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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.10%3A_The_E1_Reaction |
Many secondary and tertiary halides undergo \(\text{E}1\) elimination in competition with the \(\text{S}_\text{N}1\) reaction in neutral or acidic solutions. For example, when -butyl chloride solvolyzes in \(80\%\) aqueous ethanol at \(25^\text{o}\), it gives \(83\%\) -butyl alcohol by substitution and \(17\%\) 2-methylpropene by elimination: The ratio of substitution and elimination remains constant throughout the reaction, which means that each process has the kinetic order with respect to the concentration of -butyl halide. The \(\text{S}_\text{N}1\) and \(\text{E}1\) reactions have a common rate-determining step, namely, slow ionization of the halide. The solvent then has the choice of attacking the intermediate carbocation at the positive carbon to effect substitution, or at a \(\beta\) hydrogen to effect elimination: Factors influencing the \(\text{E}1\) reactions are expected to be similar to those for the \(\text{S}_\text{N}1\) reactions. An ionizing solvent is necessary, and for easy reaction the \(RX\) compound must have a good leaving group and form a relatively stable \(R^\oplus\) cation. Therefore the \(\text{E}1\) orders of reaction rates are \(X = I\) \(>\) \(Br\) \(>\) \(Cl\) \(>\) \(F\) and \(R\) \(>\) \(R\) \(>\) \(R\). With halides such as 2-chloro-2-methylbutane, which can give different alkenes depending on the direction of elimination, the \(\text{E}1\) reaction is like the \(\text{E}2\) reaction in tending to favor the most stable or highly substituted alkene: Another feature of \(\text{E}1\) reactions (and also of \(\text{S}_\text{N}1\) reactions) is the tendency of the initially formed carbocation to rearrange, especially if a more stable carbocation is formed thereby. For example, the very slow \(\text{S}_\text{N}1\) solvolysis of neopentyl iodide in methanoic acid leads predominantly to 2-methyl-2-butene: In this reaction, ionization results in migration of a methyl group with its bonding pair of electrons from the \(\beta\) to the \(\alpha\) carbon, thereby transforming an unstable primary carbocation to a relatively stable tertiary carbocation. Elimination of a proton completes the reaction. Rearrangements involving shifts of hydrogen (as \(H:^\ominus\)) occur with comparable ease if a more stable carbocation can be formed thereby: Rearrangements of carbocations are among the fastest organic reactions known and must be reckoned with as a possibility whenever carbocation intermediates are involved. Alcohols and ethers rarely undergo substitution or elimination unless strong acid is present. As we noted in Section 8-7D the acid is necessary to convert a relatively poor leaving group (\(HO^\ominus\), \(CH_3O^\ominus\)) into a relatively good one (\(H_2O\), \(CH_3OH\)). Thus the dehydration of alcohols to alkenes is an acid-catalyzed reaction requiring strong acids such as sulfuric or phosphoric acid: These are synthetically useful reactions for the preparation of alkenes when the alkene is less available than the alcohol. They can occur by either the \(\text{E}1\) or \(\text{E}2\) mechanism depending on the alcohol, the acid catalyst, the solvent, and the temperature. and (1977) | 3,189 | 647 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/03%3A_Distributions_Probability_and_Expected_Values/3.02%3A_Outcomes_Events_and_Probability |
We also need to introduce the idea that a function that successfully models the results of past experiments can be used to predict some of the characteristics of future results. We reason as follows: We have results from drawing many samples of a random variable from some distribution. We suppose that a mathematical representation has been found that adequately summarizes the results of these experiences. If the underlying distribution—the physical system in scientific applications—remains the same, we expect that a long series of future results would give rise to essentially the same mathematical representation. If 25% of many previous results have had a particular characteristic, we expect that 25% of a large number of future trials will have the same characteristic. We also say that there is one chance in four that the next individual result will have this characteristic; when we say this, we mean that 25% of a large number of future trials will have this characteristic, and the next trial has as good a chance as any other to be among those that do. Given a distribution, the possible outcomes must be mutually exclusive; in any given trial, the random variable can have only one of its possible values. Consequently, a discrete distribution is completely described when the probability of each of its outcomes is specified. Many distributions are comprised of a finite set of mutually exclusive possible outcomes. If each of these outcomes is equally likely, the probability that we will observe any particular outcome in the next trial is \(1/N\). We often find it convenient to group the set of possible outcomes into subsets in such a way that each outcome is in one and only one of the subsets. We say that such assignments of outcomes to subsets are , because every possible outcome is assigned to some subset; we say that such assignments are , because no outcome belongs to more than one subset. We call each such subset an . When we partition the possible outcomes into exhaustive and mutually exclusive events, we can say the same things about the probabilities of events that we can say about the probabilities of outcomes. In our discussions, the term “events” will always refer to an exhaustive and mutually exclusive partitioning of the possible outcomes. Distinguishing between outcomes and events just gives us some language conventions that enable us to create alternative groupings of the same set of real world observations. Suppose that we define a particular event to be a subset of outcomes that we denote as . If in a large number of trials, the fraction of outcomes that belong to this subset is , we say that the probability is that the outcome of the next trial will belong to this event. To express this in more mathematical notation, we write \(P\left(U\right)=F\). When we do so, we mean that the fraction of a large number of future trials that belong to this subset will be , and the next trial has as good a chance as any other to be among those that do. In a sample comprising observations, the best forecast we can make of the number of occurrences of is \(M\times P(U)\), and we call this the of in a sample of size . The idea of grouping real world observations into either outcomes or events is easy to remember if we keep in mind the example of tossing a die. The die has six faces, which are labeled with 1, 2, 3, 4, 5, or 6 dots. The dots distinguish one face from another. On any given toss, one face of the die must land on top. Therefore, there are six possible outcomes. Since each face has as good a chance as any other of landing on top, the six possible outcomes are equally probable. The probability of any given outcome is \({1}/{6}\). If we ask about the probability that the next toss will result in one of the even-numbered faces landing on top, we are asking about the probability of an event—the event that the next toss will have the characteristic that an even-numbered face lands on top. Let us call this event \(X\). That is, event \(X\) occurs if the outcome is a 2, a 4, or a 6. These are three of the six equally likely outcomes. Evidently, the probability of this event is \({3}/{6}={1}/{2}\). Having defined event \(X\) as the probability of an even-number outcome, we still have several alternative ways to assign the odd-number outcomes to events. One assignment would be to say that all of the odd-number outcomes belong to a second event—the event that the outcome is odd. The events “even outcome” and “odd outcome” are exhaustive and mutually exclusive. We could create another set of events by assigning the outcomes 1 and 3 to event \(Y\), and the outcome 5 to event \(Z\). Events \(X\), \(Y\), and \(Z\) are also exhaustive and mutually exclusive. We have a great deal of latitude in the way we assign the possible outcomes to events. If it suits our purposes, we can create many different exhaustive and mutually exclusive partitionings of the outcomes of a given distribution. We require that each partitioning of outcomes into events be exhaustive and mutually exclusive, because we want to apply the laws of probability to events. | 5,157 | 648 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.02%3A_Reaction_Mechanisms/3.2.06%3A_Steady_State_Approximation |
The steady state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction are constant. This method can only be applied when the first step of the reaction is significantly slower than subsequent step in an intermediate-forming consecutive reaction. Before discussing the steady state approximation, it must be understood that the approximation is derived to simplify the kinetic expression for product concentration, [product]. Consider the following sequential reaction: \[A \xrightarrow[]{k_1} B \xrightarrow {k_2} C \nonumber \] Calculating the [product] depends on all the rate constants in each step. For example, if the kinetic method was used to find the concentration of C, [C], at time t in the above reaction, the expression would be \[[C] = [A]_0 \left(1+ \dfrac{k_2e^{-k_1t}-k_1e^{-k_2t})}{k_1 - k_2}\right) \label{1} \] With a more complicated mechanisms, the kinetic expression becomes harder to derive. To simplify this calculation, we often use one of two approximations for determining the overall reaction rates of consecutive reactions: the steady state approximation and the . This article concerns the steady state approximation. The steady state approximation is applies to a consecutive reaction with a slow first step and a fast second step (\(k_1 \ll k_2\)). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product, such as product B in the above example. \[\dfrac{d[B]}{dt} = 0 = k_1[A] - k_2[B] \label{2} \] Thus \[[B] = \dfrac{k_1[A]}{k_2} \label{3} \] From the mechanism: \[\dfrac{d[C]}{dt} = k_2[B] = \dfrac{k_2k_1[A]}{k_2} = k_1[A] \label{4} \] Solving for \([C]\): \[[C] = [A]_0 (1- e^{-k_1t}) \label{5} \] Equation \(\ref{5}\) is much simpler to derive than Equation \(\ref{1}\), especially with a more complicated multi-step reaction mechanisms. Consider the reaction: \[A + 2B \xrightarrow[]{} C \nonumber \] a: First we use the Steady State Approximation for the intermediate (i.e., Equation \ref{2}) \[\dfrac{d[I]}{dt} = k_1[A,B] - k_{-1}[I] - k_2[I,B] = 0 \nonumber \] then we solve for the (steady-state) concentration of the intermediate \[[I] = \dfrac{k_1[A,B]}{ k_{-1} +k_2[B]} \nonumber \] Because the second step is much faster than the first step, then \(k_2 \gg k_{-1}\) then \(k_{-1} \approx 0\) for this approximation and the above equation can be simplified to \[[I] = \dfrac{k_1[A]}{k_2} \nonumber. \nonumber \] The rate law for the production of \([C]\) can be constructed directly from the second step and when the steady-state concentration of \(I\) is added, the final rate law expression is derived. \[ \begin{align*}\dfrac{d[C]}{dt} & = k_2[I,B] \\[4pt] &= \dfrac{k_1k_2[A,B]}{k_2} \\[4pt] &= k_1[A,B] \end{align*} \] b: Direct inspection of the final rate law derived above gives these parameters: In 1925, George E. Briggs and John B. S. Haldane applied the steady state approximation method to determine the rate law of the enzyme-catalyzed reaction (Figure 1). The following assumptions were made: This gives the following: \[\dfrac{d[P]}{dt} = k_2[ES] \label{6} \] where \[\dfrac{d[ES]}{dt} = 0 = k_1[E,S] - k_{-1}[ES] - k_2[ES] \label{7} \] Because \[[S] \gg [E] \label{8} \] Using the second assumption and the fact that enzyme concentration equals the initial concentration of enzyme minus the concentration of the enzyme-substrate intermediate, \[[E] = [E]_o - [ES] \label{9} \] The following equation is obtained: \[k_1[E]_o[S] = k_{-1}[ES] + k_2[ES] + k_1[ES,S] \label{10} \] From this equation, the concentration of the ES intermediate can be found: \[[ES] = \dfrac{k_1[E]_o[S]}{(k_{-1} + k_2) + k_1[S]} \label{11} \] Substitute this into Equation \(\ref{6}\) gives, \[\dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{[(k_1+k_2)/k_1]+[S]} = \dfrac{k_2[E]_0[S]}{K_M+[S]} \label{12} \] where \[K_M = \dfrac{k_{-1}+k_{2}}{k_1} \label{13} \] Because in most of the cases, only the initial \(d[P]/dt\) is measured to determine the rate of product formation, Equation \ref{12} can be rewritten as: \[v_o = \dfrac{d[P]_0}{dt} = \dfrac{k_2[E]_0[S]}{K_M+[S]} \label{14} \] Because \([E]_o = v_{max}/k_2\). Equation \(\ref{14}\) becomes the following: \[ \begin{align} v_0 &= \dfrac{d[P]_o}{dt} \\[4pt] &= \dfrac{(k_2/k_2)v_{max}[S]}{K_M+[S]} \label{15} \\ &= \dfrac{v_{max}[S]}{K_M+[S]} \label{16} \end{align} \] This equation is a useful tool to in calculating \(v_{max}\) and \(K_M\) (the ) of an enzyme by using the (1/[S] vs. 1/v ) or the (v /[S] vs. v ). Given the reaction \(A \xrightarrow[]{k_1} B \xrightarrow[]{k_2} C\) where k = 0.2 M s , k = 2000 s 1) d[A]/dt = -k 2) Because k is much larger than k , this is a steady state reaction. 3) d[C]/dt = k [B] where d[B]/dt = k [A] - k [B] = 0 so, [B] = k [A]/k Substitute this into d[C]/dt d[C]/dt = k [A] 4) d[C]/dt = 0.2M s (1M) = 0.2 s 5) [C] = [A] (1-e ) = 2M(1-e ) = 0.9 M | 5,012 | 649 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Fractional_Distillation_of_Non-ideal_Mixtures_(Azeotropes) |
Remember that a large positive deviation from produces a vapor pressure curve with a maximum value at some composition other than pure A or B. If a mixture has a high vapor pressure it means that it will have a low boiling point. The molecules are escaping easily and you won't have to heat the mixture much to overcome the intermolecular attractions completely. The implication of this is that the boiling point / composition curve will have a minimum value lower than the boiling points of either A or B. In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures. Suppose you are going to distil a mixture of ethanol and water with composition C as shown on the next diagram. It will boil at a temperature given by the liquid curve and produce a vapor with composition C . When that vapor condenses it will, of course, still have the composition C . If you reboil that, it will produce a new vapor with composition C . You can see that if you carried on with this boiling-condensing-reboiling sequence, you would eventually end up with a vapor with a composition of 95.6% ethanol. If you condense that you obviously get a liquid with 95.6% ethanol. What happens if you reboil that liquid? The liquid curve and the vapor curve meet at that point. The vapor produced will have that same composition of 95.6% ethanol. If you condense it again, it will still have that same composition. You have hit a barrier. It is impossible to get pure ethanol by distilling any mixture of ethanol and water containing less than 95.6% of ethanol. This particular mixture of ethanol and water boils as if it were a pure liquid. It has a constant boiling point, and the vapor composition is exactly the same as the liquid. It is known as a constant boiling mixture or or an . The implications of this for of dilute solutions of ethanol are obvious. The liquid collected by condensing the vapor from the top of the fractionating column cannot be pure ethanol. The best you can produce by simple fractional distillation is 95.6% ethanol. What you can get (although it isn't very useful!) from the mixture is pure water. As ethanol rich vapor is given off from the liquid boiling in the distillation flask, it will eventually lose all the ethanol to leave just water. Distilling a mixture of ethanol containing less than 95.6% of ethanol by mass lets you collect: What if you distil a mixture containing more than 95.6% ethanol? Work it out for yourself using the phase diagram, and starting with a composition to the right of the azeotropic mixture. You should find that you get: Nitric acid and water form mixtures in which particles break away to form the vapor with much more difficulty than in either of the pure liquids. You can see this from the vapor pressure / composition curve discussed further up the page. That means that mixtures of nitric acid and water can have boiling points higher than either of the pure liquids because it needs extra heat to break the stronger attractions in the mixture. In the case of mixtures of nitric acid and water, there is a maximum boiling point of 120.5°C when the mixture contains 68% by mass of nitric acid. That compares with the boiling point of pure nitric acid at 86°C, and water at 100°C. Notice the much bigger difference this time due to the presence of the new ionic interactions (see above). The phase diagram looks like this: Distilling dilute nitric acid. Start with a dilute solution of nitric acid with a composition of C and trace through what happens. The vapor produced is richer in water than the original acid. If you condense the vapor and reboil it, the new vapor is even richer in water. Fractional distillation of dilute nitric acid will enable you to collect pure water from the top of the fractionating column. As the acid loses water, it becomes more concentrated. Its concentration gradually increases until it gets to 68% by mass of nitric acid. At that point, the vapor produced has exactly the same concentration as the liquid, because the two curves meet. You produce a constant boiling mixture (or azeotropic mixture or azeotrope) and if you distil dilute nitric acid, that's what you will eventually be left with in the distillation flask. You cannot produce pure nitric acid from the dilute acid by distilling it. You cannot produce pure nitric acid from the dilute acid (<68%) by distilling it. This time you are starting with a concentration C to the right of the azeotropic mixture. The vapor formed is richer in nitric acid. If you condense and reboil this, you will get a still richer vapor. If you continue to do this all the way up the fractionating column, you can get pure nitric acid out of the top. As far as the liquid in the distillation flask is concerned, it is gradually losing nitric acid. Its concentration drifts down towards the azeotropic composition. Once it reaches that, there cannot be any further change, because it then boils to give a vapor with the same composition as the liquid. Distilling a nitric acid / water mixture containing more than 68% by mass of nitric acid gives you pure nitric acid from the top of the fractionating column and the azeotropic mixture left in the distillation flask. You can produce pure nitric acid from the concetrated acid (>68%) by distilling it. Jim Clark ( ) | 5,873 | 651 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Polyprotic_Acids_and_Bases_1 |
