pi k_BT} \right) ^{1/2} \nonumber \] and A table of definite integrals shows \[ \int_{0}^{\infty} x e^{- a x^2} dx = \dfrac{1}{2a} \nonumber \] so \[ \langle v_x \rangle = 2N \left( \dfrac{1}{2\alpha}\right) = \dfrac{N}{\alpha} \nonumber \] Substituting our definitions for \(N\) and \(\alpha\) produces \[ \langle v_x \rangle = \left( \dfrac{m}{2\pi k_BT} \right)^{1/2} \left( \dfrac{2 k_BT}{m} \right) = \left( \dfrac{2\pi k_BT}{ \pi m} \right)^{1/2} \nonumber \] This expression indicates the average speed for motion of in one direction. However, real gas samples have molecules not only with a distribution of molecular speeds and but also a random distribution of directions. Using normal vector magnitude properties (or simply using the Pythagorean Theorem), it can be seen that \[ \langle v \rangle^2 = \langle v_x \rangle^2 + \langle v_y \rangle^2 + \langle v_z \rangle^2 \nonumber \] Since the direction of travel is random, the velocity can have any component in x, y, or z directions with equal probability. As such, the average value of the x, y, or z components of velocity should be the same. And so \[ \langle v \rangle^2 = 3 \langle v_x \rangle^2 \nonumber \] Substituting this into the expression for pressure (Equation \ref{press}) yields \[ p =\dfrac{ N_{tot}m}{3V} \langle v \rangle^2 \nonumber \] All that remains is to determine the form of the distribution of velocity magnitudes the gas molecules can take. One of the first people to address this distribution was James Clerk Maxwell (1831-1879). In his 1860 paper (Maxwell, Illustrations of the dynamical theory of gases. Part 1. On the motions and collisions of perfectly elastic spheres, 1860), proposed a form for this distribution of speeds which proved to be consistent with observed properties of gases (such as their viscosities). He derived this expression based on a transformation of coordinate system from Cartesian coordinates (\(x\), \(y\), \(z\)) to (\(v\), \(\theta\), \(\phi\)). In this new coordinate system, v represents the magnitude of the velocity (or the speed) and all of the directional data is carried in the angles \(\theta\) and \(\phi\). The infinitesimal volume unit becomes \[ dx\,dy\,dz\, = v^2 \sin( \theta) \,dv\,d\theta \,d\phi \nonumber \] Applying this transformation of coordinates, and ignoring the angular part (since he was interested only in the speed) (Equation \ref{MB}) took the following form \[ f(v) = N v^2 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right) \label{MBFullN} \] This function has three basic parts to it: (\(N\)), a velocity dependence (\(v^2\)), and an exponential term that contains the kinetic energy (\(½ mv^2\)). Since the function represents the fraction of molecules with the speed \(v\), the sum of the fractions for all possible velocities must be unity. This sum can be calculated as an integral. The normalization constant ensures that \[ \int_0^{\infty} f(v) dv = 1 \nonumber \] Choosing the normalization constant as \[ N =4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } \nonumber \] yields the final form of the Maxwell distribution of molecular speeds. \[ N =4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } v^2 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right) \label{MBFull} \] At low velocities, the \(v^2\) term causes the function to increase with increasing \(v\), but then at larger values of \(v\), the exponential term causes it to drop back down asymptotically to zero. The distribution will spread over a larger range of speed at higher temperatures, but collapse to a smaller range of values at lower temperatures (Table 2.3.1). Using the Maxwell distribution as a distribution of probabilities, the average molecular speed in a sample of gas molecules can be determined. \[ \begin{align} \langle v \rangle & = \int _{-\infty}^{\infty} v \,f(v) dv \\ & = \int _{-\infty}^{\infty} v\, 4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } v^2 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right)\ dv \\ & = 4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } \int _{-\infty}^{\infty} v^3 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right)\ dv \end{align} \nonumber \] The following can be found in a table of integrals: \[ \int_0^{\infty} x^{2n+1} e^{-ax^2} dx = \dfrac{n!}{2a^{n+1}} \nonumber \] So \[\langle v \rangle = 4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } \left[ \dfrac{1}{2 \left( \dfrac{m}{2 k_B T} \right) ^2 } \right] \nonumber \] Which simplifies to \[\langle v \rangle = \left( \dfrac{8 k_BT}{\pi m} \right) ^{1/2} \nonumber \] Note: the value of \(\langle v \rangle \) is twice that of \(\langle v_x \rangle \) which was derived in an earlier example! \[\langle v \rangle = 2\langle v_x \rangle \nonumber \] What is the average value of the squared speed according to the Maxwell distribution law? \[ \begin{align} \langle v^2 \rangle & = \int _{-\infty}^{\infty} v^2 \,f(v) dv \\ & = \int _{-\infty}^{\infty} v^2\, 4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } v^2 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right)\ dv \\ & = 4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } \int _{-\infty}^{\infty} v^4 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right)\ dv \end{align} \nonumber \] A table of integrals indicates that \[ \int_0^{\infty} x^{2n} e^{-ax^2} dx = \dfrac{1 \cdot 3 \,cdot 5 \dots (2n-1)}{2^{n+1}a^n} \sqrt{\dfrac{\pi}{a}} \nonumber \] Substitution (noting that \(n = 2\)) yields \[\langle v^2 \rangle = 4\pi \sqrt{\left( \dfrac{m}{2\pi k_BT} \right) ^3 } \left[ \dfrac{1 \cdot 3}{2^3 \left( \dfrac{m}{2 k_BT} \right) ^2 } \sqrt{\dfrac{\pi}{\left( \dfrac{m}{2 k_BT} \right)}} \right] \nonumber \] which simplifies to \[\langle v^2 \rangle = \dfrac{3 k_BT}{ m} \nonumber \] : The square root of this average squared speed is called the (RMS) speed, and has the value \[v_{rms} = \sqrt{ \langle v^2 \rangle } = \left( \dfrac{3 k_BT}{ m} \right)^{1/2} \nonumber \] The entire distribution is also affected by molecular mass. For lighter molecules, the distribution is spread across a broader range of speeds at a given temperature, but collapses to a smaller range for heavier molecules (Table 2.3.2). The probability distribution function can also be used to derive an expression for the most probable speed (\(v_{mp}\)), the average (\(v_{ave}\)), and the root-mean-square (\(v_{rms}\)) speeds as a function of the temperature and masses of the molecules in the sample. The most probable speed is the one with the . That will be the speed that yields the maximum value of \(f(v)\). It is found by solving the expression \[ \dfrac{d}{dv} f(v) = 0 \nonumber \] for the value of \(v\) that makes it true. This will be the value that gives the maximum value of \(f(v)\) for the given temperature. Similarly, the average value can be found using the distribution in the following fashion \[ v_{ave} = \langle v \rangle \nonumber \] and the (RMS) speed by finding the square root of the average value of \(v^2\). Both demonstrated above. \[ v_{rms} = \sqrt{ \langle v^2 \rangle} \nonumber \] | 15,996 | 2,931 |
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After completing this section, you should be able to Make certain that you can define, and use in context, the key term below. The limitations of the system are illustrated in the examples given below. At first you might say , because it appears that two ethyl groups appear on the same side of the double bond. However, the correct answer is . The rule is that the designation or must correspond to the configuration of the carbon chain. Tracing out the seven-carbon chain in the compound shown above, you change sides as you pass through the double bond: So, the full name for this compound is -4-ethyl-3methyl-3-heptene. In cases where two or more double bonds are present, you must be prepared to assign an or designation to each of the double bonds. For example: Another use for these sequence rules will be part of the discussion of optical isomerism in Section 9.5. When each carbon in a double bond is attached to a hydrogen and and a non-hydrogen substituent, the geometric isomers can be identified by using the nomenclature discussed in the previous section. However, when a double bond is attached to three or four non-hydrogen substituents there are some examples where nomenclature is ineffective in describing the substituents orientation in geometric isomers. In these situations the rigorous IUPAC system for naming alkene isomers, called the E/Z system, is used. The E/Z system analyzes the two substituents attached to each carbon in the double bond and assigns each either a high or low priority. If the higher priority group on both carbons in the double bond the side the alkene is said to have a (from German zusammen = together). You could think of Z as Zame Zide to help memorize it. If the higher priority group on sides the alkene has an (from German entgegen = opposite). Note, if both substituents an a double bond carbon are exactly the same there is no E/Z isomerizem is possible. Also, if E/Z isomerism is possible, interchanging the substituents attached on double-bond carbon converts one isomer to the other. Substituent priority for the E,Z system is assigned using the Cahn–Ingold–Prelog (CIP) sequence rules. These are the same rules used to assign R/S configurations to chiral centers in . A brief overview of using CIP rules to determine alkene configuration is given here but CIP rules are discussed in greater detail in . The priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system First, determine the two stubstitents on each double-bond carbon separately. Rank these substituents based on which directly attached to the double-bond carbon. The substituent whose atom has a higher atomic number takes precedence over the substituent whose atom has a lower atomic number. Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins. To illustrate this, consider the molecule at the left. Is the double bond here or ? At the left end of the double bond, Br > C. But the right end of the double bond requires a careful analysis. At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins". Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH -O-CH ). This means that the isomer shown is opposite = entgegen = . And what is the name? The "name" is (E)-2-Bromo-3-(methoxymethyl)hex-2-en-4-yne. If the first atom on both substituents are the identical, then proceed along both substituent chains until the first point of difference is determined. Remember that atoms involved in multiple bonds are considered with a specific set of rules. These atoms are treated as if they have the same number of single-bond atoms as they have attached to multiply bonded atoms. An easy example which shows the necessity of the E/Z system is the alkene, 1-bromo-2-chloro-2-fluoro-1-iodoethene, which has four different substituents attached to the double bond. The figure below shows that there are two distinctly different geometric isomers for this molecule neither of which can be named using the system. Consider the left hand structure. On the double bond carbon on the left, the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at carbon on the right. The attached atoms are Cl and F, with Cl having the higher atomic number and the higher priority. When considering the relative positions of the higher priority groups, the higher priority group is "down" on the left double bond carbon and "down" at right double bond carbon. Since the two higher priority groups are both on the side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E). The Figure above shows the two isomers of 2-butene. You should recognize them as and . Let's analyze them to see whether they are or . Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is ( )-2-butene. Now look at the right hand structure (the isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on sides of the double bond, they are entgegen = opposite. Therefore, this is ( )-2-butene. In simple cases, such as 2-butene, corresponds to and to . However, that is a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the / system is that it will always work. In contrast, the / system breaks down with many ambiguous cases. The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene. It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as or . On the other hand, the / system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the ( ) isomer. Similarly, the right hand structure is ( ). Consider the molecule shown at the left. This is 2-bromo-2-butene -- ignoring the geometric isomerism for now. or ? This molecule is clearly . The two methyl groups are on the same side. More rigorously, the "parent chain" is . or ? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group -- because the other group is -H. On the right, the methyl is the low priority group -- because the other group is -Br. That is, the high priority groups are -CH (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = . This example should convince you that and are not synonyms. and / are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine / or / separately, as needed. If the compound contains more than one double bond, then each one is analyzed and declared to be or . The configuration at the left hand double bond is ; at the right hand double bond it is . Thus this compound is (1 ,4 )-1,5-dichloro-1,4-hexadiene. Consider the compound below This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end. What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH -CH (an ethyl group) and -CH=CH (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the -isomer; the compound is ( )-1-chloro-2-ethyl-1,3-butadiene. The configuration about double bonds is undoubtedly best specified by the / notation when there is no ambiguity involved. Unfortunately, many compounds cannot be described adequately by the / system. Consider, for example, configurational isomers of 1-fluoro-1-chloro-2-bromo-2-iodo-ethene, 9 and 10. There is no obvious way in which the / system can be used: A system that is easy to use and which is based on the sequence rules already described for the , system works as follows: The isomer is designated as the isomer in which the top priority groups are on the same side ( is taken from the German word zusammen- together). The isomer has these groups on opposite sides ( , German for entgegen across). Two further examples show how the nomenclature is used: Which of the following sets has a higher ranking? a) -CH or -CH Br b) -Br or -Cl c) -CH=CH or -CH=O a) -CH Br b) -Br c) -CH=O Place the following sets of substituents in each group in order of lowest priority (1 ) to highest priority (4 ) a) -CH(CH ) , -CH CH , -C(CH ) , -CH b) -NH , -F, -Br, -CH c) -SH, -NH , -F, -H a) (lowest priority) -CH < -CH CH < -CH(CH ) < -C(CH ) (highest priority) b) (lowest priority) -CH < -NH < -F < -Br (highest priority) c) (lowest priority) -H < -NH < -F < -SH (highest priority) Label the following alkenes as E, Z, or neither. The higher priority group is highlighted in red. a) E b) E c) E d) Z e) neither (both isopropyls on the right have the same priority) Name the following alkenes. The higher priority group is highlighted in red. a) (Z)-4-ethyl-5-methyloct-3-ene or (Z)-4-ethyl-5-methyl-3-octene b) 3-ethyl-6-methyl-4-propylhept-3-ene or 3-ethyl-6-methyl-4-propyl-3-heptene c) (E)-2-chloro-4-bromo-5-ethyl-7-methyldec-4-ene or (E)-2-chloro-4-bromo-5-ethyl-7-methyl-4-decene | 11,708 | 2,932 |
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After completing this section, you should be able to Although can be used to explain the relative stabilities of carbocations, this explanation is certainly not the only one, and is by no means universally accepted. A more common explanation, involving the concept of an inductive effect, is given below. It is a general principle in chemistry that the more a charge is dispersed, the more stable is the species carrying the charge. Put simply, a species in which a positive charge is shared between two atoms would be more stable than a similar species in which the charge is borne wholly by a single atom. In a tertiary carbocation, the positively charged carbon atom attracts the bonding electrons in the three carbon-carbon sigma ( ) bonds, and thus creates slight positive charges on the carbon atoms of the three surrounding alkyl groups (and, indeed, on the hydrogen atoms attached to them). Chemists sometimes use an arrow to represent this inductive release: These diagrams do not reflect the geometry of the carbocation. The overall charge on the carbocation remains unchanged, but some of the charge is now carried by the alkyl groups attached to the central carbon atom; that is, the charge has been dispersed. In the tertiary carbocation shown above, the three alkyl groups help to stabilize the positive charge. In a secondary carbocation, only two alkyl groups would be available for this purpose, while a primary carbocation has only one alkyl group available. Thus the observed order of stability for carbocations is as follows: tertiary > secondary > primary > methyl. The next step in understanding why Markovnikov's rule is often followed in electrophilic additions, involves understanding the structure and stability of the carboncation intermediate formed during the mechanism. Carbocations typically have three substituents which makes the carbon sp hybridized and gives the overall molecule a trigonal planar geometry. The carbocation's substituents are all in the same plane and have a bond angle of 120 between them. The carbon atom in the carbocation is electron deficient; it only has six valence electrons which are used to form three sigma covalent bonds with the substituents. The carbocation carbon has an unoccupied p orbital which is perpendicular to the plane created by the substituents. The p orbital can easily accept electron pairs during reactions making carbocations excellent Lewis acids. By being a reactive intermediate of the electrophilic addition mechanism, the stability of a carbocation has a direct effect on the reaction. The critical question now becomes, A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density will help to stabilize it. Conversely, a carbocation will be by an electron withdrawing group. Extensive experimental evidence has shown that a carbocation becomes more stable as the number of alkyl substituents increases. Carbocations can be given a designation based on the number of alkyl groups attached to the carbocation carbon. Three alkyl groups is called a tertiary (3 ) carbocation, 2 alkyl groups is called secondary (2 ), and 1 alkyl group is called primary (1 ). No alkyl groups are attached (3 hydrogen substituents) is called a methyl carbocation. The overall order of stability is as follows: Alkyl groups stabilized carbocations for two reasons. The first is through inductive effects. As discussed in , inductive effects occur when the electrons in covalent bonds are shifted towards an nearby atom with a higher electronegativity. In this case, the positively charged carbocation draws in electron density from the surrounding substituents thereby gaining stabilization by slightly reducing its positive charge. Alkyl groups are more effective at inductively donating electron density than a hydrogen because they are larger, more polarizable, and contain more bonding electrons. As more alkyl groups are attached to the carbocation more inductive electron donation occurs and the carbocation becomes more stable. The second reason alkyl groups stabilize carbocations is through hyperconjugation. As previously discussed in , hyperconjugation is an electron donation that occurs from the parallel overlap of p orbitals with adjacent hybridized orbitals participating in sigma bonds. This electron donation serves to stabilize the carbocation. As the number of alkyl substituents increases, the number of sigma bonds available for hyperconjugation increases, and the carbocation tends to become more stabilized. In the example of ethyl carbocation shown below, the p orbital from a sp hybridized carbocation carbon involved interacts with a sp hybridized orbital participating in an adjacent C-H sigma bond. Electron density from the C-H sigma bond is donated into carbocation's p orbital providing stabilization. The molecular orbital of the ethyl carbocation shows the interaction of electrons in methyl group's C-H sigma bonds with the adjacent empty p orbital from the carbocation. The interaction creates a bonding molecular orbital which extends over the three atom chain (C-C-H) involved in hyperconjugation. The expanded molecular orbital helps to stabilize the carbocation. It is not accurate to say, however, that carbocations with higher substitution are more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary. The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A. In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is S 1 or S 2. Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a carbocation: This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations. Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron groups by induction, they are electron groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in ). Consider the two pairs of carbocation species below: In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually the positive charge, because they are electron withdrawing by induction. Finally, carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
a vinylic carbocation (very unstable) In which of the structures below is the carbocation expected to be more stable? Explain. In the carbocation on the left, the positive charge is located in a position relative to the nitrogen such that the lone pair of electrons on the nitrogen can be donated to fill the empty orbital. This is not possible for the carbocation species on the right. Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms. When considering the possibility that a nucleophilic substitution reaction proceeds an S 1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an S 1 mechanism must be considered unlikely, and the reaction probably proceeds by an S 2 mechanism. In the next chapter we will see several examples of biologically important S 1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure. State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning. a) 1 (tertiary vs. secondary carbocation) b) 1 (tertiary vs. secondary carbocation) c) 2 (positive charge is further from electron-withdrawing fluorine) d) 1 (lone pair on nitrogen can donate electrons by resonance) e) 1 (allylic carbocation – positive charge can be delocalized to a second carbon) | 10,335 | 2,933 |
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Carbohydrates are the most abundant class of organic compounds found in living organisms. They originate as products of , an endothermic reductive condensation of carbon dioxide requiring light energy and the pigment chlorophyll. The formulas of many carbohydrates can be written as carbon hydrates, C (H O) , hence their name. The carbohydrates are a major source of metabolic energy, both for plants and for animals that depend on plants for food. Aside from the sugars and starches that meet this vital nutritional role, carbohydrates also serve as a structural material (cellulose), a component of the energy transport compound ATP, recognition sites on cell surfaces, and one of three essential components of DNA and RNA. Carbohydrates are called or, if they are relatively small, sugars. Several classifications of carbohydrates have proven useful, and are outlined in the following table. Glucose and other saccharides are extensively cleaved by periodic acid, thanks to the abundance of vicinal diol moieties in their structure. This oxidative cleavage, known as the is particularly useful for the analysis of selective O-substituted derivatives of saccharides, since ether functions do not react. The stoichiometry of aldohexose cleavage is shown in the following equation. The four chiral centers in glucose indicate there may be as many as sixteen (2 ) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula. on this link for a review. At the time Fischer undertook the glucose project it was not possible to establish the of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the -family. The mirror images of these configurations were then designated the -family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red. The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. . It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths. of representative aldoses may be examined by clicking on the Fischer formulas for glyceraldehyde, erythrose, threose, ribose, arabinose, allose, altrose, glucose or mannose in the above diagram. Emil Fischer made use of several key reactions in the course of his carbohydrate studies. These are described here, together with the information that each delivers. As noted above, sugars may be classified as or based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called , since the oxidant (Ag or Cu ) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars. When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an . Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an . By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below. Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration. Sodium borohydride reduction of an aldose makes the ends of the resulting chain identical, HOCH (CHOH) CH OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table. The osazone reaction was developed and used by Emil Fischer to identify aldose sugars differing in configuration only at the alpha-carbon. The upper equation shows the general form of the osazone reaction, which effects an alpha-carbon oxidation with formation of a bis-phenylhydrazone, known as an osazone. Application of the osazone reaction to D-glucose and D-mannose demonstrates that these compounds differ in configuration only at C-2. These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structurespresented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis. If a monosaccharide has a carbonyl function on one of the inner atoms of the carbon chain it is classified as a . Dihydroxyacetone may not be a sugar, but it is included as the ketose analog of glyceraldehyde. The carbonyl group is commonly found at C-2, as illustrated by the following examples (chiral centers are colored red). As expected, the carbonyl function of a ketose may be reduced by sodium borohydride, usually to a mixture of epimeric products. D-Fructose, the sweetest of the common natural sugars, is for example reduced to a mixture of D-glucitol (sorbitol) and D-mannitol, named after the aldohexoses from which they may also be obtained by analogous reduction. Mannitol is itself a common natural carbohydrate. Although the ketoses are distinct isomers of the aldose monosaccharides, the chemistry of both classes is linked due to their facile interconversion in the presence of acid or base catalysts. This interconversion, and the corresponding epimerization at sites alpha to the carbonyl functions, occurs by way of an tautomeric intermediate. , an equation illustrating these isomerizations will be displayed. Because of base-catalyzed isomerizations of this kind, the Tollens' reagent is not useful for distinguishing aldoses from ketoses or for specific oxidation of aldoses to the corresponding aldonic acids. Oxidation by HOBr is preferred for the latter conversion. Fischer's brilliant elucidation of the configuration of glucose did not remove all uncertainty concerning its structure. Two different crystalline forms of glucose were reported in 1895. Each of these gave all the characteristic reactions of glucose, and when dissolved in water equilibrated to the same mixture. This equilibration takes place over a period of many minutes, and the change in optical activity that occurs is called . These facts are summarized in the diagram below. When glucose was converted to its pentamethyl ether (reaction with excess CH I & AgOH), two different isomers were isolated, and neither exhibited the expected aldehyde reactions. Acid-catalyzed hydrolysis of the pentamethyl ether derivatives, however, gave a tetramethyl derivative that was oxidized by Tollen's reagent and reduced by sodium borohydride, as expected for an aldehyde. These reactions will be displayed above by . The search for scientific truth often proceeds in stages, and the structural elucidation of glucose serves as a good example. It should be clear from the new evidence presented above, that the open chain pentahydroxyhexanal structure drawn above must be modified. Somehow a new stereogenic center must be created, and the aldehyde must be deactivated in the pentamethyl derivative. A simple solution to this dilemma is achieved by converting the open aldehyde structure for glucose into a cyclic hemiacetal, called a , as shown in the following diagram. The linear aldehyde is tipped on its side, and rotation about the C4-C5 bond brings the C5-hydroxyl function close to the aldehyde carbon. For ease of viewing, the six-membered hemiacetal structure is drawn as a flat hexagon, but it actually assumes a chair conformation. The hemiacetal carbon atom (C-1) becomes a new stereogenic center, commonly referred to as the , and the α and β-isomers are called . We can now consider how this modification of the glucose structure accounts for the puzzling facts noted above. First, we know that hemiacetals are in equilibrium with their carbonyl and alcohol components when in solution. Consequently, fresh solutions of either alpha or beta-glucose crystals in water should establish an equilibrium mixture of both anomers, plus the open chain chain form. This will be shown above by . Note that despite the very low concentration of the open chain aldehyde in this mixture, typical chemical reactions of aldehydes take place rapidly. Second, a pentamethyl ether derivative of the pyranose structure converts the hemiacetal function to an acetal. Acetals are stable to base, so this product should not react with Tollen's reagent or be reduced by sodium borohydride. Acid hydrolysis of acetals regenerates the carbonyl and alcohol components, and in the case of the glucose derivative this will be a tetramethyl ether of the pyranose hemiacetal. This compound will, of course, undergo typical aldehyde reactions. this relationship will be displayed above. As noted above, the preferred structural form of many monosaccharides may be that of a cyclic hemiacetal. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed (five-membered) or (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown on the right. Ribose, an important aldopentose, commonly adopts a furanose structure, as shown in the following illustration. By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer. The anomeric carbon atom (colored red here) is placed on the right. . on the following diagram to see a . The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a , after the British chemist, Norman Haworth. As with the furanose ring, the anomeric carbon is placed on the right with the ring oxygen to the back of the edgewise view. In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring (beta is up & alpha is down). These Haworth formulas are convenient for displaying stereochemical relationships, but do not represent the true shape of the molecules. We know that these molecules are actually puckered in a fashion we call a chair conformation. Examples of four typical pyranose structures are shown below, both as Haworth projections and as the more representative chair conformers. The anomeric carbons are colored red. of these glucose, galactose, mannose and allose pyranose structures may be viewed by ing Here. A practice page for examining the configurations of aldohexoses may be viewed by ing Here. The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldolhexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. The formation of acetal derivatives illustrates how subtle changes may alter this selectivity. . the display will change to illustrate this. A pyranose structure for D-glucose is drawn in the rose-shaded box on the left. Acetal derivatives have been prepared by acid-catalyzed reactions with benzaldehyde and acetone. As a rule, benzaldehyde forms six-membered cyclic acetals, whereas acetone prefers to form five-membered acetals. The top equation shows the formation and some reactions of the 4,6-O-benzylidene acetal, a commonly employed protective group. A methyl glycoside derivative of this compound (see below) leaves the C-2 and C-3 hydroxyl groups exposed to reactions such as the periodic acid cleavage, shown as the last step. The formation of an isopropylidene acetal at C-1 and C-2, center structure, leaves the C-3 hydroxyl as the only unprotected function. Selective oxidation to a ketone is then possible. Finally, direct di-O-isopropylidene derivatization of glucose by reaction with excess acetone results in a change to a furanose structure in which the C-3 hydroxyl is again unprotected. However, the same reaction with D-galactose, shown in the blue-shaded box, produces a pyranose product in which the C-6 hydroxyl is unprotected. Both derivatives do not react with Tollens' reagent. This difference in behavior is attributed to the cis-orientation of the C-3 and C-4 hydroxyl groups in galactose, which permits formation of a less strained five-membered cyclic acetal, compared with the trans-C-3 and C-4 hydroxyl groups in glucose. Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules. The ring size of these cyclic monosaccharides was determined by oxidation and chain cleavage of their tetra methyl ether derivatives. To see how this was done for glucose Here. Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called . This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid. The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the . Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the and the alcohol component as the . Two examples of naturally occurring glycosides and one example of an amino derivative will be displayed above . Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials. For a discussion of the anomeric effect Here. For examples of structurally and functionally modified sugars Here. When the alcohol component of a glycoside is provided by a hydroxyl function on another monosaccharide, the compound is called a . Four examples of disaccharides composed of two glucose units are shown in the following diagram. The individual glucopyranose rings are labeled A and B, and the glycoside bonding is circled in light blue. Notice that the glycoside bond may be alpha, as in maltose and trehalose, or beta as in cellobiose and gentiobiose. Acid-catalyzed hydrolysis of these disaccharides yields glucose as the only product. Enzyme-catalyzed hydrolysis is selective for a specific glycoside bond, so an alpha-glycosidase cleaves maltose and trehalose to glucose, but does not cleave cellobiose or gentiobiose. A beta-glycosidase has the opposite activity. In order to draw a representative structure for cellobiose, one of the glucopyranose rings must be rotated by 180º, but this feature is often omitted in favor of retaining the usual perspective for the individual rings. The bonding between the glucopyranose rings in cellobiose and maltose is from the anomeric carbon in ring A to the C-4 hydroxyl group on ring B. This leaves the anomeric carbon in ring B free, so cellobiose and maltose both may assume alpha and beta anomers at that site (the beta form is shown in the diagram). Gentiobiose has a beta-glycoside link, originating at C-1 in ring A and terminating at C-6 in ring B. Its alpha-anomer is drawn in the diagram. Because cellobiose, maltose and gentiobiose are hemiacetals they are all reducing sugars (oxidized by Tollen's reagent). Trehalose, a disaccharide found in certain mushrooms, is a bis-acetal, and is therefore a non-reducing sugar. A systematic nomenclature for disaccharides exists, but as the following examples illustrate, these are often lengthy. Disaccharides made up of other sugars are known, but glucose is often one of the components. Two important examples of such mixed disaccharides will be displayed above . Lactose, also known as milk sugar, is a galactose-glucose compound joined as a beta-glycoside. It is a reducing sugar because of the hemiacetal function remaining in the glucose moiety. Many adults, particularly those from regions where milk is not a dietary staple, have a metabolic intolerance for lactose. Infants have a digestive enzyme which cleaves the beta-glycoside bond in lactose, but production of this enzyme stops with weaning. Cheese is less subject to the lactose intolerance problem, since most of the lactose is removed with the whey. Sucrose, or cane sugar, is our most commonly used sweetening agent. It is a non-reducing disaccharide composed of glucose and fructose joined at the anomeric carbon of each by glycoside bonds (one alpha and one beta). In the formula shown here the fructose ring has been rotated 180º from its conventional perspective. Here Additional Topics For a brief discussion of sweetening agents Here. For examples of some larger saccharide oligomers Here. As the name implies, polysaccharides are large high-molecular weight molecules constructed by joining monosaccharide units together by glycosidic bonds. They are sometimes called . The most important compounds in this class, cellulose, starch and glycogen are all polymers of glucose. This is easily demonstrated by acid-catalyzed hydrolysis to the monosaccharide. Since partial hydrolysis of cellulose gives varying amounts of cellobiose, we conclude the glucose units in this macromolecule are joined by beta-glycoside bonds between C-1 and C-4 sites of adjacent sugars. Partial hydrolysis of starch and glycogen produces the disaccharide maltose together with low molecular weight dextrans, polysaccharides in which glucose molecules are joined by alpha-glycoside links between C-1 and C-6, as well as the alpha C-1 to C-4 links found in maltose. Polysaccharides built from other monosaccharides (e.g. mannose, galactose, xylose and arabinose) are also known, but will not be discussed here. Over half of the total organic carbon in the earth's biosphere is in . Cotton fibers are essentially pure cellulose, and the wood of bushes and trees is about 50% cellulose. As a polymer of glucose, cellulose has the formula (C H O ) where n ranges from 500 to 5,000, depending on the source of the polymer. The glucose units in cellulose are linked in a linear fashion, as shown in the drawing below. The beta-glycoside bonds permit these chains to stretch out, and this conformation is stabilized by intramolecular hydrogen bonds. A parallel orientation of adjacent chains is also favored by intermolecular hydrogen bonds. Although an individual hydrogen bond is relatively weak, many such bonds acting together can impart great stability to certain conformations of large molecules. Most animals cannot digest cellulose as a food, and in the diets of humans this part of our vegetable intake functions as roughage and is eliminated largely unchanged. Some animals (the cow and termites, for example) harbor intestinal microorganisms that breakdown cellulose into monosaccharide nutrients by the use of beta-glycosidase enzymes. Cellulose is commonly accompanied by a lower molecular weight, branched, amorphous polymer called . In contrast to cellulose, hemicellulose is structurally weak and is easily hydrolyzed by dilute acid or base. Also, many enzymes catalyze its hydrolysis. Hemicelluloses are composed of many D-pentose sugars, with xylose being the major component. Mannose and mannuronic acid are often present, as well as galactose and galacturonic acid. Starch is a polymer of glucose, found in roots, rhizomes, seeds, stems, tubers and corms of plants, as microscopic granules having characteristic shapes and sizes. Most animals, including humans, depend on these plant starches for nourishment. The structure of starch is more complex than that of cellulose. The intact granules are insoluble in cold water, but grinding or swelling them in warm water causes them to burst. The released starch consists of two fractions. About 20% is a water soluble material called . Molecules of amylose are linear chains of several thousand glucose units joined by alpha C-1 to C-4 glycoside bonds. Amylose solutions are actually dispersions of hydrated helical micelles. The majority of the starch is a much higher molecular weight substance, consisting of nearly a million glucose units, and called . Molecules of amylopectin are branched networks built from C-1 to C-4 and C-1 to C-6 glycoside links, and are essentially water insoluble. Representative structural formulas for amylose and amylopectin will be shown above . To see an expanded structure for amylopectin on the diagram. The branching in this diagram is exaggerated, since on average, branches only occur every twenty five glucose units. Hydrolysis of starch, usually by enzymatic reactions, produces a syrupy liquid consisting largely of glucose. When cornstarch is the feedstock, this product is known as . It is widely used to soften texture, add volume, prohibit crystallization and enhance the flavor of foods. Glycogen is the glucose storage polymer used by animals. It has a structure similar to amylopectin, but is even more highly branched (about every tenth glucose unit). The degree of branching in these polysaccharides may be measured by enzymatic or chemical analysis. For examples of chemical analysis of branching Here. Cotton, probably the most useful natural fiber, is nearly pure cellulose. The manufacture of textiles from cotton involves physical manipulation of the raw material by carding, combing and spinning selected fibers. For fabrics the best cotton has long fibers, and short fibers or cotton dust are removed. Crude cellulose is also available from wood pulp by dissolving the lignan matrix surrounding it. These less desirable cellulose sources are widely used for making paper. In order to expand the ways in which cellulose can be put to practical use, chemists have devised techniques for preparing solutions of cellulose derivatives that can be spun into fibers, spread into a film or cast in various solid forms. A key factor in these transformations are the three free hydroxyl groups on each glucose unit in the cellulose chain, --[C H O(OH) ] --. Esterification of these functions leads to polymeric products having very different properties compared with cellulose itself. Cellulose Nitrate, first prepared over 150 years ago by treating cellulose with nitric acid, is the earliest synthetic polymer to see general use. The fully nitrated compound, --[C H O(ONO ) ] --, called guncotton, is explosively flammable and is a component of smokeless powder. Partially nitrated cellulose is called . Pyroxylin is soluble in ether and at one time was used for photographic film and lacquers. The high flammability of pyroxylin caused many tragic cinema fires during its period of use. Furthermore, slow hydrolysis of pyroxylin yields nitric acid, a process that contributes to the deterioration of early motion picture films in storage. , --[C H O(OAc) ] --, is less flammable than pyroxylin, and has replaced it in most applications. It is prepared by reaction of cellulose with acetic anhydride and an acid catalyst. The properties of the product vary with the degree of acetylation. Some chain shortening occurs unavoidably in the preparations. An acetone solution of cellulose acetate may be forced through a spinneret to generate filaments, called , that can be woven into fabrics. , is prepared by formation of an alkali soluble xanthate derivative that can be spun into a fiber that reforms the cellulose polymer by acid quenching. The following general equation illustrates these transformations. The product fiber is called . | 27,972 | 2,934 |
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The "phase" of a substance is the particular physical state it is in. The most common phases are solid, liquid, and gas, each easily distinguishable by their significantly different physical properties. A given substance can exist in different phases under different conditions: water can exist as solid ice, liquid, or steam, but water molecules are \(\ce{H_2O}\) regardless of the phase. Furthermore, a substance changes phase without undergoing any chemical transformation: the evaporation of water or the melting of ice occur without decomposition or modification of the water molecules. In describing the differing states of matter changes between them, we will also assume an understanding of the principles of the and the . We will also assume an understanding of the bonding, structure, and properties of individual molecules. We have developed a very clear molecular picture of the gas phase, via the Kinetic Molecular Theory. The gas particles (atoms or molecules) are very distant from one another, sufficiently so that there are no interactions between the particles. The path of each particle is independent of the paths of all other particles. We can determine many of the properties of the gas from this description; for example, the pressure can be determined by calculating the average force exerted by collisions of the gas particles with the walls of the container. To discuss liquids and solids, though, we will be forced to abandon the most fundamental pieces of the Kinetic Molecular Theory of Gases. First, it is clear that the particles in the liquid or solid phases are very much closer together than they are in the gas phase, because the densities of these "condensed" phases are of the order of a thousand times greater than the typical density of a gas. In fact, we should expect that the particles in the liquid or solid phases are essentially in contact with each other constantly. Second, since the particles in liquid or solid are in close contact, it is not reasonable to imagine that the particles do not interact with one another. Our assumption that the gas particles do not interact is based, in part, on the concept that the particles are too far apart to interact. Moreover, particles in a liquid or solid must interact, for without attractions between these particles, random motion would require that the solid or liquid dissipate or fall apart. In this study, we will pursue a model to describe the differences between condensed phases and gases and to describe the transitions which occur between the solid, liquid, and gas phases. We will find that intermolecular interactions play the most important role in governing phase transitions, and we will pursue an understanding of the variations of these intermolecular interactions for different substances. We begin by returning to our observations of . Recall that we trap an amount of gas in a cylinder fitted with a piston, and we apply a fixed pressure to the piston. We vary the temperature of the gas, and since the pressure applied to the piston is constant, the piston moves to maintain a constant pressure of the trapped gas. At each temperature, we then measure the volume of the gas. From our previous observations, we know that the volume of the gas is proportional to the absolute temperature in degrees Kelvin. Thus a graph of volume versus absolute temperature is a straight line, which can be extrapolated to zero volume at \(0 \: \text{K}\). Consider, then, trying to measure the volume for lower and lower temperatures to follow the graph. To be specific, we take exactly \(1.00 \: \text{mol}\) of butane, \(\ce{C_4H_{10}}\) at \(1 \: \text{atm}\) pressure. As we lower the temperature from \(400 \: \text{K}\) to \(300 \: \text{K}\), we observe the expected proportional decrease in the volume from \(32.8 \: \text{L}\) to \(24.6 \: \text{L}\). However, when we reach \(272.6 \: \text{K}\), the volume of the butane drops very abruptly, falling to about \(0.097 \: \text{L}\) at temperatures just slightly below \(272.6 \: \text{K}\). This is less than one-half of one percent of the previous volume! The striking change in volume is shown in the graph as a vertical line at \(272.6 \: \text{K}\). This dramatic change in physical properties at one temperature is referred to as a . When cooling butane through the temperature \(272.6 \: \text{K}\), the butane is abruptly converted at that temperature from one phase, gas, to another phase, liquid, with very different physical properties. If we reverse the process, starting with liquid butane at \(1 \: \text{atm}\) pressure and temperature below \(272.6 \: \text{K}\) and then heating, we find that the butane remains entirely liquid for temperatures below \(272.6 \: \text{K}\) and then becomes entirely gas for temperatures above \(272.6 \: \text{K}\). We refer to the temperature of the phase transition as the temperature. (We will discuss the phases present the boiling point, rather than above and below that temperature, in another section.) We now consider how the phase transition depends on a variety of factors. First, we consider capturing \(2.00 \: \text{mol}\) of butane in the cylinder initially, still at \(1 \: \text{atm}\) pressure. The volume of \(2.00 \: \text{mol}\) is twice that of \(1.00 \: \text{mol}\), by . The proportional decrease in the volume of \(2.00 \: \text{mol}\) of gas is shown in Figure 13.2 along with the previous result for \(1.00 \: \text{mol}\). Note that the phase transition is observed to occur at exactly the same temperature, \(272.6 \: \text{K}\), even though there is double the mass of butane. Consider instead then varying the applied pressure. The result for cooling \(1.00 \: \text{mol}\) of butane at a constant \(2.00 \: \text{atm}\) pressure is also shown in Figure 13.2. We observe the now familiar phase transition with a similar dramatic drop in volume. However, in this case, we find that the phase transition occurs at \(293.2 \: \text{K}\), over \(20 \: \text{K}\) higher than at the lower pressure. We can measure the boiling point temperature of butane as a function of the applied pressure, and this result is plotted in Figure 13.3. Finally, we consider varying the substance which we trap in the cylinder. In each case, we discover that the boiling point temperature depends on both what the substance is and on the applied pressure, but does not depend on the amount of the substance we trap. In Figure 13.3, we have also plotted the boiling point as a function of the pressure for several substances. It is very clear that the boiling points for different substances can be very different from one another, although the variation of the boiling point with pressure looks similar from one substance to the next. Our previous observations indicate that, for a given pressure, there is a phase transition temperature for liquid and gas: below the boiling point, the liquid is the only phase which exists, and any gas which might exist at that point will spontaneously condense into liquid. Above the boiling point, the gas is the only stable phase. However, we can also commonly observe that any liquid left in an open container will, under most conditions, eventually evaporate, even if the temperature of the liquid is well below the normal boiling point. For example, we often observe that liquid water evaporates at temperatures well below the boiling point. This observation only seems surprising in light of the discussion above. Why would liquid water spontaneously evaporate if liquid is the more stable phase below the boiling point? We clearly need to further develop our understanding of phase transitions. The tendency of a liquid to evaporate is referred to as its : a more volatile liquid evaporates more readily. To make a quantitative measure of liquid volatility, we slightly modify our previous cylinder-piston apparatus by adding a gauge to measure the pressure of gas inside the cylinder (see Figure 13.4). We begin with liquid water only in the cylinder with an applied pressure of \(1 \: \text{atm}\) at a temperature of \(25^\text{o} \text{C}\). We now pull back the piston by an arbitrary amount, and then we lock the piston in place, fixing the volume trapped inside the cylinder. We might expect to have created a vacuum in the cavity above the liquid water, and as such we might expect that the pressure inside the cylinder is small or zero. Although there was initially no gas in the container, we observe that the pressure inside the container rises to a fixed value of \(23.8 \: \text{torr}\). Clearly, the observation of pressure indicates the presence of gaseous water inside the container, arising from evaporation of some, but not all, of the liquid water. Therefore, some of the liquid water must have evaporated. On the other hand, a look inside the container reveals that there is still liquid water present. Since both a liquid phase and a gas phase are present at the same time, we say that the liquid water and the water vapor must be in . The term in this case indicates that neither the vapor nor the liquid spontaneously converts into the other phase. Rather, both phases are stable at equilibrium. Very interestingly, we can repeat this measurement by pulling the piston back to any other arbitrary position before locking it down, and, provided that there is still some liquid water present, the pressure in the container in every case rises to the same fixed value of \(23.8 \: \text{torr}\). It does not matter what volume we have trapped inside the cylinder, nor does it matter how much liquid water we started with. As long as there is still some liquid water present in the cylinder at equilibrium, the pressure of the vapor above that liquid is \(23.8 \: \text{torr}\) at \(25^\text{o} \text{C}\). Note that, in varying either the amount of liquid initially or the fixed volume of the container, the of liquid water that evaporates must be different in each case. This can be seen from the fact that the volume available for vapor must be different in varying either the volume of the container or the initial volume of the liquid. Since we observe that the pressure of the vapor is the same at a fixed temperature, the differing volumes reveal differing numbers of moles of water vapor. Clearly it is the of the vapor, not the amount, which is the most important property in establishing the equilibrium between the liquid and the vapor. We can conclude that, at a given fixed temperature, there is a single specific pressure at which a given liquid and its vapor will be in phase equilibrium. We call this the of the liquid. We can immediately observe some important features of the vapor pressure. First, for a given substance, the vapor pressure varies with the temperature. This can be found by simply increasing the temperature on the closed container in the preceding experiment. In every case, we observe that the equilibrium vapor pressure increases with increases in the temperature. The vapor pressures of several liquids at several temperatures are shown in Figure 13.5. The vapor pressure for each liquid increases smoothly with the temperature, although the relationship between vapor pressure and temperature is definitely not proportional. Second, Figure 13.5 clearly illustrates that the vapor pressure depends strongly on what the liquid substance is. These variations reflect the differing of the liquids: those with higher vapor pressures are more volatile. In addition, there is a very interesting correlation between the volatility of a liquid and the boiling point of the liquid. Without exception, the substances with high boiling points have low vapor pressures and vice versa. Looking more closely at the connection between boiling point and vapor pressure, we can find an important relationship. Looking at Figure 13.5, we discover that the vapor pressure of each liquid is equal to \(760 \: \text{torr}\) (which is equal to \(1 \: \text{atm}\)) at the boiling point for that liquid. How should we interpret this? At an applied pressure of \(1 \: \text{atm}\), the temperature of the phase transition from liquid to gas is the temperature at which the vapor pressure of the liquid is equal to \(1 \: \text{atm}\). This statement is actually true regardless of which pressure we consider: if we apply a pressure of \(0.9 \: \text{atm}\), the boiling point temperature is the temperature at which the liquid has a vapor pressure of \(0.9 \: \text{atm}\). Stated generally, the liquid undergoes phase transition at the temperature where the vapor pressure equals the applied pressure. Since the boiling point is the temperature at which the applied pressure equals the vapor pressure, then we can view Figure 13.5 in a different way. Consider the specific case of water, with vapor pressure given in Figure 13.6. To find the boiling point temperature at \(1 \: \text{atm}\) pressure, we need to find the temperature at which the vapor pressure is \(1 \: \text{atm}\). To do so, we find the point on the graph where the vapor pressure is \(1 \: \text{atm}\) and read off the corresponding temperature, which must be the boiling point. This will work at any given pressure. Viewed this way, for water Figure 13.6 gives us the vapor pressure as a function of the temperature the boiling point temperature as a function of the pressure. They are the same graph. Recall that, at the boiling point, we observe that both liquid and gas are at equilibrium with one another. This is true at every combination of applied pressure and boiling point temperature. Therefore, for every combination of temperature and pressure on the graph in Figure 13.6, we observe liquid-gas equilibrium. What happens at temperature/pressure combinations which are not on the line in Figure 13.6? To find out, we first start at a temperature-pressure combination on the graph and elevate the temperature. The vapor pressure of the liquid rises, and if the applied pressure does not also increase, then the vapor pressure will be greater than the applied pressure. We must therefore not be at equilibrium anymore. All of the liquid vaporizes, and there is only gas in the container. Conversely, if we start at a point on the graph and lower the temperature, the vapor pressure is below the applied pressure, and we observe that all of the gas condenses into the liquid. Now, what if we start at a temperature-pressure combination on the graph and elevate the applied pressure without raising the temperature? The applied pressure will be greater than the vapor pressure, and all of the gas will condense into the liquid. Figure 13.6 thus actually reveals to us what phase or phases are present at each combination of temperature and pressure: along the line, liquid and gas are in equilibrium; above the line, only liquid is present; below the line, only gas is present. When we label the graph with the phase or phases present in each region as in Figure 13.6 we refer to the graph as a . Of course, Figure 13.6 only includes liquid, gas, and liquid-gas equilibrium. We know that, if the temperature is low enough, we expect that the water will freeze into solid. To complete the phase diagram, we need additional observations. We go back to our apparatus in Figure 13.4 and we establish liquid-gas water phase equilibrium at a temperature of \(25^\text{o} \text{C}\) and \(23.8 \: \text{torr}\). If we slowly lower the temperature, the vapor pressure decreases slowly as well, as shown in Figure 13.6. If we continue to lower the temperature, though, we observe an interesting transition, as shown in the more detailed Figure 13.7. The very smooth variation in the vapor pressure shows a slight, almost unnoticeable break very near to \(0^\text{o} \text{C}\). Below this temperature, the pressure continues to vary smoothly, but along a slightly different curve. To understand what we have observed, we examine the contents of the container. We find that, at temperatures below \(0^\text{o} \text{C}\), the water in the container is now an equilibrium mixture of water vapor and solid water (ice), and there is no liquid present. The direct transition from solid to gas, without liquid, is called . For pressure-temperature combinations along this new curve below \(0^\text{o} \text{C}\), then, the curve shows the solid-gas equilibrium conditions. As before, we can interpret this two ways. The solid-gas curve gives the vapor pressure of the solid water as a function of temperature, and also gives the sublimation temperature as a function of applied pressure. Figure 13.7 is still not a complete phase diagram, because we have not included the combinations of temperature and pressure at which solid and liquid are at equilibrium. As a starting point for these observations, we look more carefully at the conditions near \(0^\text{o} \text{C}\). Very careful measurements reveal that the solid-gas line and the liquid-gas line intersect in Figure 13.7 where the temperature is \(0.01^\text{o} \text{C}\). Under these conditions, we observe inside the container that solid, liquid, and gas are all three at equilibrium inside the container. As such, this unique temperature-pressure combination is called the . At this point, the liquid and the solid have the same vapor pressure, so all three phases can be at equilibrium. If we raise the applied pressure slightly above the triple point, the vapor must disappear. We can observe that, by very slightly varying the temperature, the solid and liquid remain in equilibrium. We can further observe that the temperature at which the solid and liquid are in equilibrium varies almost imperceptibly as we increase the pressure. If we include the solid-liquid equilibrium conditions on the previous phase diagram, we get Figure 13.8, where the solid-liquid line is very nearly vertical. Each substance has its own unique phase diagram, corresponding to the diagram in Figure 13.8 for water. There are several questions raised by our observations of phase equilibrium and vapor pressure. The first we will consider is why the pressure of a vapor in equilibrium with its liquid does not depend on the volume of the container into which the liquid evaporates, or on the amount of liquid in the container, or on the amount of vapor in the container. Why do we get the same pressure for the same temperature, regardless of other conditions? To address this question, we need to understand the coexistence of vapor and liquid in equilibrium. How is this equilibrium achieved? To approach these questions, let us look again at the situation in Figure 13.4. We begin with a container with a fixed volume containing some liquid, and equilibrium is achieved at the vapor pressure of the liquid at the fixed temperature given. When we adjust the volume to a larger fixed volume, the pressure adjusts to equilibrium at exactly the same vapor pressure. Clearly, there are more molecules in the vapor after the volume is increased and equilibrium is reestablished, because the vapor exerts the same pressure in a larger container at the same temperature. Also clearly, more liquid must have evaporated to achieve this equilibrium. A very interesting question to pose here is how the liquid responded to the increase in volume, which presumably only affected the space in which the gas molecules move. How did the liquid "know" to evaporate when the volume was increased? The molecules in the liquid could not detect the increase in volume for the gas, and thus could not possibly be responding to that increase. The only reasonable conclusion is that the molecules in the liquid were always evaporating, even before the volume of the container was increased. There must be a constant movement of molecules from the liquid phase into the gas phase. Since the pressure of the gas above the liquid remains constant when the volume is constant, then there must be a constant number of molecules in the gas. If evaporation is constantly occurring, then condensation must also be occurring constantly, and molecules in the gas must constantly be entering the liquid phase. Since the pressure remains constant in a fixed volume, then the number of molecules entering the gas from the liquid must be exactly offset by the number of molecules entering the liquid from the gas. At equilibrium, therefore, the pressure and temperature inside the container are unchanging, but there is constant movement of molecules between the phases. This is called . The situation is "equilibrium" in that the observable properties of the liquid and gas in the container are not changing, but the situation is "dynamic" in that there is constant movement of molecules between phases. The dynamic processes that take place offset each other exactly so that the properties of the liquid and gas do not change. What happens when we increase the volume of the container to a larger fixed volume? We know that the pressure equilibrates at the same vapor pressure, and that therefore there are more molecules in the vapor phase. How did they get there? It must be the case that when the volume is increased, evaporation initially occurs more rapidly than condensation until equilibrium is achieved. The rate of evaporation must be determined by the number of molecules in the liquid which have sufficient kinetic energy to escape the intermolecular forces in the liquid, and according to the kinetic molecular theory, this number depends only on the temperature, not on the volume. However, the rate of condensation must depend on the frequency of molecules striking the surface of the liquid. According to the Kinetic Molecular Theory, this frequency must decrease when the volume is increased, because the density of molecules in the gas decreases. Therefore, the rate of condensation becomes smaller than the rate of evaporation when the volume is increased, and therefore there is a net flow of molecules from liquid to gas. This continues until the density of molecules in the gas is restored to its original value, at which point the rate of evaporation is matched by the rate of condensation. At this point, this pressure stops increasing and is the same as it was before the volume was increased. In the phase diagram for water in Figure 13.8, start at the point where the temperature is \(60^\text{o} \text{C}\) and the pressure is \(400 \: \text{torr}\). Slowly increase the temperature with constant pressure until the temperature is \(100^\text{o} \text{C}\). State what happens physically to the water during this heating process. In the phase diagram for water in Figure 13.8, start at the point where the temperature is \(60^\text{o} \text{C}\) and the pressure is \(400 \: \text{torr}\). Slowly lower the pressure at constant temperature until the pressure is \(80 \: \text{torr}\). State what happens physically to the water during this process. Explain why Figure 13.6 is both a graph of the boiling point of liquid water as a function of applied pressure and a graph of the vapor pressure of liquid water as a function of temperature. We observe that, when the applied pressure is less than the vapor pressure of a liquid, all of the liquid will spontaneously evaporate. In terms of dynamic equilibrium, explain why no liquid can be present under these conditions. Using arguments from the Kinetic Molecular Theory and the concept of dynamic equilibrium, explain why, at a given applied pressure, there can be one and only one temperature, the boiling point, at which a specific liquid and its vapor pressure can be in equilibrium. Using dynamic equilibrium arguments, explain why the vapor pressure of a liquid is independent of the amount of liquid present. Using dynamic equilibrium arguments, explain why the vapor pressure of a liquid is independent of the volume available for the vapor above the liquid. Using dynamic equilibrium arguments, explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces. According to Figure 13.5, the vapor pressure of phenol is much less than the vapor pressure of dimethyl ether. Which of these substances has the greater intermolecular attractions? Which substance has the higher boiling point? Explain the difference in the intermolecular attractions in terms of molecular structure. The text describes dynamic equilibrium between a liquid and its vapor at the boiling point. Describe the dynamic equilibrium between a liquid and its solid at the melting point. Using this description, explain why the melting point of a solid varies very little as the pressure increases. ; Chemistry) | 24,652 | 2,935 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO13._(pi)_Donation_Steps |
Some nucleophiles are added to carbonyls, lose a proton to drop a positive charge, and have a lone pair again. If the (former) carbonyl oxygen also has a lone pair, a potential lone-pair/lonepair repulsion problem exists. Partly for this reason, two heteroatoms bonded to one carbon often present an inherently unstable situation. These kinds of species often decompose readily via pi donation. In pi donation, a lone pair on one heteroatom is donated to the carbon shared by both heteroatoms. As a result, the other heteroatom is pushed off the carbon. This event is helped if one of the heteroatoms is already protonated, so that it comes off as a neutral species. Fill in any missing lone pairs, provide curved arrows to show pi donation, and show the resonance structures that result. Show how each of the following species might break down into two molecules via pi donation. As a result of pi donation, neutral, protic nucleophiles often replace the carbonyl oxygen entirely. In effect, the nucleophile adds twice. The lone pair that is revealed after a deprotonation step adds again to the same carbon, pushing that carbonyl oxygen out of the molecule entirely. In the case of alcohol nucleophiles, a ketal or acetal results. This kind of molecule looks like two ethers that meet at one carbon. A ketal is a "masked" carbonyl; it still contains a carbon with two bonds to oxygen. However, a ketone is no longer an electrophile like a carbonyl compound. Imines also result from pi donation. If an primary amine donates to a carbonyl, it can lose its first proton to reveal a lone pair. Once that lone pair donates, pushing off the carbonyl oxygen, a second proton can be dropped to allow the nitrogen to lose its positive charge. Imines are similar to carbonyls in that they contain a carbon-heteroatom double bond and so they are still good electrophiles. , | 1,877 | 2,936 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.03%3A_Handling_Large_and_Small_Numbers |
illustrates a common occurrence in science—results often involve very large numbers or very small fractions. The United States used 66 500 000 000 000 000 000 J (joules) of energy in 1971, and the mass of a water molecule is 0.000 000 000 000 000 000 000 029 9 g. Such numbers are inconvenient to write and hard to read correctly. (We have divided the digits into groups of three to make it easier to locate the decimal point. Spaces are used instead of commas because many countries use a comma to indicate the decimal.) There are two ways of handling this problem. We can express a quantity in larger or smaller units, as in , or we can use a better way to write small and large numbers. The latter approach involves what is called or . The position of the decimal point is indicated by a power (or exponent) of 10. For example, \[138= 13.8\cdot 10= 1.38\cdot 10\cdot 10= 1.38\cdot 10^2 \nonumber \] \[0.004\ 83= \frac{4.83}{10\cdot10\cdot10} = 4.83\cdot\frac{1}{10^3} = 4.83\cdot10^{-3} \nonumber \] A number with a negative exponent is simply the of (one divided by) the same number with the equivalent positive exponent. Therefore decimal fractions (numbers between zero and one) may be expressed using negative powers of 10. Numbers between 1 and 10 require no exponential part, and those larger than 10 involve positive exponents. By convention the power of 10 is chosen so that there is one digit to the left of the decimal point in the ordinary number. That is, we would usually write 5280 as 5.28 × 10 not as 0.528 × 10 or 52.8 × 10 . To convert a number from ordinary to scientific notation, count how many places the decimal point must be shifted to arrive at a number between 1 and 10. If these shifts are to the left, the number was large to begin with and we multiply by a large (that is, positive) power of 10. If the shift is to the right, a reciprocal (negative) power of 10 must be used. Express the following numbers in scientific notation: (a) 7563; (b) 0.0156. In this case the decimal point must be shifted places: Therefore we use an exponent of +3: \[7563= 7.563\cdot10^3 \nonumber \] Shifting the decimal point to the yields a number between 1 and 10: Therefore the exponent is –2: \[0.0156= 1.56\cdot10^{-2} \nonumber \] When working with exponential notation, it is often necessary to add, subtract, multiply, or divide numbers. When multiplying and dividing, you must remember that multiplication corresponds to addition of exponents, and division to their subtraction. Multiplication: \(\text{10}^a\cdot\text{10}^b = \text{10}^{\small(a + b\small)}\) Division: \(\frac{10^a}{10^b} = 10^{(a - b)}\) Hence \((3.0\cdot10^5)\cdot(5.0\cdot10^3) = 15.0\cdot10^{(5+3)}= 15.0\cdot10^8= 1.50\cdot10^9\) and \(\frac{3.0\cdot10^5}{5.0\cdot10^3} =0.6\cdot10^{(5- 3)} = 0.6\cdot10^2= 6.0\cdot10\) Evaluate the following, giving your answer in correct exponential notation: Addition and subtraction require that all numbers be converted to the same power of 10. (This corresponds to lining up the decimal points.) Evaluate the following, giving your answer in scientific notation: \((6.32\cdot10^2) – (1.83 \cdot10^\cdot{-1}) \) \((3.72 \cdot10^4) + (1.63\cdot10^5) – (1.7 10^3) \) First convert to the same power of 10; then add the ordinary numbers. \( \begin{align}&&6.32\cdot10^2&=& 632 \\&&–1.83\cdot10^{-1}&=& – 0.183 \\632 – 0.183&=& 631.817&=& 6.318 17\cdot10^2 \end{align}\) Convert all powers of 10 to 10 . \( \begin{align}3.72\cdot10^4 &=& 3.72 \cdot10^4 &=& 3.72 × 10^4 \\1.63\cdot10^5&=& 1.63\cdot10 \cdot10^4&=& 16.3 \cdot10^4 \\–1.7 \cdot10^3&=& –1.7 \cdot10^{-1}\cdot10^4&=& – 0.17\cdot10^4 \\ (3.72\cdot10^4) + (16.3\cdot10^4) - (0.17 \cdot10^4)&=& 19.85 \cdot10^4&=& 1.985\cdot10^5\end{align} \) Scientific notation is becoming more common every day. Many electronic pocket calculators use it to express numbers which otherwise would not fit into their displays. For example, an eight-digit calculator could not display the number 6 800 000 000.The decimal point would remain fixed on the right, and the 6 and the 8 would “overflow” to the left side. Such a number is often displayed as 6.8 09 which means 6.8 × 10 . If you use a calculator which does have scientific notation, we recommend that you express all numbers as powers of 10 before doing any arithmetic. Follow the rules in the last two examples, using your calculator to do arithmetic on the ordinary numbers. You should be able to add or subtract the powers of 10 in your head. Computers also are prone to print results in scientific notation, and they use yet another minor modification. The printed number 2.3074 E-07 means 2.3074 × 10 for example. In this case the E indicates that the number following is an exponent of 10. Again in reference to , we could express the mass of smoke collected as 3.42 × 10 g and the volume of the balloon as 1.021 926 4 × 10 cm . There is something strange about the second quantity, though. It contains a number which was copied directly from the display of an electronic calculator and has . The reliability of a quantity derived from a measurement is customarily indicated by the number of significant figures (or significant digits) it contains. For example, the three significant digits in the quantity 3.42 × 10 g tell us that a balance was used on which we could distinguish 3.42 × 10 g from 3.43 × 10 g or 3.41 × 10 g. There might be some question about the last digit, but those to the left of it are taken as completely reliable. Another way to indicate the same thing is (3.42±0.01) × 10 g. Our measurement is somewhere between 3.41 × 10 g and 3.43 × 10 g. As another example of choosing an appropriate number of significant digits, let us read the volume of liquid in a graduated cylinder. The bottom of the meniscus lies between graduations corresponding to 38 and 39 cm . We can estimate that it is at 38.5 cm , but the last digit might be off a bit-perhaps it looks like 38.4 or 38.6 cm to you. Since the third digit is in question, we should use three significant figures. The volume would be recorded as 38.5 cm . Laboratory equipment is often calibrated similarly to this graduated cylinder—you should estimate to the nearest tenth of the smallest graduation. In some ordinary numbers, for example, 0.001 23, zeros serve merely to locate the decimal point. They do not indicate the reliability of the measurement and therefore are not significant. Another advantage of scientific notation is that we can assume that all digits are significant. Thus if 0.001 23 is written as 1.23 × 10 , only the 1, 2, and 3, which indicate the reliability of the measurement, are written. The decimal point is located by the power of 10. If the rule expressed in the previous paragraph is applied to the volume of air collected in our pollution experiment, 1.021 926 4 × 10 cm , we find that the volume has eight significant digits. This implies that it was determined to ±1 cm out of about 10 million cm , a reliability which corresponds to locating a grasshopper at some point along the road from Philadelphia to New York City. For experiments as crude as ours, this is not likely. Let us see just how good the measurement was. You will recall that we , 106 in. The three significant figures imply that this might have been as large as 107 in or as small as 105 in. We can repeat the calculation with each of these quantities to see how far off the volume would be: \[\begin{equation} \begin{split} r &= \frac{1}{2} \times \text{107 in} = \text{53.5 in} \times \frac{\text{1 cm}}{\text{0.3937 in}} \\ \\ &=135.890 27 \text{cm} \end{split} \end{equation} \nonumber \] \[\begin{align} V &= \frac{4}{3} \times 3.141 59 \times (135.890 27)^3 \\ \\ &= 10 511 225 \text{cm}^3 = 1.051 122 5 \times 10^7 \text{cm}^3 \end{align} \nonumber \] or \[\begin{align} V &= \frac{4}{3} \times 3.14159 \times \left( \frac{1}{2} \times \text{105 in} \frac{\text{1 cm}}{\text{0.3937 in}}\right)^3 \\ \\ &= 9 932 759 \text{cm}^3 = 0.993 275 9 \times 10^7 \text{cm}^3 \end{align} \nonumber \] That is, the volume is between 0.99 × 10 and 1.05 × 10 cm or (1.02 ± 0.03) × 10 cm . We should our result to three significant figures, for example, 1.02 cm³, because the last digit, namely 2, is in question. Application of the Rules for Rounding Numbers can be illustrated by an example. Round each of the numbers below to three significant figures. Apply rules 1 and 2: (Note that a different result would be obtained if the digits were incorrectly rounded one at a time from the right.) Apply rules 1 and 3: 34.8 34.9 Apply rules 1 and 4: 34.7 34.8 Apply rules 1 and 5: 34.2 34.2 Apply rule 5: 34.3 34.4 To how many significant figures should we round our air-pollution results? We have already done a calculation involving multiplication and division to obtain the volume of our gas-collection balloon. It involved the following numbers: \(\frac{4}{3}\) and \(\frac{1}{2}\) The result of the calculation contained three significant figures — the same as the least-reliable number. This illustrates the general rule that Defined numbers such as π, ½ or 100 cm/1m are assumed to have an infinite number of significant figures. In the case of addition and subtraction, a different rule applies. Suppose, for example, that we weighed a smoke-collection filter on a relatively inaccurate balance that could only be read to the nearest 0.01 g. After collecting a sample, the filter was reweighed on a single-pan balance to determine the mass of smoke particles. Final mass: 2.374 g (colored digits are in question) Initial mass: –2.3 g Mass of smoke: 0.0 g Since the initial weighing could have been anywhere from 2.31 to 2.33 g, all three figures in the final result are in question. (It must be between 0.0445 and 0.0645 g). Thus there is but one significant digit, and the result is 0.05g. The rule here is Note that subtraction can drastically reduce the number of significant digits when this rule is applied. Rounding numbers is especially important if you use an electronic calculator, since these machines usually display a large number of digits, most of which are meaningless. The best procedure is to carry all digits to the end of the calculation (your calculator will not mind the extra work!) and then round appropriately. Answers to subsequent calculations in this book will be rounded according to the rules given. You may wish to go back to previous examples and round their answers correctly as well. Evaluate the following expressions, rounding the answer to the appropriate number of significant figures. When we suggested filling a surplus weather balloon to measure how much gas was pumped through our air-pollution collector, we mentioned that this would be a rather crude way to determine volume. For one thing, it would not be all that simple to measure the diameter of an 8- or 9-ft sphere reliably. Using a yardstick, we would be lucky to have successive measurements agree within half an inch or so. It was for this reason that the result was reported to the nearest inch. The degree to which repeated measurements of the same quantity yield the same result is called . Repetition of a highly precise measurement would yield almost identical results, whereas low precision implies numbers would differ by a significant percentage from each other. A highly precise measurement of the diameter of our balloon could be achieved, but it would probably not be worthwhile. We have assumed a spherical shape, but this is almost certainly not exactly correct. No matter how precisely we determine the diameter, our measurement of gas volume will be influenced by deviations from the assumed shape. When one or more of our assumptions about a measuring instrument are wrong, the of a result will be affected. An obvious example would be a foot rule divided into 11 equal inches. Measurements employing this instrument might agree very precisely, but they would not be very accurate. An important point of a different kind is illustrated in the last two paragraphs. A great many common words have been adopted into the language of science. Usually such an adoption is accompanied by an unambiguous scientific definition which does not appear in a normal dictionary. Another example is quantity, which we have defined in terms of “number × unit.” Other English words like bulk, size, amount, and so forth, may be synonymous with quantity in everyday speech, but not in science. As you encounter other words like this, try to learn and use the scientific definition as soon as possible, and avoid confusing it with the other meanings you already know. Even granting the crudeness of the measurements we have just described, they would be adequate to demonstrate whether or not an air-pollution problem existed. The next step would be to find a chemist or public health official who was an expert in assessing air quality, present your data, and convince that person to lend his or her skill and authority to your contention that something was wrong. Such a person would have available equipment whose precision and accuracy were adequate for highly reliable measurements and would be able to make authoritative public statements about the extent of the air-pollution problem. Several sites were inspired by Charles Eames' "Powers of Ten" : | 13,344 | 2,937 |
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Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid. Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy ( ) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure \(\Page {1}\)), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy (\(E_0\)) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than \(E_0\). The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than \(E_0\) has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure. To understand the causes of vapor pressure, consider the apparatus shown in Figure \(\Page {2}\). When a liquid is introduced into an evacuated chamber (part (a) in Figure \(\Page {2}\)), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with \(KE > E_0\) will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase. As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure \(\Page {2}\)). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure \(\Page {3}\). Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid. If a liquid is in an container, however, most of the molecules that escape into the vapor phase will collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure \(\Page {4}\)) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile. The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure \(\Page {4}\). Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release. Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions. A Discussing Vapor Pressure and Boiling Points. The exponential rise in vapor pressure with increasing temperature in Figure \(\Page {4}\) allows us to use natural logarithms to express the nonlinear relationship as a linear one. \[ \boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1} \] where Plotting \(\ln P\) versus the inverse of the absolute temperature (\(1/T\)) is a straight line with a slope of −Δ / . Equation \(\ref{Eq1}\), called the , can be used to calculate the \(ΔH_{vap}\) of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine \(ΔH_{vap}\) is to measure the vapor pressure of a liquid at temperatures and insert the values of \(P\) and \(T\) for these points into Equation \(\ref{Eq2}\), which is derived from the Clausius–Clapeyron equation: \[ \ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2} \] Conversely, if we know Δ and the vapor pressure \(P_1\) at any temperature \(T_1\), we can use Equation \(\ref{Eq2}\) to calculate the vapor pressure \(P_2\) at any other temperature \(T_2\), as shown in Example \(\Page {1}\). A Discussing the Clausius-Clapeyron Equation. Link: The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table: From these data, calculate the enthalpy of vaporization (Δ ) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.) vapor pressures at four temperatures Δ of mercury and vapor pressure at 160°C The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation \(\ref{Eq1}\) and find the value of Δ from the slope of the line, an alternative approach is to use Equation \(\ref{Eq2}\) to obtain Δ directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C ( ) and 120.0°C ( ) into Equation \(\ref{Eq2}\) gives \[\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \\[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \\[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \\[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber \] We can now use this value of Δ to calculate the vapor pressure of the liquid ( ) at 160.0°C ( ): \[ \ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber \] Using the relationship \(e^{\ln x} = x\), we have \[\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \\[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \\[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber \] At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect. The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr? 1896°C As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure \(\Page {4}\) are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table \(\Page {1}\) lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure increases, the boiling point of a liquid increases and vice versa. Use Figure \(\Page {4}\) to estimate the following. Data in Figure \(\Page {4}\), pressure, and boiling point corresponding boiling point and pressure Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure \(\Page {4}\) to estimate the following. 200°C 450 mmHg Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called or , generates a above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via . Eventually, a is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of . Under these conditions, a liquid exhibits a characteristic that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the . This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. are liquids with high vapor pressures, which tend to evaporate readily from an open container; have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its . ( ) | 14,787 | 2,938 |
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We often encounter cases where two of the variables , , n and are allowed to vary for a given sample of gas, and we are interested in the change in the value of the third under the new conditions. If we rearrange the ideal gas law so that , , and , the quantities that change, are on one side and the constant terms ( and for a given sample of gas) are on the other, we obtain
\( \dfrac{PV}{T}=nR = constant \tag{10.4.1} \) Thus the quantity /n is constant
\( \dfrac{P_{1}V_{1}}{n_{1}T_{1}}= \dfrac{P_{2}V_{2}}{n_{2}T_{2}} \tag{10.4.2} \) In many of these problems one considers the changes in a sample of gas where the number of moles does not change. In such a case one can write the relationship as
\( \dfrac{P_{1}V_{1}}{T_{1}}= \dfrac{P_{2}V_{2}}{T_{2}} \tag{10.4.3} \) If two of the parameters specifying the thermodynamic state of the gas sample are held constant, one recovers simple proportional relationships among the other two which were experimentally observed centuries ago by \( V \propto \frac{1}{P} \left ( at \;constant\; n,\; V \right ) \tag{10.4.4} \) expresses the idea that at constant composition (e.g. the number of moles of gas) and temperature the product PV is a constant. \( V \propto T \left (at \;constant\; n, P \right ) \tag{10.4.5} \) expresses the idea that at constant composition (e.g. the number of moles of gas) and pressure the ratio V/T is a constant. Using Boyle's law this also can be written as P/T being constant \( V \propto n \left (at \;constant\; T, P \right ) \tag{10.4.6} \) In its original formulation, Avogadro's law stated that equal volumes of gas at constant temperature and pressure held the same number of molecules Brief descriptions of these scientists work are found at the end of this section. Charles originally observed the change of volume with temperature in balloon ascents and descents that he made. Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C. How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example 5? temperature, pressure, amount, and volume in August; temperature in January volume in January Use the results from Example 5 for August as the initial conditions and then calculate the due to the change in temperature from 86°F to 14°F. Begin by constructing a table showing the initial and final conditions. Rearrange the ideal gas law to isolate those quantities that differ between the initial and final states on one side of the equation, in this case and . Equate the ratios of those terms that change for the two sets of conditions. Making sure to use the appropriate units, insert the quantities and solve for the unknown parameter. To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions: Thus we are asked to calculate the effect of a change in temperature on the volume of a fixed amount of gas at constant pressure. Both \(n\) and \(P\) are the same in both cases (\(n_i=n_f,P_i=P_f\)). Therefore, Equation 10.4.3 can be simplified to: \[\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f \notag }\] This is the relationship first noted by Charles. Solving the equation for \(V_f\), we get: \[V_f=V_i\times\dfrac{T_f}{T_i}=\rm31150\;L\times\dfrac{263\;K}{303\;K}=2.70\times10^4\;L \notag \] It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change. It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change. Exercise At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen ( = −196°C). What is the final volume of the gas in the balloon? 0.52 L Example 8 illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle ( = constant) and the relationship between volume and amount observed by Avogadro ( / = constant). We will not do so for all cases, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example 8 can be applied in such case, as we demonstrate in Example 9 (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion). Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise 5 (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)? initial volume, amount, temperature, and pressure; final temperature final pressure Follow the strategy outlined in Example 8. Prepare a table to determine which parameters change and which are held constant: Once again, two parameters are constant while one is varied, and we are asked to calculate the fourth. As before, we begin with the ideal gas law and rearrange it as necessary to get all the constant quantities on one side. In this case, because and are constant, we rearrange to obtain \( P=\left ( \dfrac{nR}{V} \right )\left ( T \right )= constant \times T \) Dividing both sides by , we obtain an equation analogous to the one in Example 6, / = / = constant. Thus the ratio of to does not change if the amount and volume of a gas are held constant. We can thus write the relationship between any two sets of values of and for the same sample of gas at the same volume as \( \dfrac{P_{i}}{T_{i}}=\dfrac{P_{f}}{T_{f}} \notag \) In this example, = 1.5 atm, = 298 K, and = 1023 K, and we are asked to find . Solving for and substituting the appropriate values, we obtain \[P_f=P_i\times\dfrac{T_f}{T_i}=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K}=5.1\;atm \notag \] This pressure is more than enough to rupture a thin sheet metal container and cause an explosion! Exercise Suppose that a fire extinguisher, filled with CO to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher? 23.4 atm We saw in Example 5 that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 10 mol of H gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude? initial pressure, temperature, amount, and volume; final pressure and temperature final volume Follow the strategy outlined in Example 6. Begin by setting up a table of the two sets of conditions: Thus all the quantities except are known. Solving for and substituting the appropriate values give \[V_f=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i}=\rm3.115\times10^4\;L\times\dfrac{0.980\; \cancel{atm}}{0.411\;\cancel{atm}}\dfrac{243\;\cancel{K}}{303\;\cancel{K}}=5.96\times10^4\;L \notag \] Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to the volume of the gas, while decreasing the temperature tends to the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both and 1/ , the variable that changes the most will have the greatest effect on . In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation. We could also have solved this problem by solving the ideal gas law for and then substituting the relevant parameters for an altitude of 23,000 ft: \( V=\dfrac{nRT}{P}= \dfrac{\left ( 1.23\times 10^{3}\; \cancel{mol}\left [ 0.082057 \; \left ( L\cdot \cancel{atm} \right )/\left ( \cancel{K} \cdot \cancel{mol} \right ) \right ]\left ( 243 \; \cancel{K} \right ) \right )}{0.411 \; \cancel{atm}} = 5.97\times 10^{4} L \) Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. Exercise A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.) 4.07 × 10 As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal. Boyle, the youngest (and 14th!) child of the Earl of Cork, was an important early figure in chemistry whose views were often at odds with accepted wisdom. Boyle’s studies of gases are reported to have utilized a very tall J-tube that he set up in the entryway of his house, which was several stories tall. He is known for the gas law that bears his name and for his book, , which was published in 1661 and influenced chemists for many years after his death. In addition, one of Boyle’s early essays on morals is said to have inspired Jonathan Swift to write . The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in . In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways ( ). A simple plot of versus gives a curve called a and reveals an relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
(\ PV = constant \tag{10.4.7} \) Dividing both sides by gives an equation illustrating the inverse relationship between and :
\( V=\dfrac{constant}{P}=constant\left ( \dfrac{1}{P} \right )\; or\; V \propto \dfrac{1}{P} \tag{10.4.8}\) where the ∝ symbol is read “is proportional to.” A plot of versus 1/ is thus a straight line whose slope is equal to the constant in . Dividing both sides of by instead of gives a similar relationship between and 1/ . The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law , after its discoverer, and can be stated as follows: Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air. The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of versus was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C ( ) or absolute zero. However, the ideal gas model does allow for reaching absolute zero, but the second law of thermodynamics, which will be discussed in Unit 7, makes this impossible. In 1783, Charles filled a balloon (“aerostatic globe”) with hydrogen (generated by the reaction of iron with more than 200 kg of acid over several days) and flew successfully for almost an hour. When the balloon descended in a nearby village, however, the terrified townspeople destroyed it. In 1804, Gay-Lussac managed to ascend to 23,000 ft (more than 7000 m) to collect samples of the atmosphere to analyze its composition as a function of altitude. In the process, he had trouble breathing and nearly froze to death, but he set an altitude record that endured for decades. (To put Gay-Lussac’s achievement in perspective, recall that modern jetliners cruise at only 35,000 ft!) We can state Charles’s and Gay-Lussac’s findings in simple terms: This relationship is often referred to as Charles’s law Charles’s law is valid for virtually all gases at temperatures well above their boiling points. not We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles. (This is the historic “Avogadro’s hypothesis” introduced in .) A logical corollary, sometimes called Avogadro’s law , describes the relationship between the volume and the amount of a gas: Stated mathematically,
\( V=\left ( constant \right )\left (n \right )or\; V \propto n \left ( at \;constant \;T \;and \;P \right )\tag{10.3.4} \) This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure ( ), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure ( ), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present ( ). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is , the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. Sketch a graph of the volume of a gas versus the pressure on the gas. What would the graph of versus look like if volume was directly proportional to pressure? What properties of a gas are described by Boyle’s law, Charles’s law, and Avogadro’s law? In each law, what quantities are held constant? Why does the constant in Boyle’s law depend on the amount of gas used and the temperature at which the experiments are carried out? Use Charles’s law to explain why cooler air sinks. 4. Use Boyle’s law to explain why it is dangerous to heat even a small quantity of water in a sealed container. For an ideal gas, is volume directly proportional or inversely proportional to temperature? What is the volume of an ideal gas at absolute zero? For a given amount of a gas, the volume, temperature, and pressure under any one set of conditions are related to the volume, the temperature, and the pressure under any other set of conditions by the equation \( \dfrac{P_{1}V_{1}}{T_{1}}= \dfrac{P_{2}V_{2}}{T_{2}} \) Derive this equation from the ideal gas law. At constant temperature, this equation reduces to one of the laws discussed in ; which one? At constant pressure, this equation reduces to one of the laws discussed in this section; which one? Predict the effect of each change on one variable if the other variables are held constant. What would the ideal gas law be if the following were true? Given the following initial and final values, what additional information is needed to solve the problem using the ideal gas law? Given the following information and using the ideal gas law, what equation would you use to solve the problem? Using the ideal gas law as a starting point, derive the relationship between the density of a gas and its molar mass. Which would you expect to be denser—nitrogen or oxygen? Why does radon gas accumulate in basements and mine shafts? Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant. Tennis balls that are made for Denver, Colorado, feel soft and do not bounce well at lower altitudes. Use the ideal gas law to explain this observation. Will a tennis ball designed to be used at sea level be harder or softer and bounce better or worse at higher altitudes? A 1.00 mol sample of gas at 25°C and 1.0 atm has an initial volume of 22.4 L. Calculate the results of each change, assuming all the other conditions remain constant. A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions? The sample is compressed to 6.0 atm at constant temperature, giving a volume of 3.99 L. Is this result consistent with Boyle’s law? A 8.60 L tank of nitrogen gas at a pressure of 455 mmHg is connected to an empty tank with a volume of 5.35 L. What is the final pressure in the system after the valve connecting the two tanks is opened? Assume that the temperature is constant. At constant temperature, what pressure in atmospheres is needed to compress 14.2 L of gas initially at 25.2 atm to a volume of 12.4 L? What pressure is needed to compress 27.8 L of gas to 20.6 L under similar conditions? One method for preparing hydrogen gas is to pass HCl gas over hot aluminum; the other product of the reaction is AlCl . If you wanted to use this reaction to fill a balloon with a volume of 28,500 L at sea level and a temperature of 78°F, what mass of aluminum would you need? What volume of HCl at STP would you need? An 3.50 g sample of acetylene is burned in excess oxygen according to the following reaction: At STP, what volume of CO (g) is produced? Calculate the density of ethylene (C H ) under each set of conditions. Determine the density of O under each set of conditions. At 140°C, the pressure of a diatomic gas in a 3.0 L flask is 635 kPa. The mass of the gas is 88.7 g. What is the most likely identity of the gas? What volume must a balloon have to hold 6.20 kg of H for an ascent from sea level to an elevation of 20,320 ft, where the temperature is −37°C and the pressure is 369 mmHg? What must be the volume of a balloon that can hold 313.0 g of helium gas and ascend from sea level to an elevation of 1.5 km, where the temperature is 10.0°C and the pressure is 635.4 mmHg? A typical automobile tire is inflated to a pressure of 28.0 lb/in. Assume that the tire is inflated when the air temperature is 20°C; the car is then driven at high speeds, which increases the temperature of the tire to 43°C. What is the pressure in the tire? If the volume of the tire had increased by 8% at the higher temperature, what would the pressure be? The average respiratory rate for adult humans is 20 breaths per minute. If each breath has a volume of 310 mL of air at 20°C and 0.997 atm, how many moles of air does a person inhale each day? If the density of air is 1.19 kg/m , what is the average molecular mass of air? Kerosene has a self-ignition temperature of 255°C. It is a common accelerant used by arsonists, but its presence is easily detected in fire debris by a variety of methods. If a 1.0 L glass bottle containing a mixture of air and kerosene vapor at an initial pressure of 1 atm and an initial temperature of 23°C is pressurized, at what pressure would the kerosene vapor ignite? Thumbnail from | 22,182 | 2,939 |
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The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system. A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure \(\Page {1}\)). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure \(\Page {1}\)) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid. The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure. The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure \(\Page {1}\), the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in . It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point. Remember that a phase diagram, such as the one in Figure \(\Page {1}\), is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error. Figure \(\Page {2}\) shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee). The phase diagram for water illustrated in Figure \(\Page {2b}\) shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure \(\Page {1}\); that is, the melting point of ice with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice. In Figure \(\Page {2b}\) point A is located at = 1 atm and = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines. Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation. Recall that pressure ( ) is the force ( ) applied per unit area ( ): \[P=\dfrac{F}{A} \nonumber \] To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is \[ F = mg \nonumber \] where is the mass and is the acceleration due to Earth’s gravity (9.81 m/s ). Thus the force is \[F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber \] If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is \[ A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber \] If the skater is gliding on one foot, the pressure exerted on the ice is \[ P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber \] The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases. Referring to the phase diagram of water in Figure \(\Page {2}\): phase diagram, temperature, and pressure physical form and physical changes Referring to the phase diagram of water in Figure \(\Page {2}\), predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm. The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid. In contrast to the phase diagram of water, the phase diagram of CO (Figure \(\Page {3}\)) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed. Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps. As the phase diagrams above demonstrate, a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature ( ), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure ( ), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in Figure \(\Page {1}\). High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa. A Discussing Phase Diagrams. | 10,098 | 2,940 |
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Often it is convenient to prepare a series of solutions of known by first preparing a single as described in . (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration. A pipet is used to measure 50.0 ml of 0.1027 HCl into a 250.00-ml volumetric flask. Distilled water is carefully added up to the mark on the flask. What is the concentration of the diluted solution? To calculate concentration, we first obtain the amount of HCI in the 50.0 ml (50.0 cm ) of solution added to the volumetric flask: \[n_{\text{HCl}}=\text{50}\text{.0 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{0}\text{.1027 mmol}}{\text{1 cm}^{\text{3}}}=\text{5}\text{.14 mmol} \nonumber \] Dividing by the new volume gives the concentration \[c_{\text{HCl}}=\dfrac{n_{\text{HCl}}}{V}=\dfrac{\text{5}\text{.14 mmol}}{\text{250}\text{.00 cm}^{\text{3}}}=\text{0.0205 mmol cm}^{\text{-3}} \nonumber \] Thus the new solution is 0.0205 . Alternatively, \[n_{\text{HCl}}=\text{50}\text{.0 mL} ~\times~ \dfrac{\text{10}^{-3}\text{L}}{\text{1 ml}} ~\times~\dfrac{\text{0}\text{.1027 mol}}{\text{1 L}} \nonumber \]
\[~=~\text{5.14}\times{10}^{-3}\text{mol} \nonumber \] \[c_{\text{HCl}}=\dfrac{n_{\text{HCl}}}{V} ~ = ~ \dfrac{\text{5}\text{.14}\times\text{10}^{-3}\text{mol}}{\text{250.00 ml}~\times~\dfrac{1 \text{ml}}{\text{10}^{-3}\text{L}}} \nonumber \] \[~ = ~ \text{0.0205 mol/L} \nonumber \] What volume of the solution of 0.316 46 KI prepared in would be required to make 50.00 ml of 0.0500 KI? Using the volume and concentration of the desired solution, we can calculate the amount of KI required. Then the concentration of the original solution (0.316 46 ) can be used to convert that amount of KI to the necessary volume. Schematically \[\begin{align} & V_{\text{new}}\xrightarrow{c_{\text{new}}}n_{\text{KI}}\xrightarrow{c_{\text{old}}}V_{\text{old}} \\ & \\ & V_{\text{old}}=\text{50}\text{.00 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{0}\text{.0500 mmol}}{\text{1 cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{1 cm}^{\text{3}}}{\text{0}\text{.316 46 mmol}}\text{ }=\text{7}\text{.90 cm}^{\text{3}} \\ \end{align} \nonumber \] Thus we should dilute a 7.90-ml aliquot of the stock solution to 50.00 ml. This could be done by measuring 7.90 ml from a buret into a 50.00-ml volumetric flask and adding water up to the mark. Note that the calculation above can be simplified, since the concentration and volume of a concentrated solution (C and V ) were used to calculate the amount of solute, and this amount was entirely transferred to the dilute solution: \[ C_{conc} \times V_{conc} = n_{conc} = n_{dil} = C_{dil} \times V_{dil} \nonumber \] So \[ C_{conc} \times V_{conc} = C_{dil} \times V_{dil} \nonumber \] So for Example \(\Page {2}\), \[( 0.316 46 M ) \times (V_{conc}) = (50.00 \text{ ml} ) \times ( 0.0500 M) \nonumber \] \[ V_{conc} = 7.90 \text{ mL} \nonumber \] , which will be diluted to 50.00 mL as before. Note that the calculated volume will have the same dimensions as the input volume, and dimensional analysis tells us that in this case we don't need to convert to liters, since L cancels when we divide M (mol/L) by M (mol/L). | 3,438 | 2,942 |
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The following sections are concerned with , , and . These seemingly unrelated subjects are discussed together because many of the calculations involving them are almost identical in form. The same is true of and of the calculations involving and . In each case one quantity is defined as the ratio of two others. The first quantity serves as a conversion factor relating the other two. A summary of the relationships and conversion factors we have encountered so far is given in Table \(\Page {1}\). An incredible variety of problems can be solved using in Table \(\Page {1}\). Sometimes only one factor is needed, but quite often several are applied in sequence, as in . In solving such problems, it is necessary first to think your way through, perhaps by writing down a road map showing the relationships among the quantities given in the problem. Then you can apply conversion factors, making sure that the units cancel, and calculate the result. The examples in these sections should give you some indication of the broad applications of the problem-solving techniques we have developed here. Once you have mastered these techniques, you will be able to do a great many useful computations which are related to problems in the chemical laboratory, in everyday life, and in the general environment. You will find that the same type of calculations, or more complicated problems based on them, will be encountered again and again throughout your study of chemistry and other sciences. \[\large\underset{\text{Gasoline}}{ \text{2C}_{\text{8}}\text{H}_{\text{12}}} \text{ + }\underset{\text{Air}}{ \text{25O}_{\text{2}}}\text{ } \text{ } \rightarrow \underset{\begin{smallmatrix} \text{Carbon} \\ \text{dioxide}\end{smallmatrix}}{ \text{16CO}_{\text{2}}} \text{ + }\underset{\text{Water}}{ \text{18H}_{\text{2}}\text{O}} \nonumber \] There are a great many circumstances in which you may need to use a balanced equation. For example, you might want to know how much air pollution would occur when 100 metric tons of coal were burned in an electric power plant, how much heat could be obtained from a kilogram of natural gas, or how much vitamin C is really present in a 300-mg tablet. In each instance someone else would probably have determined what reaction takes place, but you would need to use the balanced equation to get the desired information. | 2,376 | 2,943 |
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Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm or lb/ft . Now we will see how representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the volume of a given of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions. When we are referring to the same object or sample of material, it is often useful to be able to convert one into another. For example, in our discussion of fossil-fuel reserves we find that 318 Pg (3.18 × 10 g) of coal, 28.6 km (2.68 × 10 m ) of petroleum, and 2.83 × 10 km (2.83 × 10 m ) of natural gas (measured at normal atmospheric pressure and 15°C) are available. But none of these quantities tells us what we really want to know ― how much could be released by burning each of these reserves? Only by converting the mass of coal and the volumes of petroleum and natural gas into their equivalent energies can we make a valid comparison. When this is done, we find that the coal could release 7.2 × 10 J, , the petroleum 1.1 × 10 J, and the gas 1.1 × 10 J of heat energy. Thus the reserves of coal are more than three times those of the other two fuels combined. It is for this reason that more attention is being paid to the development of new ways for using coal resources than to oil or gas. Conversion of one kind of quantity into another is usually done with what can be called a , but the conversion factor is based on a mathematical (D = m / V) or mathematical equation that relates parameters. Since we have not yet discussed energy or the units (joules) in which it is measured, an example involving the more familiar quantities mass and volume will be used to illustrate the way conversion factors are employed. The same principles apply to finding how much energy would be released by burning a fuel, and that problem will be encountered later. For helpful context about the above discussion, check out the following Crash Course Chemistry video: Suppose we have a rectangular solid sample of gold which measures 3.04 cm × 8.14 cm × 17.3 cm. We can easily calculate that its volume is 428 cm but how much is it worth? The price of gold is about 5 dollars per gram, and so we need to know the mass rather than the volume. It is unlikely that we would have available a scale or balance which could weigh accurately such a large, heavy sample, and so we would have to determine the mass of gold equivalent to a volume of 428 cm . This can be done by manipulating the equation which defines density, ρ = m / V. If we multiply both sides by V, we obtain \[V \times \rho =\dfrac{m}{V}\times V = m\label{1} \] \[m = V \times \rho \nonumber \] or \[mass = \text{volume} \times \text{ density } \nonumber \] Taking the density of gold from , we can now calculate \[\text{Mass}= m =V \rho =\text{428 cm}^{3}\times \dfrac{\text{10}\text{0.32 g}}{\text{1 cm}^{3}}=8.27\times \text{10}^{3}\text{g}=\text{8}\text{.27 kg} \nonumber \] This is more than 18 lb of gold. At the price quoted above, it would be worth over 40 000 dollars! The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. \(\ref{1}\) are multiplied by 1/ρ, we have \[\dfrac{\text{1}}{\rho }\times m=V \rho \times \dfrac{\text{1}}{\rho }=V \nonumber \] \[V=m \times \dfrac{\text{1}}{\rho }\label{2} \] Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter? A mistake sometimes made by beginning students is to confuse density with , which also may have units of g/cm . By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function \[ C = \dfrac{m}{V} \nonumber \] In this case, \(V\) refers to the volume of a solution, which contains both a solute and solvent. Given a concentration of an alloy is 10 g gold in 100 cm of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to calculate the volume of gold in 20 g of the alloy as follows: \[20 \text{g} \times \dfrac{\text{100 cm^3}}{\text{10 g}} = 200 \text{ cm}^{3} \nonumber \] It is only possible to calculate the volume of gold if the density of the alloy is known, so that the volume of alloy represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function. The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood. A solution of ethanol with a concentration of 0.1754 g / cm has a density of 0.96923 g / cm and a freezing point of -9 ° F . What is the volume of ethanol (D = 0.78522 g / cm at 25 °C) in 100 g of the solution? The volume of 100 g of solution is \[ V = m \div D = 100 \text{ g} \div 0.96923 \text{ g} \text{ cm}^{3} = 103.17 \text{ cm}^{3} \nonumber \] The mass of ethanol in this volume is \[ m = V \times C = 103.17 \text{ cm}^{3} \times 0.1754 \text{ g /} \text{ cm}^{3} = 18.097 \text{ g} \nonumber \] \[ \text{The volume of ethanol } = m \div D = 18.097 \text{ g} \div 0.78522 \text{ g / } \text{cm}^{3} = 23.05 \text{cm}^{3} \nonumber \] Note that we cannot calculate the volume of ethanol by \[\dfrac {\dfrac{0.96923 g}{cm^3} \times 100 cm^3}{\dfrac {0.78522 g}{cm^3}} \normalsize = 123.4 \text{cm}^{3} \nonumber \] even though this equation is dimensionally correct. Note: Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions. Find the volume occupied by a 4.73-g sample of benzene. The is 0.880 g cm . Using Eq. (2), \[\text{Volume = }V\text{ = }m\text{ }\times \text{ }\dfrac{\text{1}}{\rho }\text{ = 4}\text{.73 g }\times \text{ }\dfrac{\text{1 cm}^{\text{3}}}{\text{0}\text{.880 g}}\text{ = 5}\text{.38 cm}^{\text{3}} \nonumber \] Note: Note that taking the reciprocal of \(\Large\tfrac{\text{0}\text{.880 g}}{\text{1 cm}^{3}}\) simply inverts the fraction ― 1 cm goes on top, and 0.880 g goes on the bottom. The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because the mathematical formula defining density relates it to mass and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. \(\ref{1}\) and \(\ref{2}\)], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate converstion factors by unit cancellation, as the following example shows: A student weighs 98.0 g of mercury. If the density of mercury is 13.6 g/cm , what volume does the sample occupy? We know that volume is related to mass through density. Therefore \[ V = m \times \text{ conversion factor} \nonumber \] Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel: \[V=m\times \dfrac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \dfrac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3} \nonumber \] If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error: \(V=\text{98}\text{.0 g}\times \dfrac{\text{13.6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\;\) (no cancellation!) It is clear that square grams per cubic centimeter are not the units we want. Using a conversion factor is very similar to using a unity factor — we know the conversion factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of the relationship [ie. the definition of density as defined by Eqs. \(\ref{1}\) and \(\ref{2}\) includes the relationships between density, mass, and volume], because it is has a value of one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing. A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below: \[\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume or }m\overset{\rho }{\longleftrightarrow}V\text{ } \nonumber \] This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written \[\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity} \nonumber \] As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution. Black ironwood has a density of 67.24 lb/ft . If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm). The road map \[V\xrightarrow{\rho }m\text{ } \nonumber \] tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation: \[ \text{Mass} = m = 47.3 \text{cm}^{3} \times \dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}} \nonumber \] Since the volume units are different, we need a unity factor to get them to cancel: \[m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \dfrac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}} \nonumber \] We now have the mass in pounds, but we want it in grams, so another unity factor is needed: \[m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{0.9 g} \nonumber \] In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique. | 11,963 | 2,944 |
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We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries. As shown in Table \(\Page {4}\), the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns (n − 1)d valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg . The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble. All the possible group 12 dihalides (MX ) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl ). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO and H SO . All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands. | 3,360 | 2,945 |
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We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. As you have learned, the molecules of a gas are stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses At a given temperature, heavier molecules move more slowly than lighter molecules. During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of \(\ce{^{235}U}\). Naturally occurring uranium is only 0.720% \(\ce{^{235}U}\), whereas most of the rest (99.275%) is \(\ce{^{238}U}\), which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound \(UF_6\) (boiling point = 56°C). isotopic content of naturally occurring uranium and atomic masses of U and U ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation 6.8.1:
\[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043\] Thus passing UF containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% UF . In this case, 0.990 = (0.00720)(1.0043) , which can be rearranged to give
\[1.0043^n=\dfrac{0.990}{0.00720}=137.50\] Thus at least a thousand effusion steps are necessary to obtain highly enriched U. Figure \(\Page {2}\) shows a small part of a system that is used to prepare enriched uranium on a large scale. Helium consists of two isotopes: He (natural abundance = 0.000134%) and He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant He by a process of gaseous effusion. a. ratio of effusion rates = 1.15200; one step gives 0.000154% He; b. 96 steps Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}\] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}\] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}\] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation 6.8.4 shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure \(\Page {3}\) for some common gases. Because all gases have the same average kinetic energy, according to the , molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure \(\Page {4}\). The average distance traveled by a molecule between collisions is the . The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 m (about 6 million miles). The the gas, the the mean free path. Calculate the rms speed of a sample of -2-butene (C H ) at 20°C. compound and temperature rms speed Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 6.8.5 to calculate the rms speed of the gas. To use Equation 6.8.4, we need to calculate the molar mass of -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C H , so its molar mass is 56.11 g/mol. Thus \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}=361\;m/s\] Calculate the rms speed of a sample of radon gas at 23°C. 1.82 × 10 m/s (about 410 mi/h) The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. In Section 10.8, we explore how the behavior of real gases differs from that of ideal gases. Graham’s law of Diffusion and Effusion: is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The of a molecule is the average distance it travels between collisions. | 8,421 | 2,946 |
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Up to this point we have been considering what would happen if a single Li atom and a single H atom were combined. When a large number of atoms of each kind combine, the result is somewhat different. Electrons are again transferred, and ions are formed, but the ions no longer pair off in twos. Instead, under the influence of their mutual attractions and repulsions, they collect together in much larger aggregates, eventually forming a three-dimensional array like that shown in the Figure On the macroscopic level a crystal of solid lithium hydride is formed. The formation of such an results in a lower potential energy than is possible if the ions only group into pairs. It is easy to see from the figure of the crystal lattice why this should be so. In an ion pair each Li ion is close to only one H ion, whereas in the crystal lattice it is close to no less than ions of opposite charge. Conversely each H ion is surrounded by six Li ions. In the crystal lattice therefore, more opposite charges are brought closer together than is possible for separate ion pairs and the potential energy is lower by an additional 227 kJ mol . The arrangement of the ions in a crystal of LiH corresponds to the lowest possible energy. If there were an alternative geometrical arrangement bringing even more ions of opposite charge even closer together than that shown in the figure, the Li ions and H ions would certainly adopt it. | 1,445 | 2,947 |
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The focus of this chapter is on analytical techniques that use a measurement of potential, current, or charge to determine an analyte’s concentration or to characterize an analyte’s chemical reactivity. Collectively we call this area of analytical chemistry because its originated from the study of the movement of electrons in an oxidation–reduction reaction. Despite the difference in instrumentation, all electrochemical techniques share several common features. Before we consider individual examples in greater detail, let’s take a moment to consider some of these similarities. As you work through the chapter, this overview will help you focus on similarities between different electrochemical methods of analysis. You will find it easier to understand a new analytical method when you can see its relationship to other similar methods. To understand electrochemistry we need to appreciate five important and interrelated concepts: (1) the electrode’s potential determines the analyte’s form at the electrode’s surface; (2) the concentration of analyte at the electrode’s surface may not be the same as its concentration in bulk solution; (3) in addition to an oxidation–reduction reaction, the analyte may participate in other chemical reactions; (4) current is a measure of the rate of the analyte’s oxidation or reduction; and (5) we cannot control simultaneously current and potential. The material in this section—particularly the five important concepts—draws upon a vision for understanding electrochemistry outlined by Larry Faulkner in the article “Understanding Electrochemistry: Some Distinctive Concepts,” , , 262–264. See also, Kissinger, P. T.; Bott, A. W. “Electrochemistry for the Non-Electrochemist,” , , , 51–53. In we introduced the ladder diagram as a tool for predicting how a change in solution conditions affects the position of an equilibrium reaction. Figure 11.1.1
, for example, shows a ladder diagram for the Fe /Fe and the Sn /Sn equilibria. If we place an electrode in a solution of Fe and Sn and adjust its potential to +0.500 V, Fe is reduced to Fe but Sn is not reduced to Sn . In we introduced the Nernst equation, which provides a mathematical relationship between the electrode’s potential and the concentrations of an analyte’s oxidized and reduced forms in solution. For example, the Nernst equation for Fe and Fe is \[E=E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}-\frac{R T}{n F} \ln \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=\frac{0.05916}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \label{11.1}\] where is the electrode’s potential and \(E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ}\) is the standard-state reduction potential for the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\). Because it is the potential of the electrode that determines the analyte’s form at the electrode’s surface, the concentration terms in Equation \ref{11.1} are those of Fe and Fe at the electrode's surface, not their concentrations in bulk solution. This distinction between a species’ surface concentration and its bulk concentration is important. Suppose we place an electrode in a solution of Fe and fix its potential at 1.00 V. From the ladder diagram in Figure 11.1.1
, we know that Fe is stable at this potential and, as shown in Figure 11.1.2
a, the concentration of Fe is the same at all distances from the electrode’s surface. If we change the electrode’s potential to +0.500 V, the concentration of Fe at the electrode’s surface decreases to approximately zero. As shown in Figure 11.1.2
b, the concentration of Fe increases as we move away from the electrode’s surface until it equals the concentration of Fe in bulk solution. The resulting concentration gradient causes additional Fe from the bulk solution to diffuse to the electrode’s surface. We call the region of solution that contains this concentration gradient in Fe the diffusion layer. We will have more to say about this in . Figure 11.1.1
and Figure 11.1.2
shows how the electrode’s potential affects the concentration of Fe and how the concentration of Fe varies as a function of distance from the electrode’s surface. The reduction of Fe to Fe , which is governed by Equation \ref{11.1}, may not be the only reaction that affects the concentration of Fe in bulk solution or at the electrode’s surface. The adsorption of Fe at the electrode’s surface or the formation of a metal–ligand complex in bulk solution, such as Fe(OH) , also affects the concentration of Fe . The reduction of Fe to Fe consumes an electron, which is drawn from the electrode. The oxidation of another species, perhaps the solvent, at a second electrode is the source of this electron. Because the reduction of Fe to Fe consumes one electron, the flow of electrons between the electrodes—in other words, the current—is a measure of the rate at which Fe is reduced. One important consequence of this observation is that the current is zero when the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\) is at equilibrium. The rate of the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\) is the change in the concentration of Fe as a function of time. If a solution of Fe and Fe is at equilibrium, the current is zero and the potential is given by Equation \ref{11.1}. If we change the potential away from its equilibrium position, current flows as the system moves toward its new equilibrium position. Although the initial current is quite large, it decreases over time, reaching zero when the reaction reaches equilibrium. The current, therefore, changes in response to the applied potential. Alternatively, we can pass a fixed current through the electrochemical cell, forcing the reduction of Fe to Fe . Because the concentrations of Fe decreases and the concentration of Fe increases, the potential, as given by Equation \ref{11.1}, also changes over time. In short, if we choose to control the potential, then we must accept the resulting current, and we must accept the resulting potential if we choose to control the current. Electrochemical measurements are made in an electrochemical cell that consists of two or more electrodes and the electronic circuitry needed to control and measure the current and the potential. In this section we introduce the basic components of electrochemical instrumentation. The simplest electrochemical cell uses two electrodes. The potential of one electrode is sensitive to the analyte’s concentration, and is called the or the . The second electrode, which we call the , completes the electrical circuit and provides a reference potential against which we measure the working electrode’s potential. Ideally the counter electrode’s potential remains constant so that we can assign to the working electrode any change in the overall cell potential. If the counter electrode’s potential is not constant, then we replace it with two electrodes: a whose potential remains constant and an that completes the electrical circuit. Because we cannot control simultaneously the current and the potential, there are only three basic experimental designs: (1) we can measure the potential when the current is zero, (2) we can measure the potential while we control the current, and (3) we can measure the current while we control the potential. Each of these experimental designs relies on , which states that the current, , passing through an electrical circuit of resistance, , generates a potential, . \[E = i R\nonumber\] Each of these experimental designs uses a different type of instrument. To help us understand how we can control and measure current and potential, . To do so the analyst observes a change in the current or the potential and manually adjusts the instrument’s settings to maintain the desired experimental conditions. It is important to understand that modern electrochemical instruments provide an automated, electronic means for controlling and measuring current and potential, and that they do so by using very different electronic circuitry than that described here. This point bears repeating: It is important to understand that the experimental designs in Figure 11.1.3
, Figure 11.1.4
, and Figure 11.1.5
do not represent the electrochemical instruments you will find in today’s analytical labs. For further information about modern electrochemical instrumentation, see this chapter’s additional resources. To measure the potential of an electrochemical cell under a condition of zero current we use a . Figure 11.1.3
shows a schematic diagram for a manual potentiometer that consists of a power supply, an electrochemical cell with a working electrode and a counter electrode, an ammeter to measure the current that passes through the electrochemical cell, an adjustable, slide-wire resistor, and a tap key for closing the circuit through the electrochemical cell. Using Ohm’s law, the current in the upper half of the circuit is \[i_{\text {upper}}=\frac{E_{\mathrm{PS}}}{R_{a b}} \nonumber\] where is the power supply’s potential, and is the resistance between points and of the slide-wire resistor. In a similar manner, the current in the lower half of the circuit is \[i_{\text {lower}}=\frac{E_{\text {cell}}}{R_{c b}} \nonumber\] where is the potential difference between the working electrode and the counter electrode, and is the resistance between the points and of the slide-wire resistor. When = = 0, no current flows through the ammeter and the potential of the electrochemical cell is \[E_{\mathrm{coll}}=\frac{R_{c b}}{R_{a b}} \times E_{\mathrm{PS}} \label{11.2}\] To determine we briefly press the tap key and observe the current at the ammeter. If the current is not zero, then we adjust the slide wire resistor and remeasure the current, continuing this process until the current is zero. When the current is zero, we use Equation \ref{11.2} to calculate . Using the tap key to briefly close the circuit through the electrochemical cell minimizes the current that passes through the cell and limits the change in the electrochemical cell’s composition. For example, passing a current of 10 A through the electrochemical cell for 1 s changes the concentrations of species in the cell by approximately 10 moles. Modern potentiometers use operational amplifiers to create a high-impedance voltmeter that measures the potential while drawing a current of less than 10 A. A , a schematic diagram of which is shown in Figure 11.1.4
, allows us to control the current that flows through an electrochemical cell. The current from the power supply through the working electrode is \[i=\frac{E_{\mathrm{PS}}}{R+R_{\mathrm{cell}}} \nonumber\] where is the potential of the power supply, is the resistance of the resistor, and is the resistance of the electrochemical cell. If >> , then the current between the auxiliary and working electrodes \[i=\frac{E_{\mathrm{PS}}}{R} \approx \text{constant} \nonumber\] maintains a constant value. To monitor the working electrode’s potential, which changes as the composition of the electrochemical cell changes, we can include an optional reference electrode and a high-impedance potentiometer. A , a schematic diagram of which is shown in Figure 11.1.5
, allows us to control the working electrode’s potential. The potential of the working electrode is measured relative to a constant-potential reference electrode that is connected to the working electrode through a high-impedance potentiometer. To set the working electrode’s potential we adjust the slide wire resistor that is connected to the auxiliary electrode. If the working electrode’s potential begins to drift, we adjust the slide wire resistor to return the potential to its initial value. The current flowing between the auxiliary electrode and the working electrode is measured with an ammeter. Modern potentiostats include waveform generators that allow us to apply a time-dependent potential profile, such as a series of potential pulses, to the working electrode. Because interfacial electrochemistry is such a broad field, let’s use Figure 11.1.6
to organize techniques by the experimental conditions we choose to use (Do we control the potential or the current? How do we change the applied potential or applied current? Do we stir the solution?) and the analytical signal we decide to measure (Current? Potential?). At the first level, we divide interfacial electrochemical techniques into static techniques and dynamic techniques. In a static technique we do not allow current to pass through the electrochemical cell and, as a result, the concentrations of all species remain constant. Potentiometry, in which we measure the potential of an electrochemical cell under static conditions, is one of the most important quantitative electrochemical methods and is discussed in detail in . Dynamic techniques, in which we allow current to flow and force a change in the concentration of species in the electrochemical cell, comprise the largest group of interfacial electrochemical techniques. Coulometry, in which we measure current as a function of time, is covered in . Amperometry and voltammetry, in which we measure current as a function of a fixed or variable potential, is the subject of . | 13,399 | 2,948 |
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In an equilibrium method the analytical signal is determined by an equilibrium reaction that involves the analyte or by a steady-state process that maintains the analyte’s concentration. When we determine the concentration of iron in water by measuring the absorbance of the orange-red \(\text{Fe(phen)}_3^{2+}\) complex, the signal depends upon the concentration of \(\text{Fe(phen)}_3^{2+}\), which, in turn, is determined by the complex’s formation constant. In the flame atomic absorption determination of Cu and Zn in tissue samples, the concentration of each metal in the flame remains constant because each step in the process of atomizing the sample is in a steady-state. In a the analytical signal is determined by the rate of a reaction that involves the analyte or by a nonsteady-state process. As a result, the analyte’s concentration changes during the time in which we monitor the signal. In many cases we can choose to complete an analysis using either an equilibrium method or a kinetic method by changing when we measure the analytical signal. For example, one method for determining the concentration of nitrite, \(\text{NO}_2^-\), in groundwater utilizes the two-step diazotization re-action shown in Figure 13.1.1