According to Brønsted and Lowry an is a proton donor and a base is a proton acceptor. This idea of proton donor and proton acceptor is important in understanding monoprotic and polyprotic acids and bases because monoprotic corresponds to the transfer of one proton and polyprotic refers to the transfer of more than one proton. Therefore, a is an acid that can donate proton, while can donate proton. Similarly, a can only accept proton, while a can accept . One way to display the differences between monoprotic and polyprotic acids and bases is through titration, which clearly depicts the equivalence points and acid or base dissociation constants. The acid dissociation constant, signified by \(K_a\), and the base dissociation constant, \(K_b\), are equilibrium constants for the dissociation of weak acids and weak bases. The larger the value of either \(K_a\) or \(K_b\) signifies a stronger acid or base, respectively. Here is a list of important equations and constants when dealing with \(K_a\) and \(K_b\): For the general equation of a weak acid, \[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \label{1} \] you need to solve for the \(K_a\) value. To do that you use \[K_a = \dfrac{[H_3O^+,A^-]}{[HA]} \label{2} \] Another necessary value is the \(pK_a\) value, and that is obtained through \(pK_a = {-logK_a}\) The procedure is very similar for weak bases. The general equation of a weak base is \[BOH \rightleftharpoons B^+ + OH^- \label{3} \] Solving for the \(K_b\)value is the same as the \(K_a\) value. You use the formula \[K_b = \dfrac{[B^+,OH^-]}{[BOH]} \label{4} \] The \(pK_b\) value is found through \(pK_b = {-logK_b}\) The \(K_w\) value is found with\(K_w = {[H3O^+]}{[OH^-]}\). \[K_w = 1.0 \times 10^{-14} \label{5} \] are acids that can release only one proton per molecule and have one equivalence point. Here is a table of some common monoprotic acids: are bases that can only react with one proton per molecule and similar to monoprotic acids, only have one equivalence point. Here is a list of some common monoprotic bases: What is the pH of the solution that results from the addition of 200 mL of 0.1 M CsOH(aq) to 50 mL of 0.2M HNO (aq)? (pKa= 3.14 for HNO ) \[\dfrac{0.1 mol}{L}*200 mL* \dfrac{1 L}{1000 mL} = {0.02 mol CsOH} \nonumber \] \[\dfrac{0.2 mol}{L}*50 mL* \dfrac{1 L}{1000 mL} = {0.01 mol HNO_2} \nonumber \] Then do an for \[CsOH + HNO_2 \rightleftharpoons H_2O + CsNO_2 \nonumber \] yielding \([CsOH]= [OH^-]= 0.01M\) Then to find pH first we find pOH \(pOH = {-log[OH^-] = -log[\dfrac{0.01}{0.25}] = 1.4}\) Then \(pH = {14 - pOH}\), plugging in pH = 14 - 1.4 = 12.6 So far, we have only considered monoprotic acids and bases, however there are various other substances that can donate or accept more than proton per molecule and these are known as . Polyprotic acids and bases have multiple dissociation constants, such as \(K_{a1}\), \(K_{a2}\), \(K_{a3}\) or \(K_{b1}\), \(K_{b2}\), and \(K_{b3}\), and equivalence points depending on the number of times dissociation occurs. are acids that can lose several protons per molecule. They can be further categorized into and , those which can donate two and three protons, respectively. The best way to demonstrate polyprotic acids and bases is with a titration curve. A titration curve displays the multiple acid dissociation constants (\(K_a\)) as portrayed below. Here is a list of some common polyprotic acids: are bases that can attach several protons per molecule. Similar to polyprotic acids, polyprotic bases can be categorized into and . Here is a list of some common polyprotic bases: For a 4.0 M H PO solution, calculate (a) [H O ] (b) [HPO 2- ] and (c) [PO 3 ]. \[H_3PO_4 + H_2O \rightleftharpoons H_3O^+ + H_2PO_4^- \nonumber \] (a) Using you get: \[K_{a1} = \dfrac{[H_3O^+,H_2PO_4^-]}{[H_3PO_4]} \nonumber \] So, \(x^2\) = .0284 \(x\) = 0.17 M (b) From part (a), \(x\) = [H PO ] = [H O ] = 0.17 M (c) To determine [H O ] and [H PO ], it was assumed that the second ionization constant was insignificant. The new equation is as follows: \(H_2PO_4^- + H_2O \rightleftharpoons H_3O^+ + HPO_4^{2-}\) Using ICE Tables again: \(K_{a2} = [HPO_4^{2-}] = 6.3 \times 10^{-8}\) The polyprotic acid H SO can ionize two times ( \(K_{a1}>>1\), \(K_{a2} = 1.1 * 10^-2\)). If we start with 9.50*10 M solution of H SO , what are the final concentrations of H SO , HSO , SO , and H O . The equation for the first ionization is \(H_2SO_4 + H_2O \rightleftharpoons H_3O^+ + HSO_4^-\). This equation goes to completion because H SO is a strong acid and \(K_{a1}>>1\). So since the reaction goes to completion, doing an you get [H 0 ] = 9.50*10 M and [HSO ] = 9.50*10 M (after the first ionization). The equation of the second ionization is \(HSO_4- + H_2O \rightleftharpoons H_3O^+ + SO_4^2-\). Using the equation \(K_{a2} = \dfrac{[H_3O^+,SO_4^2-^-]}{[HSO_4^-]}\), \(K_{a2} = 1.1 * 10^-2\), and an to get \(x^2 + .0.0205x - 0.0001045 = 0\). Then you use the quadratic equation to solve for X, to get \(x\) = 0.004226. Now we need to solve for the necessary concentrations \([H_2S0_4]\) = 0 (because the first ionization reaction went to completion) \([HS0_4^-]\) = \(k_{a1}\) - \(k_{a2}\) = 9.50*10 M - 0.004226 M = 5.27*10 M \([SO_4^2-]\) = \(k_{a2}\) = .004226 M \([H_3O^+]\) = \(k_{a1}\) + \(k_{a2}\) = 9.50*10 M + 0.004226 M = 1.37*10 M Assuming that the [H 0 ] is the same for all the ionizations. In fact, the pH is dominated by only the first ionization, but the later ionizations do contribute very slightly. | 5,594 | 653 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/19%3A_d-Block_Metal_Chemistry_-_General_Considerations/19.05%3A_Characteristic_Properties_-_A_General_Perspective/19.5B%3A_Paramagnetism |
In CHEM1902 (C10K) we introduced the formula used to relate the magnetic moment to the number of unpaired spins in a transition metal complex. During the laboratory session you will carry out a measurement of the magnetic susceptibility which is a measure of the force exerted by the magnetic field on a unit mass of the sample under investigation. This is related to the number of unpaired electrons per unit weight and hence per mole and in the simplest picture we consider that this is solely dependent on the presence of unpaired electrons.
For a Ti(III) complex with 1 unpaired electron this corresponds to: We will see later that while the spin-only approximation works in many cases, for a more complete analysis it is necessary to consider the contribution made by the orbital motion of the electron as well. | 839 | 654 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/09%3A_Molecular_Geometry_and_Covalent_Bonding_Models/9.04%3A_Delocalized_Bonding_and_Molecular_Orbitals |
delocalized In , we described the electrons in isolated atoms as having certain spatial distributions, called , each with a particular . Just as the positions and energies of electrons in can be described in terms of (AOs), the positions and energies of electrons in can be described in terms of molecular orbitals (MOs) —a spatial distribution of electrons that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory , is a approach to bonding. Molecular orbital theory is a delocalized bonding approach that explains the colors of compounds, their stability, and resonance. Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins. We begin our discussion of molecular orbitals with the simplest molecule, H , formed from two isolated hydrogen atoms, each with a 1 electron configuration. As we explained in electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called . (For more information on wave functions, see ) The 1 atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the of the two H 1 wave functions, and the other produced by taking their :
\( \begin{matrix}
MO(1)= & AO(atom\; A) & +& AO(atomB) \\
MO(1)= & AO(atom\; A) & -&AO(atomB)
\end{matrix} \tag{6.5.1} \) The molecular orbitals created from are called linear combinations of atomic orbitals (LCAOs) . A molecule must have as many molecular orbitals as there are atomic orbitals. Adding two atomic orbitals corresponds to interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is . The molecular orbital corresponding to the sum of the two H 1 orbitals is called a σ combination (pronounced “sigma one ess”) (part (a) and part (b) in ). In a sigma (σ) orbital, , the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1 denotes the atomic orbitals from which the molecular orbital was derived: \( \sigma _{1s}^{*} \) \( \sigma _{1s}^{*} \)
\( \sigma _{1s} \approx 1s\left ( A \right ) + 1s\left ( B \right ) \tag{6.5.2} \) Conversely, subtracting one atomic orbital from another corresponds to interference between two waves, which reduces their intensity and causes a in the internuclear electron probability density (part (c) and part (d) in ). The resulting pattern contains a where the electron density is zero. The molecular orbital corresponding to the difference is called \( \sigma _{1s}^{*} \) (“sigma one ess star”). In a sigma star (σ*) orbital , there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:
\( \sigma _{1s}^{\star } \approx 1s\left ( A \right ) - 1s\left ( B \right ) \tag{6.5.3} \) A molecule must have as many molecular orbitals as there are atomic orbitals. The electron density in the σ molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ orbital represents a bonding molecular orbital. . In contrast, electrons in the \( \sigma _{1s}^{\star } \) orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the \( \sigma _{1s}^{\star } \) orbital is an antibonding molecular orbital . Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not. Because electrons in the σ orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ molecular orbital has a energy than either of the hydrogen 1 atomic orbitals. Conversely, electrons in the \( \sigma _{1s}^{\star } \) orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1 atomic orbitals. Consequently, the \( \sigma _{1s}^{\star } \) molecular orbital has a energy than either of the hydrogen 1 atomic orbitals. The σ (bonding) molecular orbital is relative to the 1 atomic orbitals, and the \( \sigma _{1s}^{\star } \) (antibonding) molecular orbital is . The relative energy levels of these orbitals are shown in the energy-level diagram in A bonding molecular orbital is lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is higher in energy (less stable). To describe the bonding in a homonuclear diatomic molecule such as H , we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram ( ). We fill the orbitals according to the Pauli principle and Hund’s rule: each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds. In the Lewis electron structures described in , the number of electron pairs holding two atoms together was called the . In the molecular orbital approach, bond order is defined as one-half the number of bonding electrons:
\( bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \tag{6.5.4} \) To calculate the bond order of H , we see from that the σ (bonding) molecular orbital contains two electrons, while the \( \sigma _{1s}^{\star } \) (antibonding) molecular orbital is empty. The bond order of H is therefore
\( \dfrac{2-0}{2}=1 \tag{5.3.5} \) This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in to describe the bonding in other pairs of atoms and ions where = 1, such as the H ion, the He ion, and the He molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule. Part (a) in shows the energy-level diagram for the H ion, which contains two protons and only one electron. The single electron occupies the σ bonding molecular orbital, giving a (σ ) electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is (1-0)/2=1/2 Because the bond order is greater than zero, the H ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H . With a bond order of only 1/2 the bond in H should be weaker than in the H molecule, and the H–H bond should be longer. As shown in , these predictions agree with the experimental data. Part (b) in is the molecular orbital energy-level diagram for He . This ion has a total of three valence electrons. Because the first two electrons completely fill the σ molecular orbital, the Pauli principle states that the third electron must be in the \( \sigma _{1s}^{\star} \) antibonding orbital, giving a \( \left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1} \) electron configuration. This electron configuration gives a bond order of (2-1)/2=1/2. As with H , the He ion should be stable, but the He–He bond should be weaker and longer than in H . In fact, the He ion can be prepared, and its properties are consistent with our predictions ( ). Finally, we examine the He molecule, formed from two He atoms with 1 electron configurations. Part (c) in is the molecular orbital energy-level diagram for He . With a total of four valence electrons, both the σ bonding and \( \sigma _{1s}^{\star } \) antibonding orbitals must contain two electrons. This gives a \( \left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1} \) electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He molecule has no net bond and is not a stable species. Experiments show that the He molecule is actually stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions. In molecular orbital theory, . Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H ion. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons. Use a molecular orbital energy-level diagram, such as those in , to predict the bond order in the He ion. Is this a stable species? chemical species molecular orbital energy-level diagram, bond order, and stability Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for the system. Determine the total number of valence electrons in the He ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and predict whether the species is stable. Two He 1 atomic orbitals combine to give two molecular orbitals: a σ bonding orbital at lower energy than the atomic orbitals and a \( \sigma _{1s}^{\star } \) antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram: The He ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He as being formed from two He ions, each of which has a single valence electron in the 1 atomic orbital. We can now fill the molecular orbital diagram: The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ ) orbital, giving a (σ ) electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. So the bond order is \( \frac{2-0}{2} =1 \) He is therefore predicted to contain a single He–He bond. Thus it should be a stable species. Exercise Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H ion. Is this a stable species? H has a valence electron configuration of \( \left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2} \) with a bond order of 0. It is therefore predicted to be unstable. So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals. A molecular orbital diagram that can be applied to any with two identical alkali metal atoms (Li and Cs , for example) is shown in part (a) in , where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σ bonding molecular orbital and a σ antibonding molecular orbital. Because each alkali metal (M) has an valence electron configuration, the M molecule has two valence electrons that fill the σ bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li , Na , K , Rb , and Cs ). The general features of these M diagrams are identical to the diagram for the H molecule in . Experimentally, all are found to be stable in the gas phase, and some are even stable in solution. Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be ), in which each metal atom has an valence electron configuration, resemble the diagram for the He molecule in part (c) in As shown in part (b) in , this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σ bonding orbital and the antibonding orbital and give a bond order of 0. Thus Be , Mg , Ca , Sr , and Ba are all expected to be unstable, in agreement with experimental data. Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na ion. chemical species molecular orbital energy-level diagram, valence electron configuration, bond order, and stability Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system. Determine the total number of valence electrons in the Na ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and predict whether the species is stable. Because sodium has a [Ne]3 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1 atomic orbitals. The Na ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ molecular orbital, a half-filled σ and a \( \left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1} \) electron configuration. The bond order is (2-1) 2=1/2 With a fractional bond order, we predict that the Na ion exists but is highly reactive. Exercise Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca ion. Ca has a \( \left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1} \) electron configurations and a bond order of 1/2 and should exist. Atomic orbitals other than orbitals can also interact to form molecular orbitals. Because individual , , and orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall from that for each subshell, for example, there are , , and orbitals ( ). All have the same energy and are therefore degenerate, but they have different spatial orientations.\( \sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \tag{5.3.6}\) Just as with orbitals, we can form molecular orbitals from orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two atomic orbitals in part (a) in , it is the mathematical of their wave functions that results in interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a \( \sigma _{np_{z}} \) because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the -axis):
\( \sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \tag{6.5.6}\) The other possible combination of the two orbitals is the mathematical sum: \( \sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \tag{6.5.7}\) In this combination, shown in part (b) in , the positive lobe of one atomic orbital overlaps the negative lobe of the other, leading to interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a \( \sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \) . Whenever orbitals combine, (more stable) than the atomic orbitals from which it was derived, and (less stable). Overlap of atomic orbital lobes with the produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals. The remaining orbitals on each of the two atoms, and , do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in , we see that we have two pairs of orbitals: the two orbitals lying in the plane of the page, and two orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the orbital on one atom can interact with only the orbital on the other, and the orbital on one atom can interact with only the on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital . The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital .
\( \pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \tag{6.5.8}\)
\( \pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \tag{6.5.9}\) The two orbitals can also combine using side-to-side interactions to produce a bonding \( \pi _{np_{y}} \) molecular orbital and an antibonding \( \pi _{np_{y}}^{\star } \) molecular orbital. Because the and atomic orbitals interact in the same way (side-to-side) and have the same energy, the \( \pi _{np_{x}} \) and \( \pi _{np_{y}} \)molecular orbitals are a degenerate pair, as are the \( \pi _{np_{x}}^{\star } \) and \( \pi _{np_{y}}^{\star } \) molecular orbitals. is an energy-level diagram that can be applied to two identical interacting atoms that have three atomic orbitals each. There are six degenerate atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals. Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an atomic orbital on one atom with an atomic orbital on another. As shown in , the sum of the two atomic wave functions ( + ) produces a σ bonding molecular orbital. Their difference ( − ) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N , O , and F . When we draw a molecular orbital diagram for a molecule, there are four key points to remember: The number of molecular orbitals is always equal to the total number of atomic orbitals we started with. We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F . We use the diagram in part (a) in ; the = 1 orbitals (σ and σ ) are located well below those of the = 2 level and are not shown. As illustrated in the diagram, the σ and σ molecular orbitals are much lower in energy than the molecular orbitals derived from the 2 atomic orbitals because of the large difference in energy between the 2 and 2 atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2 orbitals on each F is \( \sigma _{2p_{z}} \) and the next most stable are the two degenerate orbitals, \( \pi _{2p_{x}} \) and \( \pi _{2p_{y}} \) For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the \( \sigma ^{\star }_{2p_{z}} \) orbital is higher in energy than either of the degenerate \( \pi _{2p_{x}}^{\star } \) and \( \pi _{2p_{y}}^{\star } \) orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy. Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ and σ orbitals, 2 fill the \( \sigma _{2p_{z}} \) orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F is predicted to have a stable F–F single bond, in agreement with experimental data. We now turn to a molecular orbital description of the bonding in O . It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O , we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in . We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ and σ orbitals, 2 more to fill the \( \sigma _{2p_{z}} \) orbital, and 4 to fill the degenerate \( \pi _{2p_{x}}^{\star } \) and \( \pi _{2p_{y}}^{\star } \) orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data: the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K. \( \sigma ^{\star }_{2p_{z}} \) \( \left ( \pi ^{\star }_{np_{x}},\; \pi ^{\star }_{np_{y}} \right ) \) None of the other bonding models can predict the presence of two unpaired electrons in O . Chemists had long wondered why, unlike most other substances, liquid O is attracted into a magnetic field. As shown in , it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed. Because the O molecule has two unpaired electrons, it is paramagnetic. Consequently, it is attracted into a magnetic field, which allows it to remain suspended between the poles of a powerful magnet until it evaporates. The magnetic properties of O are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H O, CO , and N in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H O, CO , and N , have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O with organic compounds to give H O, CO , and N would require that at least one of the electrons on O change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a . Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke! For period 2 diatomic molecules to the left of N in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the \( \sigma _{2p_{z}} \) molecular orbital is slightly in energy than the degenerate \( \pi ^{\star }_{np_{x}} \) and \( \pi ^{\star }_{np_{y}} \) orbitals. The difference in energy between the 2 and 2 atomic orbitals increases from Li to F due to increasing nuclear charge and poor screening of the 2 electrons by electrons in the 2 subshell. The bonding interaction between the 2 orbital on one atom and the 2 orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ orbital and increases the energy of the \( \sigma _{2p_{z}} \) orbital. Thus for Li , Be , B , C , and N , the \( \sigma _{2p_{z}} \) orbital is higher in energy than the \( \sigma _{3p_{z}} \) orbitals, as shown in Experimentally, it is found that the energy gap between the and atomic orbitals as the nuclear charge increases ( ). Thus for example, the \( \sigma _{2p_{z}} \) molecular orbital is at a lower energy than the \( \pi _{2p_{x,y}} \) pair. \( \sigma _{2p_{z}} \) \( \pi _{2p_{x,y}} \) \( \sigma _{2p_{z}} \) \( \pi _{2p_{x,y}} \) Completing the diagram for N in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N versus 141.2 pm in F ), and the bond energy is much greater for N than for F (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N bond is much shorter and stronger than the F bond, consistent with what we would expect when comparing a triple bond with a single bond. Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S , a bright blue gas at high temperatures. chemical species molecular orbital energy-level diagram, bond order, and number of unpaired electrons Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S . Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S . Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule. Calculate the bond order and describe the bonding. Sulfur has a [Ne]3 3 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in and , we need to know how close in energy the 3 and 3 atomic orbitals are because their energy separation will determine whether the \( \pi _{3p_{x,y}} \) or the \( \sigma _{3p_{z}} \) molecular orbital is higher in energy. Because the – energy gap as the nuclear charge increases ( ), the \( \sigma _{3p_{z}} \) molecular orbital will be lower in energy than the \( \pi _{3p_{x,y}} \) pair. The molecular orbital energy-level diagram is as follows: Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through \( \pi _{3p_{x}} \) and \( \pi _{3p_{y}} \), leaving 2 electrons to occupy the degenerate \( \pi ^{\star }_{3p_{x}} \) and \( \pi ^{\star }_{3p_{y}} \) pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S is \( \left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2} \) with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond. Exercise Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O ). \( \left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4} \) bond order of 1; no unpaired electrons Diatomic molecules with two different atoms are called heteronuclear diatomic molecules . When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χ > χ ), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in . The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond. The bonding molecular orbitals are closer in energy to the atomic orbitals of the more electronegative B atom. Consequently, the electrons in the bonding orbitals are not shared equally between the two atoms. On average, they are closer to the B atom, resulting in a polar covalent bond. A molecular orbital energy-level diagram is always skewed toward the more electronegative atom. Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O with N at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O to produce NO , which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals. Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO ( ) shows that the general pattern is similar to that for the O molecule (see ). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2 atomic orbitals, the 11th electron must occupy one of the degenerate π orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N and O molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot. \( \left ( \pi ^{\star }_{2p_{x}},\; \pi ^{\star }_{2p_{y}} \right ) \) Molecular orbital theory can also tell us something about the of NO. As indicated in the energy-level diagram in , NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, NO is easily oxidized to the NO cation, which is isoelectronic with N and has a bond order of 3, corresponding to an N≡O triple bond. Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from that the 1 orbital of atomic hydrogen is closest in energy to the 3 orbitals of chlorine. Consequently, the filled Cl 3 atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1 and Cl 3 orbitals. Of the three orbitals, only one, designated as 3 , can interact with the H 1 orbital. The 3 and 3 atomic orbitals have no net overlap with the 1 orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3 , 3 , and 3 orbitals do not change when HCl forms, they are called nonbonding molecular orbitals . A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3 than to the H 1 atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give \( H^{\delta +} -- Cl^{\delta -} \) Electrons in nonbonding molecular orbitals have no effect on bond order. Use a “skewed” molecular orbital energy-level diagram like the one in to describe the bonding in the cyanide ion (CN ). What is the bond order? chemical species “skewed” molecular orbital energy-level diagram, bonding description, and bond order Calculate the total number of valence electrons in CN . Then place these electrons in a molecular orbital energy-level diagram like in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and describe the bonding in CN . The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like fills the five lowest-energy orbitals, as shown here: Because χ > χ , the atomic orbitals of N (on the right) are lower in energy than those of C. The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N . Exercise Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl ). What is the bond order? All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1. Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods. A is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the of bonding is a approach. Molecular orbitals are constructed using , which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals. A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely contains at least one node perpendicular to the internuclear axis. A (bonding) or a (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a (bonding) and a (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane. The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an . The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the , defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist. Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered , molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called , using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry. What is the distinction between an atomic orbital and a molecular orbital? How many electrons can a molecular orbital accommodate? Why is the molecular orbital approach to bonding called a approach? How is the energy of an electron affected by interacting with more than one positively charged atomic nucleus at a time? Does the energy of the system increase, decrease, or remain unchanged? Why? Constructive and destructive interference of waves can be used to understand how bonding and antibonding molecular orbitals are formed from atomic orbitals. Does constructive interference of waves result in increased or decreased electron probability density between the nuclei? Is the result of constructive interference best described as a bonding molecular orbital or an antibonding molecular orbital? What is a “node” in molecular orbital theory? How is it similar to the nodes found in atomic orbitals? What is the difference between an orbital and a σ orbital? How are the two similar? Why is a σ molecular orbital lower in energy than the two atomic orbitals from which it is derived? Why is a σ* molecular orbital higher in energy than the two atomic orbitals from which it is derived? What is meant by the term in molecular orbital theory? How is the bond order determined from molecular orbital theory different from the bond order obtained using Lewis electron structures? How is it similar? What is the effect of placing an electron in an antibonding orbital on the bond order, the stability of the molecule, and the reactivity of a molecule? How can the molecular orbital approach to bonding be used to predict a molecule’s stability? What advantages does this method have over the Lewis electron-pair approach to bonding? What is the relationship between bond length and bond order? What effect do antibonding electrons have on bond length? on bond strength? Draw a diagram that illustrates how atomic orbitals can form both σ and π molecular orbitals. Which type of molecular orbital typically results in a stronger bond? What is the minimum number of nodes in σ, π, σ*, and π*? How are the nodes in bonding orbitals different from the nodes in antibonding orbitals? It is possible to form both σ and π molecular orbitals with the overlap of a orbital with a orbital, yet it is possible to form only σ molecular orbitals between and orbitals. Illustrate why this is so with a diagram showing the three types of overlap between this set of orbitals. Include a fourth image that shows why and orbitals cannot combine to form a π molecular orbital. Is it possible for an orbital on one atom to interact with an orbital on another atom to produce molecular orbitals? Why or why not? Can the same be said of and orbitals on adjacent atoms? What is meant by in molecular orbital theory? Is it possible for σ molecular orbitals to form a degenerate pair? Explain your answer. Why are bonding molecular orbitals lower in energy than the parent atomic orbitals? Why are antibonding molecular orbitals higher in energy than the parent atomic orbitals? What is meant by the ? Atomic orbitals on different atoms have different energies. When atomic orbitals from nonidentical atoms are combined to form molecular orbitals, what is the effect of this difference in energy on the resulting molecular orbitals? If two atomic orbitals have different energies, how does this affect the orbital overlap and the molecular orbitals formed by combining the atomic orbitals? Are the Al–Cl bonds in AlCl stronger, the same strength, or weaker than the Al–Br bonds in AlBr ? Why? Are the Ga–Cl bonds in GaCl stronger, the same strength, or weaker than the Sb–Cl bonds in SbCl ? Why? What is meant by a molecular orbital, and how is it formed? How does the energy of a nonbonding orbital compare with the energy of bonding or antibonding molecular orbitals derived from the same atomic orbitals? Many features of molecular orbital theory have analogs in Lewis electron structures. How do Lewis electron structures represent How does electron screening affect the energy difference between the 2 and 2 atomic orbitals of the period 2 elements? How does the energy difference between the 2 and 2 atomic orbitals depend on the effective nuclear charge? For σ versus π, π versus σ*, and σ* versus π*, which of the resulting molecular orbitals is lower in energy? The energy of a σ molecular orbital is usually lower than the energy of a π molecular orbital derived from the same set of atomic orbitals. Under specific conditions, however, the order can be reversed. What causes this reversal? In which portion of the periodic table is this kind of orbital energy reversal most likely to be observed? Is the \( \sigma _{2p_{z}} \) molecular orbital stabilized or destabilized by interaction with the molecular orbital in N ? in O ? In which molecule is this interaction most important? Explain how the Lewis electron-pair approach and molecular orbital theory differ in their treatment of bonding in O . Why is it crucial to our existence that O is paramagnetic? Will NO or CO react more quickly with O ? Explain your answer. How is the energy-level diagram of a heteronuclear diatomic molecule, such as CO, different from that of a homonuclear diatomic molecule, such as N ? How does molecular orbital theory describe the existence of polar bonds? How is this apparent in the molecular orbital diagram of HCl? An atomic orbital is a region of space around an atom that has a non-zero probability for an electron with a particular energy. Analogously, a molecular orbital is a region of space in a molecule that has a non-zero probability for an electron with a particular energy. Both an atomic orbital and a molecular orbital can contain two electrons. No. Because an orbital on one atom is perpendicular to an orbital on an adjacent atom, the net overlap between the two is zero. This is also true for and orbitals on adjacent atoms. Use a qualitative molecular orbital energy-level diagram to describe the bonding in S . What is the bond order? How many unpaired electrons does it have? Use a qualitative molecular orbital energy-level diagram to describe the bonding in F . What is the bond order? How many unpaired electrons does it have? If three atomic orbitals combine to form molecular orbitals, how many molecular orbitals are generated? How