[Method 4500-NO B in , American Public Health Association: Washington, DC, 20th Ed., 1998]. The final product, which is a reddish-purple azo dye, absorbs visible light at a wavelength of 543 nm. Because neither reaction in Figure 13.1.1
is rapid, the absorbance—which is directly proportional to the concentration of nitrite—is measured 10 min after we add the last reagent, a lapse of time that ensures that the concentration of the azo dyes reaches the steady-state value required of an equilibrium method. We can use the same set of reactions as the basis for a kinetic method if we measure the solution’s absorbance during this 10-min development period, obtaining information about the reaction’s rate. If the measured rate is a function of the concentration of \(\text{NO}_2^-\), then we can use the rate to determine its concentration in the sample [Karayannis, M. I.; Piperaki, E. A.; Maniadaki, M. M. , , 13–23]. There are many potential advantages to a kinetic method of analysis, perhaps the most important of which is the ability to use chemical reactions and systems that are slow to reach equilibrium. In this chapter we examine three techniques that rely on measurements made while the analytical system is under kinetic control: chemical kinetic techniques, in which we measure the rate of a chemical reaction; radiochemical techniques, in which we measure the decay of a radioactive element; and flow injection analysis, in which we inject the analyte into a continuously flowing carrier stream, where its mixes with and reacts with reagents in the stream under conditions controlled by the kinetic processes of convection and diffusion. | 2,908 | 2,949 |
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Let’s begin by choosing a simple quantitative problem that requires a single measurement: What is the mass of a penny? You probably recognize that our statement of the problem is too broad. For example, are we interested in the mass of a United States penny or of a Canadian penny, or is the difference relevant? Because a penny’s composition and size may differ from country to country, let’s narrow our problem to pennies from the United States. There are other concerns we might consider. For example, the United States Mint produces pennies at two locations (Figure 4.1.1
). Because it seems unlikely that a penny’s mass depends on where it is minted, we will ignore this concern. Another concern is whether the mass of a newly minted penny is different from the mass of a circulating penny. Because the answer this time is not obvious, let’s further narrow our question and ask “What is the mass of a circulating United States Penny?” A good way to begin our analysis is to gather some preliminary data. Table 4.1.1
shows masses for seven pennies collected from my change jar. In examining this data we see that our question does not have a simple answer. That is, we can not use the mass of a single penny to draw a specific conclusion about the mass of any other penny (although we might reasonably conclude that all pennies weigh at least 3 g). We can, however, characterize this data by reporting the spread of the individual measurements around a central value. One way to characterize the data in Table 4.1.1
is to assume that the masses of individual pennies are scattered randomly around a central value that is the best estimate of a penny’s expected, or “true” mass. There are two common ways to estimate central tendency: the mean and the median. The , \(\overline{X}\), is the numerical average for a data set. We calculate the mean by dividing the sum of the individual values by the size of the data set \[\overline{X} = \frac {\sum_{i = 1}^n X_i} {n} \nonumber\] where \(X_i\) is the measurement, and is the size of the data set. What is the mean for the data in Table 4.1.1
? To calculate the mean we add together the results for all measurements \[3.080 + 3.094 + 3.107 + 3.056 + 3.112 + 3.174 + 3.198 = 21.821 \text{ g} \nonumber\] and divide by the number of measurements \[\overline{X} = \frac {21.821 \text{ g}} {7} = 3.117 \text{ g} \nonumber\] The mean is the most common estimate of central tendency. It is not a robust estimate, however, because a single extreme value—one much larger or much smaller than the remainder of the data—influences strongly the mean’s value [Rousseeuw, P. J. , , 1–20]. For example, if we accidently record the third penny’s mass as 31.07 g instead of 3.107 g, the mean changes from 3.117 g to 7.112 g! An estimate for a statistical parameter is robust if its value is not affected too much by an unusually large or an unusually small measurement. The , \(\widetilde{X}\), is the middle value when we order our data from the smallest to the largest value. When the data has an odd number of values, the median is the middle value. For an even number of values, the median is the average of the /2 and the ( /2) + 1 values, where is the size of the data set. When = 5, the median is the third value in the ordered data set; for = 6, the median is the average of the third and fourth members of the ordered data set. What is the median for the data in ? To determine the median we order the measurements from the smallest to the largest value \(3.056 \quad 3.080 \quad 3.094 \quad 3.107 \quad 3.112 \quad 3.174 \quad 3.198\) Because there are seven measurements, the median is the fourth value in the ordered data; thus, the median is 3.107 g. As shown by Example 4.1.1
and Example 4.1.2
, the mean and the median provide similar estimates of central tendency when all measurements are comparable in magnitude. The median, however, is a more robust estimate of central tendency because it is less sensitive to measurements with extreme values. For example, if we accidently record the third penny’s mass as 31.07 g instead of 3.107 g, the median’s value changes from 3.107 g to 3.112 g. If the mean or the median provides an estimate of a penny’s expected mass, then the spread of individual measurements about the mean or median provides an estimate of the difference in mass among pennies or of the uncertainty in measuring mass with a balance. Although we often define the spread relative to a specific measure of central tendency, its magnitude is independent of the central value. Although shifting all measurements in the same direction by adding or subtracting a constant value changes the mean or median, it does not change the spread. There are three common measures of spread: the range, the standard deviation, and the variance. at the end of the chapter asks you to show that this is true. The , , is the difference between a data set’s largest and smallest values. \[w = X_\text{largest} - X_\text{smallest} \nonumber\] The range provides information about the total variability in the data set, but does not provide information about the distribution of individual values. The range for the data in is \[w = 3.198 \text{ g} - 3.056 \text{ g} = 0.142 \text{ g} \nonumber\] The , , describes the spread of individual values about their mean, and is given as \[s = \sqrt{\frac {\sum_{i = 1}^{n} (X_i - \overline{X})^{2}} {n - 1}} \label{4.1}\] where \(X_i\) is one of the individual values in the data set, and \(\overline{X}\) is the data set's mean value. Frequently, we report the relative standard deviation, , instead of the absolute standard deviation. \[s_r = \frac {s} {\overline{X}} \nonumber\] The percent relative standard deviation, % , is \(s_r \times 100\). The relative standard deviation is important because it allows for a more meaningful comparison between data sets when the individual measurements differ significantly in magnitude. Consider again the data in . If we multiply each value by 10, the absolute standard deviation will increase by 10 as well; the relative standard deviation, however, is the same. Report the standard deviation, the relative standard deviation, and the percent relative standard deviation for the data in ? To calculate the standard deviation we first calculate the difference between each measurement and the data set’s mean value (3.117), square the resulting differences, and add them together to find the numerator of Equation \ref{4.1} \[\begin{align*}
(3.080-3.117)^2 = (-0.037)^2 = 0.001369\\
(3.094-3.117)^2 = (-0.023)^2 = 0.000529\\
(3.107-3.117)^2 = (-0.010)^2 = 0.000100\\
(3.056-3.117)^2 = (-0.061)^2 = 0.003721\\
(3.112-3.117)^2 = (-0.005)^2 = 0.000025\\
(3.174-3.117)^2 = (+0.057)^2 = 0.003249\\
(3.198-3.117)^2 = (+0.081)^2 = \underline{0.006561}\\
0.015554
\end{align*}\] For obvious reasons, the numerator of Equation \ref{4.1} is called a sum of squares. Next, we divide this sum of squares by – 1, where is the number of measurements, and take the square root. \[s = \sqrt{\frac {0.015554} {7 - 1}} = 0.051 \text{ g} \nonumber\] Finally, the relative standard deviation and percent relative standard deviation are \[s_r = \frac {0.051 \text{ g}} {3.117 \text{ g}} = 0.016 \nonumber\] \[\% s_r = (0.016) \times 100 = 1.6 \% \nonumber\] It is much easier to determine the standard deviation using a scientific calculator with built in statistical functions. Many scientific calculators include two keys for calculating the standard deviation. One key calculates the standard deviation for a data set of samples drawn from a larger collection of possible samples, which corresponds to Equation \ref{4.1}. The other key calculates the standard deviation for all possible samples. The latter is known as the population’s standard deviation, which we will cover later in this chapter. Your calculator’s manual will help you determine the appropriate key for each. Another common measure of spread is the , which is the square of the standard deviation. We usually report a data set’s standard deviation, rather than its variance, because the mean value and the standard deviation share the same unit. As we will see shortly, the variance is a useful measure of spread because its values are additive. What is the variance for the data in ? Solution The variance is the square of the absolute standard deviation. Using the standard deviation from Example 4.1.3
gives the variance as \[s^2 = (0.051)^2 = 0.0026 \nonumber\] The following data were collected as part of a quality control study for the analysis of sodium in serum; results are concentrations of Na in mmol/L. \(140 \quad 143 \quad 141 \quad 137 \quad 132 \quad 157 \quad 143 \quad 149 \quad 118 \quad 145\) Report the mean, the median, the range, the standard deviation, and the variance for this data. This data is a portion of a larger data set from Andrew, D. F.; Herzberg, A. M. , Springer-Verlag:New York, 1985, pp. 151–155. : To find the mean we add together the individual measurements and divide by the number of measurements. The sum of the 10 concentrations is 1405. Dividing the sum by 10, gives the mean as 140.5, or \(1.40 \times 10^2\) mmol/L. : To find the median we arrange the 10 measurements from the smallest concentration to the largest concentration; thus \(118 \quad 132 \quad 137 \quad 140 \quad 141 \quad 143 \quad 143 \quad 145 \quad 149 \quad 157\) The median for a data set with 10 members is the average of the fifth and sixth values; thus, the median is (141 + 143)/2, or 142 mmol/L. : The range is the difference between the largest value and the smallest value; thus, the range is 157 – 118 = 39 mmol/L. : To calculate the standard deviation we first calculate the absolute difference between each measurement and the mean value (140.5), square the resulting differences, and add them together. The differences are \(–0.5 \quad 2.5 \quad 0.5 \quad –3.5 \quad –8.5 \quad 16.5 \quad 2.5 \quad 8.5 \quad –22.5 \quad 4.5\) and the squared differences are \(0.25 \quad 6.25 \quad 0.25 \quad 12.25 \quad 72.25 \quad 272.25 \quad 6.25 \quad 72.25 \quad 506.25 \quad 20.25\) The total sum of squares, which is the numerator of Equation \ref{4.1}, is 968.50. The standard deviation is \[s = \sqrt{\frac {968.50} {10 - 1}} = 10.37 \approx 10.4 \nonumber\] : The variance is the square of the standard deviation, or 108. | 10,390 | 2,952 |
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Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be useful to review and be familiar with the following concepts: To illustrate this concept, consider the element vanadium, which \(\ce{V^{2+}}\) and \(\ce{V^{3+}}\)). The oxidizing \[ \ce{V \rightarrow V^{2+} + 2e^{-}} \label{1}\] The vanadium in the \( \ce{V^{2+}}\) ion has an oxidation state of +2. Removal of another electron gives the \(\ce{V^{3+}}\) ion: \[ \ce{V^{2+} \rightarrow V^{3+} + e^{-}} \label{2}\] The vanadium in the \(\ce{V^{3+} }\) ion has an oxidation state of +3. Removal of another electron forms the ion \(\ce{VO2+}\): \[ \ce{V^{3+} + H_2O \rightarrow VO^{2+} + 2H^{+} + e^{-}} \label{3}\] The vanadium in the \(\ce{VO^{2+}}\) is now in an oxidation state of +4. Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations \ref{1} and \ref{2}), but not for the ion in Equation \ref{3}). The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth electron to form another the \(\ce{VO_2^{+}}\) ion with the vanadium in a +5 oxidation state. \[ \ce{VO^{2+} + H_2O \rightarrow VO_2^{+} + 2H^{+} + e^{-}}\] Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of zero. If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is common for nonmetals such as sulfur: \[ \ce{S + 2e^- \rightarrow S^{2-}} \] Here the sulfur has an oxidation state of -2. The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state. Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations. Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules provide a simpler method. The reasons for the exceptions : Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H . The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1. Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0). : Peroxides include hydrogen peroxide, H O . This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero. Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it. : The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2. : Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is -1, and work the correct state out using fluorine or oxygen as a reference. An example of this situation is given below. What is the oxidation state of chromium in Cr ? For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion) What is the oxidation state of chromium in CrCl ? This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let equal the oxidation state of chromium: The oxidation state of chromium is +3. What is the oxidation state of chromium in Cr(H O) ? This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia. The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr . The oxidation state is +3. What is the oxidation state of chromium in the dichromate ion, Cr O ? The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present. 2n + 7(-2) = -2 n = +6 What is the oxidation state of copper in CuSO ? Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulfur) with variable oxidation states. In cases like these, some chemical intuition is useful. Here are two ways of approaching this problem: You will have come across names like iron(II) sulfate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe and Fe ions. This can also be extended to negative ions. Iron(II) sulfate is FeSO . SO sulfate The sulfite ion is SO . The oxidation state of the sulfur is +4. This ion is more properly named the sulfate(IV) ion. The ending indicates that the sulfur is in a negative ion. FeSO is properly named iron(II) sulfate(VI), and FeSO is iron(II) sulfate(IV). Because of the potential for confusion in these names, the older names of sulfate and sulfite are more commonly used in introductory chemistry courses. This is the most common function of oxidation states. Remember: In each of the following examples, we have to decide whether the reaction is a redox reaction, and if so, which species have been oxidized and which have been reduced. This is the reaction between magnesium and hydrogen chloride: \[ \ce{Mg + 2HCl -> MgCl2 +H2} \nonumber\] Assign each element its oxidation state to determine if any change states over the course of the reaction: The oxidation state of magnesium has increased from 0 to +2; the element has been oxidized. The oxidation state of hydrogen has decreased—hydrogen has been reduced. The chlorine is in the same oxidation state on both sides of the equation—it has not been oxidized or reduced. The reaction between sodium hydroxide and hydrochloric acid is: \[ NaOH + HCl \rightarrow NaCl + H_2O\] The oxidation states are assigned: None of the elements are oxidized or reduced. This is not a redox reaction. The reaction between chlorine and cold dilute sodium hydroxide solution is given below: \[ \ce{2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O} \nonumber\] It is probable that the elemental chlorine has changed oxidation state because it has formed two ionic compounds. Chlorine is the only element to have changed oxidation state. However, its transition is more complicated than previously-discussed examples: it is both oxidized and reduced. The NaCl chlorine atom is reduced to a -1 oxidation state; the NaClO chlorine atom is oxidized to a state of +1. This type of reaction, in which a single substance is both oxidized and reduced, is called a disproportionation reaction. Oxidation states can be useful in working out the stoichiometry for titration reactions when there is insufficient information to work out the complete ionic equation. Each time an oxidation state changes by one unit, one electron has been transferred. If the oxidation state of one substance in a reaction decreases by 2, it has gained 2 electrons. Another species in the reaction must have lost those electrons. Any oxidation state decrease in one substance must be accompanied by an equal oxidation state increase in another. Ions containing cerium in the +4 oxidation state are oxidizing agents, capable of oxidizing molybdenum from the +2 to the +6 oxidation state (from Mo to MoO ). Cerium is reduced to the +3 oxidation state (Ce ) in the process. What are the reacting proportions? The oxidation state of the molybdenum increases by 4. Therefore, the oxidation state of the cerium must decrease by 4 to compensate. The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion. Here is a more common example involving iron(II) ions and manganate(VII) ions: A solution of potassium manganate(VII), KMnO , acidified with dilute sulfuric acid oxidizes iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction. The oxidation state of the manganese in the manganate(VII) ion is +7, as indicated by the name (but it should be fairly straightforward and useful practice to figure it out from the chemical formula) In the process of transitioning to manganese(II) ions, the oxidation state of manganese decreases by 5. Every reactive iron(II) ion increases its oxidation state by 1. Therefore, there must be five iron(II) ions reacting for every one manganate(VII) ion. The left-hand side of the equation is therefore written as: MnO + 5Fe + ? The right-hand side is written as: Mn + 5Fe + ? The remaining atoms and the charges must be balanced using some intuitive guessing. In this case, it is probable that the oxygen will end up in water, which must be balanced with hydrogen. It has been specified that this reaction takes place under acidic conditions, providing plenty of The fully balanced equation is displayed below: \[ MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} \nonumber\] Jim Clark ( ) | 10,126 | 2,953 |
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John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. This theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. Before discussing the atomic theory, this article explains the theories that Dalton used as a basis for his theory: the law of conservation of mass and the law of constant composition. The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, . The law of conservation of mass was formulated by Antoine Lavoisier (1743-1794) as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air. Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, the illustration below shows that the burning of word does follow the law of conservation of mass. Scientists did not take into account the gases that play a critical role in this reaction. Joseph Proust (1754-1826) formulated the law of constant composition (also called the ). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law; 31 grams of H O and 8 grams of H O are made up of the same percent of hydrogen and oxygen. 1. Each chemical element is composed of extremely small particles that are indivisible and cannot be seen by the naked eye, called . Atoms can neither be created nor destroyed. Pictured below is a helium atom. The purple and red dots represent the neutrons and protons in the nucleus. The black area around the nucleus represent the electron cloud. The following sections discuss this further. 2. All atoms of an element are alike in mass and other properties, but the atoms of one element differ from all other elements. For example, gold and silver have different atomic masses and different properties. ld): Courtesy of Chris Ralph that released this image into the public domain. 3. For each compound, different elements combine in a simple numerical ratio. The illustration below describes this rule. The second equation for the reaction is incorrect because half of an atom does not exist. Atomic theory can be used to answers the questions presented above. A pure gold necklace is made up of atoms. A pure gold necklace and a pure silver necklace are different because they have different atoms. Pure gold mixed with pure copper forms rose gold. The gold and copper atoms combine in a simple numerical ratio. The first rule was proven incorrect when scientists divided atoms in a process called . The second rule was proven incorrect by the discovery that not all atoms of the same element have the same mass; there are different . . It correctly explains the law of conservation of mass: if atoms of an element are indestructible, then the same atom must be present after a chemical reaction as before and, and the mass must constant. Dalton’s atomic theory also explains the law of constant composition: if all the atoms of an element are alike in mass and if atoms unite in fixed numerical ratios, the percent composition of a compound must have a unique value without regards to the sample analyzed. The atomic theory led to the creation of the law of multiple proportions. The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers. The illustration of the third rule of the atomic theory correctly depicts this law. The first cathode-ray tube (CRT) was invented by Michael Faraday (1791-1867). Cathode rays are a type of radiation emitted by the negative terminal, the cathode, and were discovered by passing electricity through nearly-evacuated glass tubes. The radiation crosses the evacuated tube to the positive terminal, the anode. Cathode rays produced by the CRT are invisible and can only be detected by light emitted by the materials that they strike, called phosphors, painted at the end of the CRT to reveal the path of the cathode rays. These phosphors showed that cathode rays travel in straight lines and have properties independent of the cathode material (whether it is gold, silver, etc.). Another significant property of cathode rays is that they are deflected by magnetic and electric fields in a manner that is identical to negatively charged material. Due to these observations, J.J. Thompson (1856-1940) concluded that cathode rays are negatively charged particles that are located in all atoms. It was George Stoney who first gave the term electrons to the cathode rays. The below figures depict the way that the cathode ray is effected by magnetics. The cathode ray is always attracted by the positive magnet and deflected by the negative magnets. After Thompson discovered the electron, he proposed the plum pudding model of an atom, which states that the electrons float in positively-charged material. This model was named after the plum-pudding dessert. In 1909, Ernest Rutherford (1871-1937) performed a series of experiments studying the inner structure of atoms using alpha particles. Rutherford knew that alpha particles are significantly more massive than electrons and positively charged. Using the plum-pudding model for reference, Rutherford predicted that particles in an alpha beam would largely pass through matter unaffected, with a small number of particles slightly deflected. The particles would only be deflected if they happened to come into contact with electrons. According to the plum pudding model, this would be very unlikely. In order to test his hypothesis, Rutherford shot a beam of alpha particles at a thin piece of gold foil. Around the gold foil Rutherford placed sheets of zinc sulfide. These sheets produced a flash of light when struck by an alpha particle. However, this experiment produced results that contradicted Rutherford's hypothesis. Rutherford observed that the majority of the alpha particles went through the foil; however, some particles were slightly deflected, a small number were greatly deflected, and another small number were thrown back in nearly the direction from which they had come. Figure 10 shows Rutherford's prediction based off of the plum-pudding model (pink) and the observed large deflections of the alpha particles (gold). To account for these observations, Rutherford devised a model called the nuclear atom. In this model, the positive charge is held in an extremely small area called the nucleus, located in the middle of the atom. Outside of the nucleus the atom is largely composed of empty space. This model states that there were positive particles within the nucleus, but failed to define what these particles are. Rutherford discovered these particles in 1919, when he conducted an experiment that scattered alpha particles against nitrogen atoms. When the alpha particles and nitrogen atoms collided protons were released. In 1933, James Chadwick (1891-1974) discovered a new type of radiation that consisted of particles. It was discovered that these neutral atoms come from the nucleus of the atom. This last discovery completed the atomic model. : When 32.0 grams (g) of methane are burned in 128.0 g of oxygen, 88.0 g of carbon dioxide and 72.0 g of water are produced. Which law is this an example of? (a) Law of Definite Proportions (b) Law of Conservation of Mass or (c) Law of Multiple Proportions. The answer is (b) Law of Conservation of Mass. The number of grams of reactants (32.0 g of methane and 128.0 g of oxygen = 160.0 g total) is equal to the number of grams of product (88.0 g of carbon dioxide and 72.0 g of water = 160.0 g total). : 8.00 grams (g) of methane are burned in 32.00 g of oxygen. The reaction produces 22.00 g of carbon dioxide and an unmeasured mass of water. What mass of water is produced? The answer is 18.00 g of water. Because the only products are water and carbon dioxide, their total mass must equal the total mass of the reactants, methane and oxygen. 8.00 g of methane + 32.00 g of oxygen = 40.00 total g of reactants. Because the total mass of the reactants equals the total mass of the products, the total mass of the products is also 40.00 g. Thus, 40.00 total g of products = 22.00 g carbon dioxide + unknown mass water. 40.00 total g of products - 22.00 g carbon dioxide = 18.00 g water. : Two experiments using sodium and chlorine are performed. In the first experiment, 4.36 grams (g) sodium are reacted with 32.24 g of chlorine, using up all the sodium. 11.08 g of sodium chloride was produced in the first experiment. In the second experiment, 4.20 g of chlorine reacted with 20.00 g of sodium, using up all the chlorine. 6.92 g of of sodium chloride was produced in the second experiment. Show that these results are consistent with the law of constant composition. To solve, determine the percent of sodium in each sample of sodium chloride. There is 4.36 g sodium for every 11.08 g of sodium chloride in the first experiment. The amount of sodium in the sodium chloride for the second experiment must be found. This is found by subtracted the known amount of reacted chlorine (4.20 g) from the amount of sodium chloride (6.92 g). 6.92 g sodium chloride - 4.20 g chlorine = 2.72 g sodium. Thus, the percent of sodium in each sample is represented below:
% Na = (4.36 g Na)/(11.08 g NaCl) x 100% = 39.4% Na % Na = (2.72 g Na)/(6.92 g NaCl) x 100% = 39.3%
The slight difference in compositions is due to significant figures: each percent has an uncertainty of .01% in either direction. The two samples of sodium chloride have the same composition. : 36.0 grams (g) of wood are burned in oxygen. The products of this reaction weigh 74.4 g. (a) What mass of oxygen is needed in this reaction? (b) What mass of oxygen is needed to burn 8.00 lb of wood? 1 lb = 453.59237 g. 36.0 g wood x (1lb)/(453.59237 g) = .0793664144 lb wood
38.4 g oxygen x (1 lb)/(453.59237 g) = .0846575087 lb oxygen
Now two ratios equal to each other can be set up to determine the unknown mass of oxygen.
(0.0793664144 lb wood)/(.0846575087 lb oxygen) = (8.00 lb wood)/(unknown mass oxygen)
Solving reveals that it requires 8.53 lb of oxygen to burn 8.00 lb of wood. : A sample of methane contains only carbon and hydrogen, with 3.00 grams (g) of carbon for every 1.00 g of hydrogen. How much hydrogen should be present in a different, 50.0 g same of methane? The answer is 12.5 g of hydrogen. If there are 3.00 g of carbon present for every 1.00 g of hydrogen, we can assume the smallest whole number combination of these elements in that ratio to be 4.00 g of methane:
50.0 g methane x (1.00 g hydrogen)/(4.00 g methane) = 12.5 g of hydrogen. | 11,490 | 2,955 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Oxidation_of_Alkenes_with_Potassium_Manganate |
The carbon-carbon double bonds in alkenes such as ethene react with potassium manganate(VII) solution (potassium permanganate solution). Alkenes react with potassium manganate(VII) solution in the cold. The colorchange depends on whether the potassium manganate(VII) is used under acidic or alkaline conditions. We'll look at the reaction with ethene. Other alkenes react in just the same way. Manganate(VII) ions are a strong oxidizing agent, and in the first instance oxidize ethene to ethane-1,2-diol (old name: ethylene glycol). Looking at the equation purely from the point of view of the organic reaction: This type of equation is quite commonly used in organic chemistry. Oxygen written in square brackets is taken to mean "oxygen from an oxidizing agent". The reason for this is that a more normal equation tends to obscure the organic change in a mass of other detail - as you will find below! The full equation depends on the conditions. . . . and then further to dark brown solid manganese(IV) oxide (manganese dioxide). This last reaction is also the one you would get if the reaction was done under neutral conditions. You will notice that there are neither hydrogen ions nor hydroxide ions on the left-hand side of the equation. You might possibly remember that further up the page it says that potassium manganate(VII) is often made slightly alkaline by adding sodium carbonate solution. Where are the hydroxide ions in this? Carbonate ions react with water to some extent to produce hydrogencarbonate ions and hydroxide ions. \[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \] It is the presence of these hydroxide ions that gives sodium carbonate solution its pH in the 10 - 11 region. If an organic compound reacts with dilute alkaline potassium manganate(VII) solution in the cold to give a green solution followed by a dark brown precipitate, then it may contain a carbon-carbon double bond. But equally it could be any one of a large number of other compounds all of which can be oxidized by manganate(VII) ions under alkaline conditions. The situation with acidified potassium manganate(VII) solution is even worse because it has a tendency to break carbon-carbon bonds. It reacts destructively with a large number of organic compounds and is rarely used in organic chemistry. You could use alkaline potassium manganate(VII) solution if, for example, all you had to do was to find out whether a hydrocarbon was an alkane or an alkene - in other words, if there was nothing else present which could be oxidized. It isn't a useful test. Bromine water is far more clear cut. The diols, such as ethane-1,2-diol, which are the products of the reaction with cold dilute potassium manganate(VII), are themselves quite easily oxidized by manganate(VII) ions. That means that the reaction p at this point unless the potassium manganate(VII) solution is very dilute, very cold, and preferably not under acidic conditions. If you are using hot concentrated acidified potassium manganate(VII) solution, what you finally end up with depends on the arrangement of groups around the carbon-carbon double bond. The formula below represents a general alkene. In organic chemistry, the symbol R is used to represent hydrocarbon groups or hydrogen in a formula when you don't want to talk about specific compounds. If you use the symbol more than once in a formula (as here), the various groups are written as R , R , etc. In this particular case, the double bond is surrounded by four such groups, and these can be any combination of same or different - so they could be 2 hydrogens, a methyl and an ethyl, or 1 hydrogen and 3 methyls, or 1 hydrogen and 1 methyl and 1 ethyl and 1 propyl, or any other combination you can think of. In other words, this formula represents every possible simple alkene: The acidified potassium manganate(VII) solution oxidizes the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds. The products are known as because they contain the carbonyl group, C=O. Carbonyl compounds can also react with potassium manganate(VII), but how they react depends on what is attached to the carbon-oxygen double bond. So we need to work through all the possible combinations. Carbonyl compounds which have two hydrocarbon groups attached to the carbonyl group are called . Ketones aren't that easy to oxidize, and so there is no further action. (But see note in red below.) If the groups attached either side of the original carbon-carbon double bond were the same, then you would end up with a single ketone. If they were different, then you would end up with a mixture of two. For example: In this case, you would end up with two identical molecules called propanone. On the other hand, if one of the methyl groups in the original molecule was replaced by an ethyl group, you would get a mixture of two different ketones - propanone and butanone. What would you get if there was a methyl and an ethyl group on both sides of the original carbon-carbon double bond? Again, you would get a single ketone formed - in this case, butanone. If you aren't sure about this, draw the structures and see. This last section is a gross over-simplification. In practice, ketones are oxidized by potassium manganate(VII) solution under these conditions. The reaction is untidy and results in breaking carbon-carbon bonds either side of the carbonyl group. Potassium manganate(VII) is such a devastating oxidizing agent that it is rarely used in organic chemistry. For example, suppose the first stage of the reaction was: In this case, the first product molecule has a methyl group and a hydrogen attached to the carbonyl group. This is a different sort of compound known as an aldehyde. Aldehydes are readily oxidized to give carboxylic acids, containing the -COOH group. So this time, the reaction will go on a further step to give ethanoic acid, CH COOH. The acid structure has been turned around slightly to make it look more like the way we normally draw acids, but the net effect is that an oxygen has been slotted in between the carbon and hydrogen. The overall effect of the potassium manganate(VII) on this kind of alkene is therefore: Obviously, if there was a hydrogen atom attached to both carbons at the ends of the carbon-carbon double bond, you would get two carboxylic acid molecules formed - which might be the same or different, depending on whether the alkyl groups were the same or different. Play around with this until you are happy about it. Draw a number of alkenes, all of which have a hydrogen attached at both ends of the carbon-carbon double bond. Vary the alkyl groups - sometimes the same on each end of the double bond, sometimes different. oxidize them to form the acids, and see what you get. You might have expected that this would produce methanoic acid, as in the equation: But it doesn't! That's because methanoic acid is also easily oxidized by potassium manganate(VII) solution. In fact, it oxidizes it all the way to carbon dioxide and water. So the equation in a case like this might be, for example: The exact nature of the other product (in this example, propanone) will vary depending on what was attached to the right-hand carbon in the carbon-carbon double bond. If there were two hydrogens at both ends of the double bond (in other words, if you had ethene), then all you would get would be carbon dioxide and water. Working back from the results helps you to work out the structure of the alkene. For example, the alkene C H has three structural isomers: Work out which of these would give each of the following results if they were treated with hot concentrated potassium manganate(VII) solution. Acids are produced when there is a hydrogen atom attached to at least one of the carbons in the carbon-carbon double bond. Since in C there is only one product, the alkene must be symmetrical around the double bond. That's but-2-ene. If you have got two hydrogens at one end of the bond, this will produce carbon dioxide. A is 2-methylpropene, because the other molecule is a ketone. B must be but-1-ene because it produces carbon dioxide and an acid. Think about both ends of the carbon-carbon double bond separately, and then combine the results afterwards. Jim Clark ( ) | 8,298 | 2,957 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/11%3A_Electrochemical_Methods/11.01%3A_Overview_of_Electrochemistry |
The focus of this chapter is on analytical techniques that use a measurement of potential, current, or charge to determine an analyte’s concentration or to characterize an analyte’s chemical reactivity. Collectively we call this area of analytical chemistry because its originated from the study of the movement of electrons in an oxidation–reduction reaction. Despite the difference in instrumentation, all electrochemical techniques share several common features. Before we consider individual examples in greater detail, let’s take a moment to consider some of these similarities. As you work through the chapter, this overview will help you focus on similarities between different electrochemical methods of analysis. You will find it easier to understand a new analytical method when you can see its relationship to other similar methods. To understand electrochemistry we need to appreciate five important and interrelated concepts: (1) the electrode’s potential determines the analyte’s form at the electrode’s surface; (2) the concentration of analyte at the electrode’s surface may not be the same as its concentration in bulk solution; (3) in addition to an oxidation–reduction reaction, the analyte may participate in other chemical reactions; (4) current is a measure of the rate of the analyte’s oxidation or reduction; and (5) we cannot control simultaneously current and potential. The material in this section—particularly the five important concepts—draws upon a vision for understanding electrochemistry outlined by Larry Faulkner in the article “Understanding Electrochemistry: Some Distinctive Concepts,” , , 262–264. See also, Kissinger, P. T.; Bott, A. W. “Electrochemistry for the Non-Electrochemist,” , , , 51–53. In we introduced the ladder diagram as a tool for predicting how a change in solution conditions affects the position of an equilibrium reaction. Figure 11.1.1
, for example, shows a ladder diagram for the Fe /Fe and the Sn /Sn equilibria. If we place an electrode in a solution of Fe and Sn and adjust its potential to +0.500 V, Fe is reduced to Fe but Sn is not reduced to Sn . In we introduced the Nernst equation, which provides a mathematical relationship between the electrode’s potential and the concentrations of an analyte’s oxidized and reduced forms in solution. For example, the Nernst equation for Fe and Fe is \[E=E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}-\frac{R T}{n F} \ln \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=\frac{0.05916}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \label{11.1}\] where is the electrode’s potential and \(E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ}\) is the standard-state reduction potential for the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\). Because it is the potential of the electrode that determines the analyte’s form at the electrode’s surface, the concentration terms in Equation \ref{11.1} are those of Fe and Fe at the electrode's surface, not their concentrations in bulk solution. This distinction between a species’ surface concentration and its bulk concentration is important. Suppose we place an electrode in a solution of Fe and fix its potential at 1.00 V. From the ladder diagram in Figure 11.1.1
, we know that Fe is stable at this potential and, as shown in Figure 11.1.2
a, the concentration of Fe is the same at all distances from the electrode’s surface. If we change the electrode’s potential to +0.500 V, the concentration of Fe at the electrode’s surface decreases to approximately zero. As shown in Figure 11.1.2
b, the concentration of Fe increases as we move away from the electrode’s surface until it equals the concentration of Fe in bulk solution. The resulting concentration gradient causes additional Fe from the bulk solution to diffuse to the electrode’s surface. We call the region of solution that contains this concentration gradient in Fe the diffusion layer. We will have more to say about this in . Figure 11.1.1
and Figure 11.1.2
shows how the electrode’s potential affects the concentration of Fe and how the concentration of Fe varies as a function of distance from the electrode’s surface. The reduction of Fe to Fe , which is governed by Equation \ref{11.1}, may not be the only reaction that affects the concentration of Fe in bulk solution or at the electrode’s surface. The adsorption of Fe at the electrode’s surface or the formation of a metal–ligand complex in bulk solution, such as Fe(OH) , also affects the concentration of Fe . The reduction of Fe to Fe consumes an electron, which is drawn from the electrode. The oxidation of another species, perhaps the solvent, at a second electrode is the source of this electron. Because the reduction of Fe to Fe consumes one electron, the flow of electrons between the electrodes—in other words, the current—is a measure of the rate at which Fe is reduced. One important consequence of this observation is that the current is zero when the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\) is at equilibrium. The rate of the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\) is the change in the concentration of Fe as a function of time. If a solution of Fe and Fe is at equilibrium, the current is zero and the potential is given by Equation \ref{11.1}. If we change the potential away from its equilibrium position, current flows as the system moves toward its new equilibrium position. Although the initial current is quite large, it decreases over time, reaching zero when the reaction reaches equilibrium. The current, therefore, changes in response to the applied potential. Alternatively, we can pass a fixed current through the electrochemical cell, forcing the reduction of Fe to Fe . Because the concentrations of Fe decreases and the concentration of Fe increases, the potential, as given by Equation \ref{11.1}, also changes over time. In short, if we choose to control the potential, then we must accept the resulting current, and we must accept the resulting potential if we choose to control the current. Electrochemical measurements are made in an electrochemical cell that consists of two or more electrodes and the electronic circuitry needed to control and measure the current and the potential. In this section we introduce the basic components of electrochemical instrumentation. The simplest electrochemical cell uses two electrodes. The potential of one electrode is sensitive to the analyte’s concentration, and is called the or the . The second electrode, which we call the , completes the electrical circuit and provides a reference potential against which we measure the working electrode’s potential. Ideally the counter electrode’s potential remains constant so that we can assign to the working electrode any change in the overall cell potential. If the counter electrode’s potential is not constant, then we replace it with two electrodes: a whose potential remains constant and an that completes the electrical circuit. Because we cannot control simultaneously the current and the potential, there are only three basic experimental designs: (1) we can measure the potential when the current is zero, (2) we can measure the potential while we control the current, and (3) we can measure the current while we control the potential. Each of these experimental designs relies on , which states that the current, , passing through an electrical circuit of resistance, , generates a potential, . \[E = i R\nonumber\] Each of these experimental designs uses a different type of instrument. To help us understand how we can control and measure current and potential, . To do so the analyst observes a change in the current or the potential and manually adjusts the instrument’s settings to maintain the desired experimental conditions. It is important to understand that modern electrochemical instruments provide an automated, electronic means for controlling and measuring current and potential, and that they do so by using very different electronic circuitry than that described here. This point bears repeating: It is important to understand that the experimental designs in Figure 11.1.3
, Figure 11.1.4
, and Figure 11.1.5
do not represent the electrochemical instruments you will find in today’s analytical labs. For further information about modern electrochemical instrumentation, see this chapter’s additional resources. To measure the potential of an electrochemical cell under a condition of zero current we use a . Figure 11.1.3
shows a schematic diagram for a manual potentiometer that consists of a power supply, an electrochemical cell with a working electrode and a counter electrode, an ammeter to measure the current that passes through the electrochemical cell, an adjustable, slide-wire resistor, and a tap key for closing the circuit through the electrochemical cell. Using Ohm’s law, the current in the upper half of the circuit is \[i_{\text {upper}}=\frac{E_{\mathrm{PS}}}{R_{a b}} \nonumber\] where is the power supply’s potential, and is the resistance between points and of the slide-wire resistor. In a similar manner, the current in the lower half of the circuit is \[i_{\text {lower}}=\frac{E_{\text {cell}}}{R_{c b}} \nonumber\] where is the potential difference between the working electrode and the counter electrode, and is the resistance between the points and of the slide-wire resistor. When = = 0, no current flows through the ammeter and the potential of the electrochemical cell is \[E_{\mathrm{coll}}=\frac{R_{c b}}{R_{a b}} \times E_{\mathrm{PS}} \label{11.2}\] To determine we briefly press the tap key and observe the current at the ammeter. If the current is not zero, then we adjust the slide wire resistor and remeasure the current, continuing this process until the current is zero. When the current is zero, we use Equation \ref{11.2} to calculate . Using the tap key to briefly close the circuit through the electrochemical cell minimizes the current that passes through the cell and limits the change in the electrochemical cell’s composition. For example, passing a current of 10 A through the electrochemical cell for 1 s changes the concentrations of species in the cell by approximately 10 moles. Modern potentiometers use operational amplifiers to create a high-impedance voltmeter that measures the potential while drawing a current of less than 10 A. A , a schematic diagram of which is shown in Figure 11.1.4
, allows us to control the current that flows through an electrochemical cell. The current from the power supply through the working electrode is \[i=\frac{E_{\mathrm{PS}}}{R+R_{\mathrm{cell}}} \nonumber\] where is the potential of the power supply, is the resistance of the resistor, and is the resistance of the electrochemical cell. If >> , then the current between the auxiliary and working electrodes \[i=\frac{E_{\mathrm{PS}}}{R} \approx \text{constant} \nonumber\] maintains a constant value. To monitor the working electrode’s potential, which changes as the composition of the electrochemical cell changes, we can include an optional reference electrode and a high-impedance potentiometer. A , a schematic diagram of which is shown in Figure 11.1.5
, allows us to control the working electrode’s potential. The potential of the working electrode is measured relative to a constant-potential reference electrode that is connected to the working electrode through a high-impedance potentiometer. To set the working electrode’s potential we adjust the slide wire resistor that is connected to the auxiliary electrode. If the working electrode’s potential begins to drift, we adjust the slide wire resistor to return the potential to its initial value. The current flowing between the auxiliary electrode and the working electrode is measured with an ammeter. Modern potentiostats include waveform generators that allow us to apply a time-dependent potential profile, such as a series of potential pulses, to the working electrode. Because interfacial electrochemistry is such a broad field, let’s use Figure 11.1.6
to organize techniques by the experimental conditions we choose to use (Do we control the potential or the current? How do we change the applied potential or applied current? Do we stir the solution?) and the analytical signal we decide to measure (Current? Potential?). At the first level, we divide interfacial electrochemical techniques into static techniques and dynamic techniques. In a static technique we do not allow current to pass through the electrochemical cell and, as a result, the concentrations of all species remain constant. Potentiometry, in which we measure the potential of an electrochemical cell under static conditions, is one of the most important quantitative electrochemical methods and is discussed in detail in . Dynamic techniques, in which we allow current to flow and force a change in the concentration of species in the electrochemical cell, comprise the largest group of interfacial electrochemical techniques. Coulometry, in which we measure current as a function of time, is covered in . Amperometry and voltammetry, in which we measure current as a function of a fixed or variable potential, is the subject of . | 13,399 | 2,958 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.04%3A_Calculating_Entropy_Changes |
Entropy changes are fairly easy to calculate so long as one knows initial and final state. For example, if the initial and final volume are the same, the entropy can be calculated by assuming a reversible, isochoric pathway and determining an expression for \(\frac{dq}{T}\). That term can then be integrated from the initial condition to the final conditions to determine the entropy change. If the initial and final temperatures are the same, the most convenient reversible path to use to calculate the entropy is an isothermal pathway. As an example, consider the isothermal expansion of an ideal gas from \(V_1\) to \(V_2\). As was derived in Chapter 3, \[ dq = nRT \dfrac{dV}{V} \nonumber \] So \(dq/T\) is given by \[\dfrac{dq}{T} = nR\dfrac{dV}{V} \nonumber \] and so \[ \Delta S = \int \dfrac{dq}{T} = nR \int_{V_1}^{V_2} \dfrac{dV}{V} = nR \ln \left( \dfrac{V_2}{V_1} \right) \label{isothermS} \] Calculate the entropy change for 1.00 mol of an ideal gas expanding isothermally from a volume of 24.4 L to 48.8 L. Recognizing that this is an isothermal process, we can use Equation \ref{isothermS} \[ \begin{align*} \Delta S &= nR \ln \left( \dfrac{V_2}{V_1} \right) \\ &= (1.00 \, mol) (8.314 J/(mol \, K)) \ln \left( \dfrac{44.8\,L}{22.4\,L } \right) \\ &= 5.76 \, J/K \end{align*} \nonumber \] For changes in which the initial and final pressures are the same, the most convenient pathway to use to calculate the entropy change is an isobaric pathway. In this case, it is useful to remember that \[dq = nC_pdT \nonumber \] So \[\dfrac{dq}{T} = nC_p \dfrac{dT}{T} \nonumber \] Integration from the initial to final temperature is used to calculate the change in entropy. If the heat capacity is constant over the temperature range \[ \int_{T_1}^{T_2} \dfrac{dq}{T} = nC_p \int_{T_1}^{T_2} \dfrac{dT}{T} = nC_p \ln \left( \dfrac{T_2}{T_1} \right) \nonumber \] If the temperature dependence of the heat capacity is known, it can be incorporated into the integral. For example, if \(C_p\) can be expressed as \[C_p =a + bT + \dfrac{c}{T^2} \nonumber \] \(\Delta S\) takes the form \[ \int_{T_1}^{T_2} \dfrac{dq}{T} = n \int_{T_1}^{T_2} \dfrac{a + bT + \dfrac{c}{T^2}}{T} dT \nonumber \] which simplifies to \[ \Delta S = n \int_{T_1}^{T_2} \left( \dfrac{a}{T} + bT + \dfrac{c}{T^3} \right) dT \nonumber \] or \[ \Delta S = n \left[ a \ln \left( \dfrac{T_2}{T_1} \right) + b(T_2-T_1) - \dfrac{c}{2} \left( \dfrac{1}{T_2^2} -\dfrac{1}{T_1^2} \right) \right] \nonumber \] Similarly to the cast of constant pressure, it is fairly simple to calculate \(\Delta S\). Since \[ dq = nC_Vdt \nonumber \] is given by \[\dfrac{dq}{T} = nC_V \dfrac{dT}{T} \nonumber \] And so for changes over which \(C_V\) is independent of the temperature \(\Delta S\) is given by \[ \Delta S = nC_v \ln \left( \dfrac{T_2}{T_1} \right) \nonumber \] The easiest pathway for which to calculate entropy changes is an adiabatic pathway. Since \(dq = 0\) for an adiabatic change, then \(dS = 0\) as well. The entropy change for a phase change at constant pressure is given by \[ \Delta S = \dfrac{q}{T} = \dfrac{\Delta H_{phase}}{T} \label{phase} \] The enthalpy of fusion for water is 6.01 kJ/mol. Calculate the entropy change for 1.0 mole of ice melting to form liquid at 273 K. This is a phase transition at constant pressure (assumed) requiring Equation \ref{phase}: \[\begin{align*} \Delta S &= \dfrac{(1\,mol)(6010\, J/mol)}{273\,K} \\ &= 22 \,J/K \end{align*} \] | 3,460 | 2,962 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Electrophilic_Addition_of_Hydrogen_Halides_II |
Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond. All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other. For example, with ethene and hydrogen chloride, you get chloroethane: With but-2-ene you get 2-chlorobutane: What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately. The alkenes react with gaseous hydrogen halides at room temperature. If the alkene is also a gas, you can simply mix the gases. If the alkene is a liquid, you can bubble the hydrogen halide through the liquid. Alkenes will also react with concentrated solutions of the gases in water. A solution of hydrogen chloride in water is, of course, hydrochloric acid. A solution of hydrogen bromide in water is hydrobromic acid - and so on. There are, however, problems with this. The water will also get involved in the reaction and you end up with a mixture of products. Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions. When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow. This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be. Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example: There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions. Alkenes react because the electrons in the \(pi\) bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this. Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes. The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride. The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction: The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster. In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond. If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product. This is in line with Markovnikov's Rule When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already. In this case, the hydrogen becomes attached to the CH group, because the CH group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH group are totally irrelevant. Unlike the other hydrogen halides, hydrogen bromide can add to a carbon-carbon double bond either way around - depending on the conditions of the reaction. If the hydrogen bromide and alkene are entirely pure, then the hydrogen bromide adds on according to Markovnikov's Rule. For example, with propene you would get 2-bromopropane. That is exactly the same as the way the other hydrogen halides add. Oxygen from the air tends to react slowly with alkenes to produce some organic peroxides, and so you do not necessarily have to add them separately. This is therefore the reaction that you will tend to get unless you take care to exclude all air from the system. In this case, the addition is the other way around, and you get 1-bromopropane: This is sometimes described as an anti-Markovnikov addition or as the peroxide effect. Organic peroxides are excellent sources of free radicals. In the presence of these, the hydrogen bromide reacts with alkenes using a different (faster) mechanism. For various reasons, this doesn't happen with the other hydrogen halides. This reaction can also happen in this way in the presence of ultra-violet light of the right wavelength to break the hydrogen-bromine bond into hydrogen and bromine free radicals. Jim Clark ( ) | 5,583 | 2,963 |
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Since the fallout from the Bhopal tragedy, efforts have centered on avoiding storage of methyl isocyanate (MIC). The final product, Sevin, is no longer being manufactured by a two-step process. The process now consists of no longer reacting naphthol with MIC (one step), but sequentially with phosgene and methyl amine. The basis for the Bhopal tragedy was the reaction shown in Figure \(\Page {1}\). Polyurethane is a polymer composed of carbamate (urethane) linkages. Polyurethanes are traditionally formed by reacting a di- or polyisocyanate with a polyol (alcoholic polymer such as PEG, polyethylene glycol). Both the isocyanates and polyols used to make polyurethanes contain, on average, two or more functional groups (either on termini or within the molecule, hence being telechelic, i.e., a di-end-functional polymer where both ends possess the same functionality) per molecule. Recent efforts have been dedicated to minimizing the use of isocyanates to synthesize polyurethanes because isocyanates are toxic. Non-isocyanate-based polyurethanes (NIPUs), especially made from soybean oils, have recently been targeted as a new greener class of polyurethanes. Shown in Figure \(\Page {2}\) is a molecular representation of polyurethane linkages (highlighted in blue). The carbamate linkages are composed of a central carbonyl moiety that has two heteroatoms attached to it – a nitrogen and an oxygen. From an organic perspective, it is an ester/amide hybridized molecular system. Interestingly, telechelic monomers such as adipic acid chloride (left telechelic monomer) and hexamethylene diamine (right telechelic monomer) such as shown in Figure \(\Page {3}\). What is immediately noticeable is that the final polymer is a blend of two distinct monomers; thus, the final properties can be tailored by judicious (and discrete) choice of the monomers. More specifically, prepolymers (> 10-12 monomer units) of each monomer may be coupled to provide distinct segments having specific properties. For example, polyethylene glycol (PEG) is a polymer that is hydrophilic, soft, rubbery, and flows well. However, the phenyl-based di-isocyanate segment is much more rigid, tough, and non-stretchable. Therefore, the overall final physical and thermal properties of the polyurethane can be tuned. The opportunity to enhance the polymerization reactivity inherent for polyurethanes can be catalyzed by a non-nucleophilic base (such as DABCO – diazabicyclooctane) shown in Figure \(\Page {4}\) In Figure \(\Page {4}\), DABCO is able to abstract a proton from an alcohol (say ethylene glycol) to allow for the nucleophilic reactivity at the cumulated carbon of the isocyanate. An additional “greener” approach to forming isocyanates, an important class of starting material in the formation of many highly important materials, is through the Curtius Rearrangement (RAR) Reaction, an example of which is shown in Figure \(\Page {5}\). The Curtius Rearrangement is a thermal or photochemical decomposition starting from carboxylic azides (left structure in reaction above) to an isocyanate (first product in reaction above). In the above reaction, the solvent also plays the role of a reactant as shown by N atom insertion into cyclohexane via a radicaloid mechanism. These intermediates may be isolated, or their reaction or hydrolysis products can be obtained. The reaction sequence that includes the subsequent reaction with water that leads to amines is called the Curtius Reaction. This reaction is similar to the Schmidt Reaction, shown below in Figure \(\Page {6}\) ( ), with acids that differs in that the acyl azide is prepared from the acyl halide and an azide salt. For the Curtius RAR, the following steps take place: Starting Reagent 1 undergoes an electrophilic attack by an anionic azide molecule to produce an azide whose mesomerism is shown in brackets 2. It then undergoes decomposition under the appropriate stressor (heat, pressure, etc.), which leads to the isocyanate 3 with the loss of nitrogen (denitrogenation). This can react with water to yield a carbamate-like molecule (carbamic acid, urethane-like) which can spontaneously decompose thru decarboxylation to a primary amine 4. In the presence of an alcohol or an amine, 3 can yield an ester, 5, and an amide, 6, respectively. | 4,310 | 2,964 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.09%3A_Enzymes |
Virtually all biochemical reactions are catalyzed by proteins called enzymes. The catalytic power and specificity of enzymes is extraordinarily high. The reactions that they catalyze are generally enhanced in rate many orders of magnitude, often as much as \(10^7\), over the nonenzymatic process. Consequently enzymatic reactions may occur under much milder conditions than comparable laboratory reactions. For example, the simple hydrolysis of an amide proceeds at a practical rate only on heating the amide in either strongly acidic or strongly basic aqueous solution, and even then reaction may not be complete for several hours. In contrast, hydrolysis of amide or peptide bonds catalyzed by typical proteolytic enzymes, such as trypsin, chymotrypsin, or carboxypeptidase A, occurs rapidly as physiological temperatures and physiological pH.\(^7\) It is one of the remarkable attributes of many enzymes that they catalyze reactions that otherwise would require strongly acidic or basic conditions. Enzymes are strictly catalysts, however, and affect only the of reaction, not the position of equilibrium; they lower the energy of the transition state, not the energies of the reactants or products (see Figure 4-4). Many enzymes appear to be tailor-made for one specific reaction involving only one reactant, which is called the . Others can function more generally with different reactants (substrates). But there is no such thing as a universal enzyme that does all things for all substrates. However, nothing seems to be left to chance; even the equilibration of carbon dioxide with water is achieved with the aid of an enzyme known as carbonic anhydrase.\(^8\) Clearly, the scope of enzyme chemistry is enormous, yet the structure and function of relatively few enzymes are understood in any detail. We can give here only a brief discussion of the mechanisms of enzyme action - first some general principles then some specific examples. An enzyme usually catalyzes a single chemical operation at a very specific position, which means that only a small part of the enzyme is intimately involved. The region of the enzyme structure where key reactions occur as the result of association of the substrate with the enzyme is called the . The initial association of the enzyme \(\left( \ce{E} \right)\) and the substrate \(\left( \ce{S} \right)\) is formation of an enzyme-substrate complex \(\left( \ce{ES} \right)\): \[\ce{E} + \ce{S} \rightleftharpoons \ce{ES}\] Complexation could occur in many different ways, but for the intimate complexation required for catalysis, the enzyme must have, or must be able to assume, a shape complementary to that of the substrate. Originally, it was believed that the substrate fitted the enzyme somewhat like a key in a lock; this concept has been modified in recent years to the theory, whereby the enzyme can adapt to fit the substrate by undergoing conformational changes (Figure 25-18). Alternatively, the substrate may be similarly induced to fit the enzyme. The complementarity is three-dimensional, an important factor in determining the specificity of enzymes to the structure and stereochemical configuration of the substrates. Detailed structures for the active sites of enzymes are difficult to obtain and have been worked out only for a few enzymes that have been studied extensively by both chemical and x-ray methods. Very revealing information has been obtained by x-ray diffraction studies of complexes between the enzyme and , which are molecules similar to actual substrates and complex with the enzyme at the active site, but do not react further. These substances often reaction of the normal substrate by associating strongly with the enzyme at the active site and not moving onward to products. The x-ray studies of enzymes complexed with nonsubstrates show that the active site generally is a cleft or cavity in the folded structure of the enzyme that is largely hydrophobic in character. The enzyme-substrate complex can be inferred to be held together largely by van der Waals attractive forces between like groups ( ), hydrogen-bonding, and by electrostatic attraction between ionic or polar groups. To achieve a stereospecific catalyzed reaction, there must be at least three points of such interactions to align properly the substrate within the cavity of the enzyme. The reaction of the \(\ce{ES}\) complex may convert the substrate to product \(\left( \ce{P} \right)\) directly, and simultaneously free the enzyme \(\left( \ce{E} \right)\) to react with more of the substrate: \[\ce{ES} \rightarrow \ce{P} + \ce{E}\] However, the reaction between enzyme and substrate often is much more complex. In many cases, the substrate becomes covalently bound to the enzyme. Then, in a subsequent step, or steps, the enzyme-bound substrate \(\left( \ce{ES'} \right)\) reacts to give products and regenerate the active enzyme \(\left( \ce{E} \right)\): \[\ce{ES} \rightarrow \ce{ES'} \rightarrow \ce{P} + \ce{E}\] The considerable detail to which we now can understand enzyme catalysis is well illustrated by what is known about the action of carboxypeptidase A. This enzyme ( and Table 25-3) is one of the digestive enzymes of the pancreas that specifically hydrolyze peptide bonds at the \(\ce{C}\)-terminal end. Both the amino-acid sequence and the three-dimensional structure of carboxypeptidase A are known. The enzyme is a single chain of 307 amino-acid residues. The chain has regions where it is associated as an \(\alpha\) helix and others where it is associated as a \(\beta\)-pleated sheet. The prosthetic group is a zinc ion bound to three specific amino acids and one water molecule near the surface of the molecule. The amino acids bound ot zinc are His 69, His 196, and Glu 72; the numbering refers to the position of the amino acid along the chain, with the amino acid at the \(\ce{N}\)-terminus being number 1. The zinc ion is essential for the activity of the enzyme and is implicated, therefore, as part of the active site. X-ray studies\(^9\) of carboxypeptidase complexed with glycyltyrosine (with which it reacts only slowly) provide a detailed description of the active site, which is shown schematically in Figure 25-19a and is explained below. 1. The tyrosine carboxylate group of the substrate is associated by electrostatic attraction with the positively charged side chain of arginine 145 \(\left( \ce{W} \right)\): 2. The tyrosine side chain of the substrate associates with a nonpolar pocket in the enzyme \(\left( \ce{X} \right)\).
3. Hydrogen bonding possibly occurs between the substrate tyrosine amide unshared pair and the side-chain \(\ce{HO}\) groups of the enzyme tyrosine 248 \(\left( \ce{Y} \right)\).
4. The glycyl carbonyl oxygen in the substrate probably is coordinated with the zinc ion \(\left( \ce{Z} \right)\), displacing the water molecule coordinated to the zinc in the uncomplexed enzyme.
5. A side-chain carboxylate anion of glutamic acid 270 is so situated with respect to the reaction center that it could well function as a nucleophile by attacking the glycine carbonyl carbon. The arrangement of the enzyme-substrate complex suggests a plausible reaction mechanism analogous to nonenzymatic mechanisms of amide hydrolysis ( ). The carboxyl group of Glu 270 can add to the amide carbonyl to form a tetrahedral intermediate that then rapidly dissociates to release the terminal amino acid, leaving the rest of the substrate bound to the enzyme as a mixed anhydride which can be symbolized as \(\ce{E(CO)OCOR_1}\). Reaction of the acyl-enzyme intermediate with water will release the peptide, minus the terminal amino acid, and regenerates the enzyme. This postulated sequence of events may leave you wondering why the enzyme speeds up the hydrolysis, especially because the sequence proceeds through an energetically unfavorable reaction, the formation of a from an and a : How, then, can we possibly expect that the enzyme could make the anhydride route be faster than the simple water route? The answer is the way in which interactions in the enzyme-substrate complex can stabilize the transition state for the anhydride-forming reaction. Thus for carboxypeptidase the zinc can act as a strong electrophile to facilitate attack on the amide carbonyl. Hydrogen bonding of the amide nitrogen to Tyr 248 both will facilitate attack on the carbonyl group and assist in the breaking of the \(\ce{C-N}\) bond. Furthermore, the nonpolar environment of the alkyl side-chains of the enzyme will increase the nucleophilicity of the \(\ce{-CO_2^-}\) group that forms the anhydride (see Section 8-7F). Inspection of Figure 25-20 shows qualitatively that if the energy of the transition state for formation of the anhydride is lowered greatly, the overall rate will be determined by the rate of hydrolysis of the anhydride! In this circumstance (assuming the anhydride hydrolysis is uncatalyzed), the efficiency of the enzyme in breaking the peptide bond is as great as it can be, at least by this particular pathway of peptide hydrolysis. Such efficiencies have been established for other enzymes. Mechanisms similar to the one described for carboxypeptidase appear to operate in the hydrolysis of amide and ester bonds catalyzed by a number of proteinases and esterases. The substrate, here generalized as \(\ce{RCOX}\), acts to acylate the enzyme, which subsequently reacts with water to give the observed products: In the acylation step a nucleophilic group on one of the amino-acid side chains at the active site behaves as the nucleophile. As we have seen in , the nucleophile of carboxypeptidase is the free carboxyl group of glutamic acid 270. In several other enzymes (chymotrypsin, subtilisin, trypsin, elastase, thrombin, acetylcholinesterase), it is the hydroxyl group of a residue: This raises another question. Why is the serine hydroxyl an effective nucleophile when water and other hydroxylic compounds clearly are not similarly effective? Apparently, the nucleophilicity of the serine \(\ce{-CH_2OH}\) is enhanced by acid-base catalysis involving proton transfers between acidic and basic side-chain functions in the vicinity of the active site. The serine is believed to transfer its \(\ce{OH}\) proton to an amphoteric\(^{10}\) site \(\ce{B-A-H}\) on the enzyme at the same instant that the proton of \(\ce{B-A-H}\) is transferred to another base \(\ce{B'}^\ominus\) (Equation 25-8). These proton transfers are, of course, reversible: \(\tag{25-8}\) Loss of its proton makes the serine hydroxyl oxygen a much more powerful nucleophile, and even though the equilibrium of Equation 25-8 must lie far to the left at physiological pH, it can increase greatly the reactivity of the serine hydroxyl. In chymotrypsin and subtilisin, this network system, as it is called, is made up of a specific aspartic acid residue, acting as \(\ce{B'}^\ominus\), and a specific histidine residue (acting as the amphoteric \(\ce{B-A-H}\)): \(^7\)The slowness with which amide bonds are hydrolyzed in the presence of either strong acids or strong bases, and their susceptibility to hydrolysis under the influence of enzymes, clearly is a key advantage in the biological functioning of peptides. Amide hydrolysis in neutral solution has a favorable, but not large, equilibrium constant. Therefore it does not take a great deal of biochemical energy either to form or to hydrolyze peptide bonds. The resistance to ordinary hydrolysis provides needed stability for proteins, and yet when it is necessary to break down the peptide bonds of proteins, as in digestion, this can be done smoothly and efficiently with the aid of the proteolytic enzymes. \(^8\)Many enzymes are named by adding the suffix to a word, or words, descriptive of the type of enzymatic activity. Thus, hydrolyze esters, hydrolyze proteins, achieve reductions, and achieve synthesis of polypeptide chains, nucleic acid chains, and other molecules. \(^9\)W. N. Lipscomb, , 81 (1970); E. T. Kaiser and B. L. Kaiser, , 219 (1972). Lipscomb received the 1976 Nobel Prize in chemistry for structural work on boranes. \(^{10}\)Amphoteric means that a substance can act either as an acid or as a base. and (1977) | 12,232 | 2,965 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods/1.5E%3A_Hot_Filtration |
A hot filtration is generally used in some crystallization, when a solid contains impurities that are insoluble in the crystallization solvent. It is also necessary in crystallization when charcoal is used to remove highly colored impurities from a solid, as charcoal is so fine that it cannot be removed by decanting. A hot filtration is performed by first pouring a few \(\text{mL}\) of solvent through a funnel containing a "fluted filter paper". A fluted filter paper has many indentations and high surface area, which allows for a fast filtration. The funnel is allowed to get hot, while the mixture to be filtered is brought to a boil. The boiling mixture is then poured through the filter paper in portions (Figures 1.81b+d). It is best to use a ring clamp to secure the filtration funnel, although the funnel could also be simply placed atop the flask. If not using a ring clamp, it is recommended to place a bent paper clip between the flask and funnel to allow for displaced air to escape the bottom flask as liquid drains (Figure 1.81c+d). Without a ring clamp, the setup is more prone to tipping and so using a ring clamp is considerably safer. A hot filtration is used for filtering solutions that will crystallize when allowed to cool. It is therefore during filtration through contact with hot solvent vapors, or crystals may prematurely form on the filter paper or in the stem of the funnel (Figure 1.82). Crystallization on the filter paper can clog the setup and cause a loss of yield (as the filter paper will be later thrown away). Crystallization in the stem hinders filtration, and can act as a plug on the bottom of the funnel. An advantage of filtration is that the boiling solvent in the filter flask helps to dissolve crystals that prematurely form in the stem of the funnel. With hot filtration, it is advised to use a short-stemmed funnel (Figure 1.83a) or stemless funnel if available, instead of a long-stemmed funnel (Figure 1.83b), as material is less likely to crystallize in a short or absent stem. As it is essential that a solution filters quickly before it has a chance to cool off in the funnel, a " " (Figure 1.84b+c) is commonly used instead of the quadrant-folded filter paper sometimes used with gravity filtration (Figure 1.84a). The greater number of bends on the fluted filter paper translate into increased surface area and quicker filtration. The folds also create space between the filter paper and glass funnel, allowing for displaced air to more easily exit the flask as liquid drains. Hot filtration is often used with crystallization, and this procedure should be inserted after the dissolution step, but before setting aside the solution to slowly cool. | 2,722 | 2,966 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/01%3A_The_Basics/1.01%3A_The_System_and_the_Surroundings |
The Zeroth Law of Thermodynamics deals with the temperature of a system. And while it may seem intuitive as to what terms like “temperature” and “system” mean, it is important to define these terms. The easiest terms to define are the ones used to describe the system of interest and the surroundings, both of which are subsets of the universe. As it turns out, there can be several types of systems, depending on the nature of the boundary that separates the system from the surroundings, and specifically whether or not it allows to the transmittance of matter or energy across it. Further, systems can be (consisting of only a single phase of matter, and with uniform concentration of all substances present throughout) or (containing multiple phases and/or varying concentrations of the constituents throughout.) A very important variable that describes a system is its composition, which can be specified by the number of moles of each component or the concentration of each component. The number of moles of a substance is given by the ratio of the number of particles to Avogadro’s number \[ n =\dfrac{N}{N_A} \nonumber \] where \(n\) is the number of moles, \(N\) is the number of particles (atoms, molecules, or formula units) and \(N_A\) is Avogadro’s number (N = 6.022 x 10 mol ). Other important variables that are used to describe a system include the important variables of pressure, temperature, and volume. Other variables may also be important, but can often be determined if these are known. Oftentimes, knowing a small number of state variables is all that is required to determine all of the other properties of a system. The relationship that allows for the determination of these properties from the values of a couple of state variables is called an . Variables that describe a system can be either (independent of the amount of any given substance present in the system) or (dependent on the amount of substance present in the system.) Temperature and color are examples of intensive variables, whereas volume and mass are examples of extensive variables. The value of intensive properties is that they can be conveniently tabulated for various substances, whereas extensive properties would be specific to individual systems. Oftentimes it is the case that the ratio of two extensive variables results in an intensive variable (since the amount of substance cancels out.) An example of this is density, which is the ratio of mass and volume. Another example is molar volume (\(V_m\)) which is the ratio of volume and number of moles of substance. For a given substance, the molar volume is inversely proportional to the density of the substance. In a homogeneous system, an intensive variable will describe not just the system as a whole, but also any subset of that system. However, this may not be the case in a heterogeneous system! | 2,885 | 2,967 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria/9.05%3A_Degree_of_Dissociation |
Reactions such as the one in the previous example involve the dissociation of a molecule. Such reactions can be easily described in terms of the fraction of reactant molecules that actually dissociate to achieve equilibrium in a sample. This fraction is called the . For the reaction in the previous example \[A(g) \rightleftharpoons 2 B(g) \nonumber \] the degree of dissociation can be used to fill out an ICE table. If the reaction is started with \(n\) moles of \(A\), and is the fraction of \(A\) molecules that dissociate, the ICE table will look as follows. The mole fractions of \(A\) and \(B\) can then be expressed by \[ \begin{align*} \chi_A &= \dfrac{n(1-\alpha)}{n(1-\alpha)+2n\alpha} \\[4pt] &= \dfrac{1 -\alpha}{1+\alpha} \\[4pt] \chi_B &= \dfrac{2 \alpha}{1+\alpha} \end{align*} \] Based on these mole fractions \[ \begin{align} K_x &= \dfrac{\left( \dfrac{2 \alpha}{1+\alpha}\right)^2}{\dfrac{1 -\alpha}{1+\alpha}} \\[4pt] &= \dfrac{4 \alpha^2}{1-\alpha^2} \end{align} \nonumber \] And so \(K_p\), which can be expressed as \[K_p = K_x(p_{tot})^{\sum \nu_i} \label{oddEq} \] is given by \[ K_p = \dfrac{4 \alpha^2}{(1-\alpha^2)} (p_{tot}) \nonumber \] Based on the values given below, find the equilibrium constant at 25 C and degree of dissociation for a system that is at a total pressure of 1.00 atm for the reaction \[N_2O_4(g) \rightleftharpoons 2 NO_2(g) \nonumber \] First, the value of \(K_p\) can be determined from \(\Delta G_{rxn}^o\) via an application of Hess' Law. \[ \begin{align*} \Delta G_{rxn}^o &= 2 \left( 51.3 \, kJ/mol \right) - 99.8 \,kJ/mol &= 2.8\, kJ/mol \end{align*} \] So, using the relationship between thermodynamics and equilibria \[ \begin{align*} \Delta G_f^o &= -RT \ln K_p \\[4pt] 2800\, kJ/mol &= -(8.314 J/(mol\,K) ( 298 \,K) \ln K_p \\[4pt] K_p &= 0.323 \,atm \end{align*} \] The degree of dissociation can then be calculated from the ICE tables at the top of the page for the dissociation of \(N_2O_4(g)\): \[ \begin{align*} K_p &= \dfrac{4 \alpha^2}{1-\alpha^2} (p_{tot}) \\[4pt] 0.323 \,atm & = \dfrac{4 \alpha^2}{1-\alpha^2} (1.00 \,atm) \end{align*} \] Solving for \(\alpha\), \[ \alpha = 0.273 \nonumber \] : since represents the fraction of N O molecules dissociated, it be a positive number between 0 and 1. Consider the gas-phase reaction \[A + 2B \rightleftharpoons 2C \nonumber \] A reaction vessel is initially filled with 1.00 mol of A and 2.00 mol of B. At equilibrium, the vessel contains 0.60 mol C and a total pressure of 0.890 atm at 1350 K. Let’s build an ICE table! From the equilibrium measurement of the number of moles of C, x = 0.30 mol. So at equilibrium, The total number of moles at equilibrium is 2.70 mol. From these data, the mole fractions can be determined. \[ \begin{align*} \chi_A &= \dfrac{0.70\,mol}{2.70\,mol} = 0.259 \\[4pt] \chi_B &= \dfrac{1.40\,mol}{2.70\,mol} = 0.519 \\[4pt] \chi_C &= \dfrac{0.60\,mol}{2.70\,mol} = 0.222 \end{align*} \] So \(K_x\) is given by \[ K_x = \dfrac{(0.222)^2}{(0.259)(0.519)^2} = 0.7064 \nonumber \] And \(K_p\) is given by Equation \ref{oddEq}, so \[K_p = 0.7604(0.890 \,atm)^{-1} = 0.792\,atm^{-1} \nonumber \] The thermodynamic equilibrium constant is unitless, of course, since the pressures are all divided by 1 atm. So the actual value of \(K_p\) is 0.794. This value can be used to calculate \(\Delta G_{rxn}^o\) using \[ \Delta G_{rxn}^o = -RT \ln K_p \nonumber \] so \[ \begin{align*} \Delta G_{rxn}^o &= - (8.314 \, J/(mol\,K))( 1350\, K) \ln (0.792) \\[4pt] &= 2590 \, J/mol \end{align*} \] | 3,552 | 2,968 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photoreceptors/Chemistry_of_Vision/Cis-Trans_Isomerization_of_Retinal |
Vitamin A, trans-retinol, is carried to the rods in the eyes from storage in the liver. First it is converted to cis-retinol by a process of isomerization, which means that the trans isomer is converted to a cis isomer. The molecule must break the pi bond, rotate on the single bond, and reform the pi bond. The cis-retinol, an alcohol, is then oxidized to cis-retinal, an aldehyde. Photochemical events in vision involve the protein opsin and the cis/trans isomers of retinal. The cis-retinal fits into a receptor site of opsin. Upon absorption of a photon of light in the visible range, cis-retinal can isomerize to all-trans-retinal. In the cis-retinal, the hydrogens (light gray in the molecular model on the left) are on the same side of the double bond (yellow in the molecular model). In the trans-retinal, the hydrogens are on opposite sides of the double bond. In fact, all of the double bonds are in the trans-configuration in this isomer: the hydrogens, or hydrogen and -CH3, are always on opposite sides of the double bonds (hence, the name "all-trans-retinal"). Note how the shape of the molecule changes as a result of this isomerization. The molecule changes from an overall bent structure to one that is more or less linear. All of this is the result of trigonal planar bonding (120 bond angles) about the double bonds. This photochemical reaction is best understood in terms of molecular orbitals, orbital energy, and electron excitation. In cis-retinal, absorption of a photon promotes a p electron in the pi bond to a higher-energy orbital. This excitation "breaks" the pi component of the double bond and is temporarily converted into a single bond. This means the molecule can now rotate around this single bond, which it does by swiveling through 180 . The double bond then reforms and locks the molecule back into position in a trans configuration of the all-trans-retinal. This isomerization occurs in a few picoseconds (10-12 s) or less. Energy from light is crucial for this isomerization process: absorption of a photon leads to breaking the double bond and consequent isomerization about half the time (in the dark is almost never happens. | 2,187 | 2,969 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Electrophilic_Addition_of_Halogens_to_Alkenes |
Halogens can act as electrophiles to attack a double bond in . Double bond represents a region of electron density and therefore functions as a nucleophile. How is it possible for a halogen to obtain positive charge to be an electrophile? As halogen molecule, for example Br approaches a double bond of the alkene, electrons in the double bond repel electrons in bromine molecule causing polarization of the halogen bond. This creates a dipolar moment in the halogen molecule bond. Heterolytic bond cleavage occurs and one of the halogens obtains positive charge and reacts as an electrophile. The reaction of the addition is not regioselective but stereoselective.Stereochemistry of this addition can be explained by the mechanism of the reaction.In the first step electrophilic halogen with a positive charge approaches the double carbon bond and 2 p orbitals of the halogen, bond with two carbon atoms and create a cyclic ion with a halogen as the intermediate step. In the second step, halogen with the negative charge attacks any of the two carbons in the cyclic ion from the back side of the cycle as in the . Therefore stereochemistry of the product is addition. \[\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X}\] Halogens that are commonly used in this type of the reaction are: \(Br\) and \(Cl\). In thermodynamical terms \(I\) is too slow for this reaction because of the size of its atom, and \(F\) is too vigorous and explosive. Solvents that are used for this type of electrophilic halogenation are inert (e.g., CCl ) can be used in this reaction. Because halogen with negative charge can attack any carbon from the opposite side of the cycle it creates a mixture of steric products.Optically inactive starting material produce optically inactive achiral products ( ) or a . Before constructing the mechanism let us summarize conditions for this reaction. We will use Br in our example for halogenation of ethylene. chemistry In the first step of the addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a intermediate cycle with the double bond. In the second step, bromide anion attacks any carbon of the bridged bromonium ion from the back side of the cycle. Cycle opens up and two halogens are in the position . Hallogens can act as electrophiles due to polarizability of their covalent bond.Addition of halogens is stereospecific and produces vicinial dihalides with anti addition.Cis starting material will give mixture of enantiomers and trans produces a meso compound. 1.What is the mechanism of adding Cl to the cyclohexene? 2.A reaction of Br molecule in an inert solvent with alkene follows? a) syn addition b) anti addition c) Morkovnikov rule 3) 4) Key: 1. 2. b 3. enantiomer 4. | 2,778 | 2,970 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.01%3A_Prelude_to_Thermodynamics |
In the sections on we indicate that is a form of and show how the quantity of heat energy absorbed or released by a chemical change can be related to the corresponding chemical equation. We also state the law of conservation of energy, and arguments in other sections have often been based on the idea that energy can neither be created nor destroyed. The law of conservation of energy is the first of three important laws involving energy and matter, which were discovered over a century ago. These laws were originally based on the movement or transfer (dynamics) of heat (thermo), and the law of conservation of energy is therefore referred to as the . and the name change to the quantity of heat absorbed by a chemical or physical change under conditions of constant pressure. You may wonder just how heat energy could be absorbed or given off when atoms and molecules change position and structure during a chemical reaction, but we have not yet developed theories of chemical bonding, molecular structure, intermolecular forces, and molecular motion to the point where a satisfactory explanation can be given. We are in a position to investigate what can happen to molecules when matter absorbs or releases heat. One result of this study will be a clearer understanding of enthalpy. At the same time we will begin to appreciate what contribute to . This gives us a solid basis for discussing several aspects of what is probably the most important problem facing our technological society today― . The first law of thermodynamics (the law of conservation of energy) states that when heat energy is supplied to a substance, that energy cannot disappear-it must still be present in the atoms or molecules of the substance. Some of the added energy makes the atoms or molecules move faster. This is called translational energy. In the case of molecules, which can rotate and vibrate, some of the added energy increases the rotational and vibrational energies of the molecules. You can investigate vibrations of the ethane molecule above in the Jmol. Finally, any atom or molecule will have a certain electronic energy which depends on how close its electron clouds are to positively charged nuclei. The total of translational, rotational, vibrational, and electronic energies is the internal energy of an atom or molecule. When chemical reactions occur, the of the products is usually different from that of the reactants, and the difference appears as heat energy in the surroundings. If the reaction is carried out in a closed container (bomb calorimeter, for example), the increase in internal energy of the atoms and molecules is exactly equal to the heat energy absorbed from the surroundings. If the internal energy decreases, the energy of the surroundings must increase; i.e., heat energy is given off. When a chemical reaction occurs at constant pressure, as in a coffee-cup calorimeter, there is a change in potential energy of the atmosphere (given by Δ ) as well as a change in heat energy of the surroundings. Because the heat energy absorbed can be measured more easily than Δ , it is convenient to define the enthalpy as the internal energy plus the increased potential energy of the atmosphere. Thus the enthalpy increase equals the heat absorbed at constant pressure. Enthalpy changes for a variety of reactions may be calculated from . They may also be estimated by summing the bond enthalpies of all bonds broken and subtracting the bond enthalpies of all bonds formed. Because the dissociation enthalpy for the same type of bond varies from one molecule to another, the second method is not as accurate as the first. However, it has the advantage that enthalpy changes for reactions of a particular compound can be estimated even if the compound has not yet been synthesized. The enthalpy change for a reaction depends on the relative strengths of the bonds broken and formed and on the relative number of bonds broken and formed. A good fuel is a substance which can combine with oxygen from the air, forming more bonds and/or stronger bonds than were originally present. The fossil fuels, coal, petroleum, and natural gas consist mainly of carbon and hydrogen. When they burn in air, strong O—H and bonds are formed in the resulting H O and CO molecules. The supply of fossil fuels is limited, and they constitute a nonrenewable resource. Coal supplies ought to last another century or two, but petroleum and natural-gas supplies will be essentially depleted in half a century or less. During the next few decades it will be possible to gasify or liquefy coal to extend our supply of gaseous and liquid fuels. Conservation of these fuels can also make a major contribution toward continuing their use. Eventually, however, it will be necessary to develop nuclear or solar energy or some unknown source of energy if we are to continue our current energy-intensive way of life.
| 4,935 | 2,972 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/11%3A_Fluids/11.09%3A_Essential_Skills_6 |
Essential Skills 3 in , introduced the common, or base-10, logarithms and showed how to use the properties of exponents to perform logarithmic calculations. In this section, we describe natural logarithms, their relationship to common logarithms, and how to do calculations with them using the same properties of exponents. Many natural phenomena exhibit an exponential rate of increase or decrease. Population growth is an example of an exponential rate of increase, whereas a runner’s performance may show an exponential decline if initial improvements are substantially greater than those that occur at later stages of training. Exponential changes are represented logarithmically by , where is an irrational number whose value is approximately 2.7183. The , abbreviated as ln, is the power to which must be raised to obtain a particular number. The natural logarithm of is 1 (ln = 1). Some important relationships between base-10 logarithms and natural logarithms are as follows: According to these relationships, ln 10 = 2.303 and log 10 = 1. Because multiplying by 1 does not change an equality, Substituting any value for 10 gives Other important relationships are as follows: Entering a value , such as 3.86, into your calculator and pressing the “ln” key gives the value of ln , which is 1.35 for = 3.86. Conversely, entering the value 1.35 and pressing “ ” key gives an answer of 3.86. Hence Calculate the natural logarithm of each number and express each as a power of the base . What number is each value the natural logarithm of? Like common logarithms, natural logarithms use the properties of exponents. We can compare the properties of common and natural logarithms: \[ \dfrac{10^{a}}{10^{b}}=10^{a-b} \notag \] \[ \dfrac{e^{a}}{e^{b}}=e^{a-b} \notag \] \[ log \left (\dfrac{1}{a} \right )=-log\left ( a \right ) \notag \] \[ ln \left (\dfrac{1}{x} \right )=-ln\left ( x \right ) \notag \] The number of significant figures in a number is the same as the number of digits after the decimal point in its logarithm. For example, the natural logarithm of 18.45 is 2.9151, which means that is equal to 18.45. Calculate the natural logarithm of each number. The answers obtained using the two methods may differ slightly due to rounding errors. Calculate the natural logarithm of each number. | 2,338 | 2,973 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Nucleophilic_Addition_Reactions_and_Reduction |
The sp-hybrid carbon atoms of the triple-bond render alkynes more electrophilic than similarly substituted alkenes. As a result, alkynes sometimes undergo addition reactions initiated by bonding to a nucleophile. This mode of reaction, illustrated below, is generally not displayed by alkenes, unless the double-bond is activated by electronegative substituents, F C=CF , or by conjugation with an electron withdrawing group. HC≡CH + KOC H in C H OH at 150 ºC → H C=CH-OC H HC≡CH + HCN + NaCN (catalytic) → H C=CH-CN The smallest and most reactive nucleophilic species is probably an electron. Electron addition to a functional group is by definition a reduction, and we noted earlier that alkynes are reduced by solutions of sodium in liquid ammonia to trans-alkenes. To understand how this reduction occurs we first need to identify two distinct reactions of sodium with liquid ammonia (boiling point -78 ºC). In the first, sodium dissolves in the pure liquid to give a deep blue solution consisting of very mobile and loosely bound electrons together with solvated sodium cations (first equation below). For practical purposes, we can consider such solutions to be a source of "free electrons" which may be used as powerful reducing agents. In the second case, ferric salts catalyze the reaction of sodium with ammonia, liberating hydrogen and forming the colorless salt sodium amide (second equation). This is analogous to the reaction of sodium with water to give sodium hydroxide, but since ammonia is 10 times weaker an acid than water, the reaction is less violent. The usefulness of this reaction is that sodium amide, NaNH , is an exceedingly strong base (18 powers of ten stronger than sodium hydroxide), which may be used to convert very weak acids into their conjugate bases. Returning to the reducing capability of the blue electron solutions, we can write a plausible mechanism for the reduction of alkynes to trans-alkenes, as shown below. Isolated carbon double-bonds are not reduced by sodium in liquid ammonia, confirming the electronegativity difference between sp and sp hybridized carbons. | 2,128 | 2,974 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Metal_Lattices/2._Metal_Packing%3A_Layers |
For comparison, maybe there is another set of atoms, also in a simple square layer. Suppose they are well-separated from each other; maybe they are far enough apart that you could fit an extra atom between each pair if you wanted to. If the free electron is in the same place -- the middle of the nearest hole -- you can see that it is much farther from the nucleus in this case. The force of attraction is much lower in this case, and the overall energy is not as low. That first case, with atoms packed more tightly together, may be preferable, because of the stronger interaction between the metal nucleus and the free electron. For reasons like this, understanding the packing efficiency in a crystal can be very important. , | 742 | 2,975 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.07%3A_The_Amount_of_Substance-_Moles |
According to the , atoms are the units of chemical reactions. The formula HgBr indicates that each molecule of this substance contains one mercury and two bromine atoms. Therefore, if we ask how much bromine is required to make a given quantity of mercury (II) bromide, the answer is two bromine atoms for each mercury atom or two bromine atoms per molecule. In other words, how much substance we have depends in a very important way on how many atoms or molecules are present. So far, we've dealt with mass ratios. Is there a way to change masses of atoms into numbers of atoms, so it is easy to see of one element will react with another, just by looking at the number of atoms that are needed? As we see below, there seems to be no fundamental connection between the number of atoms or molecules in the chemical equations, and typical measures of "how much": 1 atom 1 molecule 1 molecule "How much?" in the above sense of the quantity of atoms or molecules present is not the same thing as "how much" in terms of volume or mass. It takes 3.47 cm Br ( ) to react with a 1 cm sample of Hg( ). That same 1 cm Hg( ) would weigh 13.59 g, but only 10.83 g Br ( ) would be needed to react with it. In terms of volume, more bromine than mercury is needed, while in terms of mass, less bromine than mercury is required. In the atomic sense, however, there are exactly as many bromine atoms as mercury atoms and twice as much bromine as mercury. Luckily, the International System of Measurements (IUPAC) has a measure of that reflects the number of atoms present, and it is called the For perspective, 1 mole of salt cubes (like those seen below) would form a cube . For additional perspective, it would take the fastest marathoner in the world just about 2 hours to run the length of one side. A mole is a huge number... Find out more about it on the next page! | 1,880 | 2,976 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.01%3A_Prelude_to_Atoms_and_Reactions |
Two very important things that chemists (and scientists in general) do include making quantitative measurements, and communicating the results of experiments as clearly and unambiguously as possible. We will now deal with another important activity of chemists—the use of their imaginations to devise theories or models to interpret their observations and measurements. Such theories or models are useful in suggesting new observations or experiments that yield additional data. They also serve to summarize existing information and aid in its recall. , first proposed in modern form by John Dalton, is one of the most important and useful ideas in chemistry. It interprets observations of the every-day world in terms of particles called atoms and molecules. events—those which humans can observe or experience with their unaided senses—are interpreted by means of objects—those so small that a special instrument or apparatus must be used to detect them. (Perhaps the term really ought to be used, because most atoms and molecules are much too small to be seen even under a microscope.) In any event, chemists continually try to explain the . The contrasting properties of solids, liquids, and gases, for example, may be ascribed to differences in spacing between and speed of motion of the constituent atoms or molecules. In the , the atomic theory and was used to explain the . The theory also agreed with . An important aspect of the atomic theory is the assignment of relative masses ( ) to the elements. Atoms and molecules are extremely small. Therefore, when calculating how much of one substance is required to react with another, chemists use a unit called the . One mole contains 6.022 × 10 of whatever kind of microscopic particles one wishes to consider. Referring to 2 mol Br specifies a certain number of Br molecules in the same way that referring to 10 gross of pencils specifies a certain number of pencils. The quantity which is measured in the units called moles is known as the . The somewhat unusual number 6.022 × 10 , also referred to as the , which specifies how many particles are in a mole, has been chosen so that the mass of 1 mol of atoms of any element is the atomic weight of that element expressed in grams. Similarly, the mass of a mole of molecules is the molecular weight expressed in grams. The molecular weight is obtained by summing atomic weights of all atoms in the molecule. This choice for the mole makes it very convenient to obtain –simply add the units grams per mole to the atomic or molecular weight. Using molar mass and the Avogadro constant, it is possible to determine the masses of individual atoms or molecules and to find how many atoms or molecules are present in a macroscopic sample of matter. A table of atomic weights and the molar masses which can be obtained from it can also be used to obtain the empirical if we know the percentage by weight of each element present. The opposite calculation, determination of weight percent from a chemical formula, is also possible. Once formulas for reactants and products are known, a can be written to describe any chemical change. Balancing an equation by adjusting the coefficients applied to each formula depends on the postulate of the atomic theory which states that atoms are neither created, destroyed, nor changed into atoms of another kind during a chemical reaction. | 3,414 | 2,977 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Complete_vs._Incomplete_Combustion_of_Alkanes |
This page deals briefly with the combustion of alkanes and cycloalkanes. In fact, there is very little difference between the two. Complete combustion (given sufficient oxygen) of any hydrocarbon produces and . It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number! For example, with propane (\(\ce{C3H8}\)), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be: \[\ce{ C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O} \nonumber\] Counting the oxygens leads directly to the final version: \[\ce{ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O} \nonumber\] With butane (\(\ce{C4H10}\)), you can again balance the carbons and hydrogens as you write the equation down. \[\ce{ C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O} \nonumber\] Counting the oxygen atoms leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" \(\ce{O2}\) molecules on the left. \[\ce{ C_4H_{10} + 6 1/2 O_2 \rightarrow 4CO_2 + 5H_2O} \nonumber\] If that offends you, double everything: \[\ce{ 2C_4H_{10} + 13 O_2 \rightarrow 8CO_2 + 10 H_2O} \nonumber\] The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules do not vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas. Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas. Oxygen is carried around the blood by hemoglobin. Unfortunately carbon monoxide also binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation. Jim Clark ( ) | 3,003 | 2,979 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO19._Conjugate_Addition |
are structures that contain alternating double and single bonds (or, in some cases, a double bond that is next to an atom with either a lone pair or a vacant orbital). Conjugated systems are usually at lower energy than regular double bonds because the electrons involved in bonding are delocalized; they are spread out over a greater area and thus can have a longer wavelength. For example, the -bonding system for 3-butene-2-one (or methyl vinyl ketone) is described by orbitals involving both the carbonyl group and the alkene group. These two groups become linked together so that there is not longer an independent carbonyl nor an independent alkene, but one "enone" (a term taken from the words alkene and ketone). Because of that extra stability, it might not be surprising that conjugated carbonyls are often a little slower to react than regular carbonyls. The surprise is that conjugated carbonyls can sometimes give additional products in which addition does not take place at the carbonyl. The product shown above is called a conjugate addition product, or a 1,4-addition product. In conjugate addition, the nucleophile does not donate to the carbonyl, but instead donates to an atom that is involved in conjugation with the carbonyl. This additional electrophilic position is sometimes called a "vinylogous" position (from the word vinyl, which refers to that CH=CH unit next to the carbonyl). Draw a mechanism with curved arrows for the conjugate addition shown above. Regular additions to carbonyls are sometimes called 1,2-additions, whereas conjugate additions are called 1,4-additions. Show why. Remember that we can look at another resonance structure of a carbonyl, one that emphasizes the electron-poverty of the carbonyl carbon. It's not a good Lewis structure because of the lack of an octet on carbon, but it does reinforce the idea that there is at least some positive charge at that carbon because it is less electronegative than oxygen. Extending that idea, we can draw an additional resonance structure in a conjugated system. That third structure suggests there may be some positive charge two carbons away from the carbonyl, on the β position on the double bond. The idea that there are two electrophilic positions in an enone is reinforced by the picture of the LUMO (the lowest-energy "empty" frontier orbital, the virtual place where an additional electron would probably go). When a lone pair is donated to an electrophile, the electrons are most likely to be donated into the LUMO. Having two possible products of a reaction can be confusing. How do you know which one will result? Often, you don't know. Frequently, both products result, so there is a mixture of commpounds. However, one product often predominates. In conjugate addition, there are a few different factors that may tilt the reaction in one direction or another. Possibly the simplest reason is steric effects. Maybe one of the electrophilic positions is more crowded than the other, and the nucleophile can access that position more easily. , | 3,060 | 2,980 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.07%3A_Electron_Waves_in_the_Hydrogen_Atom |
An electron in an atom differs in two ways from the hypothetical . First, in an atom the electron occupies all three dimensions of ordinary space. This permits the shapes of the electron waves to be more complicated. Second, the electron is not confined in an atom by the solid walls of a box. Instead, the electrostatic force of attraction between the positive nucleus and the negative electron prevents the latter from escaping. In 1926 Erwin Schrödinger (1887 to 1961) devised a mathematical procedure by which the electron waves and energies for this more complicated situation could be obtained. A solution of the is beyond the scope of a general chemistry text. However, a great many chemical phenomena can be better understood if one is familiar with Schrödinger’s results, and we shall consider them in some detail. The distribution of electron density predicted by the solution of Schrödinger’s equation for a hydrogen atom which has the minimum possible quantity of energy is illustrated in Figure 1. A number of general characteristics of the behavior of electrons in atoms and molecules may be observed from this figure. First of all, the hydrogen atom does not have a well-defined boundary. The number of dots per unit area is greatest near the nucleus of the atom at the center of the diagram (where the two axes cross). Electron density decreases as distance from the nucleus increases, but there are a few dots at distances as great as 200 pm (2.00 Å) from the center. Thus as one gets closer and closer to the nucleus of an atom, electron density builds up slowly and steadily from a very small value to a large one. Another way of stating the same thing is to say that the electron cloud becomes more dense as the center of the atom is approached. A second characteristic evident from Figure 1 is the of the electron cloud. In this two-dimensional diagram it appears to be approximately ; in three dimensions it would be (Figure 2). The following image is a three-dimensional dot-density diagram for the orbital of the hydrogen atom shown in Figure 1. This lowest-energy orbital, called the 1s orbital, appears spherically symmetric. In two dimensions, this can be illustrated more readily by drawing a circle (or in three dimensions, a sphere) which contains a large percentage (say 75 or 90 percent) of the dots, as has been done in the figure above. Since such a sphere or circle encloses most of (but not all) the electron density, it is about as close as one can come to drawing a boundary which encloses the atom. in two and three dimensions are easier to draw quickly than are dot-density diagrams. Therefore chemists use them a great deal. | 2,686 | 2,981 |
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When speaking of solubility or miscibility, or when doing quantitative experiments involving solutions, it is necessary to know the exact composition of a solution. This is invariably given in terms of a telling us how much solute is dissolved in a unit quantity of solvent or solution. The ratio can be a ratio of masses, of amounts of substances, or of volumes, or it can be some of these. For example, was defined as the : \[c_{\text{solute}}=\dfrac{n_{\text{solute}}}{V_{\text{solute}}} \nonumber \] The two simplest measures of the composition are the , which is the , and the , which is the . If we indicate the solute by and the solvent by , the mass fraction and the mole fraction are defined by \[w_{A}=\dfrac{m_{A}}{m_{A}\text{ + }m_{B}}\text{ and }x_{A}=\dfrac{n_{A}}{n_{A}\text{ + }n_{B}} \nonumber \] A solution is prepared by dissolving 18.65 g naphthalene, C H in 89.32 g benzene, C H . Find (a) the mass fraction and (b) the mole fraction of the naphthalene. The mass fraction is easily calculated from the masses: \[w_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\dfrac{m_{\text{C}_{\text{10}}\text{H}_{\text{8}}}}{m_{\text{C}_{\text{10}}\text{H}_{\text{8}}}\text{ + }m_{\text{C}_{\text{6}}\text{H}_{\text{6}}}}=\dfrac{\text{18}\text{.65 g}}{\text{18}\text{.65 g + 89}\text{.32 g}}=\text{0}\text{.1727} \nonumber \] It is sometimes useful to distinguish mass of solute and mass of solution for purposes of calculation. In such a case we can write = 0.1727 g C H per g solution In order to calculate the mole fraction, we must first calculate the amount of each substance. Since \[M_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\text{128}\text{.18 g mol}^{-\text{1}}\text{ and }M_{\text{C}_{\text{6}}\text{H}_{\text{6}}}=\text{78}\text{.11 g mol}^{-\text{1}}\text{ } \nonumber \] we find \[n_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\dfrac{\text{18}\text{.65 g}}{\text{128}\text{.18 g mol}^{-\text{1}}}=\text{0}\text{.1455} \nonumber \] \[n_{\text{C}_{\text{6}}\text{H}_{\text{6}}}=\dfrac{\text{89}\text{.32 g}}{\text{78}\text{.11 g mol}^{-\text{1}}}=\text{1}\text{.144} \nonumber \] Thus \[x_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\dfrac{n_{\text{C}_{\text{10}}\text{H}_{\text{8}}}}{n_{\text{C}_{\text{10}}\text{H}_{\text{8}}}\text{ + }n_{\text{C}_{\text{6}}\text{H}_{\text{6}}}}=\dfrac{\text{0}\text{.1455 mol}}{\text{0}\text{.1455 mol + 0}\text{.144 mol}}=\text{0}\text{.1455}=\text{0}\text{.1128} \nonumber \] or The mass fraction is useful because it does not require that we know the exact chemical nature of both solute and solvent. Thus if we dissolve 10 g crude oil in 10 g gasoline, we can calculate = 0.5 even though the solute and solvent are both mixtures of alkanes and have no definite molar mass. By contrast, the mole fraction is useful when we want to know the nature of the solution on the microscopic level. In the above example, for instance, we know that for every 100 mol of solution, 11.28 mol is naphthalene. On the molecular level this means that out of each 100 molecules in the solution, 11.28 will, on the average, be naphthalene molecules. The mass fraction of a solution is often encountered in other disguises. The (strictly speaking, the percentage) of a solution is often defined by the formula \[\text{Weight percentage of }A=\dfrac{m_{A}}{m_{A}\text{ + }m_{B}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ } \nonumber \] This definition is really the same as that of the mass fraction because the percent sign means “divided by 100.” Thus 100% is merely a synonym for 100/100, that is, the number 1, and we can write \[ w_{\text{C}_{10} \text{H}_8} = 0.1727 \times 1 = 0.1727 \times 100\% = 17.27\% \nonumber \] When the mass fraction is very small, it is often expressed in (ppm) or parts ppb . These symbols can be handled in much the same way as a percentage if you remember how they are related to unity: 1 = 100% = 10 ppm = 10 ppb In other words \(\begin{align}
1\% = \tfrac{1}{100} & = 10^{-2} \\
\text{1 ppm} & = {10}^{-6} \\
\text{1 ppb} & = {10}^{-9}
\end{align}\) A 1-kg sample of water from Lake Powell, Utah, is found to contain 10 ng mercury. Walleyed pike caught in the lake contain 0.427 ppm mercury. (a) What is the mass fraction of mercury (in ppb) in water from Lake Powell? (b) If you ate 2 lb of walleyed pike caught from the lake, what mass of mercury would you ingest?