many molecular orbitals result from the combination of four atomic orbitals? From five? If two atoms interact to form a bond, and each atom has four atomic orbitals, how many molecular orbitals will form? Sketch the possible ways of combining two 1 orbitals on adjacent atoms. How many molecular orbitals can be formed by this combination? Be sure to indicate any nodal planes. Sketch the possible ways of combining two 2 orbitals on adjacent atoms. How many molecular orbitals can be formed by this combination? Be sure to indicate any nodal planes. If a diatomic molecule has a bond order of 2 and six bonding electrons, how many antibonding electrons must it have? What would be the corresponding Lewis electron structure (disregarding lone pairs)? What would be the effect of a one-electron reduction on the bond distance? What is the bond order of a diatomic molecule with six bonding electrons and no antibonding electrons? If an analogous diatomic molecule has six bonding electrons and four antibonding electrons, which has the stronger bond? the shorter bond distance? If the highest occupied molecular orbital in both molecules is bonding, how will a one-electron oxidation affect the bond length? Qualitatively discuss how the bond distance in a diatomic molecule would be affected by adding an electron to Explain why the oxidation of O decreases the bond distance, whereas the oxidation of N the N–N distance. Could Lewis electron structures be employed to answer this problem? Draw a molecular orbital energy-level diagram for Na . What is the bond order in this ion? Is this ion likely to be a stable species? If not, would you recommend an oxidation or a reduction to improve stability? Explain your answer. Based on your answers, will Na , Na , or Na be the most stable? Why? Draw a molecular orbital energy-level diagram for Xe , showing only the valence orbitals and electrons. What is the bond order in this ion? Is this ion likely to be a stable species? If not, would you recommend an oxidation or a reduction to improve stability? Explain your answer. Based on your answers, will Xe , Xe , or Xe be most stable? Why? Draw a molecular orbital energy-level diagram for O and predict its valence electron configuration, bond order, and stability. Draw a molecular orbital energy-level diagram for C and predict its valence electron configuration, bond order, and stability. If all the orbitals in the valence shells of two atoms interact, how many molecular orbitals are formed? Why is it not possible to form three π orbitals (and the corresponding antibonding orbitals) from the set of six orbitals? Draw a complete energy-level diagram for B . Determine the bond order and whether the molecule is paramagnetic or diamagnetic. Explain your rationale for the order of the molecular orbitals. Sketch a molecular orbital energy-level diagram for each ion. Based on your diagram, what is the bond order of each species? The diatomic molecule BN has never been detected. Assume that its molecular orbital diagram would be similar to that shown for CN in but that the ( \sigma _{2p_{z}} \) molecular orbital is higher in energy than the \( \pi _{2p_{x,y}} \) molecular orbitals. Of the species BN, CO, C , and N , which are isoelectronic? Of the species CN , NO , B , and O , which are isoelectronic? The bond order is 1, and the ion has no unpaired electrons. The number of molecular orbitals is always equal to the number of atomic orbitals you start with. Thus, combining three atomic orbitals gives three molecular orbitals, and combining four or five atomic orbitals will give four or five molecular orbitals, respectively. Combining two atomic orbitals gives two molecular orbitals, a σ (bonding) orbital with no nodal planes, and a σ* (antibonding) orbital with a nodal plane perpendicular to the internuclear axis. Sodium contains only a single valence electron in its 3 atomic orbital. Combining two 3 atomic orbitals gives two molecular orbitals; as shown in the diagram, these are a σ (bonding) orbital and a σ* (antibonding) orbital. Although each sodium atom contributes one valence electron, the +1 charge indicates that one electron has been removed. Placing the single electron in the lowest energy molecular orbital gives a \( \sigma ^{1}_{3s} \) electronic configuration and a bond order of 0.5. Consequently, Na should be a stable species. Oxidizing Na by one electron to give Na would remove the electron in the σ molecular orbital, giving a bond order of 0. Conversely, reducing Na by one electron to give Na would put an additional electron into the σ molecular orbital, giving a bond order of 1. Thus, reduction to Na would produce a more stable species than oxidation to Na . The Na ion would have two electrons in the bonding σ molecular orbital and one electron in the σ antibonding molecular orbital, giving a bond order of 0.5. Thus, Na is the most stable of the three species. BN and C are isoelectronic, with 12 valence electrons, while N and CO are isoelectronic, with 14 valence electrons. by Harvard Natural Sciences Lecture Demos on YouTube | 53,211 | 656 |
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A is defined as a homogeneous mixture of two or more components existing in a single phase. In this description, the focus will be on liquid solutions because within the realm of biology and chemistry, liquid solutions play an important role in multiple processes. Without the existence of solutions, a cell would not be able to carry out and other signaling cascades necessary for cell growth and development. Chemists, therefore, have studied the processes involved in solution chemistry in order to further the understanding of the solution chemistry in nature. The mixing of solutions is driven by , opposed to being driven by . While an by definition does not have interactions between particles, an ideal solution assumes there are interactions. Without the interactions, the solution would not be in a liquid phase. Rather, ideal solutions are defined as having an enthalpy of mixing or enthalpy of solution equal to zero (ΔH or ΔH = 0). This is because the interactions between two liquids, A-B, is the average of the A-A interactions and the B-B interactions. In an ideal solution the average A-A and B-B interactions are identical so there is no difference between the average A-B interactions and the A-A/B-B interactions. Since in biology and chemistry the average interactions between A and B are not always equivalent to the interactions of A or B alone, the enthalpy of mixing is not zero. Consequently, a new term is used to describe the of molecules in solution. , \(a_1\), is the effective concentration that takes into account the deviation from ideal behavior, with the activity of an ideal solution equal to one. An activity coefficient, \( \gamma_1\), is utilized to convert from the solute’s mole fraction, \(x_1\), (as a unit of concentration, mole fraction can be calculated from other concentration units like molarity, molality, or percent by weight) to activity, \(a_1\). \[ a_1=\gamma_1x_1 \tag{1}\] The Debye-Hückel formula is used to calculate the activity coefficient. \[ \log \gamma_\pm = - \dfrac{1.824 \times 10^6} { \left( \epsilon T \right)^{3/2}} | z_+ z_- | \sqrt I \tag{2}\] This form of the Debye-Hückel equation is used if the solvent is water at 298 K. \[ \log \gamma_\pm = - 0.509 | z_+ z_- | \sqrt I \tag{3}\] Consider a solution of 0.01 M MgCl (aq) with an ionic strength of 0.030 M. What is the mean activity coefficient? \( \log \gamma_\pm = - \displaystyle \frac{1.824 \times 10^6} { \left( \epsilon T \right)^{3/2}} | z_+ z_- | \sqrt I \) \( \log \gamma_\pm = - 0.509 | z_+ z_- | \sqrt I \) \( \log \gamma_\pm = - \displaystyle \frac{1.824 \times 10^6} { \left( 78.54 \cdot 298 \mathrm {K} \right)^{3/2}} | 2 \cdot 1 | \sqrt {0.0030} \) \( \log \gamma_\pm = - 0.509 | 2 \cdot 1 | \sqrt {0.0030} \; m \) \( \gamma_\pm = 0.67 \) \( \gamma_\pm = 0.67 \) | 2,830 | 657 |
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The rate at which molecules collide which is the frequency of collisions is called the , Z, which has units of collisions per unit of time. Given a box of molecules A and B, the collision frequency between molecules A and B is defined by the following equation: \[Z=N_{A}N_{B}\sigma_{AB}\sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}} \nonumber \] where: In order for a successful collision to occur, the reactant molecules must collide with enough kinetic energy to break original bonds and form new bonds to become the product molecules. This energy is called the activation energy for the reaction; it is also often referred to as the energy barrier. The fraction of collisions with enough energy to overcome the activation barrier is given by: where: The fraction of successful collisions is directly proportional to the temperature and inversely proportional to the activation energy. The rate constant of the gas-phase reaction is proportional to the product of the and the . As stated above, sufficient kinetic energy is required for a successful reaction; however, they must also collide properly (see below). Compare the following equation to the : where The collision frequency equation can thus be given as follows: \[Z = N_A\rho_{AB} \sqrt{\frac{8k_BT}{\pi\mu_{AB}}} \nonumber \] where: Reactions with more complex reactants and greater needs for collision orientation specificity, such as that below, have smaller steric factors (and therefore lower chances of success). \[H_2 + C_2H_4 \rightarrow C_2H_6 \nonumber \] The opposite holds true for simpler reactions; they have a relatively larger steric factors: \[H^. + H^. \rightarrow H_2 \nonumber \] The collision theory is used to predict the reaction rates for gas-phase reactions. It is a rough approximation due to the complications of the steric factor; furthermore, some of the assumptions do not hold in practical scenarios. For example, in real life molecules are not perfect spheres. Finally, the concepts of collision frequency can be applied in the laboratory: Although the collision theory deals with gas-phase reactions, its concepts can also be applied to reactions that take place in solvents; however, the properties of the solvents (for example: solvent cage) will affect the rate of reactions. Ultimately, collision theory illustrates how reactions occur; it can be used to approximate the rate constants of reactions, and its concepts can be directly applied in the laboratory. | 2,473 | 658 |
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\(S\) can be written as \(S(U,\bar{X})\), where \(\bar{X}\) is a vector of all independent internal extensive variable (e.g. all but one U , and all other X ). Because is monotonic in and continuous, we can invert to \(U(S,\bar{X})\). This relation is fully equivalent to the fundamental relation. Because of the shape of \(S(U,\bar{X})\) or \(U(S,\bar{X})\), as shown in the figure, maximizing the entropy at constant U is equivalent to minimizing the energy at constant S. This is the familiar version from mechanics, where system properties are usually formulated in terms of energies, instead of entropies. Working for now with \(U\) for a simple system, \(U(S,\vec{X}\) we can write \[du=\left(\dfrac{\partial U}{\partial S}\right)_{\vec{X}}dS + \left(\dfrac{\partial U}{\partial \vec{X}}\right)_S \cdot d\vec{X}\] with appropriate constraints n each \(\dfrac{\partial U}{\partial X_i}\) derivative \[dU = TdS + \vec{I} \cdot d\vec{X}\] where \(T \equiv \left( \dfrac{\partial U}{\partial S} \right)_{\vec{X}} > 0\) by . By construction, \(T\) and the \(\{I\}\) are intensive variables. For example, \[ U \rightarrow \lambda U \] and \[ S \rightarrow \lambda S \Rightarrow T \rightarrow \left( \dfrac{\partial \lambda U}{\partial \lambda S} \right) =T\] We consider in detail the properties of the energy derivative \(T\), and then briefly by analogy other intensive variables \(\{I_i\}\). Let all the \(\vec{X}\) (such as \(dV\), \(dn_i\), \(dM\), etc.) equal to zero: no mechanical macroscopic variables are being altered except for energy. It then follows that \(dU = dq\) because \(dw = 0\). Therefore \[dq= TdS\] for small (quasistatic) charges in heat, the change in system entropy is linearly proportional to the heat increment. Thus as we add energy to the uncontrollable degrees of freedom of our system, entropy increases, in accord with the notion that entropy is disorder. Furthermore, we can rewrite this as \[ dS = \dfrac{dq}{T}.\] When \(T\) is larger, the entropy increases less for a given heat input. What is this quantity \(T\)? Consider a closed composite system \(\{S\}\) of two subsystems \(\{S_1\}\) and \(\{S_2\}\) separated by a diathermal wall. A diathermal wall allows only heat flow, so \(dX=0\) again. At equilibrium, \[ dS = 0 = \left( \dfrac{\partial S_1}{\partial U_1} \right)_x dU_1 + \left(\dfrac{\partial S_2}{\partial U_2} \right)_x dU_2\] according to P2 or \[dS = \dfrac{1}{T_1} dU_1 + \dfrac{1}{T_2} dU_2.\] But \(dU = 0\) for a closed system by P1, from which follows that \[dU_2 = -dU_1\] or \[ dS = \left ( \dfrac{1}{T_2} - \dfrac{1}{T_2} \right) dU_1.\] At equilibrium, \(dS = 0\) for any variation of \(dU_1\), which can only be true if \[ \left ( \dfrac{1}{T_2} - \dfrac{1}{T_2} \right) = 0 \Rightarrow T_1=T_2\] Thus, \(T\) . This is the most straightforward definition of temperature: the thing that becomes equal when heat stops flowing from one place to another. We can thus identify the intensive variable as the temperature of the system. Temperature is always guaranteed to be positive by P3 because entropy is a monotonically increasing function of energy. Finally, if \(T = (\partial U/ \partial S)_X\), we can rewrite the third postulate as \[ \lim_{T \rightarrow 0} S =0,\] more commonly known as the As all the energy is removed from a system by lowering its temperature, the system becomes completely ordered. It is worth noting that there are systems (glasses), where reaching this limit takes an inordinate amount of time. A very general principle of quantum mechanics guarantees that the third law holds even in those cases, if we can actually get the system to equilibrium: a coordinate or spin Hamiltonian always has a single groundstate of \(A_1\) symmetry. This is the state any system reaches as \(T \rightarrow 0\). In practice, this state may just not be reachable even approximately in glasses, and heuristic replacements of the third law have been developed for this case, which is really a non-equilibrium case. \[\Delta S _{closed} > 0\] always by postulate P2 \[ds = \dfrac{dq}{T}\] by P2 for a quasistatic process when no work is done \[T>0\] always by postulate P3 \[T_1=T_2\] for two systems in thermal equilibrium \[\lim _{T \rightarrow 0} S =0\] always by P3, difficult to reach even approximately in some cases Thus \(T\) and \(S\) have all the intuitive characteristics of temperature and disorder, and we can take them as representing temperature and disorder. The latter can be justified even more deeply by making use of statistical mechanics in later chapters, where the second postulate follows from microscopic properties of the system. A note on units: \(TS\) must have units of energy. It would be convenient to let \(T\) have units of energy (as an “energy per unit size of the system”) and to let \(S\) be unitless, but for historical reasons, \(T\) has arbitrary units of Kelvin and S has units of Joules/Kelvin to compensate. The more complex a composite system becomes, the more extensive variables it requires beyond \(U\), leading to additional intensive variables. For example: \(V\) (volume) leads to an energy change \[ dU_v = \left( \dfrac{\partial U}{\partial V} \right)_{\vec{X}} dV \equiv -PdV.