The mass fraction of mercury is by definition \[w_{\text{Hg}}=\dfrac{m_{\text{Hg}}}{m_{\text{solution}}} \nonumber \] Therefore \[w_{\text{Hg}}=\dfrac{\text{10 ng}}{\text{1 kg}}=\dfrac{\text{10 }\times \text{ 10}^{-\text{9}}\text{ g}}{\text{1 }\times \text{ 10}^{\text{3}}\text{ g}}=\text{1 }\times \text{ 10}^{-\text{11}} \nonumber \] In ppb \[w_{\text{Hg}}=\text{1 }\times \text{ 10}^{-\text{11}}\text{ }\times \text{ 1}=\text{1 }\times \text{ 10}^{-\text{11}}\text{ }\times \text{ }10^{\text{9}}\text{ ppb}=\text{0}\text{0.01 ppb} \nonumber \] Assuming the mercury to be uniformly distributed throughout the walleyed pike, we have \[w_{\text{Hg}} = 0.427 \text{ ppm} = 0.427 \times 10^{-7} \nonumber \] By definition and \[w_{\text{Hg}}=\dfrac{m_{\text{Hg}}}{m_{\text{walleyed pike}}} \nonumber \] and \[m_{\text{Hg}}=w_{\text{Hg}}\text{ }\times \text{ }m_{\text{walleyed pike}}=\text{4}\text{0.27 }\times \text{ 10}^{-\text{7}}\text{ }\times \text{ 2}\text{.0 lb} \nonumber \] \[=\text{4}\text{.27 }\times \text{ 10}^{-\text{7}}\text{ }\times \text{ 2}\text{.0 lb }\times \dfrac{\text{1 kg}}{\text{2}\text{.2 lb}}\text{ }\times \text{ }\dfrac{\text{10}^{\text{3}}\text{ g}}{\text{1 kg}}=\text{3}\text{.9 }\times \text{ 10 g}=\text{390 }\mu \text{g} \nonumber \] From this example you can see that from nearly the same mass of fish, almost 40 000 times as much mercury would be obtained as from the water. Indeed, if you ate 2 lb of walleyed pike every day, you would exceed the minimum dosage (300 μg for a 70-kg human) at which symptoms of mercury poisoning can appear. Fortunately, most of us do not eat fish every day, nor are we as gluttonous as the example suggests. Nevertheless, the much higher mass fraction of mercury in fish than in water shows that very small quantities of mercury in the environment can be magnified many times in living systems. This process of will be discussed more thoroughly in the sections on Biochemistry. | 6,339 | 2,982 |
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After having some familiarity with , we can discuss atoms containing more than one electron. The diagrams shown here give a visual representation of the electrons in multi-electron atoms, using a different color for each electron. Use the buttons on the jmols to toggle electrons on or off. The following list explains the rules for predicting the electron configurations for atoms. By knowing the configuration of the previous element on the periodic table and by using these rules, determining the electron configuration for an atom having more than one electron is straightforward and simple. ( ). may be built up from that of the element preceding it in the periodic system by adding one proton (and an appropriate number of neutrons) to the nucleus and one extranuclear electron. . Each time an electron is added, it occupies the available subshell of . The appropriate shell may be determined from a diagram such as Figure 1 which arranges the subshells in order of increasing energy. Once a subshell becomes filled, the subshell of the next higher energy starts to fill. . No more than two electrons can occupy a single . When two electrons occupy the same orbital, they must be of opposite spin ( ). Hund’s rule. When electrons are added to a subshell where more than one orbital of the same energy is available, their spins remain parallel and they occupy different orbitals. Electron pairing does not occur until it is required by lack of another empty orbital in the subshell. | 1,514 | 2,984 |
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In the following sections we are going to study the way in which matter can both absorb energy and emit it in the form of such as light. The pattern in which matter absorbs or emits radiation is called its . In the past, and still to this day, studies of the spectrum of a substance have furnished important clues to the structure of matter. At the same time, the spectrum of a substance is often a very useful way of characterizing and hence identifying and analyzing that substance. Many of the properties of electromagnetic radiation can be explained if light is thought of as periodically varying electric and magnetic fields (electromagnetic waves). Such waves can be characterized by their frequency or their wavelength λ, and their speed of propagation is always λ = = 2.998 × 10 m s . Some properties of light are more easily explained in terms of particles called photons. The energy of a photon is given by = , where = 6.626 × 10 J s and is called Planck’s constant. When any element is heated to a high temperature or excited in a discharge tube, . Niels Bohr was able to predict the wavelengths of the lines in the spectrum of hydrogen by means of a theory which assigned the single electron to specific energy levels and hence to orbits of specific radius. Absorption of an appropriate quantity of energy can raise the hydrogen atom from a lower to a higher energy level, while emission of electromagnetic radiation corresponds to a change from a higher to a lower energy level. Although is quantitatively accurate only for hydrogen, his idea of energy levels is useful for all other atoms and even for molecules. , energy levels arise because of different speeds and kinds of molecular vibrations and rotations as well as because electrons are moved farther from or closer to positively charged nuclei. In organic compounds some groups of atoms vibrate at much the same frequency no matter what molecule they are in. The energy levels of such vibrations usually differ by roughly the energies of infrared photons, and many organic functional groups can be identified by the characteristic frequencies at which they absorb infrared radiation. When , band spectra occur. Some of the energy of each absorbed photon goes to excite an electron, but varying amounts also increase vibrational and rotational energies. Thus photons are absorbed over a broad range of frequencies and wavelengths. The most convenient theory by which the electronic energies of molecules can be predicted is . It assumes that electrons in a molecule occupy orbitals which are not confined to a single atom but rather extend over the entire molecule. Bonding molecular orbitals involve constructive interference between two electron waves, while antibonding molecular orbitals involve destructive interference. An electron occupying an antibonding MO is higher in energy than it would be if the atoms were not bonded together, and so antibonding electrons cancel the effect of bonding electrons. This explains why molecules such as He or Ne do not form. Molecular-orbital theory is especially useful in dealing with molecules for which resonance structures must be drawn. Because molecular orbitals can be delocalized over several atoms, there is no need for several resonance structures in the case of molecules like O and C H . The greater the extent of , the smaller the separation between molecular energy levels and the longer the wavelength at which absorption of ultraviolet or visible light can occur. Thus compounds containing long chains of alternating single and double bonds or having several benzene rings connected together often absorb visible light and are colored. | 3,703 | 2,985 |
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may involve proton transfers and other bond-breaking and bond-making processes, as well as electron transfers, and therefore the equations involved are much more difficult to deal with than those describing . In order to be able to recognize redox reactions, we need a method for keeping a careful account of all the electrons. This is done by assigning to each atom before and after the reaction. For example, in NO the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. This arbitrary assignment corresponds to the nitrogen’s having lost its original five valence electrons to the electronegative oxygens. In NO , on the other hand, the nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more electron than it had in NO . This arbitrarily assigned gain of one electron corresponds to reduction of the nitrogen atom on going from NO to NO . As a general rule, reduction corresponds to a lowering of the oxidation number of some atom. Oxidation corresponds to increasing the oxidation number of some atom. Applying the oxidation number rules to the following equation, we have
Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO does not really have a +5 charge which can be reduced to +4 in NO . Instead, there are covalent bonds and electron-pair sharing between nitrogen and oxygen in both species, and nitrogen has certainly not lost its valence electrons entirely to oxygen. Even though this may (and indeed should) make you suspicious of the validity of oxidation numbers, they are undoubtedly a useful tool for spotting electron-transfer processes. So long as they are used for that purpose only, and not taken to mean that atoms in covalent species actually have the large charges oxidation numbers often imply, their use is quite valid. The general rules for oxidation numbers are seen below, taken from the following page in the Analytical Chemistry Core Textbook: : : : : Identify the redox reactions and the reducing and oxidizing agents from the following: Solution: The appropriate oxidation numbers are The oxidation numbers show that no redox has occurred. This is an acid-base reaction because a proton, but no electrons, has been transferred. H S has been oxidized, losing two electrons to form elemental S. Since H S donates electrons, it is the reducing agent. HClO accepts these electrons and is reduced to Cl . Since it accepts electrons, HClO is the oxidizing agent. | 2,626 | 2,986 |
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In a classic paper published in 1913, the young Niels Bohr, then working with Rutherford in Manchester, England, proceeded to show how could be explained in terms of a very simple model of the hydrogen atom. The model was based on the nuclear view of atomic structure which had just been proposed by Rutherford. Bohr’s model is shown in Figure \(\Page {1}\). An of charge – and mass \(m\) moves around a heavy of charge + . Ordinarily the electron would move in a straight line, but the attraction of the nucleus bends its path so that it moves with a constant velocity \(u\) in a perfect circle of radius around the nucleus. The situation and the mathematics are very similar to that of a planet arcing round the sun. The major difference is that instead of the force of gravity there is an electrostatic force of attraction \(F\) between the proton and the electron described by : \[F=k\frac{e^{2}}{r^{2}} \label{1} \] where \(k\) has the value 8.9876 × 10 J m C . Expressions for both the kinetic and potential energies of the electron can be derived using Equation \(\ref{1}\) and the principles of elementary physics. Such a derivation can be found in most introductory physics texts. The two expressions are \[E_{k}=\frac{1}{2}mu^{2}=\frac{1}{2}k\frac{e^{2}}{r} \label{2} \] and \[E_{p}=-k\frac{e^{2}}{r} \label{3} \] If these are added together, we obtain a simple formula for the total energy of the electron: \[\begin{align} E &=E_{k}+E_{p} \\[4pt] &=-\frac{1}{2}k\frac{e^{2}}{r} \label{4} \end{align} \] If we now insert the known values of \(e\) and \(k\), we have the result \[ \begin{align} E &= -\frac{1}{2}\times\frac{8.9876 \times 10^{9} \text{ J m C}^{-2}\times(1.6022 \times 10^{-19} \text{ C})^{2}}{r} \\[4pt] &=-\frac{1.1536 \times 10^{-28} \text{ J M} }{r} \label{5} \end{align} \] From Equation \(\ref{5}\) we see that the total energy of the electron is very negative for an orbit with a small radius but increases as the orbit gets larger. In addition to suggesting the planetary model just described, Bohr also made two further postulates which enabled him to explain the spectrum of hydrogen. The first of these was the suggestion that an electron of high energy circling the nucleus at a large radius can lose some of that energy and assume an orbit of lower energy closer to the nucleus. The energy lost by the electron is emitted as a photon of light of frequency \(\nu\) given by Planck’s formula \[\Delta E = h\upsilon \label{6} \] where \(ΔE\) is the energy lost by the electron. Bohr’s second postulate was that are possible to the electron in a hydrogen atom. This enabled him to explain why it is that only light of a few particular frequencies can be emitted by the hydrogen atom. Since only a limited number of orbits are allowed, when an electron shifts from an outer to an inner orbit, the photon which emerges cannot have just any frequency but only that frequency corresponding to the energy difference between two allowed orbits. According to Bohr’s theory two of the allowed orbits in the hydrogen atom have radii of 52.918 and 211.67 pm. Calculate the energy, the frequency, and the wavelength of the photon emitted when the electron moves from the outer to the inner of these two orbits. Labeling the outer orbit 2 and the inner orbit 1, we first calculate the energy of each orbit from Equation \(\ref{5}\): \[E_{2}=-\dfrac{1.1536\times10^{-28} \text{Jm}}{211.67\times10^{-12} \text{m}}=-0.54500 \text{aJ} \nonumber \] \[E_{1}=-\dfrac{1.1536\times10^{-28} \text{Jm}}{52.918\times10^{-12} \text{m}}=-2.1780\, \text{aJ} \nonumber \] Thus \[ \Delta E = – 0.545 00 \text{aJ} – (– 2.1780 \text{aJ}) = 1.6330\, \text{aJ} \nonumber \] Using Equation \(\ref{6}\), we now have \[\upsilon=\dfrac{\Delta E}{h}=\dfrac{1.6330\times10^{-18}\text{ J}}{6.6262\times10^{-34} \text{ J s}}=2.4645\times10^{15} \text{ s}^{-1}=2.4645 \text{ PHz} \nonumber \] Finally \( \lambda = \dfrac{c}{\upsilon} = 1.2164 \times 10^{-7} \text{ m} = 121.64 \text{ nm} \). In order to predict the right frequencies for the lines in the hydrogen spectrum, Bohr found that he had to assume that the quantity \(mur\) (called the angular momentum by physicists) needed to be a multiple of \(h/2π\). In other words the condition restricting the orbits to only certain radii and certain energies was found to be \[mur=\frac{nh}{2\pi} \qquad \label{7} \] where could have the value 1, 2, 3, etc. By manipulating both Equations \(\ref{7}\) and \(\ref{2}\), it is possible to show that this restriction on the angular momentum restricts the radii of orbits to those given by the expression \[ r = \frac{n^{2}h^{2}}{4\pi^{2}mke^{2}} \qquad n = 1, 2, 3, \cdots \label{8} \] If the known values of and are inserted, this formula reduces to the convenient form \[ r = n^2 \times 52.918 \text{ pm} \qquad n = 1, 2, 3, \cdots \label{9} \] Bohr’s postulate thus restricts the electron to orbits for which the radius is 52.9 pm, 2 × 52.9 pm, 3 × 52.9 pm, and so on. If we substitute Equation \(\ref{8}\) into Equation \(\ref{4}\), we arrive at a general expression for the energy in terms of n: \[E=-\frac{1}{2}k\frac{e^{2}}{r}=-\frac{1}{2}ke^{2}\times\frac{4\pi^{2}mke^{2}}{n^{2}h^{2}} =\frac{2\pi^{2}k^{2}e^{4}m}{n^{2}h^{2}} \label{10} \] Again substituting in the known values for all the constants, we obtain \[E=-\frac{2.1800 \text{ aJ}}{n^{2}} \label{11} \] The integer \(n\) thus determines how far the electron is from the nucleus and how much energy it has, just as the principal quantum number \(n\) described previously. Using Equation \(\ref{10}\) or \(\ref{11}\) find the ionization energy of the hydrogen atom. The ionization energy of the hydrogen atom corresponds to the energy difference between the electron in its innermost orbit ( = 1) and the electron when completely separated from the proton. For the completely separated electron r = ∞ (infinity) and so does . Thus \[E_{1}=-\dfrac{2.1800 \text{aJ}}{1^{2}}=-2.1800\text{aJ} \nonumber \] and \[E_{\infty}=-\dfrac{2.1800 \text{aJ}}{\infty^{2}}=0.0000\text{aJ} \nonumber \] The energy difference is thus \[\Delta E=E_{infty}-E_{1}=2.1800 \text{aJ} \nonumber \] which is the ionization energy per atom. On a molar basis the ionization energy is the Avogadro constant times this quantity; namely, \[2.1800 \times 10^{-18} \text{J} \times 6.0221 \times 10^{23} \text{mol}^{-1} = 1312.8 \text{ kJ mol}^{-1}\ n\nonumber \] : In an atom, the electronic configuration of lowest energy is called the while other configurations are called . We can now derive Rydberg’s experimental formula from Bohr’s theory. Suppose an electron moves from an outer orbit for which the quantum number is to an inner orbit of quantum number . The energy lost by the electron and emitted as a photon is then given by \[\Delta E=E_{2}-E_{1}= – 2.1800 \text{ aJ} \left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right) = 2.1800 \text{ aJ} \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \label{12} \] However, \[\Delta E=h\upsilon=\frac{hc}{\lambda} \label{13} \] where \(λ\) is the wavelength of the photon. Combining Eqs. \(\ref{2}\) and \(\ref{3}\), we obtain \[\frac{1}{\lambda}=\frac{2.1800 \text{ aJ}}{hc}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \nonumber \] or \[\frac{1}{\lambda}=1.0975\times10^{7}m^{-1} \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \label{14} \] This expression is of exactly the same form as that found experimentally by Rydberg with a value for R of 1.0974 × 10 m , very close to the experimental value of 1.097 094 × 10 m . Even better agreement can be obtained if allowance is made for the fact that the nucleus is not stationary but that the electron and nucleus revolve around a common center of gravity. Calculate the wavelength of the light emitted when the electron in a hydrogen atom drops from the = 3 to the = 1 orbit. In what region of the spectrum does this spectral line lie? To what series does it belong? From Equation \(\ref{14}\) we find \[\dfrac{1}{\lambda}=1.0975\times10^{7}m^{-1}\times\left(1-\dfrac{1}{9}\right)=9.7547\times10^{6}m^{-1} \nonumber \] giving \(\lambda = 102.51 \text{ nm} \). This is the second line in the Lyman series and lies in the far ultraviolet. The experimentally determined wavelength is 102.573 nm. Bohr’s success with the hydrogen atom soon led to attempts both by him and by others to extend the same model to other atoms. On the qualitative level these attempts met with some success, and a general picture of electrons occupying orbits in successive levels and sub-levels, similar to that shown in Figure 5.2, began to emerge. On the quantitative level, however, all attempts to calculate accurate values for the energies of the electrons in their quantized orbitals were dismal failures. It was not until Schrödinger’s introduction of wave mechanics in 1926 that these difficulties could be resolved. Suddenly, it seemed, everything fell into place. Since then virtually every line in the spectrum of every element has been accounted for theoretically. As a result, we now have a very exact, though mathematically rather complex, picture of the behavior of electrons in both the ground state and in excited states of atoms. In particular, the study of atomic spectra has allowed us to determine the ionization energies of all the elements very accurately. Details of the spectra of polyelectronic atoms are complex, and so we will consider only one example: sodium. Excited states of sodium may be obtained by increasing the energy of the atom so that the 3 valence electron occupies the 3 , 3 , 4 , 4 , 4 , 4 or some other orbital. By contrast with the hydrogen atom, however, a sodium atom has other electrons which shield the valence electron from nuclear charge, and this shielding is different for each different orbital shape ( , , , , etc.). Consequently the energy of an excited sodium atom whose electron configuration is 1 2 2 4 is as that of an excited sodium atom whose configuration is 1 2 2 4 . Different shielding of the outermost (4 or 4 ) electron results in a different energy. Because of this, four formulas are needed to describe the energy of the sodium atom—one for each of the orbital shapes available to the outermost electron: \[\begin{align*} E_{ns} & =\frac{2.1800 \text{ aJ}}{(n-a_{s})^{2}}\\E_{np} & =\frac{2.1800 \text{ aJ}}{(n-a_{p})^{2}}\\ E_{nd} &= \frac{2.1800 \text{ aJ}}{(n-a_{d})^{2}}\\ E_{nf} & = \frac{2.1800 \text{ aJ}}{(n-a_{f})^{2}} \end{align*} \nonumber \] In all these equations represents the principal quantum number. It must be 3 or greater since the electron is in the 3 orbital to begin with. The different shielding requires a different correction for each type of orbital: = 1.36; = 0.87; = 0.012; and = 0.001. Because there are four different sets of energy levels, the number of transitions between levels (and hence the number of lines in the spectrum) is larger for sodium than for hydrogen. Early spectroscopists were able to distinguish four different types of lines, which they labeled the , , , and series. It is from the abbreviation of these terms that we have obtained the modern symbols , , , and . As most readers will know, when almost any sodium compound is held in a Bunsen burner, it imparts a brilliant yellow color to the flame. This yellow color corresponds to the most prominent line in the sodium spectrum. Its wavelength is 589 nm. On the atomic scale this line is caused by the sodium atom moving from an excited state (in which the valence electron is in a 3 orbital) to the ground state (in which the electron is in a 3 orbital). Using the above equations we can obtain approximate values for the two energies involved in the transition: \[E_{3p}=\frac{2.1800 \text{ aJ}}{(3-0.87)^{2}}=-0.4805\, \text{ aJ} \nonumber \] and \[E_{3s}=\dfrac{2.1800 \text{ aJ}}{(3-1.36)^{2}}=-0.8105\, \text{ aJ}. \nonumber \] Thus \(\Delta E= 0.3300 \text{ aJ}\) giving \(\lambda = \dfrac{hc}{\Delta E}=602 \text{ nm}\) This agrees approximately with the experimental result. Another feature of the sodium spectrum deserves mention. Careful observation reveals that the yellow color of sodium is actually due to two closely spaced lines (a ). One has a wavelength of 588.995 nm, and the other is at 589.592 nm. When the electron is in a 3 orbital, its spin can be aligned in two ways with respect to the axis of the orbital. The small difference in energy between these two orientations results in two slightly different wavelengths | 12,527 | 2,987 |
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Early scientists explored the relationships among the pressure of a gas ( ) and its temperature ( ), volume ( ), and amount ( ) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the . As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal. The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure \(\Page {1}\). In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure \(\Page {2}\)). A simple plot of \(V\) versus \(P\) gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows: \[PV = \rm constant \label{10.3.1} \] Dividing both sides by \(P\) gives an equation illustrating the inverse relationship between \(P\) and \(V\): or \[V \propto \dfrac{1}{P} \label{10.3.3} \] where the ∝ symbol is read “is proportional to.” A plot of versus 1/ is thus a straight line whose slope is equal to the constant in Equations \(\ref{10.3.1}\) and \(\ref{10.3.3}\). Dividing both sides of Equation \(\ref{10.3.1}\) by instead of gives a similar relationship between and 1/ . The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: This law in practice is shown in Figure \(\Page {2}\). At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air. The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure \(\Page {3}\)).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure \(\Page {3}\) that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure \(\Page {3}\), plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis. The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K). We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure \(\Page {3}\) is often referred to as Charles’s law and is stated mathematically as \[V ={\rm const.}\; T \label{10.3.4} \] or \[V \propto T \label{10.3.5} \] with not Charles’s law is valid for virtually all gases at temperatures well above their boiling points. We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure \(\Page {4}\)). This is the historic “Avogadro’s hypothesis.” A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: Stated mathematically, This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures. For a sample of gas, The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure \(\Page {5}\). Volume with increasing temperature or amount, but with increasing pressure. The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure ( ), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure ( ), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present ( ). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is , the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | 7,633 | 2,988 |
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By comparison with , and have microscopic structures in which the constituent particles are very close together. The volume occupied by a given amount of a solid or liquid is much less than that of the corresponding gas. Consequently solids and liquids collectively are called condensed phases. The properties of solids and liquids are much more dependent on intermolecular forces and on atomic, molecular, or ionic sizes and shapes than are the properties of gases. NaCl crystal lattice CO Lattice Quartz Gallium Crystal Bismuth Crystal Despite their greater variation with changes in molecular structure, some properties of condensed phases are quite general. In a solid, for example, microscopic particles are arranged in a regular, repeating crystal . Above you can see microscopic images of this lattice in NaCl and CO . On a macroscopic scale the 3 crystals on the right each show unique, repeating shapes due to their microscopic lattices. There are only a limited number of different ways such a lattice can form, and so it is worth spending some time to see what they are, which we will do later in the chapter. Similarly, useful generalizations can be made regarding the properties of liquids and about changes of phase - when a solid melts, a liquid vaporizes, and so on. The liquid phase, where microscopic particles are close together but can still move past one another, provides an ideal medium for chemical reactions. Reactant molecules can move toward one another because they are not held in fixed locations as in a solid, and a great many more collisions between molecules are possible because they are much closer together than in a gas. Such collisions lead to breaking of some bonds and formation of new ones, that is, to chemical reactions. This molecular intimacy without rigidity, combined with ease of handling of liquids in the laboratory, leads chemists to carry out many reactions in the liquid phase. Usually such reactions involve solutions of reactants in liquid solvents. Consequently we shall examine some general properties of as well. | 2,090 | 2,989 |
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We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO , as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure \(\Page {1}\). Phase changes are accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is . Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is . The energy change associated with each common phase change is shown in Figure \(\Page {1}\). Δ is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state. Previously, we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar (\(ΔH_{fus}\)), the energy required to convert from a solid to a liquid, a process known as fusion (or melting), as well as the normal boiling points and enthalpies of vaporization (\(ΔH_{vap}\)) of selected compounds are listed in Table \(\Page {1}\). The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid). Less energy is needed to allow molecules to move past each other than to separate them totally. The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimation. The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (Δ ). Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO (dry ice); iodine (Figure \(\Page {2}\)); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in Figure \(\Page {1}\), the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same ; this is an application of . \[ΔH_{sub} =ΔH_{fus} +ΔH_{vap} \label{Eq1} \] Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the (CFCs) and the hydrofluorocarbons (HCFCs). The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C. The processes on the right side of Figure \(\Page {1}\)—freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve. Figure \(\Page {3}\) shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat ( ) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a rate than seen in the other phases because the heat capacity of steam is than that of ice or water. Thus . In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils. The temperature of a sample does not change during a phase change. If heat is added at a constant rate, as in Figure \(\Page {3}\), then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure \(\Page {3}\), the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion. A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form. The cooling curve, a plot of temperature versus cooling time, in Figure \(\Page {4}\) plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure \(\Page {3}\), the cooling curve is an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system. Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a ) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO (dry ice) into the cloud from an airplane. Solid CO sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO sublimes, it absorbs heat from the cloud, often with the desired results. A Discussing the Thermodynamics of Phase Changes. If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol. mass, volume, initial temperature, density, specific heats, and \(ΔH_{fus}\) final temperature Substitute the given values into the general equation relating heat gained (by the ice) to heat lost (by the tea) to obtain the final temperature of the mixture. When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by \[q=mC_sΔT \nonumber \] where \(q\) is heat, \(m\) is mass, \(C_s\) is the specific heat, and \(ΔT\) is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C. The amount of heat gained by the ice cube as it melts is determined by its enthalpy of fusion in kJ/mol: \[q=nΔH_{fus} \nonumber \] For our 50.0 g ice cube: \[\begin{align*} q_{ice} &= 50.0 g⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \\[4pt] &= 16.7\, kJ \end{align*} \nonumber \] Thus, when the ice cube has just melted, it has absorbed 16.7 kJ of heat from the tea. We can then substitute this value into the first equation to determine the change in temperature of the tea: \[q_{tea} = - 16,700 J = 500 mL⋅\dfrac{1.00\: g}{1\: mL}⋅4.184 J/(g•°C) ΔT \nonumber \] \[ΔT = - 7.98 °C = T_f - T_i \nonumber \] \[T_f = 12.02 °C \nonumber \] This would be the temperature of the tea when the ice cube has just finished melting; however, this leaves the melted ice still at 0.0°C. We might more practically want to know what the final temperature of the mixture of tea will be once the melted ice has come to thermal equilibrium with the tea. To determine this, we can add one more step to the calculation by plugging in to the general equation relating heat gained and heat lost again: \[\begin{align*} q_{ice} &= - q_{tea} \\[4pt] q_{ice} &= m_{ice}C_sΔT = 50.0g⋅4.184 J/(g•°C)⋅(T_f - 0.0°C) \\[4pt] &= 209.2 J/°C⋅T_f \end{align*} \nonumber \] \[q_{tea} = m_{tea}C_sΔT = 500g⋅4.184 J/(g•°C)⋅(T_f - 12.02°C) = 2092 J/°C⋅T_f - 25,150 J \nonumber \] \[209.2 J/°C⋅T_f = - 2092 J/°C⋅T_f + 25,150 J \nonumber \] \[2301.2 J/°C⋅T_f = 25,150 J \nonumber \] \[T_f = 10.9 °C \nonumber \] The final temperature is in between the initial temperatures of the tea (12.02 °C) and the melted ice (0.0 °C), so this answer makes sense. In this example, the tea loses much more heat in melting the ice than in mixing with the cold water, showing the importance of accounting for the heat of phase changes! Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example \(\Page {1}\) 200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow. Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of , or . All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are . Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always . The conversion of a solid to a liquid is called . The energy required to melt 1 mol of a substance is its enthalpy of fusion (Δ ). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (Δ ). The direct conversion of a solid to a gas is . The amount of energy needed to sublime 1 mol of a substance is its and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called . Heating curves relate temperature changes to phase transitions. A , a liquid at a temperature and pressure at which it should be a gas, is not stable. A is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a , a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a of the same or another substance can induce crystallization. | 17,271 | 2,990 |
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(where X= Br or Cl, we will discuss further in detail later why other members of the halogen family Flourine and Iodine are not used in halogenation of benzenes) ) where X= Br or Cl. In the following examples, the halogen we will look at is Bromine. In the example of bromine, in order to make bromine electrophillic enough to react with benzene, we use the aid of an aluminum halide such as aluminum bromide. With aluminum bromide as a Lewis acid, we can mix Br with AlBr to give us Br . The presence of Br is a much better electrophile than Br alone. Bromination is acheived with the help of AlBr (Lewis acid catalysts) as it polarizes the Br-Br bond. The polarization causes polarization causes the bromine atoms within the Br-Br bond to become more electrophillic. The presence of Br compared to Br alone is a much better electrophille that can then react with benzene. As the bromine has now become more electrophillic after activation of a catalyst, an electrophillic attack by the benzene occurs at the terminal bromine of Br-Br-AlBr This allows the other bromine atom to leave with the AlBr as a good leaving group, AlBr -. After the electrophilic attack of bromide to the benzene, the hydrogen on the same carbon as bromine substitutes the carbocation in which resulted from the attack. Hence it being an electrophilic aromatic SUBSTITUTION. Since the by-product aluminum tetrabromide is a strong nucleophile, it pulls of a proton from the Hydrogen on the same carbon as bromine. In the end, AlBr was not consumed by the reaction and is regenerated. It serves as our catalyst in the halogenation of benzenes. The electrophillic bromination of benzenes is an exothermic reaction. Considering the exothermic rates of aromatic halogenation decreasing down the periodic table in the Halogen family, Flourination is the most exothermic and Iodination would be the least. Being so exothermic, a reaction of flourine with benzene is explosive! For iodine, electrophillic iodination is generally endothermic, hence a reaction is often not possible. Similar to bromide, chlorination would require the aid of an activating presence such as Alumnium Chloride or Ferric Chloride. The mechanism of this reaction is the same as with Bromination of benzene. What reagents would you need to get the given product? What product would result from the given reagents? What is the major product given the reagents below? Draw the formatin of Cl from AlCl and Cl . Draw the mechanism of the reaction between Cl and a benzene. 1. Cl and AlCl3 or Cl and FeCl 2. No Reaction 3. 4. 5. | 2,594 | 2,991 |
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The following set of experiments and class exercises introduce students to the importance of sampling on the quality of analytical results. The following experiments describe homemade sampling devices for collecting samples in the field. The following experiments introduce students to methods for extracting analytes from their matrix. The following papers provides a general introduction to the terminology used in describing sampling. Further information on the statistics of sampling is covered in the following papers and textbooks. The process of collecting a sample presents a variety of difficulties, particularly with respect to the analyte’s integrity. The following papers provide representative examples of sampling problems. The following texts and articles provide additional information on methods for separating analytes and inter- ferents. | 868 | 2,992 |
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We have already noted the importance of in the chemical laboratory, in the natural environment, and in the human body. Many reactions in aqueous solutions involve weak acids or bases or slightly soluble substances, and in such cases one or more equilibria are achieved in solution. Furthermore, the equilibrium state is usually reached almost instantaneously, and so we can use the equilibrium law to calculate the concentrations and amounts of substance of different species in solution. Such information enables us to understand, predict, and control what will happen in solution, and it has numerous practical applications. Equilibrium constants may be used to obtain information about reactions in solution, and in many cases the results of equilibrium calculations will be applied to practical problems. Acid-base reactions in aqueous solutions are intimately related to water’s ability to act as both a weak acid and a weak base, . In any aqueous solution at 25°C, the equilibrium constant of water (k ) is as follows: \[\text{K}_w={[\text{ H}_{3}\text{O}^{+},\text{OH}^{-}]}=\frac{1.00 \times{10}^{-14}}{\text{1 mol L}^{-\text{1}}} \nonumber \] \[[\text{H}_{3}\text{O}^{+},\text{OH}^{-}]={K_{w}}=\text{1.00}\times\text{10}^{-14} \text{ mol}^{2} \text{ L}^{-2} \nonumber \] and concentrations of H O and OH can vary from roughly 10 to 10 mol/L. This makes it convenient to define as: \[\begin{align}\text{pH}=-\text{log}\frac{[\text{ H}_{3}\text{O}^{+}]}{\text{1 mol L}^{1}}\text{ } \\\\\text{pOH}=-\text{log}\frac{[\text{ OH}^{-}]}{\text{1 mol L}^{-1}} \\\end{align} \nonumber \] Since molecules of a strong acid transfer their protons to water molecules completely, [H O ] (and hence pH) can be obtained directly from the stoichiometric concentration of the solution. Similarly [OH ] and pOH may be obtained from the stoichiometric concentration of a strong base. In the case of and , proton-transfer reactions proceed to only a limited extent and a dynamic equilibrium is set up. In such cases an acid constant (equilibrium constant for acids) or a base constant as well as the stoichiometric concentration of weak acid or base are required to calculate [H O ], [OH ], pH, or pOH. and for a are related, and their product is always . Often it is necessary or desirable to restrict the pH of an aqueous solution to a narrow range. This can be accomplished by means of a buffer solution―one which contains a conjugate weak acid-weak base pair. If a small amount of strong base is added to a buffer, the OH ions are consumed by the conjugate weak acid, so they have little influence on pH. Similarly, a small amount of strong acid can be consumed by the conjugate weak base in a buffer. To a good approximation the [H O ] in a buffer solution depends only on for the weak acid and the stoichiometric concentrations of the weak acid and weak base. for acid-base on are conjugate acid-base pairs, each member of which is a different color. An indicator changes from the color of the conjugate acid to the color of the conjugate base as pH increases from approximately p – 1 to p + 1. For titrations involving only strong acids and strong bases, several indicators are usually capable of signaling the endpoint because there is a large jump in within ± 0.05 L of the exact stoichiometric volume of titrant. In the case of titrations which involve a weak acid or a weak base, a buffer solution is involved and the jump in pH is smaller. Consequently greater care is required in selection of an appropriate indicator. A dynamic equilibrium is set up when a solid compound is in contact with a saturated solution. In the case of an ionic solid, the equilibrium constant for such a process is called . can be determined by measurement of the solubility of a compound, and it is useful in predicting whether the compound will precipitate when ionic solutions are mixed. The , in which an increase in the concentration of one ion decreases the concentration of the other ion of an insoluble compound, can be interpreted quantitatively using solubility products. It is also true that removal of one ion of an insoluble compound from solution will increase the concentration of the other ion, and hence the solubility. It is for this reason that often dissolve in acidic solutions―protonation of the anion effectively reduces its concentration to the point where the solubility product is not exceeded. | 4,445 | 2,993 |
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. In addition to these individual basis; please contact What is the main difference between the VSEPR model and Lewis electron structures? What are the differences between molecular geometry and Lewis electron structures? Can two molecules with the same Lewis electron structures have different molecular geometries? Can two molecules with the same molecular geometry have different Lewis electron structures? In each case, support your answer with an example. How does the VSEPR model deal with the presence of multiple bonds? Three molecules have the following generic formulas: AX , AX E, and AX E . Predict the molecular geometry of each, and arrange them in order of increasing X–A–X angle. Which has the smaller angles around the central atom—H S or SiH ? Why? Do the Lewis electron structures of these molecules predict which has the smaller angle? Discuss in your own words why lone pairs of electrons occupy more space than bonding pairs. How does the presence of lone pairs affect molecular geometry? When using VSEPR to predict molecular geometry, the importance of repulsions between electron pairs decreases in the following order: LP–LP, LP–BP, BP–BP. Explain this order. Draw structures of real molecules that separately show each of these interactions. How do multiple bonds affect molecular geometry? Does a multiple bond take up more or less space around an atom than a single bond? a lone pair? Straight-chain alkanes do not have linear structures but are “kinked.” Using -hexane as an example, explain why this is so. Compare the geometry of 1-hexene to that of -hexane. How is molecular geometry related to the presence or absence of a molecular dipole moment? How are molecular geometry and dipole moments related to physical properties such as melting point and boiling point? What two features of a molecule’s structure and bonding are required for a molecule to be considered polar? Is COF likely to have a significant dipole moment? Explain your answer. When a chemist says that a molecule is , what does this mean? What are the general physical properties of polar molecules? Use the VSPER model and your knowledge of bonding and dipole moments to predict which molecules will be liquids or solids at room temperature and which will be gases. Explain your rationale for each choice. Justify your answers. The idealized molecular geometry of BrF is square pyramidal, with one lone pair. What effect does the lone pair have on the actual molecular geometry of BrF ? If LP–BP repulsions were than BP–BP repulsions, what would be the effect on the molecular geometry of BrF ? Which has the smallest bond angle around the central atom—H S, H Se, or H Te? the largest? Justify your answers. Which of these molecular geometries results in a molecule with a net dipole moment: linear, bent, trigonal planar, tetrahedral, seesaw, trigonal pyramidal, square pyramidal, and octahedral? For the geometries that do not always produce a net dipole moment, what factor(s) will result in a net dipole moment? To a first approximation, the VSEPR model assumes that multiple bonds and single bonds have the same effect on electron pair geometry and molecular geometry; in other words, VSEPR treats multiple bonds like single bonds. Only when considering fine points of molecular structure does VSEPR recognize that multiple bonds occupy more space around the central atom than single bonds. Physical properties like boiling point and melting point depend upon the existence and magnitude of the dipole moment of a molecule. In general, molecules that have substantial dipole moments are likely to exhibit greater intermolecular interactions, resulting in higher melting points and boiling points. The term “polar” is generally used to mean that a molecule has an asymmetrical structure and contains polar bonds. The resulting dipole moment causes the substance to have a higher boiling or melting point than a nonpolar substance. Give the number of electron groups around the central atom and the molecular geometry for each molecule. Classify the electron groups in each species as bonding pairs or lone pairs. Give the number of electron groups around the central atom and the molecular geometry for each species. Classify the electron groups in each species as bonding pairs or lone pairs. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons. Give the number of electron groups around the central atom and the molecular geometry for each molecule. For structures that are not linear, draw three-dimensional representations, clearly showing the positions of the lone pairs of electrons. What is the molecular geometry of ClF ? Draw a three-dimensional representation of its structure and explain the effect of any lone pairs on the idealized geometry. Predict the molecular geometry of each of the following. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles. Predict whether each molecule has a net dipole moment. Justify your answers and indicate the direction of any bond dipoles. Of the molecules Cl C=Cl , IF , and SF , which has a net dipole moment? Explain your reasoning. Of the molecules SO , XeF , and H C=Cl , which has a net dipole moment? Explain your reasoning. four electron groups, pyramidal molecular geometry The idealized geometry is T shaped, but the two lone pairs of electrons on Cl will distort the structure, making the F–Cl–F angle than 180°. Cl C=CCl : Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl C=CCl does not have a net dipole moment. IF : In this structure, the individual I–F bond dipoles cannot cancel one another, giving IF a net dipole moment. SF : The S–F bonds are quite polar, but the individual bond dipoles cancel one another in an octahedral structure. Thus, SF has no net dipole moment. Predict whether each compound is purely covalent, purely ionic, or polar covalent. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond. Classify each species as having 0%–40% ionic character, 40%–60% ionic character, or 60%–100% ionic character based on the type of bonding you would expect. Justify your reasoning. If the bond distance in HCl (dipole moment = 1.109 D) were double the actual value of 127.46 pm, what would be the effect on the charge localized on each atom? What would be the percent negative charge on Cl? At the actual bond distance, how would doubling the charge on each atom affect the dipole moment? Would this represent more ionic or covalent character? Calculate the percent ionic character of HF (dipole moment = 1.826 D) if the H–F bond distance is 92 pm. Calculate the percent ionic character of CO (dipole moment = 0.110 D) if the C–O distance is 113 pm. Calculate the percent ionic character of PbS and PbO in the gas phase, given the following information: for PbS, = 228.69 pm and µ = 3.59 D; for PbO, = 192.18 pm and µ = 4.64 D. Would you classify these compounds as having covalent or polar covalent bonds in the solid state? Arrange , , and in order of increasing strength of the bond formed to a hydrogen atom. Explain your reasoning. What atomic orbitals are combined to form , , , and ? What is the maximum number of electron-pair bonds that can be formed using each set of hybrid orbitals? Why is it incorrect to say that an atom with hybridization will form only three bonds? The carbon atom in the carbonate anion is hybridized. How many bonds to carbon are present in the carbonate ion? Which orbitals on carbon are used to form each bond? If hybridization did not occur, how many bonds would N, O, C, and B form in a neutral molecule, and what would be the approximate molecular geometry? How are hybridization and molecular geometry related? Which has a stronger correlation—molecular geometry and hybridization or Lewis structures and hybridization? In the valence bond approach to bonding in BeF , which step(s) require(s) an energy input, and which release(s) energy? The energies of hybrid orbitals are intermediate between the energies of the atomic orbitals from which they are formed. Why? How are lone pairs on the central atom treated using hybrid orbitals? Because nitrogen bonds to only three hydrogen atoms in ammonia, why doesn’t the nitrogen atom use hybrid orbitals instead of hybrids? Using arguments based on orbital hybridization, explain why the CCl ion does not exist. Species such as NF and OF are unknown. If 3 atomic orbitals were much lower energy, low enough to be involved in hybrid orbital formation, what effect would this have on the stability of such species? Why? What molecular geometry, electron-pair geometry, and hybridization would be expected for each molecule? Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH . How does your diagram compare with that for methane? What is the molecular geometry? Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH . How does your diagram compare with that for methane? What is the molecular geometry? Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each molecule. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each species. What is the hybridization of the central atom in each of the following? What is the hybridization of the central atom in each of the following? What is the hybridization of the central atom in PF ? Is this ion likely to exist? Why or why not? What would be the shape of the molecule? What is the hybridization of the central atom in SF ? Is this ion likely to exist? Why or why not? What would be the shape of the molecule? The promotion and hybridization process is exactly the same as shown for CH in the chapter. The only difference is that the C atom uses the four singly occupied hybrid orbitals to form electron-pair bonds with only H atoms, and an electron is added to the fourth hybrid orbital to give a charge of 1–. The electron-pair geometry is tetrahedral, but the molecular geometry is pyramidal, as in NH . , trigonal planar , pyramidal , trigonal planar The central atoms in CF , CCl , IO , and SiH are all sp hybridized. The phosphorus atom in the PF ion is hybridized, and the ion is octahedral. The PF ion is isoelectronic with SF and has essentially the same structure. It should therefore be a stable species. | 11,194 | 2,995 |
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Bonding in metals and semiconductors can be described using band theory, in which a set of molecular orbitals is generated that extends throughout the solid. The primary learning objective of this Module is to describe the electrical properties of solid using band theory. To explain the observed properties of metals, a more sophisticated approach is needed than the electron-sea model commonly described. The molecular orbital theory used to explain the delocalized π bonding in polyatomic ions and molecules such as NO , ozone, and 1,3-butadiene can be adapted to accommodate the much higher number of atomic orbitals that interact with one another simultaneously in metals. In a 1 mol sample of a metal, there can be more than 10 orbital interactions to consider. In our molecular orbital description of metals, however, we begin by considering a simple one-dimensional example: a linear arrangement of n metal atoms, each containing a single electron in an s orbital. We use this example to describe an approach to metallic bonding called band theory, which assumes that the valence orbitals of the atoms in a solid interact, generating a set of molecular orbitals that extend throughout the solid. If the distance between the metal atoms is short enough for the orbitals to interact, they produce bonding, antibonding, and nonbonding molecular orbitals. The left portion of Figure \(\Page {1}\) shows the pattern of molecular orbitals that results from the interaction of ns orbitals as n increases from 2 to 5. As we saw previously, the lowest-energy orbital is the completely bonding molecular orbital, whereas the highest-energy orbital is the completely antibonding molecular orbital. Molecular orbitals of intermediate energy have fewer nodes than the totally antibonding molecular orbital. The energy separation between adjacent orbitals decreases as the number of interacting orbitals increases. For n = 30, there are still discrete, well-resolved energy levels, but as n increases from 30 to a number close to Avogadro’s number, the spacing between adjacent energy levels becomes almost infinitely small. The result is essentially a continuum of energy levels, as shown on the right in Figure \(\Page {1}\), each of which corresponds to a particular molecular orbital extending throughout the linear array of metal atoms. The levels that are lowest in energy correspond to mostly bonding combinations of atomic orbitals, those highest in energy correspond to mostly antibonding combinations, and those in the middle correspond to essentially nonbonding combinations. The continuous set of allowed energy levels shown on the right in Figure \(\Page {1}\) is called an energy band. The difference in energy between the highest and lowest energy levels is the bandwidth and is proportional to the strength of the interaction between orbitals on adjacent atoms: the stronger the interaction, the larger the bandwidth. Because the band contains as many energy levels as molecular orbitals, and the number of molecular orbitals is the same as the number of interacting atomic orbitals, the band in Figure \(\Page {1}\) contains n energy levels corresponding to the combining of s orbitals from n metal atoms. Each of the original s orbitals could contain a maximum of two electrons, so the band can accommodate a total of 2n electrons. Recall, however, that each of the metal atoms we started with contained only a single electron in each s orbital, so there are only n electrons to place in the band. Just as with atomic orbitals or molecular orbitals, the electrons occupy the lowest energy levels available. Consequently, only the lower half of the band is filled. This corresponds to filling all of the bonding molecular orbitals in the linear array of metal atoms and results in the strongest possible bonding. The previous example was a one-dimensional array of atoms that had only s orbitals. To extrapolate to two- or three-dimensional systems and atoms with electrons in p and d orbitals is straightforward in principle, even though in practice the mathematics becomes more complex, and the resulting molecular orbitals are more difficult to visualize. The resulting energy-level diagrams are essentially the same as the diagram of the one-dimensional example in Figure \(\Page {1}\), with the following exception: they contain as many bands as there are different types of interacting orbitals. Because different atomic orbitals interact differently, each band will have a different bandwidth and will be centered at a different energy, corresponding to the energy of the parent atomic orbital of an isolated atom. Because the 1s, 2s, and 2p orbitals of a period 3 atom are filled core levels, they do not interact strongly with the corresponding orbitals on adjacent atoms. Hence they form rather narrow bands that are well separated in energy (Figure \(\Page {2}\)). These bands are completely filled (both the bonding and antibonding levels are completely populated), so they do not make a net contribution to bonding in the solid. The energy difference between the highest level of one band and the lowest level of the next is the band gap. It represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals. Because they extend farther from the nucleus, the valence orbitals of adjacent atoms (3s and 3p in Figure \(\Page {2}\)) interact much more strongly with one another than do the filled core levels; as a result, the valence bands have a larger bandwidth. In fact, the bands derived from the 3s and 3p atomic orbitals are wider than the energy gap between them, so the result is overlapping bands. These have molecular orbitals derived from two or more valence orbitals with similar energies. As the valence band is filled with one, two, or three electrons per atom for Na, Mg, and Al, respectively, the combined band that arises from the overlap of the 3s and 3p bands is also filling up; it has a total capacity of eight electrons per atom (two electrons for each 3s orbital and six electrons for each set of 3p orbitals). With Na, therefore, which has one valence electron, the combined valence band is one-eighth filled; with Mg (two valence electrons), it is one-fourth filled; and with Al, it is three-eighths filled, as indicated in Figure \(\Page {2}\). The partially filled valence band is absolutely crucial for explaining metallic behavior because it guarantees that there are unoccupied energy levels at an infinitesimally small energy above the highest occupied level. Band theory can explain virtually all the properties of metals. Metals conduct electricity, for example, because only a very small amount of energy is required to excite an electron from a filled level to an empty one, where it is free to migrate rapidly throughout the crystal in response to an applied electric field. Similarly, metals have high heat capacities (as you no doubt remember from the last time a doctor or a nurse placed a stethoscope on your skin) because the electrons in the valence band can absorb thermal energy by being excited to the low-lying empty energy levels. Finally, metals are lustrous because light of various wavelengths can be absorbed, causing the valence electrons to be excited into any of the empty energy levels above the highest occupied level. When the electrons decay back to low-lying empty levels, they emit light of different wavelengths. Because electrons can be excited from many different filled levels in a metallic solid and can then decay back to any of many empty levels, light of varying wavelengths is absorbed and reemitted, which results in the characteristic shiny appearance that we associate with metals. For a solid to exhibit metallic behavior, Without a set of delocalized orbitals, there is no pathway by which electrons can move through the solid. Band theory explains the correlation between the valence electron configuration of a metal and the strength of metallic bonding. The valence electrons of transition metals occupy either their valence ns, (n − 1)d, and np orbitals (with a total capacity of 18 electrons per metal atom) or their ns and (n − 1)d orbitals (a total capacity of 12 electrons per metal atom). These atomic orbitals are close enough in energy that the derived bands overlap, so the valence electrons are not confined to a specific orbital. Metals with 6 to 9 valence electrons (which correspond to groups 6–9) are those most likely to fill the valence bands approximately halfway. Those electrons therefore occupy the highest possible number of bonding levels, while the number of antibonding levels occupied is minimal. Not coincidentally, the elements of these groups exhibit physical properties consistent with the presence of the strongest metallic bonding, such as very high melting points. In contrast to metals, electrical insulators are materials that conduct electricity poorly because their valence bands are full. The energy gap between the highest filled levels and the lowest empty levels is so large that the empty levels are inaccessible: thermal energy cannot excite an electron from a filled level to an empty one. The valence-band structure of diamond, for example, is shown in Figure \(\Page {3a}\). Because diamond has only 4 bonded neighbors rather than the 6 to 12 typical of metals, the carbon 2s and 2p orbitals combine to form two bands in the solid, with the one at lower energy representing bonding molecular orbitals and the one at higher energy representing antibonding molecular orbitals. Each band can accommodate four electrons per atom, so only the lower band is occupied. Because the energy gap between the filled band and the empty band is very large (530 kJ/mol), at normal temperatures thermal energy cannot excite electrons from the filled level into the empty band. Thus there is no pathway by which electrons can move through the solid, so diamond has one of the lowest electrical conductivities known. What if the difference in energy between the highest occupied level and the lowest empty level is intermediate between those of electrical conductors and insulators? This is the case for silicon and germanium, which have the same structure as diamond. Because Si–Si and Ge–Ge bonds are substantially weaker than C–C bonds, the energy gap between the filled and empty bands becomes much smaller as we go down group 14 (part (b) and part (c) of Figure \(\Page {2}\)). Consequently, thermal energy is able to excite a small number of electrons from the filled valence band of Si and Ge into the empty band above it, which is called the conduction band. Exciting electrons from the filled valence band to the empty conduction band causes an increase in electrical conductivity for two reasons: Consequently, Si is a much better electrical conductor than diamond, and Ge is even better, although both are still much poorer conductors than a typical metal (Figure \(\Page {4}\)). Substances such as Si and Ge that have conductivities between those of metals and insulators are called semiconductors. Many binary compounds of the main group elements exhibit semiconducting behavior similar to that of Si and Ge. For example, gallium arsenide (GaAs) is isoelectronic with Ge and has the same crystalline structure, with alternating Ga and As atoms; not surprisingly, it is also a semiconductor. The electronic structure of semiconductors is compared with the structures of metals and insulators in Figure \(\Page {5}\). Because thermal energy can excite electrons across the band gap in a semiconductor, increasing the temperature increases the number of electrons that have sufficient kinetic energy to be promoted into the conduction band. The electrical conductivity of a semiconductor therefore increases rapidly with increasing temperature, in contrast to the behavior of a purely metallic crystal. In a metal, as an electron travels through the crystal in response to an applied electrical potential, it cannot travel very far before it encounters and collides with a metal nucleus. The more often such encounters occur, the slower the net motion of the electron through the crystal, and the lower the conductivity. As the temperature of the solid increases, the metal atoms in the lattice acquire more and more kinetic energy. Because their positions are fixed in the lattice, however, the increased kinetic energy increases only the extent to which they vibrate about their fixed positions. At higher temperatures, therefore, the metal nuclei collide with the mobile electrons more frequently and with greater energy, thus decreasing the conductivity. This effect is, however, substantially smaller than the increase in conductivity with temperature exhibited by semiconductors. For example, the conductivity of a tungsten wire decreases by a factor of only about two over the temperature range 750–1500 K, whereas the conductivity of silicon increases approximately 100-fold over the same temperature range. These trends are illustrated in Figure \(\Page {6}\). Doping is a process used to tune the electrical properties of commercial semiconductors by deliberately introducing small amounts of impurities. If an impurity contains more valence electrons than the atoms of the host lattice (e.g., when small amounts of a group 15 atom are introduced into a crystal of a group 14 element), then the doped solid has more electrons available to conduct current than the pure host has. As shown in Figure \(\Page {7a}\), adding an impurity such as phosphorus to a silicon crystal creates occasional electron-rich sites in the lattice. The electronic energy of these sites lies between those of the filled valence band and the empty conduction band but closer to the conduction band. Because the atoms that were introduced are surrounded by host atoms, and the electrons associated with the impurity are close in energy to the conduction band, those extra electrons are relatively easily excited into the empty conduction band of the host. Such a substance is called an n-type semiconductor, with the n indicating that the added charge carriers are negative (they are electrons). If the impurity atoms contain fewer valence electrons than the atoms of the host (e.g., when small amounts of a group 13 atom are introduced into a crystal of a group 14 element), then the doped solid has fewer electrons than the pure host. Perhaps unexpectedly, this also results in increased conductivity because the impurity atoms generate holes in the valence band. As shown in Figure \(\Page {7b}\), adding an impurity such as gallium to a silicon crystal creates isolated electron-deficient sites in the host lattice. The electronic energy of these empty sites also lies between those of the filled valence band and the empty conduction band of the host but much closer to the filled valence band. It is therefore relatively easy to excite electrons from the valence band of the host to the isolated impurity atoms, thus forming holes in the valence band. This kind of substance is called a with the p standing for positive charge carrier (i.e., a hole). Holes in what was a filled band are just as effective as electrons in an empty band at conducting electricity. The electrical conductivity of a semiconductor is roughly proportional to the number of charge carriers, so doping is a precise way to adjust the conductivity of a semiconductor over a wide range. The entire semiconductor industry is built on methods for preparing samples of Si, Ge, or GaAs doped with precise amounts of desired impurities and assembling silicon chips and other complex devices with junctions between n- and p-type semiconductors in varying numbers and arrangements. Because silicon does not stand up well to temperatures above approximately 100°C, scientists have been interested in developing semiconductors made from diamonds, a more thermally stable material. A new method has been developed based on vapor deposition, in which a gaseous mixture is heated to a high temperature to produce carbon that then condenses on a diamond kernel. This is the same method now used to create cultured diamonds, which are indistinguishable from natural diamonds. The diamonds are heated to more than 2000°C under high pressure to harden them even further. Doping the diamonds with boron has produced p-type semiconductors, whereas doping them with boron and deuterium achieves n-type behavior. Because of their thermal stability, diamond semiconductors have potential uses as microprocessors in high-voltage applications. A crystalline solid has the following band structure, with the purple areas representing regions occupied by electrons. The lower band is completely occupied by electrons, and the upper level is about one-third filled with electrons. : band structure : variations in electrical properties with conditions : : A substance has the following band structure, in which the lower band is half-filled with electrons (purple area) and the upper band is empty. : Band theory assumes that the valence orbitals of the atoms in a solid interact to generate a set of molecular orbitals that extend throughout the solid; the continuous set of allowed energy levels is an energy band. The difference in energy between the highest and lowest allowed levels within a given band is the bandwidth, and the difference in energy between the highest level of one band and the lowest level of the band above it is the band gap. If the width of adjacent bands is larger than the energy gap between them, overlapping bands result, in which molecular orbitals derived from two or more kinds of valence orbitals have similar energies. Metallic properties depend on a partially occupied band corresponding to a set of molecular orbitals that extend throughout the solid to form a band of energy levels. If a solid has a filled valence band with a relatively low-lying empty band above it (a conduction band), then electrons can be excited by thermal energy from the filled band into the vacant band where they can then migrate through the crystal, resulting in electrical conductivity. Electrical insulators are poor conductors because their valence bands are full. Semiconductors have electrical conductivities intermediate between those of insulators and metals. The electrical conductivity of semiconductors increases rapidly with increasing temperature, whereas the electrical conductivity of metals decreases slowly with increasing temperature. The properties of semiconductors can be modified by doping, or introducing impurities. Adding an element with more valence electrons than the atoms of the host populates the conduction band, resulting in an n-type semiconductor with increased electrical conductivity. Adding an element with fewer valence electrons than the atoms of the host generates holes in the valence band, resulting in a p-type semiconductor that also exhibits increased electrical conductivity. | 18,969 | 2,996 |