\] The intensive derivative is called the pressure of the system. \(PV\) has units of Joules, so \(P\) must have units of Joules/m or N/m . Thus \(P\) certainly has the units we normally associate with pressure, or force per unit area. Usually \( \partial U/ \partial V < 0\) because squeezing a system increases its energy. Thus \(P\) is generally a positive quantity, again in accord with our intuition. Note however that there is no postulate that says \(P\) must be positive. In fact, we can bring systems to negative pressure by pulling on the system, or putting tension on it. Is \(P\) is in fact pressure? It is easy to see that it is, by applying the minimum energy principle to a flexible wall, in analogy to what was done for temperature above: \[ dU=0 = dU_1 +dU_2 \] by Postulate 1 \[ dU= T_1dS_1 - P_!dV_1 + T_2dS_2 -P_2dV_2\] by Postulate 2' \[ dU = (T_1-T_2)dS_1 - (P_1 - P_2) dV_1\] In the third line, we assume a system and process, so \(dV = 0\) and \(dS = 0\). When the energy has reached equilibrium, the equation must hold for any small perturbation of the entropy or volume of subsystem 1, which can only be satisfied if \(T_1 = T_2\) (again), and \(P_1 = P_2\). Thus, . This is the most straightforward definition of pressure: the thing that is equalized between two systems when the volume can change to whatever it wants. \(P\) is a pressure, not just in units, but agrees with our intuitive notion of what a pressure should be. \(A\) (area in surface system) \[\Rightarrow dU_A = \left( \dfrac{\partial U}{\partial A} \right)_X dA = -\Gamma dA\] where \(\Gamma\) has units of (N/m) and is therefore the surface tension. \(M\) (magnetization) \[ \Rightarrow dU_M = \left(\dfrac{\partial U}{\partial M}\right)_{\vec{X}} dM \equiv HdM \] where \(H\) is the externally applied magnetic field. \(n_i\) (mole number) \[ \Rightarrow dU_{n_i} = \left(\dfrac{\partial U}{\partial n_i}\right)_{\vec{X}} dn_1 \equiv \mu_i dn_i \] where \(\mu_i\) is the chemical potential equalized when particles are allowed to flow. \(L\) (length) \[ \Rightarrow dU_L = \left(\dfrac{\partial U}{\partial L}\right)_{\vec{X}} dL \equiv FdL \] where \(F\) is the linear tension force. Many more conjugate pairs of extensive and intensive variables are possible, but this gives the general picture. For an arbitrary variation in \(U\) we have \[ dU = TdS + \vec{I} \cdot d\vec{X},\] where \( \vec{I}\) is the vector of all intensive variables except temperature. Often, we will use \[ dU = TdS - PdV + \mu dn\] as an example, when dealing with a simple 3-dimensional 1-component system. Consider for a closed system. Because is extensive, \(S(\lambda U,\lambda \vec{X}) = \lambda S(U, \vec{X})\). This agrees with the intuitive notion that 2 identical disordered systems amount to twice as much disorder as a single one. Similarly, \( U(\lambda S,\lambda \vec{X}) = \lambda U(S, \vec{X})\). Differentiating both sides with respect to \(\lambda\) yields \[ \left( \dfrac{\partial U}{\partial \lambda S}\right)_{\vec{X}} \left( \dfrac{\partial \lambda S}{\partial \lambda }\right) + \left( \dfrac{\partial U}{\partial \lambda \vec{X}}\right)_{\vec{X}} \cdot \left( \dfrac{\partial \lambda \vec{X}}{\partial \lambda}\right) = U(S,\vec{X}) \] or \[ \left( \dfrac{\partial U}{\partial \lambda S}\right)_{\vec{X}} S + \left( \dfrac{\partial U}{\partial \lambda \vec{X}}\right)_{\vec{X}} \cdot \vec{X} =U(S,\vec{X})\] When \(\lambda = 1\), this yields \[ \left( \dfrac{\partial U}{\partial S}\right)_{\vec{X}} S + \left( \dfrac{\partial U}{\partial \vec{X}}\right)_{\vec{X}} \cdot \vec{X} = U\] or \[ U=TS + \vec{I} \cdot \vec{X}\] Thus the energy has a surprisingly simple form: it is simply a bilinear function of the intensive and extensive parameters; it is known as . The formula for energy looks like the formula for \(dU\) with the \(dS\) removed. For example, \( U=TS-PV + \mu n\) for a simple one-component system. Solving for \(S\) yields an analogous formula in the entropy representation, \[ S=\left( \dfrac{1}{T} \right) U - \left(\dfrac{\vec{I}}{T} \right) \cdot \vec{X}\] for example \[ S=\left( \dfrac{1}{T} \right) U - \left(\dfrac{P}{T} \right) V - \dfrac{\mu}{T} n \] The entropy is also a of its intensive and extensive parameters. The differential of \(U\) combined with first order homogeneity requires that not all intensive parameters be independent. For a completely arbitrary variation of \(U\), \[dU =TdS + SdT + \vec{I} \cdot d\vec{X} + \vec{X} \cdot d\vec{I}\] But we know from earlier that \[dU =TdS + \vec{I} \cdot d\vec{X} \Rightarrow SdT + \vec{X} \cdot d\vec{I} = 0\] Using this , one intensive parameter can be expressed in terms of the others. For example, consider a simple multicomponent system: \[ U = TS-PV \sum_{i=1}^r \mu_in_i \Rightarrow SdT - VdP + \sum_{i=1}^r n_id\mu_i\] \[\Rightarrow d\mu_i = \left( \dfrac{V}{n_1}\right) dP- \left( \dfrac{S}{n_1}\right) dT - \sum_{i=2}^r {\dfrac{n_i}{n_1} d\mu_i}\] One chemical potential change can be expressed in terms of pressure, temperature, and the other chemical potentials. In general, an \(r\)-component simple 3-D system has only \(2 + (r-1) = r+1\) degrees of freedom. This will be useful for multi-phase systems. For example, let two phases of the same substance be at equilibrium, and particle flow is allowed from one phase to another. Then \(\mu_1 = \mu_2\) (or particles would flow to the phase of lower chemical potential according to 3.), and to remain at equilibrium when the chemical potential changes, \(d\mu_1 = d\mu_2\). Combining the Gibbs-Duhem relations for each phase, \[ S_1dT - V_1dP =-d\mu_1 \] and \[S_2dT - V_2dP =-d\mu_2\] \[ \overset{d\mu_1=d\mu_2} {\longrightarrow} (S_1-S_2)dT=(V_1-V_2)dP\] or \[ \dfrac{dP}{dT}=\dfrac{\Delta S_{12}}{\Delta V_{12}}\] Thus letting \(d\mu_1 = d\mu_2\) traces out the , \( conditions where the two phases are at equilibrium. This is known as the equation. Often we do not know the fundamental equation \(U(S,\vec{X})\) or \(S(U,\vec{X})\) instead we know equation involving intensive variables, known as . For example, \[ U = U(S,X) \Rightarrow T=\left(\dfrac{\partial U}{\partial S}\right)_X=T(S,X).\] Similarly, the derivative with respect to any other \(X\) yields the corresponding equation of state \(I(S, X)\). These are called , and express one intensive variable in terms of all the extensive variables. There are as many equations of state as there are extensive variables for the system (e.g. \(r+2\) for a simple \(r\)-component system). Note that an equation of state does not contain the same amount of information as the original fundamental relation; it can be integrated up to a constant that depends on all extensive variables except the one involved in the derivative, but that part of (or \(S\)), if we derive equations of state from \(S(U, X)\) cannot simply be left out. If all the equations of state in normal form are known, we can reconstruct the fundamental relation by using the Euler form from 4 \[ U= T(S,\vec{X})S+\vec{I}(S,\vec{X})\cdot \vec{X}\] this is also solvable for \(S\) because of P3. If they are not known in normal form, we may also be able to obtain the fundamental relation by integrating a differential form, such as \[ dS =\left(\dfrac{1}{T} \right) dU - \left(\dfrac{I}{T}\right) \cdot d\vec{X}.\] If needed, we can compute one intensive variable from the Gibbs-Duhem relation, so we need one less equation of state (only \(r+1\) for a simple \(r\)-component system) to evaluate the fundamental relation. Finally, equations of state may also be substituted into one another, yielding equations that depend on more than one intensive variable. These are also referred to as equations of state, but they are not in normal form. Let us consider two examples of how to determine a fundamental relation. We start with the fundamental relation for a rubber band, where we can write down reasonable guesses for both equations of state needed. \[ dU = TdS + FdL \Rightarrow ds = \dfrac{1}{T} dU - \dfrac{F}{T} dL\] We need equations of state so \(T\) and \(F\) can be eliminated to yield \(S(U,L)\): a) \(F=c_1T(L-L_0)\); \(L_0\) is the relaxed length of the rubber band, and we are treating it like a linear spring once stretched. An unusual feature is that \(F\) increases with \(T\). At higher T polymer chains wrinkle into more random coils, causing shrinkage, and increasing the tension for the same length. b) \(U=c_2L_0T\), as long as \(F\) depends only linearly on \(T \Rightarrow F/T=F(L)\) only. The reason is that \[ \dfrac{\partial^2 S(U,L)}{\partial U \partial L} =\dfrac{\partial}{\partial U} \left( \dfrac{-F}{T} \right) = \dfrac{\partial}{\partial L} \left( \dfrac{1}{T} \right) =0 \] so \(\dfrac{1}{T}\) can be any single-valued function of \(U\) as long as it is independent of \(L\); for simplicity we pick \(U \sim T\), as for an ideal gas. We can now insert the two equations of state into the differential form, and integrate it \[ dS = \dfrac{c_2L_o}{U} dU -c_1(L-L_0)dL \Rightarrow\] \[ S=S_0 + c_2L_0 \ln \dfrac{U}{U_0} - \dfrac{c_1}{2} (L-L_0)^2\] The constant can be determined by invoking the third law. However, note that this can lead to singularities if the equations of state themselves are not correct at low temperature, as is the case in this example. Moreover, note that \(c_2\) most be intensive, and \(c_1^{-1}\) must be extensive so that \( is extensive. From the fundamental relation we can calculate any desired properties of the rubber band. Alternatively, we could try to obtain the fundamental relation in terms of \(U = TS + FL\), but then we would need \(T(S,L)\) and \( ( , )\) instead of \(\frac{1}{T}(U,L)\) and \(\frac{F}{T}(U,L)\), which were not available. Similarly, to plug into \(S = U/T – FL/T\), we would need \(T(U, L)\) and \(F(U, L)\); we have the former, but not the latter: the equation of state for \(F=c_1T(L-L_0)\) is in terms of another intensive variables, and not in the basic form required for the Euler form. Note that plugging \[T = U/c_2L_0\] into \(F(T, L)\) to get a \(F(U, L)\) will not help either because this does not yield an equation of state as it would have been obtained by taking the derivative of \(S\). As another example, consider the fundamental relation for an ideal monatomic gas. In this case, we will derive one of the equations of state from the others, get all three equations of state in normal form, before inserting all three to obtain the fundamental relation. The gas has 1 component, so we need \(r+1=2\) equations of state to get started: \[Pv-Rt \tag{1}\] \[u=\dfrac{3}{2}RT \tag{2}\] Here the two well-known equations of state for an ideal gas are written in terms of intensive variables \(u = U/n\) and \(V = V/n\). Again the first equation depends on two intensive variables and is not in standard form. We can bring both equations into standard form as follows: \[ P=\dfrac{R}{v}T=\dfrac{R}{v} \left(\dfrac{2u}{3R}\right) = \dfrac{2}{3}\dfrac{u}{v} = -\left( \dfrac{\partial u}{\partial v} \right)_{S,n} \tag{1'}\] \[ T=\dfrac{2u}{3R} = \left( \dfrac{\partial u}{\partial S}\right)_{V,n} \tag{2'}\] We now need \(\mu(u,n)\) as the third equation of state. Proceeding with the Gibbs-Duhem relation, \[ d\mu = -Sdt +vdP.\] We must eliminate \(S\) since we formulated \(P\) and \(T\) as a function of \(U\), not \(S\). Using the bilinear form of \(S\), \[ d\mu = -\left( \dfrac{u}{T} + \dfrac{Pv}{T} -\dfrac{\mu}{T} \right) dT + vdP.\] Next we eliminate \(P\) and \(T\) by using equations 1’ and 2’: \[d\mu = -d\mu -\dfrac{2}{3} du + \mu \dfrac{du}{u} + \dfrac{2}{3}du - \dfrac{2}{3}u\dfrac{dv}{v}.\] We then divide by on both sides, rearrange, and integrate: \[ \dfrac{d\mu}{u}- \mu \dfrac{du}{u^2} = d\left(\dfrac{\mu}{u}\right) = -\dfrac{du}{u} - \dfrac{2}{3} \dfrac{dv}{d}\] \[ \int_0^{final} d\left(\dfrac{\mu}{u}\right) = \left(\dfrac{\mu}{u}\right)-\left(\dfrac{\mu}{u}\right)_0= -\ln \dfrac{u}{u_0}-\dfrac{2}{3}\ln \dfrac{v}{v_0}\] or \[ \mu = -u \ln \dfrac{u}{u_0} - \dfrac{2}{3} u \ln \dfrac{v}{v_0} + u\left(\dfrac{\mu}{u}\right)_0\] This is the third equation of state, for the chemical potential. We now have all intensive parameters as normal form equations of state, to construct the fundamental relations \(s(u,n)\) or \(u(s, v)\). (Of course, the homogeneous first order property means that to get \(S\) and \(U\), we just multiply by \(n\).) Doing \(s\), for example, \[ s = \dfrac{1}{T} u + \dfrac{P}{T} v - \dfrac{\mu}{T}\] \[ = \dfrac{3R}{2u} u + \dfrac{2u}{3v}\dfrac{3R}{2u} v - \dfrac{3R}{2u} u \left\{ -\ln \dfrac{u}{u_0} - \dfrac{2}{3} \ln \dfrac{v}{v_0} + \left( \dfrac{\mu}{u}\right)_0 \right\}\] \[ = \dfrac{5}{2}R - \dfrac{3}{2} R \left(\dfrac{\mu}{u} \right)_0 +\dfrac{3}{2}R \ln \dfrac{u}{u_0} + R\ln \dfrac{v}{v_0}\] \[ =\dfrac{3}{2}R\ln u + R\ln v +c\] Note that this equation of state violates Postulate 3: \[\left(\dfrac{\partial U}{\partial S} \right)_V=T=\dfrac{2u}{3R}\] so \( T \rightarrow 0 \equiv u \rightarrow 0\); but \(S \rightarrow 0\) as \(T \rightarrow 0\), it approaches \(-\infty\). Thus, either \(PV=nRT\), or \(U=\dfrac{3}{2} nRT\), or both must be high-temperature approximations that fail as \(T \rightarrow 0\). At low \(T\), excluded volume effects, particle interaction, and quantum effects come into play. The ideal gas equation would have to be replaced by a more accurate equation, such as the van der Waals equation to satisfy the third law closer to \(T = 0\). In that sense, thermodynamics can point out to us when approximate equations of state break down. The first derivatives (intensive parameters) are very useful because they correspond to quantities that are equalized among equilibrated subsystems. However, the first order relationship \(dS=0\), although necessary by Postulate 2 at equilibrium, is not sufficient. The extremum in \(S\) must be a : \[ d^2S < 0\] or according to Postulate 1: \[d^2U > 0\] Extrema with \[ d^2S > 0\] or \[ d^2S = 0\] are also possible (minima, saddles, degenerate points). However, thermodynamics cannot make statements about such points without some further assumptions that go beyond the postulates. This suggests that the study of second derivative will be fruitful, to ensure that one is working near a stable equilibrium point. Three of these second derivatives encountered later are \[ \alpha =\dfrac{1}{V} \left(\dfrac{\partial V}{\partial T}\right)_{P,n_i}\] \[\kappa = -\dfrac{1}{V} \left( \dfrac{\partial V}{\partial P} \right)_T\] \[ c_p = \dfrac{T}{n} \left ( \dfrac{\partial S}{\partial T} \right)_P = \left( \dfrac{dq}{dT} \right)_P\] For a simple system, only three second derivatives are linearly independent if we exclude ones based on \(\dfrac{\partial }{\partial n_i}\). The reason is that the terms in the energy \(U = TS – PV + …\) have only three second derivatives, or \(d \mu\) is not a perfect differential. Rather than picking those three, we will usually work with the first independent set, corresponding to quantities with more obvious physical interpretations to chemists working at constant pressure and temperature. We consider the corresponding fundamental relations in the next chapter. | 20,357 | 663 |
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Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a (or, simply, complex), a species formed between a central metal ion and one or more surrounding , molecules or ions that contain at least one lone pair of electrons, such as the [Al(H O) ] ion. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu or Ru , have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions. As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu ion {[Cu(H O) ] }. Because it is a stronger base than H O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH ) (H O) ] ion. Formation of the [Cu(NH ) (H O) ] complex is accompanied by a dramatic color change, as shown in \(\Page {1}\). The solution changes from the light blue of [Cu(H O) ] to the blue-violet characteristic of the [Cu(NH ) (H O) ] ion. Figure \(\Page {1}\): The Formation of Complex Ions. An aqueous solution of \(\ce{CuSO4}\) consists of hydrated Cu ions in the form of pale blue [Cu(H O) ] (left). The addition of aqueous ammonia to the solution results in the formation of the intensely blue-violet [Cu(NH ) (H O) ] ions, usually written as [Cu(NH ) ] ion (right) because ammonia, a stronger base than H O, replaces water molecules from the hydrated Cu ion. For a more complete description, see www.youtube.com/watch?v=IQNcLH6OZK0. The replacement of water molecules from [Cu(H O) ] by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu for simplicity, we can write the equilibrium reactions as follows: The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu ion contains six H O ligands, but the complex ion that is produced contains only four \(\ce{NH_3}\) ligands, not six. \[Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{17.3.2}\] The equilibrium constant for the formation of the complex ion from the hydrated ion is called the . The equilibrium constant expression for has the same general form as any other equilibrium constant expression. In this case, the expression is as follows: The formation constant ( ) has the same general form as any other equilibrium constant expression. Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu (aq) ion is represented as Cu for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, ), the more stable the product. With = 2.1 × 10 , the [Cu(NH ) (H O) ] complex ion is very stable. The formation constants for some common complex ions are listed in . If 12.5 g of \(\ce{Cu(NO3)2•6H2O}\) is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu (aq)? mass of Cu salt and volume and concentration of ammonia solution equilibrium concentration of Cu (aq) Adding an ionic compound that contains Cu to an aqueous ammonia solution will result in the formation of [Cu(NH ) ] (aq), as shown in . We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible. The initial concentration of Cu from the amount of added copper nitrate prior to any reaction is as follows: \[12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}\] Because the stoichiometry of the reaction is four NH to one Cu , the amount of NH required to react completely with the Cu is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 10 ), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH ) ] in the first step is complete and allow some of it to dissociate into Cu and NH until equilibrium has been reached. If we define as the amount of Cu produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH ) ] is − , and the change in the concentration of ammonia is +4 , as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations. \[\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber\] Substituting the final concentrations into the expression for the formation constant ( ) and assuming that << 0.0846, which allows us to remove from the sum and difference, \[\begin{align*}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}},\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \\x&=2.1\times10^{-14}\end{align*}\] The value of indicates that our assumption was justified. The equilibrium concentration of Cu (aq) in a 1.00 M ammonia solution is therefore 2.1 × 10 M. The ferrocyanide ion {[Fe(CN) ] } is very stable, with a of 1 × 10 . Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K [Fe(CN) ]. 2 × 10 M What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a of 5.35 × 10 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt. The reaction for the dissolution of silver bromide is as follows: \[AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{17.3.4a}\] with \[K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{17.3.4b}\] The equilibrium lies far to the left, and the equilibrium concentrations of Ag and Br ions are very low (7.31 × 10 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S O ). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate. The reaction of Ag with thiosulfate is as follows: \[Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{17.3.5a}\] with \[K_f = 2.9 \times 10^{13} \label{17.3.5b}\] The magnitude of the equilibrium constant indicates that almost all Ag ions in solution will be immediately complexed by thiosulfate to form [Ag(S O ) ] . We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together: \[\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13}