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The Earth formed from the same cloud of matter that formed the Sun, but the planets acquired different compositions during the formation and evolution of the solar system. In turn, the natural history of the Earth caused parts of this planet to have differing concentrations of the elements. The mass of the Earth is approximately 5.98×1024 kg. In bulk, by mass, it is composed mostly of iron (32.1%), oxygen (30.1%), silicon (15.1%), magnesium (13.9%), sulfur (2.9%), nickel (1.8%), calcium (1.5%), and aluminum (1.4%); with the remaining 1.2% consisting of trace amounts of other elements. Figure \(\Page {1}\) illustrates the relative atomic-abundance of the chemical elements in Earth's upper continental crust, which is relatively accessible for measurements and estimation. Many of the elements shown in the graph are classified into (partially overlapping) categories: Most metals are found as types of rock in the Earth's crust. These ores contain sufficient minerals with important elements including metals that can be economically extracted from the rock. Metal ores are generally oxides, sulfides, silicates (Table \(\Page {1}\)) or "native" metals (such as native copper) that are not commonly concentrated in the Earth's crust, or "noble" metals (not usually forming compounds) such as gold (Figure \(\Page {2}\)). The ores must be processed to extract the metals of interest from the waste rock and from the ore minerals. Extractive metallurgy is a branch of metallurgical engineering wherein process and methods of extraction of metals from their natural mineral deposits are studied. The field is a materials science, covering all aspects of the types of ore, washing, concentration, separation, chemical processes and extraction of pure metal and their alloying to suit various applications, sometimes for direct use as a finished product, but more often in a form that requires further working to achieve the given properties to suit the applications. The field of ferrous and non-ferrous extractive metallurgy have specialties that are generically grouped into the categories of mineral processing, hydrometallurgy, pyrometallurgy, and electrometallurgy based on the process adopted to extract the metal. Several processes are used for extraction of same metal depending on occurrence and chemical requirements. It takes multiple steps to extract the "important" element from the ore: Most minerals are chemical compounds that contain metals and other elements. After mining, large pieces of the ore feed are broken through crushing and/or grinding. This step creates particles that are either mostly valuable or mostly waste. TConcentrating the particles of value in a form supporting separation enables the desired metal to be removed from waste products. Froth flotation is one process for separating minerals from by taking advantage of differences in their . whereby hydrophobicity differences between valuable minerals other are increased through the use of and wetting agents (Figure \(\Page {3}\)). Flotation is used for the separation of a large range of sulfides, carbonates and oxides prior to further refinement. Phosphates and coal are also upgraded (purified) by flotation technology. Other ways to concentration minerals include approached based on density (Froth Flotation) used if ore density is less than density of impurities, or via melting points, or magnetic properties (Magnetic separation). This can be used with magnetic ores (i.e Fe O ) that are passed through a magnetic field, which attracts the magnetic ore and ignore the nonmagnetic impurities. Ore is crushed and heated to a high temp using a strong blast of hot air. The process converts the ores to their oxides which can then be reduced. Consider the natural nd \(ZnCO_3\) (s \(ZnO_{(s)}\) and \(CO_{2(g)}\) \(\ref{Roast1}\) a he hot air involved in roasting sphalerite oxidizes it \(\ref{Roast2}\) to produces \(ZnO_{(s)}\) and \(SO_{2(g)}\). \[ZnCO_{3(s)} \overset{\Delta}{\longrightarrow} ZnO_{(s)} + CO_{2(g)} \label{Roast1}\] \[2 ZnS_{(s)} + 3 O_{2(g)} \overset{\Delta}{\longrightarrow} 2 ZnO_{(s)} + 2 SO_{2(g)} \label{Roast2}\] \(C_{(s)}\) and \(CO_{(g)}\) are often used as reducing agents in simultaneous reactions. Oxides of Cr, V, and Mn are reduced using Al. During the reduction process the metal oxide is heated to a temperature above its boiling point in order to vaporize it and condense as a liquid. For example, \(ZnO_{(s)}\) that is produced in the roasting process is combined with \(C_{(s)}/CO_{(g)}\) mixture and heated at a 1100 C. Two reduction reactions are viable at this temperature with either carbon (Equation \(\ref{Reduction1}\)) or carbon monoxide ((Equation \(\ref{Reduction2}\)) acting the reducing agent. \[ZnO_{(s)} +C_{(s)} \overset{\Delta}{\longrightarrow} Zn_{(g)} + CO_{(g)} \label{Reduction1}\] \[ZnO_{(s)} + CO_{(g)} \overset{\Delta}{\longrightarrow} Zn_{(g)} + CO_{2(g)} \label{Reduction2} \] The impurities contained in the metal product of the roasting/reduction process are removed. For example, the \(Zn(l)\) produced in the reduction process often contains impurities of Cd and Pb. Though of Zn(l) will work in the refining process, a more commonly used and efficient method is . The \(ZnO_{(s)}\) produced from roasting is dissolved in \(H_2SO_{4(aq)}\) and \(Zn_{(s)}\) powder is added allowing the impurities to be displaced. The solution is then electrolyzed and to get a pure metallic \(Zn_{(s)}\) as a result. A rod containing the desired pure metal and impurities is passed through a series of heating coils and cooled again. The process isolates the impurities that concentrate in the molten zones, leaving the portions behind them somewhat more pure. This process is repeated until impurities are moved to the end of the rod and cut off, resulting in an almost completely pure metal rod. is the processes of roasting an ore a high temperatures and then reducing its oxide product. Its characteristics include: The oldest, and still the most common smelting process for oxide ores involves heating them in the presence of carbon. Originally, charcoal was used, but industrial-scale smelting uses coke, a crude form of carbon prepared by pyrolysis (heating) of coal. The basic reactions are: \[MO + C \rightleftharpoons M + CO \label{2.1}\] \[MO + ½ O_2 \rightleftharpoons M + ½ CO_2 \label{2.2}\] \[MO + CO \rightleftharpoons M + CO_2 \label{2.3}\] Each of these can be regarded as a pair of in which the metal M and the carbon are effectively competing for the oxygen atom. Using reaction \(\ref{2.1}\) as an example, it can be broken down into the following two parts: \[MO \rightleftharpoons M + ½ O_2 \;\;\; ΔG^o > 0 \label{2.4}\] \[C + ½ O_2 \rightleftharpoons CO \;\;\; ΔG^o < 0 \label{2.5}\] At ordinary environmental temperatures, reaction \(\ref{2.4}\) is always spontaneous in the reverse direction (that is why ores form in the first place!), so Δ ° of Reaction \(\ref{2.4}\) will be positive. Δ ° for reaction \(\ref{2.5}\) is always negative, but at low temperatures it will not be sufficiently negative to drive \(\ref{2.4}\). The process depends on the different ways in which the free energies of reactions like \(\ref{2.4}\) and \(\ref{2.4}\) vary with the temperature. This temperature dependence is almost entirely dominated by the Δ ° term in the Gibbs function, and thus by the entropy change. The latter depends mainly on Δ , the change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always accompanied by a large increase in the system volume so Δ for this step is always positive and the reaction becomes more spontaneous at higher temperatures. The temperature dependences of the reactions that take up oxygen vary, however (Table \(\Page {1}\)). A plot of the temperature dependences of the free energies of these reactions, superimposed on similar plots for the oxygen removal reactions \(\ref{2.4}\) is called an . For a given oxide MO to be smeltable, the temperature must be high enough that reaction \(\ref{2.4}\) falls below that of at least one of the oxygen-consuming reactions. The slopes of the lines on this diagram are determined by the sign of the entropy change. Examination of the Ellingham diagram shown above illustrates why the metals known to the ancients were mainly those such as copper and lead, which can be obtained by smelting at the relatively low temperatures that were obtainable by the methods available at the time in which a charcoal fire supplied both the heat and the carbon. Thus the bronze age preceded the iron age; the latter had to await the development of technology capable of producing higher temperatures, such as the blast furnace. Smelting of aluminum oxide by carbon requires temperatures too high to be practical; commercial production of aluminum is accomplished by electrolysis of the molten ore. In pyrometallurgy, an ore is heated with a reductant to obtain the metal. Theoretically, it should be possible to obtain virtually any metal from its ore by using coke as the reductant. For example, the reduction of calcum metral from \(CaO_{(s)}\): \[\mathrm{CaO(s) + C(s)\xrightarrow{\Delta}Ca(l) + CO(g)} \label{23.2.1}\] Unfortunately, many transition metals (e.g., ) react with carbon to form stable binary carbides: \[Ti + C \rightarrow TiC.\] Consequently, more expensive reductants (such as hydrogen, aluminum, magnesium, or calcium) must be used to obtain these metals. Many metals that occur naturally as sulfides can be obtained by heating the sulfide in air, as shown for lead in the following equation: \[\mathrm{PbS(s) + O_2(g) \xrightarrow{\Delta}Pb(l) + SO_2(g)} \label{23.2.2}\] The reaction is driven to completion by the formation of \(SO_2\), a stable gas. Titanium is used in the military and aircraft industry b/c of its low density and ability to maintain its strength are high temps. Titanium is produced using the following steps: The Kroll process is slow, expensive, and proposes a health/safety risk due to the high temperature vacuum distillation. There are alternative means of producing titanium including the electrolysis of TiO pellets via the following procedure: Cu ores often contain iron sulfides and to achieve a pure(uncontaminated) Cu product, the ores undergo the following processes : Step 1: Concentration of ore by flotation (Figure \(\Page {6}\). Step 2: Conversion of Fe sulfide ores to Fe oxides through roasting (e.g., roasting of Chalcopyrite ore ) \[ 2 CuFeS)2 + 3 O)2 \overset{\Delta}{\longrightarrow} 2 FeO + 2 CuS + 2 SO_2\] : mixing with coke/charcoal and heated at high temperatures (i.e 800 C) for copper to keep Cu from remaining in sulfide form). Impurities are further removed by adding flux if not already present in ore (if gangue is acidic basic flux is added, if gangue is basic acidic flux (i.e., SiO ) is used. Smelting separates the material to separate into 2 layers ( : bottom layer containing molten sulfides of Fe and Cu and : top layer formed by reaction of oxides of Fe, Ca, and Al w/ SiO ). \[FeO_{(s)} + SiO_{2(s)} \overset{\Delta}{\longrightarrow} FeSiO_{3(l)}\] : Air is blown through Cu matte in a separate furnace. This converts the remaining iron sulfide to iron oxide followed by the formation of FeSiO (l) slag. The slag is poured off and the process is repeated resulting in blister copper with bubbles of SO (g) still present. \[2CuS_{(l)} + 3O_{2(g)} \overset{\Delta}{\longrightarrow} 2Cu_2O_{(l)} + 2SO_{2(g)}\] \[ 2Cu_2O_{(l)} + Cu_2S_{(l)} \overset{\Delta}{\longrightarrow} 6Cu(l) +SO_{2(g)} \] Blister Cu is further refined through electrolysis to achieve a high-purity Cu product. The method of refining zinc described above is an example of a hydrometallurgical process and the hydrometallurgical process eliminates the need to control gaseous emissions that are often produced in roasting. is the process of extraction and refining that involves the use of water and aqueous solutions. It is carried out at moderate temps and is generally carried out in three steps: Extractive metallurgy is the practice of removing valuable metals from an ore and refining the extracted raw metals into purer form. The field of extractive metallurgy encompasses many specialty sub-disciplines, including mineral processing, hydrometallurgy, pyrometallurgy, and electrometallurgy. Especially in hydrometallurgy, the coordination chemistry of the metals involved plays a large role in their solubility and reactivity as the ore is refined into precious metal. Many methods of extractive metallurgy may be applied to get a pure metal from its naturally occurring ore. First ores must be concentrated and separated from their earthly impurities. Concentration of ores may be achieved by froth flotation in which an ore is contained in a froth that floats at the top of a water vat. After the ore has been concentrated it must be roasted to a high temperature allowing for the desired metal to be oxidized. Once the metal is oxidized it is reduced by a reducing agent. Reduction may require the metal oxide to be heated at a high temperature. The metal must be further refined to remove all of the impurities that it may contain. Refining a metal can be accomplished by electrolysis or zone refining. Certain ores such as those of copper and titanium require special treatment before they can be refined. The extraction of iron ore is particularly important for commercial uses because of the mass production of its alloy, steel. Boundless (www.boundless.com) ) | 13,582 | 2,997 |
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In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The is the substance that loses electrons and is oxidized in the process; the is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements. Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two , one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows: \[Zn_{(s)} + Br_{2(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Br^−_{(aq)} \label{19.1}\] The half-reactions are as follows: reduction half-reaction: \[Br_{2(aq)} + 2e^− \rightarrow 2Br^−_{(aq)} \label{19.2}\] oxidation half-reaction: \[Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^− \label{19.3}\] Each half-reaction is written to show what is actually occurring in the system; Zn is the reductant in this reaction (it loses electrons), and Br is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation \(\ref{19.1}\)). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation. In any redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an . There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A uses the energy released during a spontaneous redox reaction (ΔG < 0) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). Both types contain two , which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure \(\Page {1}\)). The oxidation half-reaction occurs at one electrode (the ), and the reduction half-reaction occurs at the other (the ). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells. Electrochemical Cells: To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu ) to give copper metal and Zn ion. The balanced chemical equation is as follows: \[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{19.4}\] We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work. This same reaction can be carried out using the galvanic cell illustrated in part (a) in Figure \(\Page {3}\). To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of Cu ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of Zn ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a , a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are Na or K and NO or SO , respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to Zn ions at the zinc electrode (the anode), and Cu ions are reduced to Cu metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of Zn ions in the Zn solution increases; simultaneously, the copper strip gains mass, and the concentration of Cu ions in the Cu solution decreases (part (b) in Figure \(\Page {3}\)). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work. The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the Zn solution would increase as the zinc metal dissolves, and the total positive charge in the Cu solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the Zn solution and a flow of cations into the Cu solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained. A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (part (a) in Figure \(\Page {3}\)). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged. A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction. Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example \(\Page {1}\). A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation: \[3Sn_{(s)} + 2NO^-_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}\] For this galvanic cell, galvanic cell and redox reaction half-reactions, identity of anode and cathode, and electrode assignment as positive or negative Solution: reduction: NO (aq) + 4H (aq) + 3e → NO(g) + 2H O(l) oxidation: Sn(s) → Sn (aq) + 2e Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn . Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of MnO in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of Sn in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation: \[2MnO^−_{4(aq)} + 5Sn^{2+}_{(aq)} + 16H^+_{(aq)} \rightarrow 2Mn^{2+}_{(aq)} + 5Sn^{4+}_{(aq)} + 8H_2O_{(l)}\] For this galvanic cell, Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the Zn/Cu cell shown in part (a) in Figure \(\Page {3}\) is written as follows: A cell diagram includes solution concentrations when they are provided. Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the . One example of this type of galvanic cell is as follows: \[Pt_{(s)}\, | \, H_{2(g)} | HCl_{(aq)}\,|\, AgCl_{(s)} \,Ag_{(s)} \label{19.5}\] This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows: cathode reaction: \[AgCl_{(s)} + e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)} \label{19.6}\] anode reaction: \[\dfrac{1}{2}\,\mathrm{H_{2(g)}}\rightarrow\mathrm{H^+_{(aq)}}+\mathrm{e^-} \label{19.7}\] overall: \[\mathrm{AgCl_{(s)}}+\dfrac{1}{2}\mathrm{H_{2(g)}}\rightarrow\mathrm{Ag_{(s)}}+\mathrm{Cl^-_{(aq)}}+\mathrm{H^+_{(aq)}} \label{19.8}\] A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity. Draw a cell diagram for the galvanic cell described in Example 1. The balanced chemical reaction is as follows: \[3Sn_{(s)} + 2NO^−_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}\] galvanic cell and redox reaction cell diagram Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left. The anode is the tin strip, and the cathode is the Pt electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus Sn(s)∣Sn (aq). We could include H SO (aq) with the contents of the anode compartment, but the sulfate ion (as HSO ) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction (NO) and the Pt electrode. These are written as HNO (aq)∣NO(g)∣Pt(s), with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge, \[Sn_{(s)}\,|\,Sn^{2+}_{(aq)}\,||\,HNO_{3(aq)}\,|\,NO_{(g)}\,|\,Pt_{(s)}\] The solution concentrations were not specified, so they are not included in this cell diagram. Draw the cell diagram for the following reaction, assuming the concentration of Ag and Mg are each 1 M: \[Mg_{(s)} + 2Ag^+_{(aq)} \rightarrow Mg^{2+}_{(aq)} + 2Ag_{(s)}\] \[ Mg(s) \,|\,Mg^{2+}(aq, 1 M \,||\,Ag^+(aq, 1 M)\,|\,Ag(s)\] Cell Diagrams: Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential. | 15,789 | 2,999 |
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Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid (\(HA\)) and its conjugate base \((A^−\)) or a weak base (\(B\)) and its conjugate acid (\(BH^+\)), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH. To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of \(\ce{H^{+}}\)). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows: \[\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1} \] and the equilibrium constant expression is as follows: \[K_a=\dfrac{[\ce{H^{+}},\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2} \] Sodium acetate (\(\ce{CH_3CO_2Na}\)) is a strong electrolyte that ionizes completely in aqueous solution to produce \(\ce{Na^{+}}\) and \(\ce{CH3CO2^{−}}\) ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added \(\ce{CH_3COO^{−}}\) and some of the \(\ce{H^{+}}\) ions originally present in solution. Because \(\ce{Na^{+}}\) is a , it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which \([\ce{H^{+}}]\) is less than the initial value. Because \([\ce{H^{+}}]\) has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH. If we instead add a strong acid such as \(\ce{HCl}\) to the system, \([\ce{H^{+}}]\) increases. Once again the equilibrium is temporarily disturbed, but the excess \(\ce{H^{+}}\) ions react with the conjugate base (\(\ce{CH_3CO_2^{−}}\)), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [\(\ce{CH_3CO_2^{−}}\)] than before. In both cases, only the equilibrium composition has changed; the ionization constant \(K_a\) for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case \(\ce{CH3CO2^{−}}\), will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect. Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant. A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized. : solution concentration and pH, \(pK_a\), and percent ionization of acid; final concentration of conjugate base or strong acid added : pH and percent ionization of formic acid : : A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The \(\ce{Na^{+}}\) ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium: \[\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber \] The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated. We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so \[\begin{align*} K_a=\dfrac{[H^+,HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \\[4pt] &\approx \dfrac{x(0.100)}{0.150} \\[4pt] &\approx 10^{−3.75} \\[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber \] Rearranging and solving for \(x\), \[\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \\[4pt] &=2.7 \times 10^{−4}\\[4pt] &=[H^+] \end{align*} \nonumber \] The value of \(x\) is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover, \[K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber \] which is greater than \(1.0 \times 10^{−6}\), so again, our assumption is justified. The final pH is: \[pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber \] compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of \(\ce{H^{+}}\) ions, driving the equilibrium to the left. Because \(HCl\) is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations. \[HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber \] To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final \([\ce{HCO2^{-}}]\). We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so \[K_a=\dfrac{[H^+,HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber \] Rearranging and solving for \(x\), \[\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \\[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber \] Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows: \[\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber \] Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding \(\ce{H^{+}}\) ions drives the dissociation equilibrium to the left. A 0.225 M solution of ethylamine (\(\ce{CH3CH2NH2}\) with \(pK_b = 3.19\)) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following: 11.16 1.3% A Discussing the Common Ion Effect: Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base (\(B\)) and its conjugate acid (\(BH^+\)). The general equation for the ionization of a weak base is as follows: \[B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3} \] If the equilibrium constant for the reaction as written in Equation \(\ref{Eq3}\) is small, for example \(K_b = 10^{−5}\), then the equilibrium constant for the reverse reaction is very large: \(K = \dfrac{1}{K_b} = 10^5\). Adding a strong base such as \(OH^-\) to the solution therefore causes the equilibrium in Equation \(\ref{Eq3}\) to shift to the left, consuming the added \(OH^-\). As a result, the \(OH^-\) ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the \(OH^-\) ion concentration, the reaction will proceed to the left to counteract the stress. If the \(pK_b\) of the base is 5.0, the \(pK_a\) of its conjugate acid is \[pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber \] Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows: \[BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4} \] Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/K = 10 . If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation \(\ref{Eq4}\) shifts to the left. As a result, the \(H^+\) ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb \(H^+\) and \(OH^-\) ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution. Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on \(K\)), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure \(\Page {1}\), when \(NaOH\) is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH. A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M \(NaOH\) to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of \(NaOH\) solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the \(NaOH\) solution results in only a relatively small change in pH. The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the \(pK_a\) or \(pK_b\) of the weak acid or weak base. The procedure is analogous to that used in Example \(\Page {1}\) to calculate the pH of a solution containing known concentrations of formic acid and formate. An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is \(HA \leftrightharpoons H^+ + A^−\), for which the equilibrium constant expression is as follows: \[K_a=\dfrac{[H^+,A^-]}{[HA]} \label{Eq5} \] This equation can be rearranged as follows: \[[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6} \] Taking the logarithm of both sides and multiplying both sides by −1, \[ \begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \\[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align} \] Replacing the negative logarithms in Equation \(\ref{Eq7}\), \[pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8} \] or, more generally, \[pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9} \] Equation \(\ref{Eq8}\) and Equation \(\ref{Eq9}\) are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their \(K_a\) values. There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations: Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = \(pK_a\) − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit. If [base] = [acid] for a buffer, then pH = \(pK_a\). Changing this ratio by a factor of 10 either way changes the pH by ±1 unit. What is the pH of a solution that contains : concentration of acid, conjugate base, and \(pK_a\); concentration of base, conjugate acid, and \(pK_b\) : pH : Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH. : According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is \[pH = pK_a + \log([A−]/[HA]). \nonumber \] Inserting the given values into the equation, \[\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \\[4pt] &=3.75+\log 1.593 \\[4pt] &=3.95 \end{align*} \nonumber \] This result makes sense because the \([A^−]/[HA]\) ratio is between 1 and 10, so the pH of the buffer must be between the \(pK_a\) (3.75) and \(pK_a + 1\), or 4.75. This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation, \[\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \\[4pt] &=3.75+\log 1.593 \\[4pt] &=3.95 \end{align*} \nonumber \] This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A ]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95). In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (\(HPy^+\)). We will therefore use Equation \(\ref{Eq9}\), the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and \([acid] = [HPy^{+}] = 0.234\, M\). We also are given \(pK_b = 8.77\) for pyridine, but we need \(pK_a\) for the pyridinium ion. Recall from Equation 16.23 that the \(pK_b\) of a weak base and the \(pK_a\) of its conjugate acid are related: \[pK_a + pK_b = pK_w. \nonumber \] Thus \(pK_a\) for the pyridinium ion is \(pK_w − pK_b = 14.00 − 8.77 = 5.23\). Substituting this \(pK_a\) value into the Henderson-Hasselbalch approximation, \[\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \\[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \\[4pt] & =5.23 −0.294 \\[4pt] &=4.94 \end{align*} \nonumber \] Once again, this result makes sense: the \([B]/[BH^+]\) ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the \(pK_a\) (5.23) and \(pK_a − 1\), or 4.23. What is the pH of a solution that contains The \(pK_a\) of benzoic acid is 4.20, and the \(pK_b\) of trimethylamine is also 4.20. 4.08 9.68 A Discussing Using the Henderson Hasselbalch Equation: The Henderson-Hasselbalch approximation ((Equation \(\ref{Eq8}\)) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example \(\Page {3}\). The buffer solution in Example \(\Page {2}\) contained 0.135 M \(\ce{HCO2H}\) and 0.215 M \(\ce{HCO2Na}\) and had a pH of 3.95. : composition and pH of buffer; concentration and volume of added acid or base : final pH : The added \(\ce{HCl}\) (a strong acid) or \(\ce{NaOH}\) (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction. We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer: \[ 100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber \] \[ 100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber \] The millimoles of \(\ce{H^{+}}\) in 5.00 mL of 1.00 M \(\ce{HCl}\) is as follows: \[ 5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber \] Next, we construct a table of initial amounts, changes in amounts, and final amounts: \[\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber \] The final amount of \(H^+\) in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final \([H^+]\) and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example \(\Page {1}\) or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of \(HCO_2^−\) and \(HCO_2H\), which can be calculated using the number of millimoles (\(n\)) of each and the total volume (\(VT\)). Substituting these values into the Henderson-Hasselbalch approximation (Equation \(\ref{Eq9}\)): \[\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \\[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \\[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber \] Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So \[\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \\[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \\[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber \] Once again, this result makes sense on two levels. First, the addition of \(HCl \)has decreased the pH from 3.95, as expected. Second, the ratio of \(HCO_2^−\) to \(HCO_2H\) is slightly less than 1, so the pH should be between the \(pK_a\) and \(pK_a\) − 1. The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of \(HCO_2H\) and 21.5 mmol of \(HCO_2^−\). The number of millimoles of \(OH^-\) in 5.00 mL of 1.00 M \(NaOH\) is as follows: With this information, we can construct a table of initial amounts, changes in amounts, and final amounts. \[\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber \] The final amount of \(OH^-\) in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both \(HCO_2^−\) and \(HCO_2H\) into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels: \[\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \\[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \\[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber \] Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the \(pK_a\) and \(pK_a\) + 1, as expected for a solution with a \(HCO_2^−/HCO_2H\) ratio between 1 and 10. The buffer solution from Example \(\Page {2}\) contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94. 5.30 4.42 Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations. A Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Base to a Buffer: The results obtained in Example \(\Page {3}\) and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of \(HCl\) or \(NaOH\) solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to \(1.1 \times 10^{−4}\) M HCl). In this case, adding 5.00 mL of 1.00 M \(HCl\) would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M \(NaOH\) would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH. The most effective buffers contain concentrations of an acid and its conjugate base. A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure \(\Page {2}\) for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of \(CH_3CO_2^−\) to \(CH_3CO_2H\) from 1:1 reduces the buffer capacity of the solution. A Discussing The Buffer Region: There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure \(\Page {3}\). As indicated by the labels, the region around \(pK_a\) corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the \(pK_a\) to approximately a pH value of 1 unit greater than the \(pK_a\), which is why buffer solutions usually have a pH that is within ±1 pH units of the \(pK_a\) of the acid component of the buffer. This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water. In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to \(K_a\). In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to \(K_b\). However, we can calculate either \(K_a\) or \(K_b\) from the other because they are related by \(K_w\). Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0. Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the \(\ce{CO2}/\ce{HCO3^{−}}\) system, which dominates the buffering action of blood plasma. The acid–base equilibrium in the \(\ce{CO2}/\ce{HCO3^{−}}\) buffer system is usually written as follows: \[\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10} \] with \(K_a = 4.5 \times 10^{−7}\) and \(pK_a = 6.35\) at 25°C. In fact, Equation \(\ref{Eq10}\) is a grossly oversimplified version of the \(\ce{CO2}/\ce{HCO3^{-}}\) system because a solution of \(\ce{CO2}\) in water contains only rather small amounts of \(H_2CO_3\). Thus Equation \(\ref{Eq10}\) does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C. As shown in Equation \(\ref{Eq11}\), \(\ce{CO2}\) is in equilibrium with \(\ce{H2CO3}\), but the equilibrium lies far to the left, with an \(\ce{H2CO3}/\ce{CO2}\) ratio less than 0.01 under most conditions: \[\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11} \] with \(K′ = 4.0 \times 10^{−3}\) at 37°C. The true \(pK_a\) of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a \(K_a\) of \(2.0 \times 10^{−4}\), which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling \(\ce{H2CO3}\) from both sides give the following overall equation for the reaction of \(\ce{CO2}\) with water to give a proton and the bicarbonate ion: \[\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a} \] with \(K'=4.0 \times 10^{−3} (37°C)\) \[\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b} \] with \(K_a=2.0 \times 10^{−4} (37°C)\) \[\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c} \] with \(K=8.0 \times 10^{−7} (37°C)\) The \(K\) value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid (\(K_a\)) and the equilibrium constant (K) for the reaction of \(\ce{CO2 (aq)} \) with water to give carbonic acid. The equilibrium equation for the reaction of \(\ce{CO2}\) with water to give bicarbonate and a proton is therefore \[K=\dfrac{[\ce{H^{+}},\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13} \] The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to , \[[\ce{CO2}]=k P_{\ce{CO2}} \nonumber \] where \(k\) is the Henry’s law constant for \(\ce{CO2}\), which is \(3.0 \times 10^{−5} \;M/mmHg\) at 37°C. Substituting this expression for \([\ce{CO2}]\) in Equation \ref{eq13}, \[K=\dfrac{[\ce{H^{+}},\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber \] where \(P_{\ce{CO2}}\) is in mmHg. Taking the negative logarithm of both sides and rearranging, \[pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15} \] Thus the pH of the solution depends on both the \(\ce{CO2}\) pressure over the solution and \([\ce{HCO3^{−}}]\). Figure \(\Page {4}\) plots the relationship between pH and \([\ce{HCO3^{−}}]\) under physiological conditions for several different values of \(P_{\ce{CO2}}\), with normal pH and \([\ce{HCO3^{−}}]\) values indicated by the dashed lines. According to Equation \ref{Eq15}, adding a strong acid to the \(\ce{CO2}/\ce{HCO3^{−}}\) system causes \([\ce{HCO3^{−}}]\) to decrease as \(\ce{HCO3^{−}}\) is converted to \(\ce{CO2}\). Excess \(\ce{CO2}\) is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in \(P_{\ce{CO2}}\). Because the change in \([\ce{HCO3^{−}}]/P_{CO_2}\) is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the \(\ce{OH^{-}}\) reacts with \(\ce{CO2}\) to form \(\ce{HCO3^{−}}\), but \(\ce{CO2}\) is replenished by the body, again limiting the change in both \([\ce{HCO3^{−}}]/P_{\ce{CO2}}\) and pH. The \(\ce{CO2}/\ce{HCO3^{−}}\) buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value. If a passenger steps out of an airplane in Denver, Colorado, for example, the lower \(P_{\ce{CO2}}\) at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and \([\ce{HCO3^{-}}]\). The increase in pH and decrease in \([\ce{HCO3^{−}}]\) in response to the decrease in \(P_{\ce{CO2}}\) are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness. A Summary of the pH Curve for a Strong Acid/Strong Base Titration: Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (\(HA\)) and its conjugate weak base (\(A^−\)). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the \(K_a\) or \(K_b\)), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their \(K_a\) values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the \(CO_2/HCO_3^−\) system, which dominates the buffering action of blood plasma. | 30,633 | 3,000 |
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The process of describing each atom’s electronic structure consists, essentially, of beginning with hydrogen and adding one proton and one electron at a time to create the next heavier element in the table. Before demonstrating how to do this, however, we must introduce the concept of electron spin. The quantum numbers \(n, \ l, \ m\) are not sufficient to fully characterize the physical state of the electrons in an atom. In 1926, Otto Stern and Walther Gerlach carried out an experiment that could not be explained in terms of the three quantum numbers \(n, \ l, \ m\) and showed that there is, in fact, another quantum-mechanical degree of freedom that needs to be included in the theory. The experiment is illustrated in the figure 8.9.1. The fact that the beam splits into 2 beams suggests that the electrons in the atoms have a degree of freedom capable of coupling to the magnetic field. That is, an electron has an \(M\) arising from a degree of freedom that has no classical analog. The magnetic moment must take on only 2 values according to the Stern-Gerlach experiment. The intrinsic property that gives rise to the magnetic moment must have some analog to a , \(S\); u The implication of the Stern-Gerlach experiment is that we need to include a fourth quantum number, \(m_s\) in our description of the physical state of the electron. That is, in addition to give its principle, angular, and magnetic quantum numbers, we also need to say if it is a spin-up electron or a spin-down electron. When scientists analyzed the emission and absorption spectra of the elements more closely, they saw that for elements having more than one electron, nearly all the lines in the spectra were actually of very closely spaced lines. Because each line represents an energy level available to electrons in the atom, there are twice as many energy levels available as would be predicted solely based on the quantum numbers , , and . Scientists also discovered that applying a magnetic field caused the lines in the pairs to split farther apart. In 1925, two graduate students in physics in the Netherlands, George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin. In an external magnetic field, the electron has two possible orientations (Figure \(\Page {2}\)). These are described by a fourth quantum number ( ), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it. The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons. He developed the Pauli exclusion principle: By giving the values of , , and , we also specify a particular orbital (e.g., 1 with = 1, = 0, = 0). Because has only two possible values (+½ or −½), two electrons, , can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements. List all the allowed combinations of the four quantum numbers ( , , , ) for electrons in a 2 orbital and predict the maximum number of electrons the 2 subshell can accommodate. orbital allowed quantum numbers and maximum number of electrons in orbital For a 2 orbital, we know that = 2, = − 1 = 1, and = − , (− +1),…, ( − 1), . There are only three possible combinations of ( , , ): (2, 1, 1), (2, 1, 0), and (2, 1, −1). Because is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of ( , , , ): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½). Hence the 2 subshell, which consists of three 2 orbitals (2 , 2 , and 2 ), can contain a total of six electrons, two in each orbital. List all the allowed combinations of the four quantum numbers ( , , , ) for a 6 orbital, and predict the total number of electrons it can contain. (6, 0, 0, +½), (6, 0, 0, −½); two electrons In addition to the three quantum numbers ( , , ) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the quantum number ( ), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. The concept of electron spin has important consequences for chemistry because the implies that no orbital can contain more than two electrons (with opposite spin). | 5,305 | 3,002 |
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We have observed and defined phase transitions and phase equilibrium. We have also observed equilibrium in a variety of reaction systems. We will assume an understanding of the postulates of the and of the energetics of chemical reactions. We have developed an understanding of the concept of equilibrium, both for phase equilibrium and reaction equilibrium. As an illustration, at normal atmospheric pressure, we expect to find \(\ce{H_2O}\) in solid form below \(0^\text{o} \text{C}\), in liquid form below \(100^\text{o} \text{C}\), and in gaseous form above \(100^\text{o} \text{C}\). What changes as we move from low temperature to high temperature cause these transitions in which phase is observed? Viewed differently, if a sample of gaseous water at \(120^\text{o} \text{C}\) is cooled to below \(100^\text{o} \text{C}\), virtually all of the water vapor spontaneously condenses to form the liquid: \[\ce{H_2O} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) \: \: \text{spontaneous below } 100^\text{o} \text{C}\] By contrast, very little of liquid water at \(80^\text{o} \text{C} spontaneously converts to gaseous water: \[\ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right) \: \: \text{not spontaneous below } 100^\text{o} \text{C}\] We can thus rephrase our question as, what determines which processes are spontaneous and which are not? What factors determine what phase is "stable"? As we know, at certain temperatures and pressures, more than one phase can be stable. For example, at \(1 \: \text{atm}\) pressure and \(0^\text{o} \text{C}\), \[\ce{H_2O} \left( s \right) \rightleftharpoons \ce{H_2O} \left( l \right) \: \: \text{equilibrium at } 0^\text{o} \text{C}\] Small variations in the amount of heat applied or extracted to the liquid-solid equilibrium cause shifts towards liquid or solid without changing the temperature of the two phases at equilibrium. Therefore, when the two phases are at equilibrium, neither direction of the phase transition is spontaneous at \(0^\text{o} \text{C}\). We therefore need to understand what factors determine when two or more phases can coexist at equilibrium. This analysis leaves unanswered a series of questions regarding the differences between liquids and gases. The concept of a gas phase or a liquid phase is not a characteristic of an individual molecule. In fact, it does not make any sense to refer to the "phase" of an individual molecule. The phase is a collective property of large numbers of molecules. Although we can discuss the importance of molecular properties regarding liquid and gas phases, we have not discussed the factors which determine whether the gas phase or the liquid phase is most stable at a given temperature and pressure. These same questions can be applied to reaction equilibrium. When a mixture of reactants and products is not at equilibrium, the reaction will occur spontaneously in one direction or the other until the reaction achieves equilibrium. What determines the direction of spontaneity? What is the driving force towards equilibrium? How does the system that equilibrium has been achieved? Our goal will be to understand the driving forces behind spontaneous processes and the determination of the equilibrium point, both for phase equilibrium and reaction equilibrium. We begin by examining common characteristics of spontaneous processes, and for simplicity, we focus on processes not involving phase transitions or chemical reactions. A very clear example of such a process is mixing. Imagine putting a drop of blue ink in a glass of water. At first, the blue dye in the ink is highly concentrated. Therefore, the molecules of the dye are closely congregated. Slowly but steadily, the dye begins to diffuse throughout the entire glass of water, so that eventually the water appears as a uniform blue color. This occurs more readily with agitation or stirring but occurs spontaneously even without such effort. Careful measurements shows that this process occurs without a change in temperature, so there is no energy input or release during the mixing. We conclude that, although there is no energetic advantage to the dye molecules dispersing themselves, they do so spontaneously. Furthermore, this process is in the sense that, without considerable effort on our part, the dye molecules will never return to form a single localized drop. We now seek an understanding of how and why this mixing occurs. Consider the following rather abstract model for the dye molecules in the water. For the glass, we take a row of ten small boxes, each one of which represents a possible location for a molecule, either of water or of dye. For the molecules, we take marbles, clear for water and red for ink. Each box will accommodate only a single marble, since two molecules cannot be in the same place at the same time. Since we see a drop of dye when the molecules are congregated, we model a "drop" as three red marbles in consecutive boxes. Notice that there are only eight ways to have a "drop" of dye, assuming that the three dye "molecules" are indistinguishable from one another. Two possibilities are shown in Figure 17.1a and Figure 17.1b. It is not difficult to find the other six. a. b. c. d. By contrast, there are many more ways to arrange the dye molecules so that they do not form a drop, i.e., so that the three molecules are not together. Two possibilities are shown in Figure 17.1c and Figure 17.1d. The total number of such possibilities is 112. (The total number of all possible arrangements can be calculated as follows: there are 10 possible locations for the first red marble, 9 for the second, and 8 for the third. This gives 720 possible arrangements, but many of these are identical, since the marbles are indistinguishable. The number of duplicates for each arrangement is 6, calculated from three choices for the first marble, two for the second, and one for the third. The total number of non-identical arrangements of the molecules is \(\frac{720}{6} = 120\).) We conclude that, if we randomly place the 3 marbles in the tray of 10 boxes, the chances are only 8 out of 120 (or 1 out of 15) of observing a drop of ink. Now, in a real experiment, there are many, many times more ink molecules and many, many times more possible positions for each molecule. To see how this comes into play, consider a row of 500 boxes and 5 blue marbles. (The of ink is thus 0.01.) The total number of distinct configurations of the red marbles in these boxes is approximately \(2 \times 10^{11}\). The number of these configurations which have all five ink marbles together in a drop is 496. If the arrangements are sampled randomly, the chances of observing a drop of ink with all five molecules together are thus about one in 500 million. The possibilities are remote even for observing a partial "droplet" consisting of fewer than all five dye molecules. The chance for four of the molecules to be found together is about one in 800,000. Even if we define a droplet to be only three molecules together, the chances of observing one are less than one in 1600. We could, with some difficulty, calculate the probability for observing a drop of ink when there are \(10^{23}\) molecules. However, it is reasonably deduced from our small calculations that the probability is essentially zero for the ink molecules, randomly distributed into the water molecules, to be found together. We conclude from this that the reason why we observe ink to disperse in water is that the probability is infinitesimally small for randomly distributed dye molecules to be congregated in a drop. Interestingly, however, when we set up the real ink and water experiment, we did not randomly distribute the ink molecules. Rather, we began initially with a drop of ink in which the dye molecules were already congregated. We know that, according to our kinetic theory, the molecules are in constant random motion. Therefore, they must be constantly rearranging themselves. Since these random motions do not energetically favor any one arrangement over any other one arrangement, we can assume that all possible arrangements are equally probable. Since most of the arrangements do not correspond to a drop of ink, then we will not observe a drop. In the case above with five red marbles in 500 boxes, we expect to see a drop only once in every 500 million times we look at the "glass". In a real glass of water with a real drop of ink, the chances are very much smaller than this. We draw two very important conclusions from our model. First, the random motions of molecules make every possible arrangement of these molecules equally probable. Second, mixing occurs spontaneously simply because there are vastly many more arrangements which are mixed than which are not. The first conclusion tells us "how" mixing occurs, and the second tells us "why". On the basis of these observations, we deduce the following preliminary generalization: a spontaneous process occurs because it produces the most probable final state. There is a subtlety in our conclusion to be considered in more detail. We have concluded that all possible arrangements of molecules are equally probable. We have further concluded that mixing occurs because the final mixed state is overwhelmingly probable. Placed together, these statements appear to be openly contradictory. To see why they are not, we must analyze the statements carefully. By an "arrangement" of the molecules, we mean a specification of the location of each and every molecule. We have assumed that, due to random molecular motion, each such arrangement is equally probable. In what sense, then, is the final state "overwhelmingly probable"? Recall the system illustrated in Figure 17.1, where we placed three identical red marbles into ten spaces. We calculated before that there are 120 unique ways to do this. If we ask for the probability of the arrangement in Figure 17.1a, the answer is thus \(\frac{1}{120}\). This is also the probability for each of the other possible arrangements, according to our model. However, if we now ask instead for the probability of observing a "mixed" state (with no drop), the answer is \(\frac{112}{120}\), whereas the probability of observing an "unmixed" state (with a drop) is only \(\frac{8}{120}\). Clearly, the probabilities are not the same when considering the less specific characteristics "mixed" and "unmixed". In chemistry, we are virtually never concerned with details, such as the locations of specific individual molecules. Rather, we are interested in more general characteristics, such as whether a system is mixed or not, or what the temperature or pressure is. These properties of interest to us are . As such, we refer to a specific arrangement of the molecules as a , and each general state (mixed or unmixed, for example) as a . All microstates have the same probability of occurring, according to our model. As such, the macrostates have widely differing probabilities. We come to an important result: the probability of observing a particular macrostate (e.g., a mixed state) is proportional to the number of microstates with that macroscopic property. For example, from Figure 17.1, there are 112 arrangements (microstates) with the "mixed" macroscopic property. As we have discussed, the probability of observing a mixed state is \(\frac{112}{120}\), which is obviously proportional to 112. Thus, one way to measure the relative probability of a particular macrostate is by the number of microstates \(W\) corresponding to that macrostate. \(W\) stands for "ways", i.e., there are 112 "ways" to get a mixed state in Figure 17.1. Now we recall our conclusion that a spontaneous process always produces the outcome with greatest probability. Since \(W\) measures this probability for any substance or system of interest, we could predict, using \(W\), whether the process leading from a given initial state to a given final state was spontaneous by simply comparing probabilities for the initial and final states. For reasons described below, we instead define a function of \(W\), \[S \left( W \right) = k \text{ln} \left( W \right)\] called the , which can be used to make such predictions about spontaneity. (The \(k\) is a proportionality constant which gives \(S\) appropriate units for our calculations.) Notice that the more microstates there are, the greater the entropy is. Therefore, a macrostate with a high probability (e.g. a mixed state) has a large entropy. We now modify our previous deduction to say that a spontaneous process produces the final state of greatest entropy. (Following modifications added below, this statement forms the .) It would seem that we could use \(W\) for our calculations and that the definition of the new function \(S\) is unnecessary. However, the following reasoning shows that \(W\) is not a convenient function for calculations. We consider two identical glasses of water at the same temperature. We expect that the value of any physical property for the water in two glasses is twice the value of that property for a single glass. For example, if the enthalpy of the water in each glass is \(H_1\), then it follows that the total enthalpy of the water in the two glasses together is \(H_\text{total} = 2H_1\). Thus, the enthalpy of a system is proportional to the quantity of material in the system: if we double the amount of water, we double the enthalpy. In direct contrast, we consider the calculation involving \(W\) for these two glasses of water. The number of microstates of the macroscopic state of one glass of water is \(W_1\), and likewise the number of microstates in the second glass of water is \(W_1\). However, if we combine the two glasses of water, the number of microstates of the total system is found from the product \(W_\text{total} = W_1 \times W_1\), which does not equal \(2W_1\). In other words, \(W\) is not proportional to the quantity of material in the system. This is inconvenient, since the value of \(W\) thus depends on whether the two systems are combined or not. (If it is not clear that we should multiply the \(W\) values, consider the simple example of rolling dice. The number of states for a single die is 6, but for two dice the number is \(6 \times 6 = 36\), not \(6 + 6 = 12\).) We therefore need a new function \(S \left( W \right)\), so that, when we combine the two glasses of water, \(S_\text{total} = S_1 + S_1\). Since \(S_\text{total} = S \left( W_\text{total} \right)\), \(S_1 = S \left( W_1 \right)\), and \(W_\text{total} = W_1 \times W_1\), then our new function \(S\) must satisfy the equation \[S \left( W_1 \times W_1 \right) = S \left( W_1 \right) + S \left( W_1 \right)\] The only function \(S\) which will satisfy this equation is the logarithm function, which has the property that \(\text{ln} \left( x \times y \right) = \text{ln} \left( x \right) + \text{ln} \left( y \right)\). We conclude that an appropriate state function which measures the number of microstates in a particular macrostate the entropy equation stated previously. It is possible, though exceedingly difficult, to calculate the entropy of any system under any conditions of interest from the equation \(S = k \text{ln} \left( W \right)\). It is also possible, using more advanced theoretical thermodynamics, to determine \(S\) experimentally by measuring heat capacities and enthalpies of phase transitions. Values of \(S\) determined experimentally, often referred to as "absolute" entropies, have been tabulated for many materials at many temperatures, and a few examples are given in Table 17.1. We treat these values as observations and attempt to understand these in the context of the entropy equation. There are several interesting generalities observed in Table 17.1. First, in comparing the entropy of the gaseous form of a substance to either its liquid or solid form at the same temperature, we find that the gas always has a substantially greater entropy. This is easy to understand from the entropy equation: the molecules in the gas phase occupy a very much larger volume. There are very many more possible locations for each gas molecule and thus very many more arrangements of the molecules in the gas. It is intuitively clear that \(W\) should be larger for a gas, and therefore the entropy of a gas is greater than that of the corresponding liquid or solid. Second, we observe that the entropy of a liquid is always greater than that of the corresponding solid. This is understandable from our kinetic molecular view of liquids and solids. Although the molecules in the liquid occupy a comparable volume to that of the molecules in the solid, each molecule in the liquid is free to move through this entire volume. The molecules in the solid are relatively fixed in location. Therefore, the number of arrangements of molecules in the liquid is significantly greater than that in the solid, so the liquid has greater entropy by the entropy equation. Third, the entropy of a substance increases with increasing temperature. The temperature is, of course, a measure of the average kinetic energy of the molecules. In a solid or liquid, then, increasing the temperature increases the total kinetic energy available to the molecules. The greater the energy, the more ways there are to distribute this energy amongst the molecules. Although we have previously only referred to the range of positions for a molecule as affecting \(W\), the range of energies available for each molecule similarly affects \(W\). As a result, as we increase the total energy of a substance, we increase \(W\) and thus the entropy. Fourth, the entropy of a substance whose molecules contain many atoms is greater than that of a substance composed of smaller molecules. The more atoms there are in a molecule, the more ways there are to arrange those atoms. With greater internal flexibility, \(W\) is larger when there are more atoms, so the entropy is greater. Fifth, the entropy of a substance with a high molecular weight is greater than that of a substance with a low molecular weight. This result is harder to understand, as it arises from the distribution of the momenta of the molecules rather than the positions and energies of the molecules. It is intuitively clear that the number of arrangements of the molecules is affected by the mass of the molecules. However, even at the same temperature, the range of momenta available for a heavier molecule is greater than for a lighter one. To see why, recall that the momentum of a molecule is \(p = mv\) and the kinetic energy is \(KE = \frac{mv^2}{2} = \frac{p^2}{2m}\). Therefore, the maximum momentum available at a fixed total kinetic energy \(KE\) is \(p = \sqrt{2mKE}\). Since this is larger for larger mass molecules, the range of momenta is greater for heavier particle, thus increasing \(W\) and the entropy. We have concluded from our observations of spontaneous mixing that a spontaneous process always produces the final state of greatest probability. A few simple observations reveal that our deduction needs some thoughtful refinement. For example, we have observed that the entropy of liquid water is greater than that of solid water. This makes sense in the context of the entropy equation, since the kinetic theory indicates that liquid water has a greater value of \(W\). Nevertheless, we observe that liquid water spontaneously freezes at temperatures below \(0^\text{o} \text{C}\). This process clearly displays a decrease in entropy and therefore evidently a shift from a more probable state to a less probable state. This appears to contradict directly our conclusion. Similarly, we expect to find condensation of water droplets from steam when steam is cooled. On days of high humidity, water spontaneously liquefies from the air on cold surfaces such as the outside of a glass of ice water or the window of an air conditioned building. In these cases, the transition from gas to liquid is clearly from a higher entropy phase to a lower entropy phase, which does not seem to follow our reasoning thus far. Our previous conclusions concerning entropy and probability increases were compelling, however, and we should be reluctant to abandon them. What we have failed to take into consideration is that these phase transitions involve changes of energy and thus heat flow. Condensation of gas to liquid and freezing of liquid to solid both involve evolution of heat. This heat flow is of consequence because our observations also revealed that the entropy of a substance can be increased significantly by heating. One way to preserve our conclusions about spontaneity and entropy is to place a condition on their validity: a spontaneous process produces the final state of greatest probability and entropy the process does not involve evolution of heat. This is an unsatisfying result, however, since most physical and chemical processes involve heat transfer. As an alternative, we can force the process not to evolve heat by the system undergoing the process: no heat can be released if there is no sink to receive the heat, and no heat can be absorbed if there is no source of heat. Therefore, we conclude from our observations that a spontaneous process produces the final state of greatest probability and entropy. This is one statement of the . How can the Second Law be applied to a process in a system that is not isolated? One way to view the lessons of the previous observations is as follows: in analyzing a process to understand why it is or is not spontaneous, we must consider both the change in entropy of the system undergoing the process the effect of heat released or absorbed during the process on the entropy of the surroundings. Although we cannot prove it here, the entropy increase of a substance due to heat \(q\) at temperature \(T\) is given by \(\Delta S = \frac{q}{T}\). From another study, we can calculate the heat transfer for a process occurring under constant pressure from the enthalpy change, \(\Delta H\). By conservation of energy, the heat flow into the surroundings must be \(-\Delta H\). Therefore, the increase in the entropy of the surroundings due to heat transfer must be \(\Delta S_\text{surr} = -\frac{\Delta H}{T}\). Notice that, if the reaction is exothermic, \(\Delta H < 0\) so \(\Delta S_\text{surr} > 0\). According to our statement of the Second Law, a spontaneous process in an isolated system is always accompanied by an increase in the entropy of the system. If we want to apply this statement to a non-isolated system, we must include the surroundings in our entropy calculation. We can say then that, for a spontaneous process, \[\Delta S_\text{total} = \Delta S_\text{sys} + \Delta S_\text{surr} > 0\] Since \(\Delta S_\text{surr} = -\frac{\Delta H}{T}\), then we can write that \(\Delta S - \frac{\Delta H}{T} > 0\). This is easily rewritten to state that, for a spontaneous process: \[\Delta H - T \Delta S < 0\] This equation is really just a different form of the Second Law of Thermodynamics. However, this form has the advantage that it takes into account the effects on both the system undergoing the process and the surroundings. Thus, this new form can be applied to non-isolated systems. This equation reveals why the temperature affects the spontaneity of processes. Recall that the condensation of water vapor occurs spontaneously at temperature below \(100^\text{o} \text{C}\) but not above. Condensation is an exothermic process; to see this, consider that the reverse process, evaporation, obviously requires heat input. Therefore \(\Delta H < 0\) for condensation. However, condensation clearly results in a decrease in entropy, therefore \(\Delta S < 0\) also. Examining the above equation, we can conclude that \(\Delta H - T \Delta S < 0\) will be less than zero for condensation only if the temperature is not too high. At high temperature, the term \(-\Delta S\), which is positive, becomes larger than \(\Delta H\), so \(\Delta H - T \Delta S > 0\) for condensation at high temperatures. Therefore, condensation only occurs at lower temperatures. Because of the considerable practical utility of the above equation in predicting the spontaneity of physical and chemical processes, it is desirable to simplify the calculation of the quantity on the left side of the inequality. One way to do this is to define a new quantity \(G = H - TS\), called the . If we calculate from this definition the change in the free energy which occurs during a process at constant temperature, we get \[\Delta G = G_\text{final} - G_\text{initial} = H_\text{final} - TS_\text{final} - \left( H_\text{initial} - TS_\text{initial} \right) = \Delta H - T \Delta S\] and therefore a simplified statement of the Second Law of Thermodynamics is that \[\Delta G < 0\] for any spontaneous process. Thus, in any spontaneous process, the free energy of the system decreases. Note that \(G\) is a state function, since it is defined in terms of \(H\), \(T\), and \(S\), all of which are state functions. Since \(G\) is a state function, then \(\Delta G\) can be calculated along any convenient path. As such, the methods used to calculate \(\Delta H\) in another study can be used just as well to calculate \(\Delta G\). As we recall, the entropy of vapor is much greater than the entropy of the corresponding amount of liquid. A look back at Table 17.1 shows that, at \(25^\text{o} \text{C}\), the entropy of one mole of liquid water is \(69.9 \: \frac{\text{J}}{\text{K}}\), whereas the entropy of one mole of water vapor is \(188.8 \: \frac{\text{J}}{\text{K}}\). Our first thought, based on our understanding of spontaneous processes and entropy, might well be that a mole of liquid water at \(25^\text{o} \text{C}\) should spontaneously convert into a mole of water vapor, since this process would greatly increase the entropy of the water. We know, however, that this does not happen. Liquid water will exist in a closed container at \(25^\text{o} \text{C}\) without spontaneously converting entirely to vapor. What have we left out? The answer, based on our discussion of free energy, is the energy associated with evaporation. The conversion of one mole of liquid water into one mole of water vapor results in absorption of \(44.0 \: \text{kJ}\) of energy from the surroundings. Recall that this loss of energy from the surroundings results in a significant decrease in entropy of the surroundings. We can calculate the amount of entropy decrease in the surroundings from \(\Delta S_\text{surr} = -\frac{\Delta H}{T}\). At \(25^\text{o} \text{C}\), this gives \(\Delta S_\text{surr} = \frac{-44.0 \: \text{kJ}}{298.15 \: \text{K}} = -147.57 \: \frac{\text{J}}{\text{K}}\) for a single mole. This entropy decrease is greater than the entropy increase of the water, \(188.8 \: \frac{\text{J}}{\text{K}} - 69.9 \: \frac{\text{J}}{\text{K}} = 118.9 \: \frac{\text{J}}{\text{K}}\). Therefore, the entropy of the universe when one mole of liquid water converts to one mole of water vapor at \(25^\text{o} \text{C}\). We can repeat this calculation in terms of the free energy change: \[\begin{align} \Delta G &= \Delta H - T \Delta S \\ &= 44000 \: \frac{\text{J}}{\text{mol}} - \left( 298.15 \: \text{K} \right) \left( 118.9 \: \frac{\text{J}}{\text{K mol}} \right) \\ &= 8.55 \: \frac{\text{kJ}}{\text{mol}} > 0 \end{align}\] Since the free energy increases in the transformation of one mole of liquid water to one mole of water vapor, we predict that the transformation will not occur spontaneously. This is something of a relief, because we have correctly predicted that the mole of liquid water is stable at \(25^\text{o} \text{C}\) relative to the mole of water vapor. We are still faced with our perplexing question, however. Why does any water evaporate at \(25^\text{o} \text{C}\)? How can this be a spontaneous process? The answer is that we have to be careful about interpreting our prediction. The entropy of one mole of water at \(25^\text{o} \text{C}\) is \(188.8 \: \frac{\text{J}}{\text{K}}\). We should clarify our prediction to say that one mole of liquid water will not spontaneously evaporate to form one mole of water vapor at \(25^\text{o} \text{C}\) and \(1.00 \: \text{atm}\) pressure. This prediction is in agreement with our observation, because we have found that the water vapor formed spontaneously above liquid water at \(25^\text{o} \text{C}\) has pressure \(23.8 \: \text{torr}\), well below \(1.00 \: \text{atm}\). Assuming that our reasoning is correct, then the spontaneous evaporation of water at \(25^\text{o} \text{C}\) when water vapor is present initially must have \(\Delta G < 0\). And, indeed, as water vapor forms and the pressure of the water vapor increases, evaporation must continue as long as \(\Delta G < 0\). Eventually, evaporation stops in a closed system when we reach the vapor pressure, so we must reach a point where \(\Delta G\) is no longer less than zero, that is, evaporation stops when \(\Delta G = 0\). This is the point where we have equilibrium between liquid and vapor. We can actually determine the conditions under which this is true. Since \(\Delta G = \Delta H - T \Delta S\), then when \(\Delta G = 0\), \(\Delta H = T \Delta S\). We already know that \(\Delta H = 44.0 \: \text{kJ}\) for the evaporation of one mole of water. Therefore, the pressure of water vapor at which \(\Delta G = 0\) at \(25^\text{o} \text{C}\) is the pressure at which \(\Delta S = \frac{\Delta H}{T} = 147.6 \: \frac{\text{J}}{\text{K}}\) for a single mole of water evaporating. This is larger than the value of \(\Delta S\) for one mole and \(1.00 \: \text{atm}\) pressure of water vapor, which as we calculated was \(118.9 \: \frac{\text{J}}{\text{K}}\). Evidently, \(\Delta S\) for evaporation changes as the pressure of the water vapor changes. We therefore need to understand why the entropy of the water vapor depends on the pressure of the water vapor. Recall that 1 mole of water vapor occupies a much smaller volume at \(1.00 \: \text{atm}\) of pressure than it does at the considerably lower vapor pressure of \(23.8 \: \text{torr}\). In the larger volume at lower pressure, the water molecules have a much larger space to move in, and therefore the number of microstates for the water molecules must be larger in a larger volume. Therefore, the entropy of one mole of water vapor is larger in a larger volume at lower pressure. The entropy change for evaporation of one mole of water is thus greater when the evaporation occurs to a lower pressure. With a greater entropy change to offset the entropy loss of the surroundings, it is possible for the evaporation to be spontaneous at lower pressure. And this is exactly what we observe. To find out how much the entropy of a gas changes as we decrease the pressure, we assume that the number of microstates \(W\) for the gas molecule is proportional to the volume \(V\). This would make sense, because the larger the volume, the more places there are for the molecules to be. Since the entropy is given by \(S = k \text{ln} \left( W \right)\), then \(S\) must also be proportional to \(\text{ln} \left( V \right)\). Therefore, we can say that \[\begin{align} S \left( V_2 \right) - S \left( V_1 \right) &= R \: \text{ln} \left( V_2 \right) - R \: \text{ln} \left( V_1 \right) \\ &= R \: \text{ln} \left( \frac{V_2}{V_1} \right) \end{align}\] We are interested in the variation of \(S\) with pressure, and we remember from Boyle's Law that, for a fixed temperature, volume is inversely related to pressure. Thus, we find that \[\begin{align} S \left( P_2 \right) - S \left( P_1 \right) &= R \: \text{ln} \left( \frac{P_1}{P_2} \right) \\ &= - \left( R \: \text{ln} \left( \frac{P_2}{P_1} \right) \right) \end{align}\] For water vapor, we know that the entropy at \(1.00 \: \text{atm}\) pressure is \(188.8 \: \frac{\text{J}}{\text{K}}\) for one mole. We can use this and the equation above to determine the entropy at any other pressure. For a pressure of \(23.8 \: \text{torr} = 0.0313 \: \text{atm}\), this equation gives that \(S \left( 23.8 \: \text{torr} \right)\) is \(217.6 \: \frac{\text{J}}{\text{K}}\) for one mole of water vapor. Therefore, at this pressure, the \(\Delta S\) for evaporation of one mole of water vapor is \(217.6 \: \frac{\text{J}}{\text{K}} - 69.9 \: \frac{\text{J}}{\text{K}} = 147.6 \: \frac{\text{J}}{\text{K}}\). We can use this to calculate that for evaporation of one mole of water at \(25^\text{o} \text{C}\) and water vapor pressure of \(23.8 \: \text{torr}\) is \(\Delta G = \Delta H - T \Delta S = 44.0 \: \text{kJ} - \left( 298.15 \: \text{K} \right) \left( 147.6 \: \frac{\text{J}}{\text{K}} \right) = 0.00 \: \text{kJ}\). This is the condition we expected for equilibrium. We can conclude that the evaporation of water when no vapor is present initially is a spontaneous process with \(\Delta G < 0\), and the evaporation continues until the water vapor has reached its equilibrium vapor pressure, at which point \(\Delta G = 0\). Having developed a thermodynamic understanding of phase equilibrium, it proves to be even more useful to examine the thermodynamic description of reaction equilibrium to understand why the reactants and products come to equilibrium at the specific values that are observed. Recall that \(\Delta G = \Delta H - T \Delta S < 0\) for a spontaneous process, and \(\Delta G = \Delta H - T \Delta S = 0\) at equilibrium. From these relations, we would predict that most (but not all) exothermic processes with \(\Delta H < 0\) are spontaneous, because all such processes increase the entropy of the surroundings when they occur. Similarly, we would predict that most (but not all) processes with \(\Delta S > 0\) are spontaneous. We try applying these conclusions to the synthesis of ammonia \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\] at \(298 \: \text{K}\), for which we find that \(\Delta S^0 = -198 \: \frac{\text{J}}{\text{mol K}}\). Note that \(\Delta S^0 < 0\) because the reaction reduces the total number of gas molecules during ammonia synthesis, thus reducing \(W\), the number of ways of arranging the atoms in these molecules. \(\Delta S^0 < 0\) suggests that ammonia synthesis should not occur at all. However, \(\Delta H^0 = -92.2 \: \frac{\text{kJ}}{\text{mol}}\). Overall, we find that \(\Delta G^0 = -33.0 \: \frac{\text{kJ}}{\text{mol}}\) at \(298 \: \text{K}\), which suggests that the synthesis of ammonia is spontaneous. Given this analysis, we are now pressured to ask, if ammonia synthesis is predicted to be spontaneous, why does the reaction come to equilibrium without fully consuming all of the reactants? The answer lies in a more careful examination of the values given: \(\Delta S^0\), \(\Delta H^0\), and \(\Delta G^0\) are the values for this reaction , which means that all of the gases in the reactants and products are taken to be at \(1 \: \text{atm}\) pressure. Thus, the fact that \(\Delta G < 0\) for the synthesis of ammonia at standard conditions means that, if all three gases are present at \(1 \: \text{atm}\) pressure, the reaction will spontaneously produce an increase in the amount of \(\ce{NH_3}\). Note that this will reduce the pressure of the \(\ce{N_2}\) and \(\ce{H_2}\) and increase the pressure of the \(\ce{NH_3}\). This changes the value of \(\Delta S\) and thus of \(\Delta G\), because as we already know the entropies of all three gases depend on their pressures. As the pressure of \(\ce{NH_3}\) increases, its entropy decreases, and as the pressures of the reactant gases decrease, their entropies increase. The result is that \(\Delta S\) becomes increasingly negative. The reaction creates more \(\ce{NH_3}\) until the value of \(\Delta S\) is sufficiently negative that \(\Delta G = \Delta H - T \Delta S = 0\). From this analysis, we can say by looking at \(\Delta S^0\), \(\Delta H^0\), and \(\Delta G^0\) that, since \(\Delta G < 0\) for ammonia synthesis, reaction equilibrium results in production of more product and less reactant than at standard conditions. Moreover, the more negative \(\Delta G^0\) is, the more strongly favored are the products over the reactants at equilibrium. By contrast, the more positive \(\Delta G^0\) is, the more strongly favored are the reactants over the products at equilibrium. Thermodynamics can also provide a quantitative understanding of the equilibrium constant. Recall that the condition for equilibrium is that \(\Delta G = 0\). As noted before, \(\Delta G\) depends on the pressures of the gases in the reaction mixture, because \(\Delta S\) depends on these pressures. Though we will not prove it here, it can be shown by application of the relationship between entropy and pressure to a reaction that the relationship between \(\Delta G\) and the pressures of the gases is given by the following equation: \[\Delta G = \Delta G^0 + RT \text{ln} \left( Q \right)\] (Recall again that the superscript \(^0\) refers to standard pressure of \(1 \: \text{atm}\). \(\Delta G^0\) is the difference between the free energies of the products and reactants when all gases are at \(1 \: \text{atm}\) pressure.) In this equation, \(Q\) is a quotient of partial pressures of the gases in the reaction mixture. In this quotient, each product gas appears in the numerator with an exponent equal to its stoichiometric coefficient, and each reactant gas appears in the denominator also with its corresponding exponent. For example, for the reaction \[\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\] \[Q = \frac{P^2_{HI}}{P_{H_2} P_{I_2}}\] However, if the pressures in \(Q\) are the equilibrium partial pressures, then \(Q\) has the same value as \(K_p\), the equilibrium constant, by definition. Moreover, if the pressures are at equilibrium, we know that \(\Delta G = 0\). If we look back at the definition of \(\Delta G\), we can conclude that \[\Delta G^0 = - \left( RT \text{ln} \left( K_p \right) \right)\] This is an exceptionally important relationship, because it relates two very different observations. To understand this significance, consider first the case where \(\Delta G^0 < 0\). We have previously reasoned that, in this case, the reaction equilibrium will favor the products. From the above equation we can note that, if \(\Delta G^0 < 0\), it must be that \(K_p > 1\). Furthermore, if \(\Delta G^0\) is a large negative number, \(K_p\) is a very large number. By contrast, if \(\Delta G^0\) is a large positive number, \(K_p\) will be a very small (though positive) number much less than 1. In this case, the reactants will be strongly favored at equilibrium. Note that the thermodynamic description of equilibrium and the dynamic description of equilibrium are complementary. Both predict the same equilibrium. In general, the thermodynamic arguments give us an understanding of the conditions under which equilibrium occurs, and the dynamic arguments help us understand how the equilibrium conditions are achieved. Each possible sequence of the 52 cards in a deck is equally probable. However, when you shuffle a deck and then examine the sequence, the deck is never ordered. Explain why in terms of microstates, macrostates, and entropy. Assess the validity of the statement, "In all spontaneous processes, the system moves toward a state of lowest energy." Correct any errors you identify. In each case, determine whether spontaneity is expected at low temperature, high temperature, any temperature, or no temperature: \(\Delta H^0 > 0\), \(\Delta S^0 > 0\)