\\ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13}
\\ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{17.3.6}\] Comparing with shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 10 . The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled. If a complex ion has a large , the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts. Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl ). Calculate the solubility of AgCl in each situation: At 25°C, = 1.77 × 10 for AgCl and = 1.1 × 10 for AgCl . of AgCl, of AgCl , and KCl concentration solubility of AgCl in water and in KCl solution with and without the formation of complex ions Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10 M. If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water. \(\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10}
\\ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5}
\\ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}\) If we let equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl ] = and [Cl ] = 1.0 − . Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that << 1.0, That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10 M solution of the AgCl complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 10 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.) Calculate the solubility of mercury(II) iodide (HgI ) in each situation: = 2.9 × 10 for HgI and = 6.8 × 10 for [HgI ] . 1.9 × 10 M 1.4 M Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca and Mg , which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O POPO , or P O ) or triphosphate (P O ) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large: with However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research. Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na , but it contains fewer dissolved minerals. Figure \(\Page {2}\): An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues. Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately ( ). One of the most important metal ions for this application is Gd , which with seven unpaired electrons is highly paramagnetic. Because Gd (aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA (diethylene triamine pentaacetic acid), whose fully protonated form is shown here. The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large . A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant ( ). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | 15,208 | 664 |
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Bond-order usually predicted from the Molecular Orbital Theory . Molecular Orbital Theory (MOT) was first proposed by Friedrich Hund and Robert Mulliken in 1933. They developed an approach to covalent bond formation which is based upon the effects of the various electron fields upon each other and which e In this article text based learning approaches have been highlighted by innovative and time economic way to enhance interest of students’ who belong to paranoia zone in chemical bonding. In this pedagogical survey, I have tried to hub one time economic pedagogy by including four (04) new formulae in the field of chemical education. This article explores the results and gives implications for context based teaching, learning and assessment in a time economic way. First of all we classify the molecules or ions into the following four sets based on total number of electrons present in them. ; [Where n = Total no of electrons] Eg. H (Total electrons = 2), Therefore B.O. = n/2 = 2/2 = 1 where n = Total no of electrons, ‘I I’ indicates Mod function i.e. the value of bond order is always positive] Eg. Li (5electrons) Therefore B.O. = I 4-5 I / 2 = 1/2 = 0.5. Eg: CO (Total electrons = 6+8=14), Therefore B.O.= I 8-14 I / 2 = 3 [Where n = Total no of electrons] Eg. NO (Total electrons = 15), Therefore B.O. = 20-15/2 = 2.5 The graphical representation presented in Fig. 1 shows that bond-order gradually increases to 1 in the range (0-2) electrons then falls to zero in the range (2-4) electrons then it further rises to 1 for (4-6) electrons and once again falls to zero for (6-8) electrons then again rises to 3 in the range (8-14) electrons and then finally falls to zero for (14-20) electrons. For total no of electrons 2, 6 and 14, we may use multiple formulae, because they fall in the overlapping region in which they intersect with each other. | 1,883 | 665 |
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The Spin Quantum Number (\(m_s\)) describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -). Each orbital can only hold two electrons. One electron will have a +1/2 spin and the other will have a -1/2 spin. Electrons like to fill orbitals before they start to pair up. Therefore the first electron in an orbital will have a spin of +1/2. After all the orbitals are half filled, the electrons start to pair up. This second electron in the orbital will have a spin of -1/2. If there are two electrons in the same orbital, it will spin in opposite directions. What is the spin quantum number for Tungsten (symbol W)? Tungsten has 4 electrons in the 5d orbital. Therefore 1 electron will go into each orbital (no pairing). The 4th electron will have a +1/2 spin. What is the spin quantum number for Gold (symbol Au)? Gold has 9 electrons in the 5d orbital. Therefore the electrons will start to pair up, which means the 9th electron will pair up, giving it a -1/2 spin. What is the spin quantum number for Sulfur (symbol S)? Sulfur has 4 electrons in the 3p orbitals. The 4th electron in this orbital will be the first one to pair up with another electron, therefore giving it a -1/2 spin. | 1,401 | 666 |
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A solution is a homogeneous mixture of two or more substances and can either be in the gas phase, the liquid phase, the solid phase. The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This (\(ΔH_{solution}\)) can either be positive (endothermic) or negative (exothermic). When understanding the enthalpy of solution, it is easiest to think of a hypothetical three-step process happening between two substances. One substance is the solute, let’s call that . The other substance is the solvent, let’s call that . The first process that happens deals only with the solute, which requires breaking all intramolecular forces holding it together. This means the solute molecules separate from each other. The enthalpy of this process is called \(ΔH_1\). This since this is always an process (requiring energy to break interactions), then \(ΔH_1 > 0\). \[ \ce{A (s) ->[\text{energy in}] A (g)} \nonumber \] The second process is very similar to the first step. Much like how the solute, , needed to break apart from itself, the solvent, , also needs to overcome the intermolecular forces holding it together. This causes the solvent molecules separate from each other. The enthalpy of this process is called \(ΔH_2\). Like the first step, this reaction is always endothermic (\(ΔH_2 > 0\)) because energy is required to break the interaction between the molecules. \[ \ce{B (l) ->[\text{energy in}] B (g)} \nonumber \] At this point, let us visualize what has happened so far. The solute, , has broken from the intermolecular forces holding it together and the solvent, , has broken from the intermolecular forces holding it together as well. It is at this time that the third process happens. We also have two values \(ΔH_1\) and \(ΔH_2\). that are both greater than zero ( ). The third process is when substance and substance mix to for a solution. The separated solute molecules and the separated solvent molecules join together to form a solution. This solution will contain one mole of the solute in an infinite amount of the solvent .The enthalpy of combining these two substances to form the solution is \(ΔH_3\) and is an reaction (releasing heat since interactions are formed) with \(ΔH_3 < 0\). \[ \ce{A (g) + B (g) ->[\text{energy out}] A(sol)} \nonumber \] The enthalpy of solution can expressed as the sum of enthalpy changes for each step: \[ΔH_{solution} = ΔH_1 + ΔH_2 + ΔH_3. \label{eq1}\] So the enthalpy of solution can either be endothermic, exothermic or neither \(ΔH_{solution} = 0\)), depending on how much heat is required or release in each step. If \(ΔH_{solution} = 0\), then the solution is called an ideal solution and if \(ΔH_{solution} > 0\) or \(ΔH_{solution} < 0\), then these solutions are called non-ideal solutions. The diagrams below can be used as visuals to help facilitate the understanding of this concept. Figure \(\Page {1}\) is for an endothermic reaction, where \(ΔH_{solution} > 0.\) Figure \(\Page {2}\) is for an exothermic reaction, where \(ΔH_{solution} < 0\). Figure \(\Page {3}\) is for an ideal solution, where \(ΔH_{solution} = 0\). The enthalpy of solution depends on the strengths of intermolecular forces of the solute and solvent and solvent (Equation \ref{eq1}). If the solution is ideal, and \(ΔH_{solution} = 0\), then \[\begin{align*} ΔH_{solution} = ΔH_1 + ΔH_2 + ΔH_3 &= 0. \label{eq2} \\[4pt] ΔH_1 + ΔH_2 &= - ΔH_3 \end{align*}\] This means the forces of attraction between like (the solute-solute and the solvent-solvent) and unlike (solute-solvent) molecules are the same (Figure \(\Page {3}\)). If the solution is non-ideal, then either \(ΔH_1\) added to \(ΔH_2\) is greater than \(ΔH_3\) or \(ΔH_3\) is greater than the sum of \(ΔH_1\) and \(ΔH_2\). The first case means the forces of attraction of unlike molecules is greater than the forces of attraction between like molecules. The second case means the forces of attraction between like molecules is greater than the forces of attraction between unlike molecules (Figure \(\Page {2}\)). Table salt (\(\ce{NaCl}\)) dissolves readily in water. In solid (\(\ce{NaCl}\)), the positive sodium ions are attracted to the negative chloride ions. The same is true of the solvent, water; the partially positive hydrogen atoms are attracted to the partially negative oxygen atoms. While (\(\ce{NaCl}\)) dissolves in water, the positive sodium cations and chloride anions are stabilized by the water molecule electric dipoles. Thus, the intermolecular interactions (i.e., ionic bonds) between (\(\ce{NaCl}\)) are broken and the salt is dissolved. The overall chemical equation for this reaction is as follows: \[\ce{NaCl (s) ->[H_2O] Na^+ (aq) + Cl^- (aq)}\] Enthalpy of solution is only one part of the driving force in the formation of solutions; the other part is the . | 4,907 | 667 |
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Vanadium takes its name from the Scandinavian goddess Vanadis and was discovered in 1801 by Andrés Manuel del Rio. It was isolated in 1867 by Henry Roscoe as a silvery-white metal that is somewhat heavier than aluminum but lighter than iron. It has excellent corrosion resistance at room temperature. The history of its discovery is an interesting tale. del Rio sent his brown ore samples, containing what he thought was a new element to Paris for analysis and confirmation, along with a brief explanation that was ambiguous. The complete analysis and description of his work were lost in a shipwreck so the Paris lab saw nothing but brown powder and a brief confusing note. A second sample sent to Berlin was mislabeled lead chromate when it arrived. del Rio gave up, losing confidence in his discovery. The element was rediscovered in 1867 by Nils Sefstrôm. Vanadium has an unusually large number of stable oxidation states (+2, +3, +4, +5)each of which is characterized by a unique color in solution. The metal is used as an alloying agent for steel. It combines with nearly all non-metals in compounds. During the for manufacturing sulfuric acid, sulfur dioxide has to be converted into sulfur trioxide, which is done by passing sulfur dioxide and oxygen over a solid vanadium(V) oxide catalyst. \[ SO_2 + \dfrac{1}{2}O_2 \ce{->[V_2O_5]} SO_3\] This is a good example of the ability of transition metals and their compounds to act as catalysts because of their ability to change their oxidation state (oxidation number). The sulfur dioxide is oxidized to sulfur trioxide by the vanadium(V) oxide. In the process, the vanadium(V) oxide is reduced to vanadium(IV) oxide. \[ SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4\] The vanadium(IV) oxide is then re-oxidized by the oxygen. \[ V_2O_4 + \dfrac{1}{2} O_2 \rightarrow V_2O_5\] Although the catalyst has been temporarily changed during the reaction, at the end it is chemically the same as it started. Vanadium has oxidation states in its compounds of +5, +4, +3 and +2. This section looks at ways of changing between them. It starts with a bit of description, and then goes on to look at the reactions in terms of standard (standard electrode potentials). The usual source of vanadium in the +5 oxidation state is ammonium metavanadate, NH VO . This isn't very soluble in water and is usually first dissolved in sodium hydroxide solution. The solution can be reduced using zinc and an acid - either hydrochloric acid or sulfuric acid, usually using moderately concentrated acid. The exact vanadium ion present in the solution is very complicated, and varies with the pH of the solution. The reaction is done under acidic conditions when the main ion present is - called the dioxovanadium(V) ion. If you do the reaction in a small flask, it is normally stoppered with some cotton wool. This allows hydrogen (produced from a side reaction between the zinc and acid) to escape. At the same time it stops much air from entering. This prevents re-oxidation of the lower oxidation states of vanadium (particularly the +2 state) by oxygen in the air. The reaction is usually warmed so that the changes happen in a reasonable time. The reduction is shown in two stages. Some individual important colors are shown, but the process is one continuous change from start to finish. The reduction from +5 to +4 It is important to notice that the green color you see isn't actually another oxidation state. it is just a mixture of the original yellow of the +5 state and the blue of the +4. Be very careful with the formulae of the two vanadium ions - they are very easy to confuse! The reduction from +4 to +2. The color changes just continue. The reason for the inverted commas around the vanadium(III) ion is that this is almost certainly a simplification. The exact nature of the complex ion will depend on which acid you use in the reduction process. The simplification is probably reasonable at this level. The vanadium(II) ion is very easily oxidized. If you remove the cotton wool from the flask and pour some solution into a test tube, it turns green because of its contact with oxygen in the air. It is oxidized back to vanadium(III). If it is allowed to stand for a long time, the solution eventually turns blue as the air oxidizes it back to the vanadium(IV) state - VO ions. Adding nitric acid (a reasonably powerful oxidizing agent) to the original vanadium(II) solution also produces blue VO ions. The vanadium(II) is again oxidized back to vanadium(IV). Let's look at the first stage of the reduction - from VO to VO . The redox potential for the vanadium half-reaction is given by: The corresponding equilibrium for the zinc is: The simple principle is that if you couple two of these half-reactions together, the one with the more positive E° value will move to the right; the one with the more negative (or less positive) E° moves to the left. If you mix together zinc and VO ions in the presence of acid to provide the H ions: That converts the two equilibria into two one-way reactions. You can write these down and combine them to give the ionic equation for the reaction if you want to. Here are the E° values for all the steps of the reduction from vanadium(V) to vanadium(II): . . . and here is the zinc value again: Remember that for the vanadium reactions to move to the right (which is what we want), their E° values must be more positive than whatever you are reacting them with. In other words, for the reactions to work, zinc must always have the more negative value - and that's the case. Zinc can reduce the vanadium through each of these steps to give the vanadium(II) ion. Suppose you replaced zinc as the reducing agent by tin. How far would the set of reductions go this time? Here are the E° values again: . . . and here is the tin value: For each reduction to happen, the vanadium reaction has to have the more positive E° value because we want it to go to the right. That means that the tin must have the more negative value. The vanadium(II) oxidation state is easily oxidized back to vanadium(III) - or even higher. You will remember that the original reduction we talked about was carried out using zinc and an acid in a flask stoppered with a piece of cotton wool to keep the air out. Air will rapidly oxidize the vanadium(II) ions - but so also will the hydrogen ions present in the solution! The vanadium(II) solution is only stable as long as you keep the air out, and in the presence of the zinc. The zinc is necessary to keep the vanadium reduced. What happens if the zinc isn't there? Look at these E° values: The reaction with the more negative E° value goes to the left; the reaction with the more positive (or less negative) one to the right. That means that the vanadium(II) ions will be oxidized to vanadium(III) ions, and the hydrogen ions reduced to hydrogen. Will the oxidation go any further - for example, to the vanadium(IV) state? Have a look at the E° values and decide: In order for the vanadium equilibrium to move to the left, it would have to have the more negative E° value. It hasn't got the more negative E° value and so the reaction does not happen. In a similar sort of way, you can work out how far nitric acid will oxidize the vanadium(II). Here's the first step: The vanadium reaction has the more negative E° value and so will move to the left; the nitric acid reaction moves to the right. Nitric acid will oxidize vanadium(II) to vanadium(III). The second stage involves these E° values: The nitric acid again has the more positive E° value and so moves to the right. The more negative (less positive) vanadium reaction moves to the left. Nitric acid will certainly oxidize vanadium(III) to vanadium(IV). Will it go all the way to vanadium(V)? No, it won't! For the vanadium reaction to move to the left to form the dioxovanadium(V) ion, it would have to have the more negative (less positive) E° value. It hasn't got a less positive value, and so the reaction does not happen. You can work out the effect of any other oxidizing agent on the lower oxidation states of vanadium in exactly the same way. But don't assume that because the E° values show that a reaction is possible, it will necessarily happen. Jim Clark ( ) | 8,257 | 668 |
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Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as , the , and . An overview of the different types laws associated with the . There are two different types of spin paring configurations for an atom or ion: paramagnetic or diamagnetic. Paramagnetic and diamagnetic configurations result from the amount of d electrons in a particular atom. The energy associated with the spin pairing of these configurations relies on a factor of three things, the atom (for its electronic configuration and number of d electrons), the Crystal Field Theory (field splitting of electrons), and the type of ligand field complex (tetrahedral or octahedral). Examples of these three factors affect on spin pairing are shown ; Being means having all electrons paired and the individual magnetic effects cancel each other out. Being means having unpaired electrons and the individual magnetic effects do not cancel each other out. The unpaired electrons carry a magnetic moment that gets stronger with the number of unpaired electrons causing the atom or ion to be attracted to an external magnetic field. According to , it takes energy to pair electrons, therefore as electrons are added to an orbital, they do it in such a way that they minimize total energy; this causes the 2s orbital to be filled before the 2p orbital. When an electron can singly occupy a given orbital, in a paramagnetic state, that configuration results in high spin energy. However, when two electrons are forced to occupy the same orbital, they experience a interelectronic repulsion effect on each other which in turn increases the total energy of the orbital. The greater this repulsion effect, the greater the energy of the orbital. To calculate this repulsion effect Jorgensen and Slater founded that for any transition metal on the basis of first order perturbation theory can be solved by; \[E(S) = E(qd^n) + \left [S(S+1)- S(S+1) \right ] D\] where \(E(qd^n)\) is the weighted mean energy of the configuration, \(S\) is the spin quantum number, \(S(S+1)\) is the average value of the total spin angular momentum and \(D\) is the metal parameter. Therefore, \[\Delta E = E(S-1)-E(S) = 2SD\] It is also a general theory that spin pairing energy in the form of repulsion increase from P to D to S orbitals. An example of an element that does not follow this suit is Carbon, whose spin pairing energy increases in the opposite direction (S to D to P). For a clearer picture of how this formula works with the elements in the periodic table, see the attached table. C'dd refers to the repulsion associated with the 3dq occupation only, C', to the intracore repulsion, and C',d to the intershell repulsion between core and d electrons. For comparison, the first column shows D = E/2S, calculated from the frozen orbitals of the configuration average. In general: In the example above, the electrons can fill the d-orbitals in two different ways. The electrons can fill lower energy orbitals and pair with an existing electron there resulting in more stability (example on the right). is needed in order to force an electron to fill an orbital that is already occupied with an electron. The electrons can also fill higher energy orbitals and avoid the pairing energy (example on the left). This requires energy and reduces stability. To figure out whether the electrons pair up or go into higher energy orbital depends on the crystal field splitting energy . If the crystal field splitting energy (Δ) is greater than pairing energy, then greater stability would be obtained if the fourth and fifth electrons get paired with the ones in the lower level. If the crystal field splitting energy (\(\Delta\)) is less than the pairing energy, greater stability is obtained by keeping the electrons unpaired. In this configuration, it is evident from previous information that the configuration on the left has a than the configuration on the right due to the differing field splitting energy and max number of unpaired electrons. Weak ligands, such as \(H_2\)O and \(F^-\), produce small crystal field splitting resulting in high-spin complexes and strongly paramagnetic. There are only two unpaired electrons in the configuration on the right, which is minimum amount of electrons known as low spin. Strong ligands, such as \(NH_3\) and \(CN^-\), produce large crystal field splitting, leading to low-spin complexes and weakly paramagnetic or sometimes even diamagnetic. \(Cr^{+2}\) \(Mn^{+3}\) 244.3 (20,425) 301.6 (25,215) \(Mn^{+2}\) \(Fe^{+3}\) A) [Fe(NO ) ] B) [FeBr ] B) Br has a very small crystal field splitting energy, causing the electrons to disperse among the orbitals freely. This allows a paramagnetic state, causing this complex to have high spin energy. | 4,962 | 670 |
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Discovered independently by William Gregor and Martin Klaproth in 1795, titanium (named for the mythological Greek Titans) was first isolated in 1910. Gregor, a Cornish vicar and amateur chemist isolated an impure oxide from ilmenite (\(FeTiO_3\)) by treatment with \(HCl\) and \(H_2SO_4\).Titanium is the second most abundant transition metal on Earth (6320 ppm) and plays a vital role as a material of construction because of its: For example, when it's alloyed with 6% aluminum and 4% vanadium, titanium has half the weight of steel and up to four times the strength. Titanium is a highly corrosion-resistant metal with great tensile strength. It is ninth in abundance for elements in the earth's crust. It has a relatively low density (about 60% that of iron). It is also the tenth most commonly occurring element in the Earth's crust. That all means that titanium should be a really important metal for all sorts of engineering applications. In fact, it is very expensive and only used for rather specialized purposes. Titanium is used, for example: Titanium is very expensive because it is awkward to extract from its ores - for example, from rutile, \(TiO_2\). Whilst a biological function in man is not known, it has excellent biocompatibility--that is the ability to be ignored by the human body's immune system--and an extreme resistance to corrosion. Titanium is now the metal of choice for hip and knee replacements. Titanium cannot be extracted by reducing the ore using carbon as a cheap reducing agent, like with iron. The problem is that titanium forms a carbide, \(\ce{TiC}\), if it is heated with carbon, so you don't get the pure metal that you need. The presence of the carbide makes the metal very brittle. That means that you have to use an alternative reducing agent. In the case of titanium, the reducing agent is either sodium or magnesium. Both of these would, of course, first have to be extracted from their ores by expensive processes. The titanium is produced by reacting titanium(IV) chloride, \(\ce{TiCl4}\) - NOT the oxide - with either sodium or magnesium. That means that you first have to convert the oxide into the chloride. That in turn means that you have the expense of the chlorine as well as the energy costs of the conversion. High temperatures are needed in both stages of the reaction. Titanium is made by a batch process. In the production of iron, for example, there is a continuous flow through the Blast Furnace. Iron ore and coke and limestone are added to the top, and iron and slag removed from the bottom. This is a very efficient way of making something. With titanium, however, you make it one batch at a time. Titanium(IV) chloride is heated with sodium or magnesium to produce titanium. The titanium is then separated from the waste products, and an entirely new reaction is set up in the same reactor. This is a slow and inefficient way of doing things. Traces of oxygen or nitrogen in the titanium tend to make the metal brittle. The reduction has to be carried out in an inert argon atmosphere rather than in air; that also adds to costs. Wilhelm J. Kroll developed the process in Luxemburg around the mid 1930's and then after moving to the USA extended it to enable the extraction of Zirconium as well. Titanium ores, mainly rutile (\(\ce{TiO2}\)) and ilmentite (\(\ce{FeTiO3}\)), are treated with carbon and chlorine gas to produce titanium tetrachloride. \[\ce{TiO2 + Cl2 \rightarrow TiCl4 + CO2} \] Titanium tetrachloride is purified by distillation (Boiling point of 136.4) to remove iron chloride. Purified titanium tetrachloride is reacted with molten magnesium under argon to produce a porous “titanium sponge”. \[\ce{TiCl4 + 2Mg \rightarrow Ti + 2MgCl2}\] Titanium sponge is melted under argon to produce ingots. They can all be prepared by direct reaction of Ti with halogen gas (X ). All are readily hydrolyzed. They are all expected to be diamagnetic. Preparations: They can be prepared by reduction of TiX with H . Titanium Oxides and Aqueous Chemistry obtained from hydrolysis of TiX or Ti(III) salts. TiO reacts with acids and bases. In Acid: TiOSO formed in H SO (Titanyl sulfate) In Base: MTiO metatitanates (eg Perovskite, CaTiO and ilmenite, FeTiO ) M TiO orthotitanates. Peroxides are highly colored and can be used for Colorimetric analysis. pH <1 [TiO (OH)(H O)x]+ pH 1-2 [(O )Ti-O-Ti(O )](OH) ; x=1-6 [Ti(H O) ] -> [Ti(OH)(H O) ] + [H+] pK=1.4 TiO + 2H + e- -> Ti + H O E=0.1V TiCl is a good Lewis acid and forms adducts on reaction with Lewis bases such as; 2PEt -> TiCl (PEt )
2MeCN -> TiCl (MeCN)
bipy -> TiCl (bipy) Solvolysis can occur if ionisable protons are present in the ligand; 2NH -> TiCl (NH ) + 2HCl
4H O -> TiO .aq + 4HCl
2EtOH -> TiCl (OEt) + 2HCl TiCl has less Lewis acid strength but can form adducts also; 3pyr -> TiCl pyr \[ TiO_2 + 2Cl_2 + 2C \longrightarrow TCl_4 + 2CO \] Reduction by sodium: Jim Clark ( ) | 5,078 | 674 |
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ClapotisOcean kayakers may get unexpectedly tossed in the air by high waves called clapotis . Clapotis arise near to shore where incoming waves hit the shore, and are reflected. The reflected outgoing waves meet incoming waves to give very high wave peaks adjacent to calm regions. This YouTube video shows a clipotis on the Oregon coast, and a still image of the clapotis is shown below. Other videos show similar effects. Surprisingly, clapotis provide a model for the behavior of electrons confined to a nucleus. Like most models, the clapotis model is flawed, because ocean waves and the shore are irregular. Clapotis therefore arise irregularly (although often in the same position, as careful observation of the videos will show. But electron waves follow a fixed, regular pattern as described below. If the waves were perfectly sinusoidal and the coastline were perfectly linear, the clapotis would appear constantly at a certain distance from shore, and the distance can be predicted by the wave model shown in a YouTube video, or in the animation below. The animation is a cross section of a wave, but could also be imagined as a wave on a string. The right end is fixed, and the left end is moving up and down. The up-and-down motion creates a that moves to the right, hits the right end, and is then reflected back towards the left. The waves moving to the right and left to create a pattern, where the crests of the waves don't travel, but are located at points labeled L2 in the animation. Between the crests, there are , or points where there is no motion at all. Somehow the motion appears on both sides of the nodes without any motion at the nodes at all. As you can see in the animation, occurs when the crests of the two waves arrive at the same point at the same time, reinforcing each other, and producing a resultant crest, or which is as high as the sum of the heights of the two waves (this is called the "superposition principle"). If a crest and a trough arrive at the same point at the same time (halfway between L/2 and L/4 in the animation), occurs, and the superposition principle predicts that they will cancel each other out, giving a node. A number of discrete, "allowed" standing wave patterns are exhibited by waves on a string. The patterns are slightly different depending on whether the string is fixed or allowed to move at one end. Below are the patterns generated by a string vibrating at different frequencies, but fixed at both ends. The animation shows two "wavelengths" between the ends, corresponding to the 4th pattern from the bottom in the figure on the left. But some frequencies are not allowed (we don't want the guitar to play all tones at once!). It happens in situations like this: This model fits with the counterintuitive behavior of electrons in atoms. Specifically, electrons appear to exist in certain regions of space, but never in regions in between. Furthermore, electrons exist in a series of discrete "orbital" wavefunctions of increasing energy. The energy increases as the number of nodes increases. So the standing wave pattern goes from one node to two nodes, for example, and can't exist anywhere in between. . Electrons don't exist anywhere between the shells. A standing wave can thus be correlated with electron density in the of an atom, as shown in the figure below. Just as the vibrating string has no fixed position at the antinodes, it is impossible to specify the position (or exact motion) of electrons constrained to a nucleus. s of another standing wave are shown here. Waves are used as a model for light as well, so let's define the and of a wave a little more precisely. When you flick a string, a moves down it; if you do this continually, say once a second, you generate a travelling wave train with a of 1 s , or one per second, where the wavelength is the distance between successive peaks (or any other repeating feature) of the wave: There is a relationship between the frequency, usually denoted "ν" ("nu"), the wavelength, usually denoted "λ" (lambda) and the speed that the wave moves down the string (or through space, if it's a light wave). If we denote the speed "c" (a symbol used for the speed of light), the relationship is: Calculate the wavelength of a microwave in a microwave oven that travels at the speed of light, c = 3.0 x 10 m s and has a frequency of 2.45 GHz (2.45 x 10 s . of 12.24 cm. Solution: Rearranging (1) we have: Microwaves are waves like light waves or radio waves, but their wavelength is much longer than light, and shorter than radio. Waves of this wavelength interact with water molecules make the molecules spin faster and thereby heat up food in a microwave oven. At much the same time as Lewis was developing his theories of electronic structure, the physicist Niels Bohr was developing a similar, but more detailed, picture of the atom. Since Bohr was interested in light (energy) emitted by atoms under certain circumstances rather than the valence of elements, he particularly wanted to be able to calculate the energies of the electrons. To do this, he needed to know the exact path followed by each electron as it moved around the nucleus. He assumed paths similar to those of the planets around the sun. The figure seen here, taken from a