\(\Delta H^0 < 0\), \(\Delta S^0 > 0\)
\(\Delta H^0 > 0\), \(\Delta S^0 < 0\)
\(\Delta H^0 < 0\), \(\Delta S^0 < 0\) Using thermodynamic equilibrium arguments, explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces. Why does the entropy of a gas increase as the volume of the gas increases? Why does the entropy decrease as the pressure increases? For each of the following reactions, calculate the value of \(\Delta S^0\), \(\Delta H^0\), and \(\Delta G^0\) at \(T = 298 \: \text{K}\) and use these to predict whether equilibrium will favor products or reactants at \(T = 298 \: \text{K}\). Also calculate \(K_p\). \(2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right)\)
\(\ce{O_3} \left( g \right) + \ce{NO} \left( g \right) \rightarrow \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)\)
\(2 \ce{O_3} \left( g \right) \rightarrow 3 \ce{O_2} \left( g \right)\) Predict the sign of the entropy for the reaction \[2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)\] Give an explanation, based on entropy and the Second Law, of why this reaction occurs spontaneously. For the reaction \(\ce{H_2} \left( g \right) \rightarrow 2 \ce{H} \left( g \right)\), predict the sign of both \(\Delta H^0\) and \(\Delta S^0\). Should this reaction be spontaneous at high temperature or at low temperature? Explain. For each of the reactions listed above, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants. Using the general definition of \(\Delta G\) and the definition of \(Q\), show that for a given set of initial partial pressures where \(Q\) is larger than \(K_p\), the reaction will spontaneously create more reactants. Also show that if \(Q\) is smaller than \(K_p\), the reaction will spontaneously create more products. ; Chemistry) | 41,629 | 3,003 |
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In order to understand the nature of the leaving group, it is important to first discuss factors that help determine whether a species will be a strong base or weak base. If you remember from general chemistry, a Lewis base is defined as a species that donates a pair of electrons to form a covalent bond. The factors that will determine whether a species wants to share its electrons or not include electronegativity, size, and resonance. : In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons. In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons. The third factor to consider in determining whether or not a species will be a strong or weak base is resonance. As you may remember from general chemistry, the formation of a resonance stabilized structure results in a species that is less willing to share its electrons. Since strong bases, by definition, want to share their electrons, resonance stabilized structures are weak bases. Now that we understand how electronegativity, size, and resonance affect basicity, we can combine these concepts with the fact that weak bases make the best leaving groups. Think about why this might be true. In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base. We will now revisit electronegativity, size, and resonance, moving our focus to the leaving group, as well providing actual examples. As Electronegativity Increases, The Ability of the Leaving Group to Leave Increases. As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing CH to be the worst leaving group and F to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that S 2 reactions of fluoroalkanes are rarely observed. Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms. As we learned previously, resonance stabilized structures are weak bases. Therefore, leaving groups that form resonance structures upon leaving are considered to be excellent leaving groups. The following diagram shows sulfur derivatives of the type ROSO and RSO . Alkyl sulfates and sulfonates like the ones shown make excellent leaving groups. This is due to the formation of a resonance stabilized structure upon leaving. | 3,613 | 3,004 |
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The empirical and molecular formulas discussed in the preceding section are precise and informative, but they have some disadvantages. First, they are inconvenient for routine verbal communication. For example, saying “C-A-three-P-O-four-two” for Ca (PO ) is much more difficult than saying “calcium phosphate.” In addition, many compounds have the same empirical and molecular formulas but different arrangements of atoms, which differences result in very different chemical and physical properties. In such cases, it is necessary for the compounds to have different names that distinguish among the possible arrangements. Many compounds, particularly those that have been known for a relatively long time, have more than one name: a common name (sometimes several), and a systematic name, which is the name assigned by adhering to specific rules. Like the names of most elements, the common names of chemical compounds generally have historical origins, although they often appear to be unrelated to the compounds of interest. For example, the systematic name for KNO is potassium nitrate, but its common name is . In this text, a systematic nomenclature is used to assign meaningful names to the millions of known substances. Unfortunately, some chemicals that are widely used in commerce and industry are still known almost exclusively by their common names; in such cases, familiarity with the common name as well as the systematic one is required. The objective of this and the next two sections is to teach how to write the formula for a simple inorganic compound from its name—and vice versa—and introduce some frequently-encountered common names. Binary ionic compounds contain only two elements. The procedure for naming such compounds is outlined in and uses the following steps: 1. Place the ions in their proper order: cation and then anion. 2. Name the cation. a. . As noted previosuly, these metals are usually in Groups 1–3, 12, and 13. The name of the cation of a metal that forms only one cation is the same as the name of the metal (with the word ion added if the cation is by itself). For example, Na is the sodium ion, Ca is the calcium ion, and Al is the aluminum ion. b. . As shown in , many metals can form more than one cation. This behavior is observed for most transition metals, many actinides, and the heaviest elements of Groups 13–15. In such cases, the positive charge on the metal is indicated by a roman numeral in parentheses immediately following the name of the metal. Thus Cu is copper(I) (read as “copper one”), Fe is iron(II), Fe is iron(III), Sn is tin(II), and Sn is tin(IV). An older system of nomenclature for such cations is still widely used, however. The name of the cation with the higher charge is formed from the root of the element’s Latin name with the suffix -ic attached, and the name of the cation with the lower charge has the same root with the suffix -ous. The names of Fe , Fe , Sn , and Sn are therefore ferric, ferrous, stannic, and stannous, respectively. Even though this text uses the systematic names with roman numerals, it is important to recognize these common names because they are still often used. For example, on the label of dental fluoride rinse, the compound chemists call tin(II) fluoride is usually listed as stannous fluoride. Some examples of metals that form more than one cation are listed in along with the names of the ions. Note that the simple Hg cation does not occur in chemical compounds. Instead, all compounds of mercury(I) contain a dimeric cation, Hg , in which the two Hg atoms are bonded together. c. . The names of the common polyatomic cations that are relatively important in ionic compounds (such as, the ammonium ion) are in 3. Name the anion. a. . Monatomic anions are named by adding the suffix -ide to the root of the name of the parent element; thus, Cl is chloride, O is oxide, P is phosphide, N is nitride (also called azide), and C is carbide. Because the charges on these ions can be predicted from their position in the periodic table, it is not necessary to specify the charge in the name. Examples of monatomic anions are in b. . Polyatomic anions typically have common names that must be memorized; some examples are in . Polyatomic anions that contain a single metal or nonmetal atom plus one or more oxygen atoms are called oxoanions (or oxyanions). In cases where only two oxoanions are known for an element, the name of the oxoanion with more oxygen atoms ends in -ate, and the name of the oxoanion with fewer oxygen atoms ends in -ite. For example, NO is nitrate and NO is nitrite. The halogens and some of the transition metals form more extensive series of oxoanions with as many as four members. In the names of these oxoanions, the prefix per- is used to identify the oxoanion with the most oxygen (so that ClO is perchlorate and ClO is chlorate), and the prefix hypo- is used to identify the anion with the fewest oxygen (ClO is chlorite and ClO is hypochlorite). The relationship between the names of oxoanions and the number of oxygen atoms present is diagrammed in Differentiating the oxoanions in such a series is no trivial matter; for example, the hypochlorite ion is the active ingredient in laundry bleach and swimming pool disinfectant, but compounds that contain the perchlorate ion can explode if they come into contact with organic substances. 4. Write the name of the compound as the name of the cation followed by the name of the anion. It is not necessary to indicate the number of cations or anions present per formula unit in the name of an ionic compound because this information is implied by the charges on the ions. The charge of the ions must be considered when writing the formula for an ionic compound from its name, however. Because the charge on the chloride ion is −1 and the charge on the calcium ion is +2, for example, consistent with their positions in the periodic table, arithmetic indicates that calcium chloride must contain twice as many chloride ions as calcium ions to maintain electrical neutrality. Thus, the formula is CaCl . Similarly, calcium phosphate must be Ca (PO ) because the cation and the anion have charges of +2 and −3, respectively. The best way to learn how to name ionic compounds is to work through a few examples, referring to and as needed. Cations are always named before anions. Most transition metals, many actinides, and the heaviest elements of groups 13–15 can form more than one cation. Write the systematic name (and the common name if applicable) for each ionic compound. : empirical formula : name : If only one charge is possible for the cation, give its name, consulting or if necessary. If the cation can have more than one charge ( ), specify the charge using roman numerals. If the anion does not contain oxygen, name it according to step 3a, using and if necessary. For polyatomic anions that contain oxygen, use and the appropriate prefix and suffix listed in step 3b. Beginning with the cation, write the name of the compound. : a. Lithium is in group 1, so we know that it forms only the Li cation, which is the lithium ion. Similarly, chlorine is in group 7, so it forms the Cl anion, which is the chloride ion. Because we begin with the name of the cation, the name of this compound is lithium chloride, which is used medically as an antidepressant drug. b. The cation is the magnesium ion, and the anion, which contains oxygen, is sulfate. Because we list the cation first, the name of this compound is magnesium sulfate. hydrated form of magnesium sulfate (MgSO ·7H O) is sold in drugstores as Epsom salts, a harsh but effective laxative. c. The cation is the ammonium ion (from ), and the anion is phosphate. C The compound is therefore ammonium phosphate, which is widely used as a fertilizer. It is not necessary to specify that the formula unit contains three ammonium ions because three are required to balance the negative charge on phosphate. d. The cation is a transition metal that often forms more than one cation ( ). We must therefore specify the positive charge on the cation in the name: copper(I) or, according to the older system, cuprous. The anion is oxide. C The name of this compound is copper(I) oxide or, in the older system, cuprous oxide. Copper(I) oxide is used as a red glaze on ceramics and in antifouling paints to prevent organisms from growing on the bottoms of boats. Write the systematic name (and the common name if applicable) for each ionic compound. : Write the formula for each compound. : systematic name : formula : Identify the cation and its charge using the location of the element in the periodic table and and form cations with different charges, use the appropriate roman numeral or suffix to indicate its charge. Identify the anion using and . Beginning with the cation, write the compound’s formula and then determine the number of cations and anions needed to achieve electrical neutrality. : a. Calcium is in group 2, so it forms only the Ca ion. Dihydrogen phosphate is the H PO − ion ( ). Two H PO ions are needed to balance the positive charge on Ca , to give Ca(H PO ) . A hydrate of calcium dihydrogen phosphate, Ca(H PO ) ·H O, is the active ingredient in baking powder. b. Aluminum, near the top of group 13 in the periodic table, forms only one cation, Al ( ). B Sulfate is SO ( ). To balance the electrical charges, we need two Al cations and three SO anions, giving Al (SO ) . Aluminum sulfate is used to tan leather and purify drinking water. c. Because chromium is a transition metal, it can form cations with different charges. The roman numeral tells us that the positive charge in this case is +3, so the cation is Cr . B Oxide is O . Thus two cations (Cr ) and three anions (O ) are required to give an electrically neutral compound, Cr O . This compound is a common green pigment that has many uses, including camouflage coatings. Write the formula for each compound. : Ionic compounds are named according to systematic procedures, although common names are widely used. Systematic nomenclature enables chemists to write the structure of any compound from its name and vice versa. Ionic compounds are named by writing the cation first, followed by the anion. If a metal can form cations with more than one charge, the charge is indicated by roman numerals in parentheses following the name of the metal. Oxoanions are polyatomic anions that contain a single metal or nonmetal atom and one or more oxygen atoms. As with ionic compounds, the system for naming covalent compounds enables chemists to write the molecular formula from the name and vice versa. This and the following section describe the rules for naming simple covalent compounds, beginning with inorganic compounds and then turning to simple organic compounds that contain only carbon and hydrogen. When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C H . The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C H . The formula unit is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane has the empirical formula C H , but it contains two C H formula units, giving a molecular formula of C H . Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na and Cl , then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg and Cl , for example, two Cl ions are needed to balance the two positive charges on each Mg ion, giving an empirical formula of MgCl . Similarly, the formula for the ionic compound that contains Na and O ions is Na O. Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions. Nomenclature of Metals: Nomenclature of Transition Metals: An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl , a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows: . When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg and O . Using the absolute values of the charges on the ions as subscripts gives the formula Mg O : This simplifies to its correct empirical formula MgO. The empirical formula has one Mg ion and one O ion. Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. : ions or elements : empirical formula for binary ionic compound : to write the empirical formula. Check to make sure the empirical formula is electrically neutral. a. Using the absolute values of the charges on the ions as the subscripts gives Ga3As3: Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga and As have charges of equal magnitude but opposite signs. One Ga ion balances the charge on one As ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices. b. Because Eu has a charge of +3 and O has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion: The subscript for Eu is 2 (from O ), and the subscript for O is 3 (from Eu ), giving Eu O ; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu O is neutral. Europium oxide is responsible for the red color in television and computer screens. c. Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca . Chlorine lies in group 17, so it should gain one electron to form Cl . Two Cl ions are needed to balance the charge on one Ca ion, which leads to the empirical formula CaCl . We could also cross charges, using the absolute value of the charge on Ca as the subscript for Cl and the absolute value of the charge on Cl as the subscript for Ca: The subscripts in CaCl cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter. Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. : Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH ) and the methylammonium (CH NH ) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in . The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K and NO ions combine in a 1:1 ratio to form KNO (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca and SO form CaSO (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH and NO form NH NO (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca and PO ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca (PO ) ; the parentheses around PO in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca P O gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO ions. Write the empirical formula for the compound formed from each ion pair. : ions : empirical formula for ionic compound : : Write the empirical formula for the compound formed from each ion pair. : Polyatomics: Many ionic compounds occur as hydrates, compounds that contain specific ratios of loosely bound water molecules, called waters of hydration. Waters of hydration can often be removed simply by heating. For example, calcium dihydrogen phosphate can form a solid that contains one molecule of water per Ca(H PO ) unit and is used as a leavening agent in the food industry to cause baked goods to rise. The empirical formula for the solid is Ca(H PO ) ·H O. In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula CuSO ·5H O. When heated, all five water molecules are lost, giving a white solid with the empirical formula CuSO . Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, CaSO ·½H2O is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas CaSO ·2H O is the less dense, flakier gypsum, a mineral used in drywall panels for home construction. When a cast would set, a mixture of plaster of Paris and water crystallized to give solid CaSO ·2H O. Similar processes are used in the setting of cement and concrete. —covalent compounds that contain only two elements—are named using a procedure similar to that used for simple ionic compounds, but prefixes are added as needed to indicate the number of atoms of each kind. The procedure, diagrammed in consists of the following steps: To demonstrate steps 1 and 2a, HCl is named hydrogen chloride (because hydrogen is to the left of chlorine in the periodic table), and PCl is phosphorus pentachloride. The order of the elements in the name of BrF , bromine trifluoride, is determined by the fact that bromine lies below fluorine in Group 17. Start with the element at the far left in the periodic table and work to the right. If two or more elements are in the same group, start with the bottom element and work up. Write the name of each binary covalent compound. molecular formula name of compound Write the name of each binary covalent compound. Write the formula for each binary covalent compound. name of compound formula List the elements in the same order as in the formula, use to identify the number of each type of atom present, and then indicate this quantity as a subscript to the right of that element when writing the formula. Write the formula for each binary covalent compound. The structures of some of the compounds in Examples 3.6.3 and 3.6.4 are shown in along with the location of the “central atom” of each compound in the periodic table. It may seem that the compositions and structures of such compounds are entirely random, but this is not true. After mastering the material discussed later on this course, one is able to predict the compositions and structures of compounds of this type with a high degree of accuracy. Nomenclature of Nonmetals: For our purposes at this point in the text, we can define an acid as a substance with at least one hydrogen atom that can dissociate to form an anion and an H+ ion (a proton) in aqueous solution, thereby forming an acidic solution. We can define bases as compounds that produce hydroxide ions (OH ) and a cation when dissolved in water, thus forming a basic solution. Solutions that are neither basic nor acidic are neutral. We will discuss the chemistry of acids and bases in more detail later, but in this section we describe the nomenclature of common acids and identify some important bases so that you can recognize them in future discussions. Pure acids and bases and their concentrated aqueous solutions are commonly encountered in the laboratory. They are usually highly corrosive, so they must be handled with care. The names of acids differentiate between (1) acids in which the H ion is attached to an oxygen atom of a polyatomic anion (these are called oxoacids, or occasionally oxyacids) and (2) acids in which the H ion is attached to some other element. In the latter case, the name of the acid begins with hydro- and ends in -ic, with the root of the name of the other element or ion in between. Recall that the name of the anion derived from this kind of acid always ends in -ide. Thus hydrogen chloride (HCl) gas dissolves in water to form hydrochloric acid (which contains H and Cl ions), hydrogen cyanide (HCN) gas forms hydrocyanic acid (which contains H and CN ions), and so forth ( ). Examples of this kind of acid are commonly encountered and very important. For instance, your stomach contains a dilute solution of hydrochloric acid to help digest food. When the mechanisms that prevent the stomach from digesting itself malfunction, the acid destroys the lining of the stomach and an ulcer forms. Acids are distinguished by whether the H ion is attached to an oxygen atom of a polyatomic anion or some other element. If an acid contains one or more H ions attached to oxygen, it is a derivative of one of the common oxoanions, such as sulfate (SO ) or nitrate (NO ). These acids contain as many H ions as are necessary to balance the negative charge on the anion, resulting in a neutral species such as H SO and HNO . The names of acids are derived from the names of anions according to the following rules: The relationship between the names of the oxoacids and the parent oxoanions is illustrated in , and some common oxoacids are in . Name and give the formula for each acid. : anion : parent acid : Refer to and to find the name of the acid. If the acid is not listed, use the guidelines given previously. : Neither species is listed in or , so we must use the information given previously to derive the name of the acid from the name of the polyatomic anion. Name and give the formula for each acid. : Nomenclature of Acids: Many organic compounds contain the carbonyl group, in which there is a carbon–oxygen double bond. In carboxylic acids, an –OH group is covalently bonded to the carbon atom of the carbonyl group. Their general formula is RCO H, sometimes written as RCOOH: where R can be an alkyl group, an aryl group, or a hydrogen atom. The simplest example, HCO H, is formic acid, so called because it is found in the secretions of stinging ants (from the Latin , meaning “ant”). Another example is acetic acid (CH CO H), which is found in vinegar. Like many acids, carboxylic acids tend to have sharp odors. For example, butyric acid (CH CH CH CO H), is responsible for the smell of rancid butter, and the characteristic odor of sour milk and vomit is due to lactic acid [CH CH(OH)CO H]. Some common carboxylic acids are shown in Table . Although carboxylic acids are covalent compounds, when they dissolve in water, they dissociate to produce H ions (just like any other acid) and RCO ions. Note that only the hydrogen attached to the oxygen atom of the CO group dissociates to form an H ion. In contrast, the hydrogen atom attached to the oxygen atom of an alcohol does not dissociate to form an H ion when an alcohol is dissolved in water. The reasons for the difference in behavior between carboxylic acids and alcohols will be discussed in . Only the hydrogen attached to the oxygen atom of the CO group dissociates to form an H ion. We will present more comprehensive definitions of bases in later chapters, but virtually every base you encounter in the meantime will be an ionic compound, such as sodium hydroxide (NaOH) and barium hydroxide [Ba(OH) ], that contain the hydroxide ion and a metal cation. These have the general formula M(OH)n. It is important to recognize that alcohols, with the general formula ROH, are covalent compounds, not ionic compounds; consequently, they do not dissociate in water to form a basic solution (containing OH ions). When a base reacts with any of the acids we have discussed, it accepts a proton (H ). For example, the hydroxide ion (OH ) accepts a proton to form H O. Thus bases are also referred to as proton acceptors. Concentrated aqueous solutions of ammonia (NH ) contain significant amounts of the hydroxide ion, even though the dissolved substance is not primarily ammonium hydroxide (NH OH) as is often stated on the label. Thus aqueous ammonia solution is also a common base. Replacing a hydrogen atom of NH3 with an alkyl group results in an amine (RNH ), which is also a base. Amines have pungent odors—for example, methylamine (CH NH ) is one of the compounds responsible for the foul odor associated with spoiled fish. The physiological importance of amines is suggested in the word vitamin, which is derived from the phrase vital amines. The word was coined to describe dietary substances that were effective at preventing scurvy, rickets, and other diseases because these substances were assumed to be amines. Subsequently, some vitamins have indeed been confirmed to be amines. Metal hydroxides (MOH) yield OH ions and are bases, alcohols (ROH) do not yield OH or H ions and are neutral, and carboxylic acids (RCO H) yield H ions and are acids. Common acids and the polyatomic anions derived from them have their own names and rules for nomenclature. The nomenclature of acids differentiates between oxoacids, in which the H ion is attached to an oxygen atom of a polyatomic ion, and acids in which the H ion is attached to another element. Carboxylic acids are an important class of organic acids. Ammonia is an important base, as are its organic derivatives, the amines. Covalent inorganic compounds are named by a procedure similar to that used for ionic compounds, using prefixes to indicate the numbers of atoms in the molecular formula. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration. | 28,450 | 3,005 |
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\[2H_2 + O_2 \rightarrow 2H_2O\] Chemical formulas and other symbols are used to indicate the starting materials, or reactants, which by convention are written on the left side of the equation, and the final compounds, or products, which are written on the right. An arrow points from the reactant to the products. The chemical reaction for the ammonium dichromate volcano in Figure \(\Page {1}\) is \[ (NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + 4H_2O \label{4.1.1} \] \[ reactant \, \, \, \, \,\, \, \, \, \, \, \,\,\, products \] The arrow is read as “yields” or “reacts to form.” indicates that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products). The equation for this reaction is even more informative when written as follows: \[ (NH_4)_2Cr_2O_7(s) \rightarrow Cr_2O_{3\;(s)} + N_{2\;(g)} + 4H_2O_{(g)} \label{4.1.2}\] is identical to except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water. Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of and . Each side of the reaction has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms. In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. and are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms, but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr–O and N–H bonds), and new bonds are formed to create the products (here, O–H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced, and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced. Introduction to Chemical Reaction Equations: | 2,230 | 3,006 |
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Scientists search for answers to questions and solutions to problems by using a procedure called the . This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(\Page {1}\)). Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 °C, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 °C. An example of a quantitative observation was the initial observation leading to the modern theory of the dinosaurs’ extinction: iridium concentrations in sediments dating to 66 million years ago were found to be 20–160 times higher than normal. The development of this theory is a good exemplar of the scientific method in action (see Figure \(\Page {2}\) below). After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a , a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses: Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists collected additional data that either support or refute it. After a hypothesis has been formed, scientists conduct experiments to test its validity. are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in the dinosaur extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a , a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why. One example of a law, the , which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Some solid compounds do not strictly obey the law of definite proportions. The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century. Whereas a law states only what happens, a attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. Figure \(\Page {2}\) summarizes the application of the scientific method in this case. Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. : components of the scientific method : statement classification Refer to the definitions in this section to determine which category best describes each statement. Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. experiment law theory hypothesis qualitative observation quantitative observation Because scientists can enter the cycle shown in Figure \(\Page {1}\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations. It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology. Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method. Fundamental Definitions in Chemistry: | 6,843 | 3,007 |
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The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen ( H), which accounts for about 90% of all atoms. In fact, H is the raw material from which all other elements are formed. In this section, we explain why H and He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements. The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in . The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table \(\Page {1}\) for some common elements. Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH , CH , and H O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements. All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei. Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known ( ). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm , and the temperature increases to about 1.5 × 10 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun. In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium: The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two \(\gamma\) rays, and a great deal of energy: \[4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2} \] These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium. Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 10 K, the helium-4 nuclei begin to fuse, producing beryllium-8: \[2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3} \] Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24: \[_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4} \] So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star. As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 10 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei: \[_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5} \] \[_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6} \] At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40: \[_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7} \] The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 10 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known. None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called ( ). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula ( ). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120: \[_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8} \] Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past. The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of reactant and product nuclides balanced nuclear equation Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction. How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction. 19 neutrons; \(_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta\) Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. | 10,576 | 3,008 |
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All matter has physical and chemical properties. are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property). Physical properties can be extensive or intensive. vary with the amount of the substance and include mass, weight, and volume. , in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature. For example, elemental sulfur is a yellow crystalline solid that does not conduct electricity and has a melting point of 115.2°C, no matter what amount is examined ( ). Scientists commonly measure intensive properties to determine a substance’s identity, whereas extensive properties convey information about the amount of the substance in a sample. Although mass and volume are both extensive properties, their ratio is an important intensive property called (\(\rho\)). Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm ). As mass increases in a given volume, density also increases. For example, lead, with its greater mass, has a far greater density than the same volume of air, just as a brick has a greater density than the same volume of Styrofoam. At a given temperature and pressure, the density of a pure substance is a constant: \[density ={mass \over volume} \rightarrow \rho ={m \over v} \label{1.2.1}\] Pure water, for example, has a density of 0.998 g/cm at 25°C. The average densities of some common substances are in Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.” Change in which the matter's physical appearance is altered, but composition remains unchanged, e.g., a change in state of matter. are: Solid, Liquid, Gas Different Definitions of Properties: When liquid water (\(H_2O\)) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass. The combustion of magnetisum metal is a chemical change (Magnesium + Oxygen → Magnesium Oxide): \[2 Mg + O_2 \rightarrow 2 MgO\] The rusting of iron is a chemical change (Iron + Oxygen → Iron Oxide/ Rust): \[4 Fe + 3O_2 \rightarrow 2 Fe_2O_3\] Using the components of composition and properties, we have the ability to distinguish one sample of matter from the others. Different Definitions of Changes: | 3,026 | 3,009 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystallography/Physical_Properties_of_Crystals/Curie_laws |
Curie extended the notion of symmetry to include that of physical phenomena and stated that: and concludes that some symmetry elements may coexist with the phenomenon but that their presence is not necessary. On the contrary, what is necessary is the absence of certain symmetry elements: ‘asymmetry creates the phenomenon’. Noting that physical phenomena usually express relations between a cause and an effect (an influence and a response), P. Curie restated the two above propositions in the following way, now known as Curie laws, although they are not, strictly speaking, laws (Curie himself spoke about 'the principle of symmetry'): | 656 | 3,010 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.11%3A_Real_Gases-_The_Effects_of_Size_and_Intermolecular_Forces |
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases. For an ideal gas, a plot of \(PV/nRT\) versus \(P\) gives a horizontal line with an intercept of 1 on the \(PV/nRT\) axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure \(\Page {1a}\)). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure \(\Page {1b}\)). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure \(\Page {2}\) for \(N_2\). Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid. Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure \(\Page {3}\)). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of / is greater than the value predicted by the ideal gas law. Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is than expected (Figure \(\Page {4}\)). Thus as shown in Figure \(\Page {2}\), at low temperatures, the ratio of (PV/nRT\) is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the \(PV/nRT\) versus \(P\) plot for many gases. Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid). The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called equation, \[ \underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1} \] and are empirical constants that are different for each gas. The values of \(a\) and \(b\) are listed in Table \(\Page {1}\) for several common gases. The pressure term in Equation \(\ref{10.9.1}\) corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, \(n^2/V^2\) represents the concentration of the gas (\(n/V\)) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure \(\Page {4}\). The volume term corrects for the volume occupied by the gaseous molecules. The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on and , respectively. Because nonzero molecular volumes produce a measured volume that is than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is than that expected based on the ideal gas law, so the \(an^2/V^2\) term must be added to the measured pressure to correct for these effects. You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)? volume of cylinder, mass of compound, pressure, and temperature safety Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law. Obtain and values for Cl from Table \(\Page {1}\). Use the van der Waals equation (\(\ref{10.9.1}\)) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture. We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol): \[\begin{align} n &=\dfrac{m}{M} \\[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \\[4pt] &=7.052\;mol\nonumber \end{align} \nonumber \] Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure: \[\begin{align} P &=\dfrac{nRT}{V} \\[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \\[4pt] &= 43.1\;atm \end{align} \nonumber \] If chlorine behaves like an ideal gas, you have a real problem! Now let’s use the van der Waals equation with the and values for Cl from Table \(\Page {1}\). Solving for \(P\) gives \[\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\\&=\rm28.2\;atm\end{align} \nonumber \] This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation. A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the 77 atm 67 atm Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure). Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals \(a\) coefficients are relatively easy to liquefy because large coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O , N , Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions. The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek , meaning “cold,” and , meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called , which selectively destroys tissues with a minimal loss of blood by the use of extreme cold. Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure \(\Page {5}\)). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles. No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of / versus at a given temperature; for an ideal gas, / versus = 1 under all conditions. At high pressures, most real gases exhibit larger / values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit / values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the , which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold . | 12,027 | 3,011 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.09%3A_Additional_Resources |
The following set of experiments introduce students to the applications of chromatography and electrophoresis. Experiments are grouped into five categories: gas chromatography, high-performance liquid chromatography, ion-exchange chromatography, size-exclusion chromatography, and electrophoresis. The following texts provide a good introduction to the broad field of separations, including chromatography and electrophoresis. A more recent discussion of peak capacity is presented in the following papers. The following references may be consulted for more information on gas chromatography. The following references provide more information on high-performance liquid chromatography. The following references may be consulted for more information on ion chromatography. The following references may be consulted for more information on supercritical fluid chromatography. The following references may be consulted for more information on capillary electrophoresis. The application of spreadsheets and computer programs for modeling chromatography is described in the following papers. The following papers discuss column efficiency, peak shapes, and overlapping chromatographic peaks. | 1,198 | 3,012 |
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There are a few basic numerical and experimental tools with which you must be familiar. Fundamental measurements in analytical chemistry, such as mass, use base SI units, such as the kilogram. Other units, such as energy, are defined in terms of these base units. When reporting a measurement, we must be careful to include only those digits that are significant, and to maintain the uncertainty implied by these significant figures when trans- forming measurements into results. The relative amount of a constituent in a sample is expressed as a concentration. There are many ways to express concentration, the most common of which are molarity, weight percent, volume percent, weight-to-volume percent, parts per million and parts per billion. Concentrations also can be expressed using p-functions. Stoichiometric relationships and calculations are important in many quantitative analyses. The stoichiometry between the reactants and the products of a chemical reaction are given by the coefficients of a balanced chemical reaction. Balances, volumetric flasks, pipets, and ovens are standard pieces of equipment that you will use routinely in the analytical lab. You should be familiar with the proper way to use this equipment. You also should be familiar with how to prepare a stock solution of known concentration, and how to prepare a dilute solution from a stock solution. analytical balance desiccator graduated cylinder molarity parts per billion scientific notation stock solution volumetric flask weight-to-volume percent concentration dilution meniscus normality p-function significant figures tare volumetric pipet desiccant formality molality parts per million quantitative transfer SI units volume percent weight percent | 1,749 | 3,014 |
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Organic chemistry glassware is often segmented so that pieces can be arranged in a variety of ways to create setups that achieve different goals. It is important that the pieces are securely fastened in an apparatus so that flammable vapors don't escape and pieces don't fall (whereupon the glassware may break or contents may be spilled). Some chemistry labs have a lattice-work of metal bars (Figure 1.3c) secured to the benchtop that can be used for clamping apparatuses, and other labs rely on ring stands (Figure 1.3a). Ring stands should be positioned so that the apparatus is clamped directly over the heavy metal base, not in the opposite direction as the base (Figure 1.3a, not 1.3b) Metal clamps are used to connect glassware to ring stands or the metal lattice work. Two common type of clamps are " " and " " (Figure 1.4a). Although in many situations the clamps can be used interchangeably, an extension clamp must be used when clamping to a round bottomed flask (Figure 1.4b), as 3-fingered clamps do not hold well. The extension clamp should securely grasp the neck of a round bottomed flask below the glass protrusion (Figure 1.4b, not Figure 1.4c). Three-fingered clamps are generally used to hold condensers (Figure 1.3b), suction flasks, and chromatography columns (Figure 1.5). Both types of clamps often come with vinyl sleeves that may be removed if desired. The vinyl sleeves provide a gentle grasp for glassware, but should not be used with hot pieces as they may melt (or in the author's experience catch on fire!). Sometimes fire resistant sleeves are also provided with clamps as an alternative (right-most clamp in Figure 1.4a). Ring clamps (or iron rings) are also commonly used in the organic lab. They are used to hold separatory funnels (Figure 1.6a), and can be used to secure funnels when filtering or pouring liquids into narrow joints (Figure 1.6b). Furthermore, they can be used along with a wire mesh to serve as a platform for supporting flasks (Figure 1.6c). Plastic clips (sometimes called " " or " ") are also commonly used to secure the connections between joints (Figure 1.7). The clips are directional, and if they don't easily snap on, they are probably upside down. Plastic clips should not be used on any part of an apparatus that will get hot, as they may melt at temperatures above 140 ˚C (Figure 1.7b). Metal versions of these clips can be used alternatively in hot areas. Clips should not be relied upon to hold any substantial weight, as they can easily fail (especially if they have been warmed). Therefore, reaction flasks should not be held with just clips, but always supported in some more significant way (e.g. with an extension clamp attached to a ring stand). | 2,731 | 3,016 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Synthesis_of_Aldehydes_and_Ketones/Grignard_Reagents |
This page takes an introductory look at how Grignard reagents are made from halogenoalkanes (haloalkanes or alkyl halides), and introduces some of their reactions. A Grignard reagent has a formula \(\ce{RMgX}\) where \(\ce{X}\) is a halogen, and \(\ce{R}\) is an alkyl or aryl (based on a benzene ring) group. For the purposes of this page, we shall take R to be an alkyl group. Grignard \(\ce{CH3CH2MgBr}\). Grig halogenoalkane ethoxyethane Everything be perfectly dry because Grignard reagents react with water (see below). Grignard Grignard reagents react with water to produce . This is the reason that everything has to be during the preparation above. \[ CH_3CH_2MgBr + H_2O \rightarrow CH_3CH_3 + Mg(OH)Br\] oduct, \(Mg(OH)Br\), is Grignard reagents react with carbon dioxide in two stages. In the first, you get an addition of the Grignard reagent to the carbon dioxide. Grignard ethoxyethane For example: The product is then hydrolyzed (reacted with water) in the presence of a dilute acid. Typically, you would add dilute sulfuric acid or dilute hydrochloric acid to the solution formed by the reaction with the CO . A carboxylic acid is produced with one more carbon than the original Grignard reagent. The usually quoted equation is (without the red bits): Almost all sources quote the formation of a basic halide such as Mg(OH)Br as the other product of the reaction. That is actually misleading because these compounds react with dilute acids. What you end up with would be a mixture of ordinary hydrated magnesium ions, halide ions and sulfate or chloride ions - depending on which dilute acid you added. Carbonyl compounds contain the C=O double bond. The simplest ones have the form: R and R' can be the same or different, and can be an alkyl group or hydrogen. If one (or both) of the R groups are hydrogens, the compounds are called aldehydes. For example: If both of the R groups are alkyl groups, the compounds are called ketones. Examples include: The reactions between the various sorts of carbonyl compounds and Grignard reagents can look quite complicated, but in fact they all react in the same way - all that changes are the groups attached to the carbon-oxygen double bond. The reactions are essentially identical to the reaction with carbon dioxide - all that differs is the nature of the organic product. In the first stage, the Grignard reagent adds across the carbon-oxygen double bond: Dilute acid is then added to this to hydrolyse it. (I am using the normally accepted equation ignoring the fact that the Mg(OH)Br will react further with the acid.) An alcohol is formed. One of the key uses of Grignard reagents is the ability to make complicated alcohols easily. What sort of alcohol you get depends on the carbonyl compound you started with - in other words, what R and R' are. In methanal, both R groups are hydrogen. Methanal is the simplest possible aldehyde. Assuming that you are starting with CH CH MgBr and using the general equation above you get always has the form: Since both R groups are hydrogen atoms, the final product will be: A primary alcohol is formed. A primary alcohol has only one alkyl group attached to the carbon atom with the -OH group on it. You could obviously get a different primary alcohol if you started from a different Grignard reagent. The next biggest aldehyde is ethanal. One of the R groups is hydrogen and the other CH . Again, think about how that relates to the general case. The alcohol formed is: So this time the final product has one CH group and one hydrogen attached: A secondary alcohol has two alkyl groups (the same or different) attached to the carbon with the -OH group on it. You could change the nature of the final secondary alcohol by either: Ketones have two alkyl groups attached to the carbon-oxygen double bond. The simplest one is propanone. This time when you replace the R groups in the general formula for the alcohol produced you get a tertiary alcohol. A tertiary alcohol has three alkyl groups attached to the carbon with the -OH attached. The alkyl groups can be any combination of same or different. You could ring the changes on the product by The bond between the carbon atom and the magnesium is polar. Carbon is more electronegative than magnesium, and so the bonding pair of electrons is pulled towards the carbon. The carbon-oxygen double bond is also highly polar with a significant amount of positive charge on the carbon atom. The nature of this bond is described in detail elsewhere on this site. Grignard nucleophile Grignard A nucleophile is a species that attacks positive (or slightly positive) centers in other molecules or ions. Jim Clark ( ) | 4,688 | 3,019 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation_State/Oxidation_States_II |
Oxidation state is a number assigned to an element in a compound according to some rules. This number enable us to describe oxidation-reduction reactions, and balancing redox chemical reactions. When a covalent bond forms between two atoms with different electronegativities the shared electrons in the bond lie closer to the more electronegative atom: In HCl (above) the oxidation number for the hydrogen would be +1 and that of the Cl would be -1 For oxidation numbers we write the sign to distinguish them from ionic (electronic) charges Oxidation state (or oxidation numbers) do not refer to real charges on the atoms, except in the case of actual ionic substances. Oxidation numbers can be determined using the following rules: The oxidation state (OS) of an element corresponds to the number of electrons, e , that an atom loses, gains, or appears to use when joining with other atoms in compounds. In determining the OS of an atom, there are seven guidelines to follow: (Note: The sum of the OSs is equal to zero for neutral compounds and equal to the charge for polyatomic ion species.) Determine the OSs of the elements in the following reactions: Determine the OS of the bold element in each of the following: Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the OS of each): (For further discussion, see the article on ). An atom is oxidized if its oxidation number increases, the reducing agent, and an atom is reduced if its oxidation number decreases, the oxidizing agent. The atom that is oxidized is the reducing agent, and the atom that is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound). Compounds of the alkali (oxidation number +1) and alkaline earth metals (oxidation number +2) are typically ionic in nature. Compounds of metals with higher oxidation numbers (e.g., tin +4) tend to form molecular compounds MgH magnesium hydride H S dihydrogen sulfide FeF iron(II) fluoride OF oxygen difluoride Mn O manganese(III) oxide Cl O dichlorine trioxide An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Redox reactions are common and vital to some of the basic functions of life, including , respiration, combustion, and or rusting. Redox reactions are comprised of two parts, a reduced half and an oxidized half, that occur together. The reduced half gains electrons and the oxidation number decreases, while the oxidized half loses electrons and the oxidation number increases. Simple ways to remember this include the mnemonic devices meaning " " and " " and meaning " " and " ." There is no net change in the number of electrons in a redox reaction. Those given off in the oxidation half reaction are taken up by another species in the reduction half reaction. The two species that exchange electrons in a redox reaction are given special names. The ion or molecule that accepts electrons is called the ; by accepting electrons it causes the oxidation of another species. Conversely, the species that donates electrons is called the ; when the reaction occurs, it reduces the other species. In other words, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound, as in disproportionation reactions). Image by A good example of a redox reaction is the in which iron atoms in ferric oxide lose (or give up) O atoms to Al atoms, producing \(Al_2O_3\) (Figure 20.1.1). \[Fe_2O_{3(s)} + 2Al_{(s)} \rightarrow Al_2O_{3(s)} + 2Fe_{(l)}\] Another example of the redox reaction is the reaction between zinc and copper sulfate. Using the equations from the previous examples, determine what is oxidized in the following reaction. \[Zn + 2H^+ \rightarrow Zn^{2+} + H_2\] The OS of H changes from +1 to 0, and the OS of Zn changes from 0 to +2. Hence, Zn is oxidized and acts as the reducing agent. What is reduced species in this reaction? \[Zn + 2H^+ \rightarrow Zn^{2+} + H_2\] The OS of H changes from +1 to 0, and the OS of Zn changes from 0 to +2. Hence, H+ ion is reduced and acts as the oxidizing agent. reactions are among the simplest redox reactions and, as the name suggests, involves "combining" elements to form a chemical compound. As usual, oxidation and reduction occur together. The general equation for a combination reaction is given below: \[A + B \rightarrow AB \] Equation: H + O → H O
Calculation: 0 + 0 →
Explanation: In this equation both H and O are free elements; following , their OSs are 0. The product is H O, which has a total OS of 0. According to , the OS of oxygen is usually -2. Therefore, the OS of H in H O must be +1. reaction is the reverse of a combination reaction, the breakdown of a chemical compound into individual elements: \[AB \rightarrow A + B\] Consider the decomposition of water: \[H_2O \rightarrow H_2 + O_2\] Calculation: (2)(+1) + (-2) = 0 → 0 + 0
Explanation: In this reaction, water is "decomposed" into hydrogen and oxygen. As in the previous example the H O has a total OS of 0; thus, according to Rule #6 the OS of oxygen is usually -2, so the OS of hydrogen in H O must be +1. A reaction involves the "replacing" of an element in the reactants with another element in the products: \[A + BC \rightarrow AB + C\] Equation: \[Cl_2 + Na\underline{Br} \rightarrow Na\underline{Cl} + Br_2\] Calculation: (0) + ((+1) + (-1) = 0) -> ((+1) + (-1) = 0) + 0
Explanation: In this equation, Br is replaced with Cl, and the Cl atoms in Cl are reduced, while the Br ion in NaBr is oxidized. A reaction is similar to a double replacement reaction, but involves "replacing" two elements in the reactants, with two in the products: \[AB + CD \rightarrow AD + CB \] Equation: Fe + H → Fe + H
Explanation: In this equation, Fe and H trade places, and oxygen and chlorine trade places. reactions almost always involve oxygen in the form of O , and are almost always exothermic, meaning they produce heat. Chemical reactions that give off light and heat and light are colloquially referred to as "burning." \[C_xH_y + O_2 \rightarrow CO_2 + H_2O\] Although combustion reactions typically involve redox reactions with a chemical being oxidized by oxygen, many chemicals "burn" in other environments. For example, both titanium and magnesium burn in nitrogen as well: \[ 2Ti(s) + N_{2}(g) \rightarrow 2TiN(s)\] \[3 Mg(s) + N_{2}(g) \rightarrow Mg_3N_{2}(s) \] Moreover, chemicals can be oxidized by other chemicals than oxygen, such as Cl or F ; these processes are also considered combustion reactions In some redox reactions a single substance can be both oxidized and reduced. These are known as reactions, with the following general equation: \[2A \rightarrow A^{+n} + A^{-n}\] Where n is the number of electrons transferred. Disproportionation reactions do not need begin with neutral molecules, and can involve more than two species with differing oxidation states (but rarely). Disproportionation reactions have some practical significance in everyday life, including the reaction of hydrogen peroxide, H O poured over a cut. This a decomposition reaction of hydrogen peroxide, which produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced. The reaction is as follows: \[2H_2O_{2}(aq) \rightarrow 2H_2O(l) + O_{2}(g)\] Explanation: On the reactant side, H has an OS of +1 and O has an OS of -1, which changes to -2 for the product H O (oxygen is reduced), and 0 in the product O (oxygen is oxidized). | 7,921 | 3,020 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.5%3A_Boiling-Point_Elevation_and_Freezing-Point_Depression |
The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the of the solute. \[\begin{align*}\Delta{T_f} &= T_f(solvent) - T_f (solution) \\[4pt] &= K_f \times m \end{align*}\] where \(\Delta{T_f}\) is the freezing point depression, \(T_f\) (solution) is the freezing point of the solution, \(T_f\) (solvent) is the freezing point of the solvent, \(K_f\) is the freezing point depression constant, and is the molality. Nonelectrolytes are substances with no ions, only molecules. , on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also, remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution. Adding solutes to an ideal solution results in a positive \(ΔS\), an increase in entropy. Because of this, the newly altered solution's chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as . These properties are dependent on the number of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease. The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas: \[\Delta{T}_f = -K_f \times m\] \[\Delta{T}_b = K_b \times m\] where \(m\) is the solute and \(K\) values are proportionality constants; (\(K_f\) and \(K_b\) for freezing and boiling, respectively). Molality is defined as the number of moles of solute per kilogram . Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer. If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for \(K_f\) and \(K_b\) respectively, are in Table \(\Page {1}\): The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute to the vapor pressure of the solution, and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures. The boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease in the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus, occurs with a temperature increase that is quantified using \[\Delta{T_b} = K_b m\] where \(K_b\) is known as the and \(m\) is the molality of the solute. Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the pure solid solvent. Therefore, a is observed. 2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 \(^{\circ}C\). What is the molar mass of the compound? First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved. \[ \begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \\[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \\[4pt] &= 0.123 m \end{align*}\] \[ \begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \\[4pt] &= 0.00923 \; m \; solute \end{align*}\] We can now find the molecular weight of the unknown compound: \[ \begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \\[4pt] &= 216.80 \; g/mol \end{align*}\] The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water. Benzophenone has a freezing point of 49.00 C. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59 C. Find the freezing point depression constant for the solvent. \(9.80\,^oC/m\) Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, \(\ce{NaCl}\) will be ineffective. Under these conditions, \(\ce{CaCl_2}\) can be used since it dissolves to make three ions instead of two for \(\ce{NaCl}\). | 6,383 | 3,023 |
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In North America, chocolate manufacturing started in Massachusetts in 1765. Today, in the factory, the beans get cleaned, and magnets take out metallic parts, and then sand, dust, and other impurities are removed. Some starch will be changed into dextrins in the roasting process to improve flavor. Machines break the beans and grind them fine until a flowing liquid is produced, called chocolate liquor. Through hydraulic pressure, cocoa butter is reduced from 55% to approximately 10% to 24% or less, and the residue forms a solid mass called press cake. The press cake is then broken, pulverized, cooled, and sifted to produce commercial cocoa powder. The baking industry uses primarily cocoa powders with a low fat content. At the factory, chocolate is also subject to an additional refining step calcleodnching. Conching has a smoothing effect. The temperature range in this process is between 55°C and 65°C (131°F and 149°F). Sugar interacts with protein to form amino sugars, and the paste losesacids and moisture and becomes smoother. This video explains the chemical reactions related to heat, melting point, and formation of crystal structures in science360.gov/obj/video/27d9...stry-chocolate | 1,215 | 3,024 |
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Distillation of mixtures may or may not produce relatively pure samples. As distillation involves boiling a solution and condensing its vapors, the composition of the distillate is identical to the composition of the vapors. Several equations can be used to describe the composition of vapor produced from a solution. states that a compound's vapor pressure is lessened when it is part of a solution, and is proportional to its molar composition. Raoult's law is shown in Equation \ref{1}. The equation means that a solution containing \(80 \: \text{mol}\%\) compound "A" and \(20 \: \text{mol}\%\) of another miscible component would at equilibrium produce \(80\%\) as many particles of compound A in the gas phase than if compound A were in pure form. \[ P_A = P_A^o \chi_A \label{1}\] where \(P_A^o\) is the vapor pressure of a sample of pure A, and \(\chi_A\) is the mole fraction of \(A\) in the mixture. of partial pressures states that the total pressure in a closed system can be found by addition of the partial pressures of each gaseous component. Dalton's law is shown in Equation \ref{2}. \[P_\text{total} = P_A + P_B \label{2}\] A combination of Raoult's and Dalton's laws is shown in Equation \ref{3}, which describes the vapor produced by a miscible two-component system (compounds A + B). This combined law shows that the vapors produced by distillation are dependent on each component's vapor pressure and quantity (mole fraction). \[P_\text{solution} = P_A^o \chi_A + P_B^o \chi_B \label{3}\] A compound's vapor pressure reflects the temperature of the solution as well as the compound's boiling point. As temperature increases, a greater percentage of molecules have sufficient energy to overcome the intermolecular forces (IMF's) holding them in the liquid phase. Therefore, a compound's vapor pressure always increases with temperature (see Table 5.2), although not in a linear fashion. When comparing two compounds at the same temperature, the compound with weaker IMF's (the one with the lower boiling point) should more easily enter the gas phase. Therefore, at any certain temperature, (see Table 5.2). This last concept is the cornerstone of distillation theory. A compound with a lower boiling point has a greater vapor pressure than a compound with a higher boiling point A simple distillation works well to purify certain mixtures, specifically to separate a liquid from non-volatile impurities (e.g. solids or salts), or from small amounts of significantly higher or lower boiling impurities. A general guideline is that a simple distillation is capable of separating components if the . A simple distillation does not work well to purify a mixture that contains components with similar boiling points (when the difference in b.p. is \(< 100^\text{o} \text{C}\)). To demonstrate, a mixture containing \(75 \: \text{mol}\%\) 1-chlorobutane (normal b.p. \(78^\text{o} \text{C}\)) and \(25 \: \text{mol}\%\) -cymene (normal b.p. \(177^\text{o} \text{C}\), Figure 5.10b) was purified using a simple distillation to collect the first two \(\text{mL}\) of distillate. . Gas chromatograph (GC) analysis of the initial mixture and distillate allowed for quantitation, although the reported percentages were not proportional to the \(\text{mol}\%\) values because of the detector used (often higher molecular weight compounds have abnormally high integration values with mass spectrometers). A calibration curve was necessary to translate the reported percentages into accurate values (Figure 5.9). To generate the calibration curve, the GC integration values were recorded for several known mixtures of 1-chlorobutane and -cymene. For example, a sample with \(95 \: \text{mol}\%\) 1-chlorobutane (x-axis) and \(5 \: \text{mol}\%\) -cymene was reported by the GC instrument to be \(42 \: \text{mol}\%\) 1-chlorobutane (y-axis). This was recorded as the data point (95,42). GC analysis (Figure 5.10) and a calibration curve (Figure 5.9 showed the distillate to be nearly pure 1-chlorobutane (\(\sim 99.5 \: \text{mol}\%\), Table 5.3). This is an example of a mixture that is purified well through simple distillation. A second distillation was performed in a similar manner using a mixture containing \(75 \: \text{mol}\%\) ethylbenzene (normal b.p. \(136^\text{o} \text{C}\)) and \(25 \: \text{mol}\%\) -cymene (normal b.p. \(177^\text{o} \text{C}\), Figure 5.12b). . GC analysis (Figure 5.12) and a calibration curve (Figure 5.11) showed the simple distillation did increase the quantity of ethylbenzene in the mixture (from \(75 \: \text{mol}\%\) to \(90 \: \text{mol}\%\), see Table 5.4), but a significant portion of -cymene remained in the distillate. This is an example of a mixture that is not completely purified by simple distillation. The two distillations in this section both started with \(75/25 \: \text{mol}\%\) mixtures, and yet the two distillations resulted in distillates with different purity. This result can be explained by analysis of Raoult's and Dalton's laws (Equations \ref{1} and \ref{2}). The equation describing the composition of vapor produced from the \(75 \: \text{mol}\%\) 1-chlorobutane/\(25 \: \text{mol}\%\) -cymene mixture is shown in Equation \ref{4}: \[ \begin{align} P_\text{solution} &= P^o_\text{1-chlorobutane} \chi_A + P^o_\text{p-cymene} \chi_B \\[4pt] &= P^o_\text{1-chlorobutane} \left( 0.75 \right) + P^o_\text{p-cymene} \left( 0.25 \right) \label{4} \end{align} \] In this distillation, the distillate was "enriched" in 1-chlorobutane, as it changed from \(75 \: \text{mol}\%\) 1-chlorobutane in the distilling flask to \(99.5 \: \text{mol}\%\) 1-chlorobutane in the distillate. The distillate contained nearly pure 1-chlorobutane, as it had the greater contribution to the solution's vapor. This is because: In this distillation, the difference in boiling point of the components was so large (\(\Delta\) b.p. of \(99^\text{o} \text{C}\)), that the vapor pressure of -cymene (and thus its contribution to the vapor phase) was essentially insignificant during the distillation. The vastly different vapor pressures (along with the differences in mole fraction), were the reason the simple distillation resulted in a nearly pure distillate. In the distillation of the \(75 \: \text{mol}\%\) ethylbenzene/\(25 \: \text{mol}\%\) -cymene mixture, the distillate was also enriched in the lower boiling component (ethylbenzene) as it had the greater vapor pressure. However, the distillate still contained \(10 \: \text{mol}\%\) -cymene, because there was only a \(41^\text{o} \text{C}\) difference in boiling point between the two components. This meant that the higher boiling component ( -cymene) had a smaller vapor pressure, leading to a significant quantity of -cymene in the distillate. Equation (5) describes the vapor produced by a miscible two-component system (compounds A + B). This mathematical equation can be used to generate phase diagrams, or , which graphically correlate temperature to the molar composition of the liquid and vapor phases. \[P_\text{solution} = P^o_A \chi_A + P^o_B \chi_B \label{5}\] Figure 5.13 shows a generic distillation curve for a two-component system. Molar composition is on the x-axis, with the left side of the plot corresponding to a sample that is pure compound A and the right side of the plot corresponding to a sample of pure compound B. In between the two sides represent mixtures of A and B. Temperature is on the y-axis, and the boiling point of each compound are marked ("bp A" and "bp B"). You may have been previously exposed to phase diagrams for pure compounds (e.g. of water or carbon dioxide), where only a single line would be used to denote a liquid-gas equilibrium. In the distillation of mixtures, there is a difference in composition between the liquid and gas phases, which is the reason for the tear-shaped appearance of the phase diagram. The tear-shaped region represents conditions where both liquid and gas coexist, during boiling of the liquid. Imagine a \(25 \: \text{mol}\%\) A/\(75 \: \text{mol}\%\) B mixture is to be distilled, and this mixture is described by the distillation curve in Figure 5.13a. When the temperature reaches the lower line of the tear drop (temperature x and point a in Figure 5.13b), the solution begins to boil. Since liquid and gas have the same temperature during boiling, the vaporization process can be thought of as following the horizontal line from position a to b in Figure 5.13b. After vaporization, the gas condenses into the distillate, which can be thought of as following the vertical line from position b to c in Figure 5.13b. Under perfect equilibrating conditions, the distillate in this example would be \(74 \: \text{mol}\%\) A/\(26 \: \text{mol}\%\) B, and is "enriched" in A as it has the lower boiling point and thus the higher vapor pressure. In a distillation curve, the leftward horizontal "movement" followed by downward vertical "movement" represents one vaporization-condensation event (see Figure 5.13b). This is often referred to as a " ", historical terminology related to the collection of distillate onto plates or trays, and represents the purification potential of a simple distillation. Every mixture has its own distillation curve, reflective of the boiling points of the components. A mixture containing two components whose boiling points are vastly different (\(\Delta\) b.p. \(> 100^\text{o} \text{C}\)) is shown in Figure 5.14. Imagine a \(25 \: \text{mol}\%\) A/\(75 \: \text{mol}\%\) B mixture is to be distilled, and this mixture is described by the distillation curve in Figure 5.14a, where the components have significantly different boiling points. One vaporization-condensation cycle (from a to b to c in Figure 5.14b) will produce a distillate that is nearly \(100\%\) A, the compound with the much lower boiling point. When distilling a mixture with vastly different boiling points, the two components act almost independently and can be easily separated by simple distillation. . When distilling mixtures, however, the temperature does not often remain steady. This section describes why mixtures distill over a range of temperatures, and the approximate temperatures at which solutions can be expected to boil. These concepts can be understood by examination of the equation that describes a solution's vapor pressure (Equation (6)) and distillation curves. \[P_\text{solution} = P_A + P_B = P^o_A \chi_A + P^o_B \chi_B \label{6}\] Imagine an ideal solution that has equimolar quantities of compound "A" (b.p. = \(50^\text{o} \text{C}\)) and compound "B" (b.p. = \(100^\text{o} \text{C}\)). A common misconception is that this solution will boil at \(50^\text{o} \text{C}\),at the boiling point of the lower-boiling compound. The solution does boil at A's boiling point because A has a reduced partial pressure when part of a solution. In fact, the partial pressure of A at \(50^\text{o} \text{C}\) would be half its normal boiling pressure, as shown in Equation \ref{7}: \[\text{At } 50^\text{o} \text{C}: \: \: \: \: \: P_A = \left( 760 \: \text{torr} \right) \left( 0.50 \right) = 380 \: \text{torr} \label{7}\] Since the partial pressures of A and B are additive, the vapor contribution of B at \(50^\text{o} \text{C}\) is also important. The partial pressure of B at \(50^\text{o} \text{C}\) would equal its vapor pressure at this temperature multiplied by its mole fraction (0.50). Since B is below its boiling point \(\left( 100^\text{o} \text{C} \right)\), its vapor pressure would be smaller than \(760 \: \text{torr}\), and the exact value would need to be looked up in a reference book. Let's say that the vapor pressure of B at \(50^\text{o} \text{C}\) is 160 torr. At 50ºC: \[ P_B = \left( 160 \: \text{torr} \right) \left( 0.50 \right) = 80 \: \text{torr} \label{8}\] boils when the pressure of the components equals the atmospheric pressure (let's assume \(760 \: \text{torr}\) for this calculation). This equimolar solution would therefore boil at \(50^\text{o} \text{C}\), as its combined pressure is less than the external pressure of \(760 \: \text{torr}\) (Equation \ref{9}). At 50 ºC: \[\begin{align} P_\text{solution} &= P_A + P_B \\[4pt] &= \left( 380 \: \text{torr} \right) + \left( 80 \: \text{torr} \right) \\[4pt] &= 460 \: \text{torr} \label{9} \end{align}\] The initial boiling point of this solution is \(66^\text{o} \text{C}\), which is the temperature where the combined pressure matches the atmospheric pressure (Equation \ref{10}, note: all vapor pressures would have to be found in a reference book). At 66 ºC: \[\begin{align} P_\text{solution} &= P^o_A \chi_A + P^o_B \chi_B \\[4pt] &= \left( 1238 \: \text{torr} \right) \left( 0.5 \right) + \left( 282 \: \text{torr} \right) \left( 0.5 \right) \\[4pt] &= \left( 619 \: \text{torr} \right) + \left( 141 \: \text{torr} \right) = 760 \: \text{torr} \label{10} \end{align}\] These calculations demonstrate that .\(^5\). This can also be seen by examination of the distillation curve for this system, where the solution boils when the temperature reaches position a in Figure 5.15a, a temperature between the boiling point of the components. It is important to realize that mixtures of organic compounds containing similar boiling points (\(\Delta\) b.p. \(< 100^\text{o} \text{C}\)) boil . They do act as separate entities (not "A boils at \(50^\text{o} \text{C}\), then B boils at \(100^\text{o} \text{C}\)"). The mixture in this example begins boiling at \(66^\text{o} \text{C}\), but after a period of time boiling would cease if maintained at this temperature. This happens because the composition of the distilling pot changes over time. Since distillation removes more of the lower boiling A, the higher boiling B will increase percentagewise in the distilling pot. Imagine that after some material has distilled, the distilling pot is now \(42.2\%\) A and \(57.8\%\) B. This solution no longer boils at \(66^\text{o} \text{C}\), as shown in Equation \ref{11}. At 66 ºC: \[\begin{align} P_\text{solution} &= P^o_A \chi_A + P^o_B \chi_B \\[4pt] &= \left( 1238 \: \text{torr} \right) \left( 0.422 \right) + \left( 282 \: \text{torr} \right) \left( 0.578 \right) \\[4pt] &= \left( 522 \: \text{torr} \right) + \left( 163 \: \text{torr} \right) \\[4pt] &= 685 \: \text{torr} \label{11} \end{align}\] At the new composition in this example, the temperature must be raised to 70 ºC to maintain boiling, as shown in Equation \ref{12}. This can also be shown by examination of the distillation curve in Figure 5.15b, where boiling commences at temperature c, but must be raised to temperature e as the distilling pot becomes more enriched in the higher boiling component (shifts to the right on the curve). At 70 ºC: \[\begin{align} P_\text{solution} &= P^o_A \chi_A + P^o_B \chi_B \\[4pt] &= \left( 1370 \: \text{torr} \right) \left( 0.422 \right) + \left( 315 \: \text{torr} \right) \left( 0.578 \right) \\[4pt] &= \left( 578 \: \text{torr} \right) + \left( 182 \: \text{torr} \right) \\[4pt] &= 760 \: \text{torr} \label{12} \end{align} \] These calculations and analysis of distillation curves demonstrate why :\(^6\) as the composition in the distilling pot changes with time (which affects the mole fraction or the x-axis on the distillation curve), the temperature must be adjusted to compensate. These calculations also imply why a pure compound distills at a constant temperature: the mole fraction is one for a pure liquid, and the distilling pot composition remains unchanged during the distillation. The calculations and distillation curves in this section enable discussion of another aspect of distillation. The partial pressures of each component at the boiling temperatures can be directly correlated to the composition of the distillate. Therefore, the distillate at \(66^\text{o} \text{C}\) is \(81 \: \text{mol}\%\) A (\(100\% \times 619 \: \text{torr}/760 \: \text{torr}\), see Equation \ref{10}) while the distillate at \(70^\text{o} \text{C}\) is \(76 \: \text{mol}\%\) A (\(100\% \times 578 \: \text{torr}/760 \: \text{torr}\), see Equation \ref{12}). , and the distillate degrades in purity as the distillation progresses. This can also be seen in the distillation curve in Figure 5.15b, where the initial distillate composition corresponds to position d (purer), while the distillate at \(70^\text{o} \text{C}\) corresponds to position f (less pure). Pure compounds distill at a constant temperature, while most solutions containing more than one volatile component distill over a range of temperatures. There is an exception to this statement, as some mixtures of certain composition also distill at a constant temperature. These mixtures are called " ", which display non-ideal behavior and do not follow Raoult's law (Equation \ref{13}). One of the best-known azeotropes is a mixture containing \(95.6\%\) ethanol and \(4.4\%\) water. When distilling a mixture containing ethanol and water (for example when concentrating fermented grains to produce hard liquor), \(95.6\%\) is the highest percentage of ethanol possible in the distillate. It is to distill \(100\%\) pure ethanol when water is in the distilling pot, as the ethanol and water co-distill as a pure substance. For this reason, most ethanol used by chemists is the \(95\%\) version as it can be purified through distillation and is the least expensive. "Absolute ethanol" can also be purchased (which is \(> 99\%\) ethanol), but is more expensive as further methods (e.g. drying agents) are required to remove residual water before or after distillation. Many compounds form azeotropes, which are specific ratios of compounds that co-distill at a constant composition and temperature. \(^7\) lists roughly 860 known azeotropes composed of two or three components each. A selection of these azeotropic mixtures are in Table 5.5. Azeotropes form when a solution deviates from Raoult's law (Equation (13)), meaning that the vapor pressure produced from an azeotropic solution does directly correlate to its mole fraction. Deviations occur when the components are either attracted to or repelled by one another, namely if they have significantly different strengths of intermolecular forces (IMF's) to themselves (e.g. A-A or B-B) as they do with the other components (e.g. A-B). In other words, a solution will deviate from Raoult's law if the enthalpy of mixing is not zero. \[P_A = P^o_A \chi_A \label{13}\] Azeotropic mixtures come in two forms: one whose boiling point is lower than any of its constituents (called a " ") or ones whose boiling point is higher than any of its constituents (called a " "). For example, the \(61\%\) benzene/\(39\%\) methanol azeotrope is a minimum boiling azeotrope as its boiling point is \(58^\text{o} \text{C}\), which is lower than the boiling point of benzene \(\left( 80^\text{o} \text{C} \right)\) and methanol \(\left( 65^\text{o} \text{C} \right)\). Minimum boiling azeotropes are much more common than maximum boiling azeotropes. Minimum boiling azeotropes occur when the component's IMF's are stronger to themselves (A-A/B-B) than to each other (A-B). This generally occurs with components that lack an affinity for one another, for example when one component can hydrogen bond but the other cannot. In these situations, the components aggregate somewhat in solution, which decreases the solution's entropy and makes it more favorable to become a gas. These solutions therefore have a higher vapor pressure than predicted by Raoult's law, leading to a lower boiling temperature. Maximum boiling azeotropes have the opposite effect, resulting from attractions in the solution that lead to lower vapor pressures than predicted, and thus higher boiling temperatures. A distillation curve for an ethanol-water mixture is shown in Figure 5.16 (not drawn to scale in order to show detail). The minimum boiling azeotrope is represented by the intersection of the two tear-drop shapes, as indicated in Figure 5.16a. this curve demonstrates why an ethanol-water mixture cannot be distilled to produce a distillate with greater than \(95.6\%\) ethanol. Imagine distilling an ethanol/water mixture of the composition at position a in Figure 5.16b). After each vaporization-condensation event (a to b to c, then c to d to e in Figure 5.16b), the ethanol concentration increases. However, material is always funneled toward the lowest point on the curve, the azeotropic composition. Concentrations higher than \(95.6\%\) ethanol can only be distilled when there is no longer any water in the system (at which point the phase diagram in Figure 5.16 is no longer applicable). Since water forms minimum-boiling azeotropes with many organic compounds, it should be clear why water must always be carefully removed with drying agents before distillation: failure to remove trace water will result in a wet distillate. \(^3\)Recall that vapor pressure is the partial pressure of a compound formed by evaporation of a pure liquid into the headspace of a closed container. \(^4\)Data from , 50\(^\text{th}\) ed., CRC Press, Cleveland, , p. 148. \(^5\)Azeotropic solutions do not follow this same pattern, as they do not obey Raoult's law. \(^6\)Azeotropic solutions again do not follow this generalization, and instead boil at a constant temperature. \(^7\)J. A. Dean, , 15\(^\text{th}\) ed., McGraw-Hill, , Sect. 5.78 and 5.79. | 21,402 | 3,025 |
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" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Problems marked with a ♦ involve multiple concepts. 1..Carbon tetrachloride (CCl4) was used as a dry cleaning solvent until it was found to cause liver cancer. Based on the structure of chloroform given in , draw the structure of carbon tetrachloride. 2. Ammonium nitrate and ammonium sulfate are used in fertilizers as a source of nitrogen. The ammonium cation is tetrahedral. Refer to to draw the structure of the ammonium ion. 3. The white light in fireworks displays is produced by burning magnesium in air, which contains oxygen. What compound is formed? 4. Sodium hydrogen sulfite, which is used for bleaching and swelling leather and to preserve flavor in almost all commercial wines, is made from sulfur dioxide. What are the formulas for these two sulfur-containing compounds? 5. Carbonic acid is used in carbonated drinks. When combined with lithium hydroxide, it produces lithium carbonate, a compound used to increase the brightness of pottery glazes and as a primary treatment for depression and bipolar disorder. Write the formula for both of these carbon-containing compounds. 6. Vinegar is a dilute solution of acetic acid, an organic acid, in water. What grouping of atoms would you expect to find in the structural formula for acetic acid? 7. ♦ Sodamide, or sodium amide, is prepared from sodium metal and gaseous ammonia. Sodamide contains the amide ion (NH ), which reacts with water to form the hydroxide anion by removing an H ion from water. Sodium amide is also used to prepare sodium cyanide. a. Write the formula for each of these sodium-containing compounds. b. What are the products of the reaction of sodamide with water? 8. A mixture of isooctane, n-pentane, and n-heptane is known to have an octane rating of 87. Use the data in to calculate how much isooctane and n-heptane are present if the mixture is known to contain 30% n-pentane. 9. A crude petroleum distillate consists of 60% n-pentane, 25% methanol, and the remainder n-hexane by mass ( ). a. What is the octane rating? b. How much MTBE would have to be added to increase the octane rating to 93? 10. Premium gasoline sold in much of the central United States has an octane rating of 93 and contains 10% ethanol. What is the octane rating of the gasoline fraction before ethanol is added? (See .) 1. 3. MgO, magnesium oxide 5. Carbonic acid is H CO ; lithium carbonate is Li CO . 7. a. Sodamide is NaNH , and sodium cyanide is NaCN. b. Sodium hydroxide (NaOH) and ammonia (NH ). b. 52 g of MTBE must be added to 48 g of the crude distillate. | 2,693 | 3,026 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Homolytic_C-H_Bond_Dissociation_Energies_of_Organic_Molecules |
The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other. Breaking a between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another \[A-B \rightarrow A^+ + B:^-\] \[A-B \rightarrow A:^- + B^+ \] or homolytically, where one electron stays with each partner. \[A-B \rightarrow A^• + B^• \] The products of homolytic cleavage are and the energy that is required to break the bond homolytically is called the (BDE) and is a measure of the strength of the bond. The BDE for a molecule A-B is calculated as the difference in the of the products and reactants for homolysis \[BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B)\] Officially, the definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is \(D_o\). However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation , which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes. Bond dissociation energy (or enthalpy) is a and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using can be used to estimate reaction enthalpies. Consider the chlorination of methane \[CH_4 + Cl_2 \rightarrow CH_3Cl + HCl\] the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed CH → CH • + H• BDE(CH -H) Cl → 2Cl• BDE(Cl )\] H• + Cl• → HCl -BDE(HCl) CH • + Cl• → CH Cl -BDE(CH -Cl) --------------------------------------------------- \[CH_4 + Cl_2 \rightarrow CH_3Cl + HCl\] \[\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl)\] Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available. Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical. RH + Cl• → R• + HCl The energy change for this step is equal to the difference in the BDEs in RH and HCl \[\Delta H = BDE(R-H) - BDE(HCl)\] This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds. It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done. Analysis of the BDEs shown in the table above shows that there are some systematic trends: The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation. 372.4±1.7 Therefore, although C-CH bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually with more substitution. Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals. Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs. Another factor that needs to be considered is the electronegativity. The says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression \[D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2 \] where \(X_A\) and \(X_B\) are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments. There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation. . 1. Gronert, S. , , 1209 | 5,561 | 3,028 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./05.E%3A_Energy_Changes_in_Chemical_Reactions_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Palm trees grow on the coast of southern England even though the latitude is the same as that of Winnipeg, Canada. What is a plausible explanation for this phenomenon? (Hint: the Gulf Stream current is a factor.) During intense exercise, your body cannot provide enough oxygen to allow the complete combustion of glucose to carbon dioxide. Under these conditions, an alternative means of obtaining energy from glucose is used in which glucose (C H O ) is converted to lactic acid (C H O H). The equation for this reaction is as follows: ♦ During the late spring, icebergs in the North Atlantic pose a hazard to shipping. To avoid them, ships travel routes that are about 30% longer. Many attempts have been made to destroy icebergs, including using explosives, torpedoes, and bombs. How much heat must be generated to melt 15% of a 1.9 × 10 kg iceberg? How many kilograms of TNT (trinitrotoluene, C H N O ) would be needed to provide enough energy to melt the ice? (The Δ for explosive decomposition of TNT is −1035.8 kJ/mol.) Many biochemical processes occur through sequences of reactions called . The total energy released by many of these pathways is much more than the energy a cell could handle if it were all released in a single step. For example, the combustion of glucose in a single step would release enough energy to kill a cell. By using a series of smaller steps that release less energy per reaction, however, the cell can extract the maximum energy from glucose without being destroyed. Referring to Equation 5.30, calculate how many grams of glucose would need to be metabolized to raise the temperature of a liver cell from an average body temperature of 37°C to 100°C, if the cell has a volume of 5000 μm . Although the cell is only 69% water, assume that the density of the cell is 1.00 and that its is the same as that of water. ♦ During smelting, naturally occurring metal oxides are reduced by carbon at high temperature. For copper(II) oxide, this process includes the following series of chemical equations: The final products are CO and Cu. The discovery of this process led to the increasing use of ores as sources of metals in ancient cultures. In fact, between 3000 BC and 2000 BC, the smelting of copper was well established, and beads made from copper are some of the earliest known metal artifacts. ♦ The earliest known Egyptian artifacts made from tin metal date back to approximately 1400 BC. If the smelting process for SnO occurs via the reaction sequence what is Δ for the conversion of SnO to Sn (s, white) by this smelting process? How much heat was released or required if 28 g of Sn metal was produced from its ore, assuming complete reaction? ♦ An average American consumes approximately 10 kJ of energy per day. The average life expectancy of an American is 77.9 years. Several theories propose that life on Earth evolved in the absence of oxygen. One theory is that primitive organisms used fermentation processes, in which sugars are decomposed in an oxygen-free environment, to obtain energy. Many kinds of fermentation processes are possible, including the conversion of glucose to lactic acid (a), to CO and ethanol (b), and to ethanol and acetic acid (c): Reaction (a) occurs in rapidly exercising muscle cells, reaction (b) occurs in yeast, and reaction (c) occurs in intestinal bacteria. Using Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C", calculate which reaction gives the greatest energy yield (most negative ) per mole of glucose. A 70 kg person expends 85 Cal/h watching television. If the person eats 8 cups of popcorn that contains 55 Cal per cup, how many kilojoules of energy from the popcorn will have been burned during a 2 h movie? After the movie, the person goes outside to play tennis and burns approximately 500 Cal/h. How long will that person have to play tennis to work off all the residual energy from the popcorn? ♦ Photosynthesis in higher plants is a complex process in which glucose is synthesized from atmospheric carbon dioxide and water in a sequence of reactions that uses light as an energy source. The overall reaction is as follows Glucose may then be used to produce the complex carbohydrates, such as cellulose, that constitute plant tissues. Adipose (fat) tissue consists of cellular protoplasm, which is mostly water, and fat globules. Nearly all the energy stored in adipose tissue comes from the chemical energy of its fat globules, totaling approximately 3500 Cal per pound of tissue. How many kilojoules of energy are stored in 10 g of adipose tissue? How many 50 g brownies would you need to consume to generate 10 lb of fat? (Refer to Table 5.5 for the necessary caloric data.) If a moderate running pace of 5 min/km expends energy at a rate of about 400 kJ/km, how many 8 oz apples would a person have to eat to have enough energy to run 5 mi? How many 4 oz hamburgers? (Refer to Table 5.5 for the necessary caloric data.) Proteins contain approximately 4 Cal/g, carbohydrates approximately 4 Cal/g, and fat approximately 9 Cal/g. How many kilojoules of energy are available from the consumption of one serving (8 oz) of each food in the table? (Data are shown per serving.) When you eat a bowl of cereal with 500 g of milk, how many Calories must your body burn to warm the milk from 4°C to a normal body temperature of 37°C? (Assume milk has the same specific heat as water.) How many Calories are burned warming the same amount of milk in a 32°C bowl of oatmeal from 32°C to normal body temperature? In some countries that experience starvation conditions, it has been found that infants don’t starve even though the milk from their mothers doesn’t contain the number of Calories thought necessary to sustain them. Propose an explanation for this. If a person’s fever is caused by an increase in the temperature of water inside the body, how much additional energy is needed if a 70 kg person with a normal body temperature of 37°C runs a temperature of 39.5°C? (A person’s body is approximately 79% water.) The old adage “feed a fever” may contain some truth in this case. How many 4 oz hamburgers would the person need to consume to cause this change? (Refer to Table 5.5 for the necessary caloric data.) 16. Approximately 810 kJ of energy is needed to evaporate water from the leaves of a 9.2 m tree in one day. What mass of water is evaporated from the tree? 2 Solution 6. Solution Why is it preferable to convert coal to syngas before use rather than burning coal as a solid fuel? What is meant by the term ? List three greenhouse gases that have been implicated in global warming. Name three factors that determine the rate of planetary CO uptake. The structure of coal is quite different from the structure of gasoline. How do their structural differences affect their enthalpies of combustion? Explain your answer. One of the side reactions that occurs during the burning of fossil fuels is How many kilograms of CO are released during the combustion of 16 gal of gasoline? Assume that gasoline is pure isooctane with a density of 0.6919 g/mL. If this combustion was used to heat 4.5 × 10 L of water from an initial temperature of 11.0°C, what would be the final temperature of the water assuming 42% efficiency in the energy transfer? A 60 W light bulb is burned for 6 hours. If we assume an efficiency of 38% in the conversion of energy from oil to electricity, how much oil must be consumed to supply the electrical energy needed to light the bulb? (1 W = 1 J/s) How many liters of cyclohexane must be burned to release as much energy as burning 10.0 lb of pine logs? The density of cyclohexane is 0.7785 g/mL, and its Δ = −46.6 kJ/g. | 7,858 | 3,029 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.01%3A_Prelude_to_Photochemistry |
The role of light in effecting chemical change has been recognized for many years. Indeed, the connection between solar energy and the biosynthesis of plant carbohydrates from carbon dioxide and water was known by the early 1800's. Yet organic photochemistry was slow to develop as a well-understood and manageable science. Progress only became rapid following the development of spectroscopy and spectroscopic techniques for structure determination and the detection of transient species. For this reason photochemistry for many years was the domain of physical and theoretical chemists. Their work laid the foundation for modern organic photochemistry, which correlates the nature of excited electronic states of molecules with the reactions they undergo. Quite apart from the unparalleled importance of photosynthesis, photochemical reactions have a great impact on biology and technology, both good and bad. Vision in all animals is triggered by photochemical reactions. The destructive effects of ultraviolet radiation on all forms of life can be traced to photochemical reactions that alter cellular DNA, and the harmful effects of overexposure to sunlight and the resulting incidence of skin cancer are well established. The technical applications of photochemistry are manifold. The dye industry is based on the fact that many organic compounds absorb particular wavelengths of visible light, and the search for better dyes and pigments around the turn of this century was largely responsible for the development of synthetic organic chemistry. Dye chemistry has helped establish the relationship between chemical structure and color, which also is important in color printing and color photography. We cover these important applications of photochemistry only briefly in this chapter, but we hope to convey some understanding of the fundamentals involved. Most photochemical reactions can be considered to occur in three stages: 1. to produce electronically excited states.
2. involving excited electronic states.
3. whereby the products of the primary photochemical reactions are converted to stable products. We shall begin with a closer look at electronic excitation, some aspects of which were discussed in . Because transfer of electronic energy from one molecule to another is a basic process in photochemistry, we will discuss energy transfer also before giving an overview of representative photochemical reactions. The closely related phenomena of chemiluminescence and bioluminescence then will be described. Finally, there will be a discussion of several important applications of photochemistry. and (1977) | 2,655 | 3,030 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/07%3A_Mixtures_and_Solutions/7.03%3A_Chemical_Potential |
In much the same fashion as the partial molar volume is defined, the is defined for compound \(i\) in a mixture: \[ \mu_i = \left( \dfrac{\partial G}{\partial n_i} \right) _{p,T,n_j\neq i} \label{eq1} \] This particular partial molar function is of particular importance, and is called the . The chemical potential tells how the Gibbs function will change as the composition of the mixture changes. And since systems tend to seek a minimum aggregate Gibbs function, the chemical potential will point to the direction the system can move in order to reduce the total Gibbs function. In general, the total change in the Gibbs function (\(dG\)) can be calculated from \[dG = \left( \dfrac{\partial G}{\partial p} \right) _{T,n_i} dp + \left( \dfrac{\partial G}{\partial T} \right) _{p, n_i }dT + \sum_i \left( \dfrac{\partial G}{\partial n_i} \right) _{T,n_j\neq i} dn_i \nonumber \] Or, by substituting the definition for the chemical potential, and evaluating the pressure and temperature derivatives as was done in Chapter 6: \[dG = Vdp - SdT + \sum_i \mu_i dn_i \nonumber \] But as it turns out, the chemical potential can be defined as the partial molar derivative any of the four major thermodynamic functions \(U\), \(H\), \(A\), or \(G\): The last definition, in which the chemical potential is defined as the partial molar Gibbs function is the most commonly used, and perhaps the most useful (Equation \ref{eq1}). As the partial most Gibbs function, it is easy to show that \[d\mu = Vdp - SdT \nonumber \] where \(V\) is the molar volume, and \(S\) is the molar entropy. Using this expression, it is easy to show that \[\left( \dfrac{\partial \mu}{\partial p} \right) _{T} = V \nonumber \] and so at constant temperature \[ \int_{\mu^o}^{\mu} d\mu = \int_{p^o}^{p} V\,dp \label{eq5} \] So that for a substance for which the molar volume is fairly independent of pressure at constant temperature (i. e., \(\kappa_T\) is very small), therefore Equation \ref{eq5} becomes \[ \int_{\mu^o}^{\mu} d\mu = V \int_{p^o}^{p} dp \nonumber \] \[ \mu - \mu^o = V(p-p^o) \nonumber \] or \[ \mu = \mu^o + V(p-p^o) \nonumber \] Where \(p^o\) is a reference pressure (generally the standard pressure of 1 atm) and \(\mu^o\) is the chemical potential at the standard pressure. If the substance is highly compressible (such as a gas) the pressure dependence of the molar volume is needed to complete the integral. If the substance is an ideal gas \[V =\dfrac{RT}{p} \nonumber \] So at constant temperature, Equation \ref{eq5} then becomes \[ \int_{\mu^o}^{\mu} d\mu = RT int_{p^o}^{p} \dfrac{dp}{p} \label{eq5b} \] or \[ \mu = \mu^o + RT \ln \left(\dfrac{p}{p^o} \right) \nonumber \] | 2,690 | 3,032 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.01%3A_Reaction_Mechanisms |
A is a set of steps, that when taken in aggregate define a chemical pathway that connects reactants to products. An elementary reaction is one that proceeds by a single process, such a molecular (or atomic) decomposition or a molecular collision. Typically, elementary reactions only come in \[A \rightarrow products \nonumber \] and \[A + B \rightarrow products \nonumber \] form. Occasionally, an elementary step that is \[A + B + C \rightarrow products \nonumber \] (involved the simultaneous collision of three atoms or molecules) but it is generally a pair of bimolecular steps acting in rapid succession, the first forming an activated complex, and the second stabilizing that complex chemically or physically. \[A + B \rightarrow AB^* \nonumber \] \[AB^* + C \rightarrow AB + C^* \nonumber \] The wonderful property of elementary reactions is that the defines the order of the rate law for the reaction step. A valid reaction mechanism must satisfy three important criteria: For the reaction \[ A + B \xrightarrow{} C \nonumber \] is the following proposed mechanism valid? \[ A +A \xrightarrow{k_1} A_2 \nonumber \] \[ A_2 + B \xrightarrow{k_1} C + A \nonumber \] Adding both proposed reactions gives \[ \cancel{2}A + \cancel{A_2} + B\xrightarrow{} \cancel{A_2} + C + \cancel{A} \nonumber \] Canceling those species that appear on both sides of the arrow leaves \[ A + B \xrightarrow{} C \nonumber \] which is the reaction, so the mechanism is at least stoichiometrically valid. However, it would still have to be consistent with the observed kinetics for the reaction and account for any side-products that are observed. | 1,652 | 3,033 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/05%3A_Cooperativity/19%3A_Self-Assembly/19.01%3A_Micelle_Formation |
In particular, we will focus on micellar structures formed from a single species of amphiphilic molecule in aqueous solution. These are typically lipids or surfactants that have a charged or polar head group linked to one or more long hydrocarbon chains. Such amphiphiles assemble into a variety of structures, the result of which depends critically on the concentration, composition, and temperature of the system. For SDS surfactant, micelles are favored. These condense hydrophobic chains into a fluid like core and present the charged head groups to the water. The formation of micelles is observed above a (CMC). As the surfactant is dissolved, the solution is primarily monomeric at low concentration, but micelles involving 30–100 molecules suddenly appear for concentrations greater than the CMC. Reprinted from . To begin investigating this phenomenon, we can start by simplifying the equilibrium to a two-state form: \[ nA \rightleftharpoons A_n \] \(K_n\) is the equilibrium constant for assembling a micelle with \(n\) amphiphiles from solution. \(n\) is the called the . \[K_n = \dfrac{[A_n]}{[A]^n} = e^{-\Delta G^0_micelle / k_BT} \label{1}\] The total number of \(A\) molecules present is the sum of the free monomers and those monomers present in micelles: \[CTOT = [A] + n[A_n].\] The fraction of monomers present in micelles: \[ \phi_mi = \dfrac{n[A_n]}{C_{TOT}} = \dfrac{n[A_n]}{[A]+n[A_n]} = \dfrac{nK_n[A]^{n-1}}{1+nK_n[A]^{n-1}} \] This function has an inflection point at the CMC, for which the steepness of the transition increases with \(n\). Setting \(φ_{mi} = 0.5\), we obtain the CMC (\(c_0\)) as \[ c_0 = [A]_{cmc} = (nK_n)^{\dfrac{-1}{n-1}} \] Function steepens with aggregation number \(n\): Thus for large n, and cooperative micelle formation: \[ \Delta G^0_{micelle} = -RT\ln{c_0} \] Note the similarity of Equation \ref{1} to the results for fractional helicity in the helix-coil transition: \[ \dfrac{s^n}{1+s^n}\] This similarity indicates that a cooperative model exists for micelle formation in which the aggregation number reflects the number of cooperative units in the process. Cooperativity can be obtained from models that require surmounting a high nucleation barrier before rapidly adding many more molecules to reach the micelle composition.The simplest description of such a process would proceed in a step-wise growth form (a zipper model) for \(n\) copies of monomer \(A\) assembling into a single micelle \(A_n\). \[ nA \rightleftharpoons A_2 +(n+2)A \rightleftharpoons A_3 +(n-3)A \rightleftharpoons ... \rightleftharpoons A_n \] \[ K_n = \prod_{i=1}^{n-1} K_i \qquad K_i = \dfrac{k_f(i \rightarrow i+1}{k_r(i+1 \rightarrow i} \] Examples of how the energy landscape looks as a function of oligomerization number ν are shown below. However, if you remove the short-range correlation, overall we expect the shape of the energy landscape to still be two-state depending on the nucleation mechanism. This picture is overly simple though, since it is not a one-dimensional chain problem. Rather, we expect that there are equilibira connecting all possible aggregation number clusters to form larger aggregates. A more appropriate description of the free energy barrier for nucleating a micelle is similar to classical nucleation theory for forming a liquid droplet from vapor. ____________________________________________________________ | 3,401 | 3,034 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.07%3A_Osmosis_and_Diffusion |
Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean? Imagine you have a cup that has \(100 \: \text{mL}\) water, and you add \(15 \: \text{g}\) of table sugar to the water. The sugar dissolves and the mixture that is now in the cup is made up of a (the sugar) that is dissolved in the (the water). The mixture of a solute in a solvent is called a . Imagine now that you have a second cup with \(100 \: \text{mL}\) of water, and you add \(45 \: \text{g}\) of table sugar to the water. Just like the first cup, the sugar is the solute, and the water is the solvent. But now you have two mixtures of different solute concentrations. In comparing two solutions of unequal solute concentration, the solution with the higher solute concentration is , and the solution with the lower solute concentration is . Solutions of equal solute concentration are . The first sugar solution is hypotonic to the second solution. The second sugar solution is hypertonic to the first. You now add the two solutions to a beaker that has been divided by a semipermeable membrane, with pores that are too small for the sugar molecules to pass through, but are big enough for the water molecules to pass through. The hypertonic solution is one side of the membrane and the hypotonic solution on the other. The hypertonic solution has a lower water concentration than the hypotonic solution, so a concentration gradient of water now exists across the membrane. Water molecules will move from the side of water concentration to the side of concentration until both solutions are isotonic. At this point, is reached. Red blood cells behave the same way (see figure below). When red blood cells are in a hypertonic (higher concentration) solution, water flows out of the cell faster than it comes in. This results in (shriveling) of the blood cell. On the other extreme, a red blood cell that is hypotonic (lower concentration outside the cell) will result in more water flowing into the cell than out. This results in swelling of the cell and potential (bursting) of the cell. In an isotonic solution, the flow of water in and out of the cell is happening at the same rate. is the diffusion of water molecules across a semipermeable membrane from an area of concentration solution (i.e., higher concentration of water) to an area of concentration solution (i.e., lower concentration of water). Water moves into and out of cells by osmosis. A red blood cell will swell and undergo hemolysis (burst) when placed in a hypotonic solution. When placed in a hypertonic solution, a red blood cell will lose water and undergo (shrivel). Animal cells tend to do best in an isotonic environment, where the flow of water in and out of the cell is occurring at equal rates. is a way that small molecules or ions move across the cell membrane without input of energy by the cell. The three main kinds of passive transport are diffusion (or simple diffusion), osmosis, and facilitated diffusion. Simple diffusion and osmosis do not involve transport proteins. Facilitated diffusion requires the assistance of proteins. is the movement of molecules from an area of high concentration of the molecules to an area with a lower concentration. For cell transport, diffusion is the movement of small molecules across the cell membrane. The difference in the concentrations of the molecules in the two areas is called the . The kinetic energy of the molecules results in random motion, causing diffusion. In simple diffusion, this process proceeds without the aid of a transport protein. It is the random motion of the molecules that causes them to move from an area of high concentration to an area with a lower concentration. Diffusion will continue until the concentration gradient has been eliminated. Since diffusion moves materials from an area of higher concentration to the lower, it is described as moving solutes "down the concentration gradient". The end result is an equal concentration, or , of molecules on both sides of the membrane. At equilibrium, movement of molecules does not stop. At equilibrium, there is equal movement of materials in both directions. Not everything can make it into your cells. Your cells have a plasma membrane that helps to guard your cells from unwanted intruders. If the outside environment of a cell is water-based, and the inside of the cell is also mostly water, something has to make sure the cell stays intact in this environment. What would happen if a cell dissolved in water, like sugar does? Obviously, the cell could not survive in such an environment. So something must protect the cell and allow it to survive in its water-based environment. All cells have a barrier around them that separates them from the environment and from other cells. This barrier is called the , or cell membrane. The plasma membrane (see figure below) is made of a double layer of special lipids, known as . The phospholipid is a lipid molecule with a hydrophilic ("water-loving") head and two hydrophobic ("water-hating") tails. Because of the hydrophilic and hydrophobic nature of the phospholipid, the molecule must be arranged in a specific pattern as only certain parts of the molecule can physically be in contact with water. Remember that there is water outside the cell, and the inside the cell is mostly water as well. So the phospholipids are arranged in a double layer (a bilayer) to keep the cell separate from its environment. Lipids do not mix with water (recall that oil is a lipid), so the phospholipid bilayer of the cell membrane acts as a barrier, keeping water out of the cell, and keeping the cytoplasm inside the cell. The cell membrane allows the cell to stay structurally intact in its water-based environment. The function of the plasma membrane is to control what goes in and out of the cell. Some molecules can go through the cell membrane to enter and leave the cell, but some cannot. The cell is therefore not completely permeable. "Permeable" means that anything can cross a barrier. An open door is completely permeable to anything that wants to enter or exit through the door. The plasma membrane is , meaning that some things can enter the cell, and some things cannot. Molecules that cannot easily pass through the bilayer include ions and small hydrophilic molecules, such as glucose, and macromolecules, including proteins and . Examples of molecules that can easily diffuse across the plasma membrane include carbon dioxide and oxygen gas. These molecules diffuse freely in and out of the cell, along their concentration gradient. Though water is a polar molecule, it can also diffuse through the plasma membrane. The inside of all cells also contain a jelly-like substance called . Cytosol is composed of water and other molecules, including , which are proteins that speed up the cell's chemical reactions. Everything in the cell sits in the cytosol, like fruit in a Jell-o mold. The term cytoplasm refers to the cytosol and all of the organelles, the specialized compartments of the cell. The cytoplasm does not include the nucleus. As a prokaryotic cell does not have a nucleus, the is in the cytoplasm. | 7,530 | 3,035 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.02%3A_Work_and_Heat |
One of the pioneers in the field of modern thermodynamics was James P. Joule (1818 - 1889). Among the experiments Joule carried out, was an attempt to measure the effect on the temperature of a sample of water that was caused by doing work on the water. Using a clever apparatus to perform work on water by using a falling weight to turn paddles within an insulated canister filled with water, Joule was able to measure a temperature increase in the water. Thus, Joule was able to show that work and heat can have the same effect on matter – a change in temperature! It would then be reasonable to conclude that heating, as well as doing work on a system will increase its energy content, and thus it’s ability to perform work in the surroundings. This leads to an important construct of the : The capacity of a system to do work is increased by heating the system or doing work on it. The (U) of a system is a measure of its capacity to supply energy that can do work within the surroundings, making U the ideal variable to keep track of the flow of heat and work energy into and out of a system. Changes in the internal energy of a system (\(\Delta U\)) can be calculated by \[\Delta U = U_f - U_i \label{FirstLaw} \] where the subscripts \(i\) and \(f\) indicate initial and final states of the system. \(U\) as it turns out, is a state variable. In other words, the amount of energy available in a system to be supplied to the surroundings is independent on how that energy came to be available. That’s important because the manner in which energy is transferred is path dependent. There are two main methods energy can be transferred to or from a system. These are suggested in the previous statement of the first law of thermodynamics. Mathematically, we can restate the first law as \[\Delta U = q + w \nonumber \] or \[dU = dq + dw \nonumber \] where q is defined as the amount of energy that flows into a system in the form of and w is the amount of energy lost due to the system doing on the surroundings. Heat is the kind of energy that in the absence of other changes would have the effect of changing the temperature of the system. A process in which heat flows into a system is from the standpoint of the system (\(q_{system} > 0\), \(q_{surroundings} < 0\)). Likewise, a process in which heat flows out of the system (into the surroundings) is called (\(q_{system} < 0\), \(q_{surroundings} > 0\)). In the absence of any energy flow in the form or work, the flow of heat into or out of a system can be measured by a change in temperature. In cases where it is difficult to measure temperature changes of the system directly, the amount of heat energy transferred in a process can be measured using a change in temperature of the soundings. (This concept will be used later in the discussion of calorimetry). An infinitesimal amount of heat flow into or out of a system can be related to a change in temperature by \[dq = C\, dT \nonumber \] where C is the and has the definition \[ C = \dfrac{dq}{\partial T} \nonumber \] Heat capacities generally have units of (J mol K ) and magnitudes equal to the number of J needed to raise the temperature of 1 mol of substance by 1 K. Similar to a heat capacity is a which is defined per unit mass rather than per mol. The specific heat of water, for example, has a value of 4.184 J g K (at constant pressure – a pathway distinction that will be discussed later.) How much energy is needed to raise the temperature of 5.0 g of water from 21.0 °C to 25.0 °C? \[\begin{align*} q &=mC \Delta T \\[4pt] &= (5.0 \,\cancel{g}) (4.184 \dfrac{J}{\cancel{g} \, \cancel{°C}}) (25.0 \cancel{°C} - 21.0 \cancel{°C}) \\[4pt] &= 84\, J \end{align*} \] A partial derivative, like a total derivative, is a slope. It gives a magnitude as to how quickly a function changes value when one of the dependent variables changes. Mathematically, a partial derivative is defined for a function \(f(x_1,x_2, \dots x_n)\) by \[\left( \dfrac{ \partial f}{\partial x_i} \right)_{x_j \neq i} = \lim_{\Delta _i \rightarrow 0} \left( \dfrac{f(x_1+ \Delta x_1 , x_2 + \Delta x_2, \dots, x_i +\Delta x_i, \dots x_n+\Delta x_n) - f(x_1,x_2, \dots x_i, \dots x_n) }{\Delta x_i} \right) \nonumber \] Because it measures how much a function changes for a change in a given dependent variable, infinitesimal changes in the in the function can be described by \[ df = \sum_i \left( \dfrac{\partial f}{\partial x_i} \right)_{x_j \neq i} \nonumber \] So that each contribution to the total change in the function \(f\) can be considered separately. For simplicity, consider an ideal gas. The pressure can be calculated for the gas using the ideal gas law. In this expression, pressure is a function of temperature and molar volume. \[ p(V,T) = \dfrac{RT}{V} \nonumber \] The partial derivatives of p can be expressed in terms of \(T\) and \(V\) as well. \[ \left( \dfrac{\partial p}{ \partial V} \right)_{T} = - \dfrac{RT}{V^2} \label{max1} \] and \[ \left( \dfrac{\partial p}{ \partial T} \right)_{V} = \dfrac{R}{V} \label{max2} \] So that the change in pressure can be expressed \[ dp = \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \label{eq3} \] or by substituting Equations \ref{max1} and \ref{max2} \[ dp = \left( - \dfrac{RT}{V^2} \right ) dV + \left( \dfrac{R}{V} \right) dT \nonumber \] Macroscopic changes can be expressed by integrating the individual pieces of Equation \ref{eq3} over appropriate intervals. \[ \Delta p = \int_{V_1}^{V_2} \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \int_{T_1}^{T_2} \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \nonumber \] This can be thought of as two consecutive changes. The first is an (constant temperature) expansion from \(V_1\) to \(V_2\) at \(T_1\) and the second is an (constant volume) temperature change from \(T_1\) to \(T_2\) at \(V_2\). For example, suppose one needs to calculate the change in pressure for an ideal gas expanding from 1.0 L/mol at 200 K to 3.0 L/mol at 400 K. The set up might look as follows. \[ \Delta p = \underbrace{ \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV}_{\text{isothermal expansion}} + \underbrace{ \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT}_{\text{isochoric heating}} \nonumber \] or \[ \begin{align*} \Delta p &= \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV + \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT \\[4pt] &= \left[ \dfrac{R(200\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} + \left[ \dfrac{RT}{3.0 \, L/mol} \right]_{ 200\,K}^{400\,K} \\[4pt] &= R \left[ \left( \dfrac{200\,K}{3.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{200\,K}{3.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \] Alternatively, one could calculate the change as an isochoric temperature change from \(T_1\) to \(T_2\) at \(V_1\) followed by an isothermal expansion from \(V_1\) to \(V_2\) at \(T_2\): \[ \Delta p = \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT + \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV \nonumber \] \[ \begin{align*} \Delta p &= \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT + \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV \\[4pt] &= \left[ \dfrac{RT}{1.0 \, L/mol} \right]_{ 200\,K}^{400\,K} + \left[ \dfrac{R(400\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} \\[4pt] &= R \left[ \left( \dfrac{400\,K}{1.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{400\,K}{1.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \] Work can take several forms, such as expansion against a resisting pressure, extending length against a resisting tension (like stretching a rubber band), stretching a surface against a surface tension (like stretching a balloon as it inflates) or pushing electrons through a circuit against a resistance. The key to defining the work that flows in a process is to start with an infinitesimal amount of work defined by what is changing in the system. The pattern followed is always an infinitesimal displacement multiplied by a resisting force. The total work can then be determined by integrating along the pathway the change follows. What is the work done by 1.00 mol an ideal gas expanding from a volume of 22.4 L to a volume of 44.8 L against a constant external pressure of 0.500 atm? \[dw = -p_{ext} dV \nonumber \] since the pressure is constant, we can integrate easily to get total work \[ \begin{align*} w &= -p_{exp} \int_{V_1}^{V_2} dV \\[4pt] &= -p_{exp} ( V_2-V_1) \\[4pt] &= -(0.500 \,am)(44.8 \,L - 22.4 \,L) \left(\dfrac{8.314 \,J}{0.08206 \,atm\,L}\right) \\[4pt] &= -1130 \,J = -1.14 \;kJ \end{align*} \] : The ratio of gas law constants can be used to convert between atm∙L and J quite conveniently! | 8,915 | 3,036 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers_II |
Buffer solutions contain a weak acid and its salt or a weak base and its salt. The pH values of these solutions do not change much when a little bit of acid or base is added. On this page, we explore the reasons why the pH of buffer solutions resists change. The blood is a natural buffer, and so are other body fluids and plant fluids due to mixtures of weak acids and bases present in them. Buffer solutions are required for many chemical experiments; they are also useful to standardize pH meters. You have investigated how the pH varies in a strong-acid and strong-base . We illustrate the titration of a weak acid by a strong base using the following examples. During the titration process and before the equivalence point is reached, some acid has been neutralized by the strong base, and the solution contains a weak acid and its salt. The solution acts as a buffer. What is the pH of a M acid solution whose acid dissociation constant is ? (What is the pH of a M weak acid before starting the titration?) Let \(\ce{HA}\) represent the weak acid, and assume M of it is ionized. Then, the ionization and equilibrium concentration is \(\begin{array}{ccccc}
\ce{HA &\rightleftharpoons &H+ &+ &A-}\\
C_{\large\textrm a}-x &&x &&x
\end{array}\) \(K_{\large\textrm a} = \dfrac{x^2}{C_{\large\textrm a}-x}\) \(x^2 + K_{\large\textrm a}x - C_{\large\textrm a}K_{\large\textrm a} = 0\) \(x = \dfrac{-K_{\large\textrm a} + (K_{\large\textrm a}^2 + 4 C_{\large\textrm a}K_{\large\textrm a})^{1/2}}{2}\) \(\ce{pH} = -\log(x)\) The method has been fully discussed in Weak Acids and Bases equilibrium. Symbols are used here, but approximations may be applied to numerical problems. Let us make a buffer solution by mixing mL of acid \(\ce{HA}\) and mL of its salt \(\ce{NaA}\). For simplicity, let us assume both the acid and the salt solutions have the same concentration M. What is the pH of the so prepared buffer solution? The acid dissociation constant is . After mixing, the concentrations and of the acid \(\ce{HA}\) and its salt \(\ce{NaA}\) respectively are \(C_{\large\textrm a} = \dfrac{C\, V_{\large\textrm a}}{V_{\large\textrm a}+V_{\large\textrm s}}\)
\(C_{\large\textrm s} = \dfrac{C\, V_{\large\textrm s}}{V_{\large\textrm a}+V_{\large\textrm s}}\) Assume M of the acid is ionized. Then, the ionization and equilibrium of the acid is shown below, but the salt is completely dissociated. \(\begin{array}{ccccc}
\ce{HA &\rightleftharpoons &H+ &+ &A-}\\
C_{\large\textrm a}-x &&x &&x
\end{array}\) \(\begin{array}{ccccc}
\ce{NaA &\rightarrow &Na+ &+ &A-}\\
&&C_{\large\textrm s} &&C_{\large\textrm s}
\end{array}\) \(\ce{Common\: ion\: [A- ]} = x+C_{\large\textrm s}\) \(K_{\large\textrm a} = \dfrac{x(x+C_{\large\textrm s})}{C_{\large\textrm a}-x}\) \(x^2 + (K_{\large\textrm a}+C_{\large\textrm s})x - C_{\large\textrm a} K_{\large\textrm a} = 0\) \(x = \dfrac{ -(K_{\large\textrm a}+C_{\large\textrm s}) + ((K_{\large\textrm a}+C_{\large\textrm s})^2 + 4 C_{\large\textrm a} K_{\large\textrm a})^{1/2}}{2}\) \(\ce{pH} = -\log(x)\) The formulas for and the pH derived above can be used to estimate the pH of any buffer solution, regardless how little salt or acid is used compared to their counterpart. When the ratio / is between 0.1 and 10, the Henderson-Hasselbalch equation is a convenient formula to use. \(K_{\large\textrm a} = \ce{\dfrac{ [H+] [A- ]}{[HA]}}\) \(\mathrm{p\mathit K_{\large\textrm a} = pH} - \log \left(\dfrac{\ce{[A- ]}}{\ce{[HA]}}\right)\) The Henderson-Hasselbalch equation is \(\begin{align}
\ce{pH} &= \mathrm{p\mathit K_{\large\textrm a}} + \log \left(\dfrac{\ce{[A- ]}}{\ce{[HA]}}\right)\\
&= \mathrm{p\mathit K_{\large\textrm a}} + \log \left(\dfrac{ [C_{\large\textrm s}] + x}{[C_{\large\textrm a}] - x}\right)
\end{align}\) Plot the titration curve when a 10.00 mL sample of 1.00 M weak acid \(\ce{HA}\) ( = 1.0e-5) is titrated with 1.00 M \(\ce{NaOH}\). \(\begin{align}
\ce{[H+]} &= (C_{\large\textrm a}K_{\large\textrm a})^{1/2}\\
&= 0.00316\\
\ce{pH} &= 2.500
\end{align}\) the sharp increase in pH when 0.1 mL (3 drops) of basic solution is added to the solution. \(C_{\large\textrm s} = \mathrm{\dfrac{0.1\times1.0\: M}{10.1} = 0.0099\: M}\)
\(C_{\large\textrm a} = \mathrm{\dfrac{9.9\times1.0\: M}{10.1}} = 0.98\) \(\begin{array}{ccccc}
\ce{HA &\rightleftharpoons &H+ &+ &A-}\\
C_{\large\textrm a}-x &&x &&x
\end{array}\) \(\ce{[A- ]} = x + 0.0099\: (= C_{\large\textrm s})\) \(K_{\large\textrm a} = \dfrac{ x\, (x + 0.0099)}{0.98 - x} = \textrm{1e-5}\) \(x^2 + 0.0099 x = 9.8\textrm{e-}6 - 1\ce e5 x\) \(x^2 + (0.0099 + 1\ce e5)\, x - 9.8\textrm{e-}6 = 0\) \(\begin{align}
x &= \dfrac{-0.0099 + (0.00992 + 4\times 0.98\times \textrm{1e-5})^{1/2}}{2}\\
&= 0.000907
\end{align}\) \(\ce{pH} = 3.042\) that using the Henderson-Hasselbalch equation will not yield the correct solution. Do you know why? \(C_{\large\textrm s} = \mathrm{\dfrac{1.0\times1.0\: M}{11.0} = 0.0901\: M}\)
\(C_{\large\textrm a} = \mathrm{\dfrac{9.0\times1.0\: M}{11.0} = 0.818}\) \(\begin{array}{ccccc}
\ce{HA &\rightleftharpoons &H+ &+ &A-}\\
C_{\large\textrm a}-x &&x &&x+0.0901
\end{array}\) \(K_{\large\textrm a} = \dfrac{x\, (x + 0.0901)}{0.818 - x} = \textrm{1e-5}\) \(\begin{align}
x &= \dfrac{-0.0901 + (0.09012 + 4\times0.818\times1\textrm{e-}5)^{1/2}}{2}\\
&= \textrm{0.0000907 (any approximation to be made?)}
\end{align}\) \(\mathrm{pH = 4.042}\) Using the yields \(\mathrm{pH = p\mathit K_{\large\textrm a}} + \log \left(\dfrac{ 0.0901(=\ce{[A- ]})}{0.818(=\ce{[HA]})}\right) = 5 - 0.958 = \textrm{4.042 (same result)}\) \(C_{\large\textrm s} = \mathrm{\dfrac{5.0\times1.0\: M}{15.0} = 0.333\: M}\)
\(C_{\large\textrm a} = \mathrm{\dfrac{5.0\times1.0\: M}{15.0} = 0.333\: M}\)
\(\begin{array}{ccccc}
\ce{HA &\rightleftharpoons &H+ &+ &A-}\\
C_{\large\textrm a}-x &&x &&x+0.3333
\end{array}\) \(K_{\large\textrm a} = \dfrac{x\, (x + 0.333)}{0.333 - x} = \textrm{1.0e-5}\) \(x^2 + (0.333-1\textrm{e-}5)\,x - 0.333\times\textrm{1e-5} = 0\) \(\begin{align}
x &= \dfrac{-0.333 + (0.3332 + 4\times3.33\textrm{e-}6)^{1/2}}{2}\\
&= 0.0000010
\end{align}\) \(\mathrm{pH = 5.000}\) : Using the Henderson Hasselbalch approximation yields the same result \(\mathrm{pH = p\mathit K_{\large\textrm a}} + \log \left(\dfrac{0.333(=\ce{[A- ]})}{0.333(=\ce{[HA]})}\right) = 5 + 0.000 = 5.000\) \(C_{\large\textrm s} = \mathrm{\dfrac{10.0\times1.0\: M}{20.0} = 0.500\: M}\)
\(C_{\large\textrm a} = \mathrm{\dfrac{0.0\times1.0\: M}{20.0} = 0.000}\) At the equivalence point, the solution contains 0.500 M of the salt \(\ce{NaA}\), and the following equilibrium must be considered: \(\begin{array}{cccccccl}
\ce{A- &+ &H2O &\rightleftharpoons &HA &+ &OH- &}\\
C_{\large\textrm s}-x &&&&x &&x &\mathrm{Equilibrium\:\, concentrations}
\end{array}\) \(\begin{align}
K_{\large\textrm b} &= \dfrac{\ce{[HA] [OH- ]}}{\ce{[A- ]}} \dfrac{\ce{[H+]}}{\ce{[H+]}}\\
&= \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}} = \dfrac{\textrm{1e-14}}{\textrm{1e-5}} = \textrm{1e-9} = \dfrac{x^2}{C_{\large\textrm s}-x}
\end{align}\) \(x = (0.500\times\textrm{1.0e-9})^{1/2} = \textrm{2.26e-5}\) \(\textrm{pOH} = -\log x = 2.651;\: \textrm{pH} = 14 - 2.651 = 11.349\) : The calculation here illustrates the or hydrolysis of the basic salt \(\ce{NaA}\). Sketch the titration curve based on the estimates given in this example, and notice the points made along the way. \(K_{\large\textrm b} = \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}} =\: ?\) What is the pH at the equivalence point when a 0.10 M \(\ce{HF}\) solution is titrated by a 0.10 M \(\ce{NaOH}\) solution? \(\ce{[Na+]} = \ce{[F- ]} = \mathrm{\dfrac{1.0\: mL \times 0.10\: M}{11.0\: mL}}\) | 7,897 | 3,037 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.09%3A_Stoichiometric_Calculations_and_Enthalpy_Changes |
There is a growing concern about damage to the environment done by emissions from manufacturing plants. Many companies are taking steps to reduce these harmful emissions by adding equipment that will trap the pollutants. In order to know what equipment (and the quantity) to order, studies are done to measure the amount of product currently produced. Since pollution is often both particulate and thermal, energy changes need to be determined in addition to the amounts of products released. Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems. Refer again to the combustion reaction of methane. Since the reaction of \(1 \: \text{mol}\) of methane released \(890.4 \: \text{kJ}\), the reaction of \(2 \: \text{mol}\) of methane would release \(2 \times 890.4 \: \text{kJ} = 1781 \: \text{kJ}\). The reaction of \(0.5 \: \text{mol}\) of methane would release \(\frac{890,4 \: \text{kJ}}{2} = 445.2 \: \text{kJ}\). As with other stoichiometry problems, the moles of a reactant or product can be linked to mass or volume. Sulfur dioxide gas reacts with oxygen to form sulfur trioxide in an exothermic reaction, according to the following thermochemical equation. \[2 \ce{SO_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right) + 198 \: \text{kJ}\nonumber \] Calculate the enthalpy change that occurs when \(58.0 \: \text{g}\) of sulfur dioxide is reacted with excess oxygen. The calculation requires two steps. The mass of \(\ce{SO_2}\) is converted to moles. Then the moles of \(\ce{SO_2}\) is multiplied by the conversion factor of \(\left( \frac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} \right)\). \[\Delta H = 58.0 \: \text{g} \: \ce{SO_2} \times \frac{1 \: \text{mol} \: \ce{SO_2}}{64.07 \: \text{g} \: \ce{SO_2}} \times \frac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} = 89.6 \: \text{kJ}\nonumber \] The mass of sulfur dioxide is slightly less than \(1 \: \text{mol}\). Since \(198 \: \text{kJ}\) is released for every \(2 \: \text{mol}\) of \(\ce{SO_2}\) that reacts, the heat released when about \(1 \: \text{mol}\) reacts is one half of 198. The \(89.6 \: \text{kJ}\) is slightly less than half of 198. The sign of \(\Delta H\) is negative because the reaction is exothermic. | 2,314 | 3,038 |
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