physics text of the period, illustrates Bohr’s theories applied to the sodium atom. Note how the Bohr model, like that of Lewis, assumes a shell structure. There are two electrons in the innermost shell, eight electrons in the next shell, and a single electron in the outermost shell. Like Lewis’ model, Bohr’s model was only partially successful. It explained some experimental results but was quite unable to account for others. In particular it failed on the quantitative mathematical level. The worked very well for a hydrogen atom with its single electron, but calculations on atoms with more than one electron always gave the wrong answer. On a chemical level, too, certain features were inadequate. There is no evidence to suggest that atoms of sodium are ever as elongated or as flat as the one in the figure. On the contrary, the way that sodium atoms pack together in a solid suggests that they extend out uniformly in all directions; i.e., they are spherical in shape. Another weakness in the theory was that it had to a shell structure rather than explain it. After all, there is nothing in the nature of planets moving around the sun which compels them to orbit in groups of two or eight. Bohr assumed that electrons behave much like planets; so why should they form shells in this way? We usually think of electron shells in terms of their energy. That's because light energy is emitted when an electron falls from a higher shell to a lower one, and measuring light energy is the most important way of determining the energy difference between shells. When electrons change levels, they emit quanta of light called ). The energy of a photon is directly related to its frequency, or inversely related to the wavelength: The constant of proportionality h is known as and has the value 6.626 × 10 J s. Light of higher frequency has higher energy, an a shorter wavelength. Light can only be absorbed by atoms if each photon has exactly the right amount of energy to promote an electron from a lower shell to a higher one. If more energy is required than a photon possesses, it can't be supplied by bombarding the atom with more photons. So we frequently find that light of one wavelength will cause a photochemical change no matter how dim it is, while light of a neighboring wavelength will not cause a photochemical change no matter how intense it is. That's because photons must be absorbed to cause a photochemical change, and they must have exactly the energy needed to be promoted to the next shell to be absorbed. If they're not absorbed, it doesn't matter how intense the light is (how many photons there are per second). What wavelength of light is emitted by a hydrogen atom when an electron falls from the third shell, where it has E = -2.42088863 × 10 J, to the second shell, where it has E = -5.44739997 × 10 J? Solution: (-5.45 × 10 ) - (-2.42 × 10 J) = -3.03 × 10 J. Note that the energy levels get more negative (more energy is released when an electron falls into them) near the nucleus, and the difference here is negative, meaning energy is released. Taking the absolute value of the energy to calculate the energy of the photon, and rearraning equation (2): Of course the shells for electrons are three dimensional, not one dimensional like guitar strings. We can begin to visualize standing waves in more than one dimension by thinking about wave patterns on a drum skin in . Some of the wave patterns are shown below. If you look carefully, you'll see circular that don't move: Electrons exist around the nucleus in "orbitals", which are three-dimensional standing waves. Electron standing waves are quite beautiful, and we'll see more of them in the next few sections. One example is the flower like "f orbital" below. Here the red parts of the "wavefunction" represent mathematically positive (upward) parts of the standing wave, while blue parts are mathematically negative (downward) parts: | 9,266 | 675 |
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The rotational quantum number is \(J\) and the energy of the rotations is: \[E(J)= \tilde{B} J(J+1) \nonumber \] Where \(\tilde{B}\) is in units of cm and is equivalent to: \[\frac{h}{8\pi^2Ic} \nonumber \] Where \(c\) is the speed of light in units of cm . However, with the exception of the ground state (i.e., \(J=0\)), multiple eigenstates exist for a given value of \(J\). This is a form of , (\(g\)), and the number of levels per energy goes as 1, 3, 5, 7, 9 etc or: \[g(J)=2J+1 \nonumber \] Therefore, the summation to find the partition function \(q_\text{rot}(T)\) contains an extra factor \(g=2J+1\). This factor is mathematically very handy, because it is the derivative of: \[g(J)=J(J+1) \nonumber \] This makes it possible to change variables to \(X=J(J+1)\) and the integration becomes very easy. Notice that just like in the translation partition function, we approximate the summation with an integral. However, because we are dealing with far fewer levels, this is less justified in the rotational case. How justified it is depends on the gas and the temperature we consider. Roughly speaking we should be at a temperature \(T>> Θ_\text{rot}\). Look at table 18.2. When do we need to actually work out the (discrete) summation instead of making a (continuous) integration out of it? Only for the lightest gases like \(\ce{He}\) and \(\ce{H2}\) do we need to worry and then only at pretty low cryogenic circumstances. At room temperature, we can take the rotational levels as a continuous set and use an integral. Notice that the vibrational \(Θ\) values are much larger! At \(T=300\,\text{K}\), usually only a single level is occupied and we are in the discrete limit. On the surface of the sun that would be a different matter of course. The rotational constant \(B\) is directly linked to the moment of inertia: \[ I = μr^2\] Where \(μ\) is the reduced mass of the molecule: \[\dfrac{1}{μ} = \dfrac{1}{m_1} + \dfrac{1}{m_2} \nonumber \] and \(r\) is the bond length. Again we can scale the behavior of different systems to one and the same picture by introducing a characteristic temperature: \[ \Theta_{rot} = \dfrac{hb}{k} = \dfrac{\hbar^2}{2Ik_b} \nonumber \] As we did for the translations, we can calculate the moment \(\langle \epsilon \rangle\). For vibrations we get a relatively complicated function of temperature. However, for rotations, the moment is simply equal to \(NkT\) and this means that the rotational contribution to the molar \(C_v\) of a diatomic is simply \(R\). As the vibrational and electronic contributions to the heat capacity are typically negligible at room temperature, we get: Rotational energies are less than vibrational and electronic energies and, at room temperature, many rotational states will be populated. | 2,786 | 676 |
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Named for the mythological character Tantalus by Anders Ekenberg who discovered it in 1802, tantalum is a heavy, gray metal that resembles the more expensive platinum in many respects and is sometimes used as an economical substitute for that element. The metal comprises only 0.0002% of the earth's crust and most deposits lie outside the U.S. which is the chief consumer. Tantalum alloys are corrosion and wear resistant and find use in dental and surgical tools. Tantalum oxide is used in some electronic components and a composite of tantalum carbide (TaC) and graphite is one of the hardest materials ever produced. | 640 | 678 |
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Analytical titrations using redox reactions were introduced shortly after the development of acid–base titrimetry. The earliest took advantage of the oxidizing power of chlorine. In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl , HCl, and HOCl) based on its ability to oxidize indigo, a dye that is colorless in its oxidized state. In 1814, Joseph Gay-Lussac developed a similar method for determining chlorine in bleaching powder. In both methods the end point is a change in color. Before the equivalence point the solution is colorless due to the oxidation of indigo. After the equivalence point, however, unreacted indigo imparts a permanent color to the solution. The number of redox titrimetric methods increased in the mid-1800s with the introduction of MnO , Cr O , and I as oxidizing titrants, and of Fe and S O as reducing titrants. Even with the availability of these new titrants, redox titrimetry was slow to develop due to the lack of suitable indicators. A titrant can serve as its own indicator if its oxidized and reduced forms differ significantly in color. For example, the intensely purple MnO ion serves as its own indicator since its reduced form, Mn , is almost colorless. Other titrants require a separate indicator. The first such indicator, diphenylamine, was introduced in the 1920s. Other redox indicators soon followed, increasing the applicability of redox titrimetry. To evaluate a redox titration we need to know the shape of its titration curve. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H O (as pH) or M (as pM) changes as we add titrant. For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species. You may recall from Chapter 6 that the Nernst equation relates a solution’s potential to the concentrations of reactants and products participating in the redox reaction. Consider, for example, a titration in which a titrand in a reduced state, , reacts with a titrant in an oxidized state, . \[A_\textrm{red}+B_\textrm{ox} \rightleftharpoons B_\textrm{red}+A_\textrm{ox}\] where is the titrand’s oxidized form, and is the titrant’s reduced form. The reaction’s potential, , is the difference between the reduction potentials for each half-reaction. \[E_\textrm{rxn}=E_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}-E_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}\] After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are identical. \[E_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}=E_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}\] This is an important observation because we can use either half-reaction to monitor the titration’s progress. Before the equivalence point the titration mixture consists of appreciable quantities of the titrand’s oxidized and reduced forms. The concentration of unreacted titrant, however, is very small. The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction \[E_\textrm{rxn}= E^o_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}-\dfrac{RT}{nF}\ln\dfrac{[A_\textrm{red}]}{[A_\textrm{ox}]}\] Although the Nernst equation is written in terms of the half-reaction’s standard state potential, a matrix-dependent often is used in its place. See for the standard state potentials and formal potentials for selected half-reactions. After the equivalence point it is easier to calculate the potential using the Nernst equation for the titrant’s half-reaction. \[E_\textrm{rxn}= E^o_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}-\dfrac{RT}{nF}\ln\dfrac{[B_\textrm{red}]}{[B_\textrm{ox}]}\] Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 M Fe with 0.100 M Ce in a matrix of 1 M HClO . The reaction in this case is \[\textrm{Fe}^{2+}(aq)+\textrm{Ce}^{4+}(aq)\rightleftharpoons \textrm{Ce}^{3+}(aq)+\textrm{Fe}^{3+}(aq)\tag{9.15}\] In 1 M HClO , the formal potential for the reduction of Fe to Fe is +0.767 V, and the formal potential for the reduction of Ce to Ce is +1.70 V. Because the equilibrium constant for reaction 9.15 is very large—it is approximately 6 × 10 —we may assume that the analyte and titrant react completely. Step 1 Calculate the volume of titrant needed to reach the equivalence point. The first task is to calculate the volume of Ce needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that \[\textrm{moles Fe}^{2+}=\textrm{moles Ce}^{4+}\] \[M_\textrm{Fe}\times V_\textrm{Fe} = M_\textrm{Ce}\times V_\textrm{Ce}\] Solving for the volume of Ce gives the equivalence point volume as \[V_\textrm{eq} = V_\textrm{Ce} = \dfrac{M_\textrm{Fe}V_\textrm{Fe}}{M_\textrm{Ce}}=\dfrac{\textrm{(0.100 M)(50.0 mL)}}{\textrm{(0.100 M)}}=\textrm{50.0 mL}\] Step 2: alculate the potential before the equivalence point by determining the concentrations of the titrand’s oxidized and reduced forms, and using the Nernst equation for the titrand’s reduction half-reaction. Before the equivalence point, the concentration of unreacted Fe and the concentration of Fe are easy to calculate. For this reason we find the potential using the Nernst equation for the Fe /Fe half-reaction. \[E = E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} - \dfrac{RT}{nF}\log\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}=+0.767\textrm V - 0.05916\log\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}\tag{9.16}\] For example, the concentrations of Fe and Fe after adding 10.0 mL of titrant are \[\begin{align}
[\textrm{Fe}^{2+}]&=\dfrac{\textrm{initial moles Fe}^{2+} - \textrm{moles Ce}^{4+}\textrm{ added}}{\textrm{total volume}}=\dfrac{M_\textrm{Fe}V_\textrm{Fe} - M_\textrm{Ce}V_\textrm{Ce}}{V_\textrm{Fe}+V_\textrm{Ce}}\\
&\mathrm{= \dfrac{(0.100\;M)(50.0\;mL)-(0.100\;M)(10.0\;mL)}{50.0\;mL+10.0\;mL} = 6.67\times10^{-2}\;M}
\end{align}\] \[\begin{align}
[\mathrm{Fe^{3+}}]&=\mathrm{\dfrac{moles\;Ce^{4+}\;added}{total\;volume}}=\dfrac{M_\textrm{Ce}V_\textrm{Ce}}{V_\textrm{Fe} + V_\textrm{Ce}}\\
&=\dfrac{\textrm{(0.100 M)(10.0 mL)}}{\textrm{50.0 mL + 10.0 mL}}=1.67\times10^{-2}\textrm{ M}
\end{align}\] \[E = +0.767\textrm{ V} - 0.05916 \log\dfrac{6.67\times10^{-2}\textrm{ M}}{1.67\times10^{-2}\textrm{ M}}=+0.731\textrm{ V}\] Step 3: Calculate the potential after the equivalence point by determining the concentrations of the titrant’s oxidized and reduced forms, and using the Nernst equation for the titrant’s reduction half-reaction. After the equivalence point, the concentration of Ce and the concentration of excess Ce are easy to calculate. For this reason we find the potential using the Nernst equation for the Ce /Ce half-reaction. \[E=E^o_\mathrm{\large{Ce^{4+}/Ce^{3+}}}-\dfrac{RT}{nF}\log\mathrm{\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}=+ 1.70\textrm{ V} - 0.05916 \log\mathrm{\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}\tag{9.17}\] \[\begin{align}
[\textrm{Ce}^{3+}]&={\dfrac{\textrm{initial moles Fe}^{2+}}{\textrm{total volume}}}=\dfrac{M_\textrm{Fe}V_\textrm{Fe}}{V_\textrm{Fe}+V_\textrm{Ce}}\\
&=\dfrac{\textrm{(0.100 M)(50.0 mL)}}{\textrm{50.0 mL + 60.0 mL}}=4.55\times10^{-3}\textrm{ M}
\end{align}\] \[\begin{align}
[\textrm{Ce}^{4+}]&=\dfrac{\textrm{moles Ce}^{4+}\textrm{ added} - \textrm{initial moles Fe}^{2+}}{\textrm{total volume}}=\dfrac{M_\textrm{Ce}V_\textrm{Ce}-M_\textrm{Fe}V_\textrm{Fe}}{V_\textrm{Fe}+V_\textrm{Ce}}\\
&=\dfrac{\textrm{(0.100 M)(60.0 mL)}-\textrm{(0.100 M)(50.0 mL)}}{\textrm{50.0 mL + 60.0 mL}}=9.09\times10^{-3}\textrm{ M}
\end{align}\] \[E=+1.70\textrm{ V}-0.05916\log\dfrac{4.55\times10^{-2}\textrm{ M}}{9.09\times10^{-3}\textrm{ M}}=+1.66\textrm{ V}\] Step 4 Calculate the potential at the equivalence point. At the titration’s equivalence point, the potential, , in equation 9.16 and equation 9.17 are identical. Adding the equations together to gives \[2E_\textrm{eq}= E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+E^o_\mathrm{\large Ce^{4+}/Ce^{3+}}-0.05916\log\dfrac{\mathrm{[{Fe}^{2+},Ce^{3+}]}}{\mathrm{[Fe^{3+},Ce^{4+}]}}\] Because [Fe ] = [Ce ] and [Ce ] = [Fe ] at the equivalence point, the log term has a value of zero and the equivalence point’s potential is \[E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} + E^o_\mathrm{\large Ce^{4+}/Ce^{3+}}}{2}=\dfrac{\textrm{0.767 V + 1.70 V}}{2}=1.23\textrm{ V}\] Additional results for this titration curve are shown in Table 9.15 and Figure 9.36. Exercise \(\Page {1}\) Calculate the titration curve for the titration of 50.0 mL of 0.0500 M Sn with 0.100 M Tl . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is \[\textrm{Sn}^{2+}(aq)+\textrm{Tl}^{3+}(aq)\rightarrow \textrm{Sn}^{4+}(aq)+\textrm{Tl}^+(aq)\] to review your answer to this exercise. To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching a redox titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.100 M Fe with 0.100 M Ce in a matrix of 1 M HClO . This is the same example that we used in developing the calculations for a redox titration curve. You can review the results of that calculation in Table 9.15 and Figure 9.36. We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 50.0 mL. Next, we draw our axes, placing the potential, , on the -axis and the titrant’s volume on the -axis. To indicate the equivalence point’s volume, we draw a vertical line corresponding to 50.0 mL of Ce . Figure 9.37a shows the result of the first step in our sketch. Before the equivalence point, the potential is determined by a redox buffer of Fe and Fe . Although we can easily calculate the potential using the Nernst equation, we can avoid this calculation by making a simple assumption. You may recall from Chapter 6 that a redox buffer operates over a range of potentials that extends approximately ±(0.05916/ ) unit on either side of . The potential is at the buffer’s lower limit \[\textrm E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}-0.05916\] when the concentration of Fe is 10× greater than that of Fe . The buffer reaches its upper potential \[\textrm E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+0.05916\] when the concentration of Fe is 10× smaller than that of Fe . The redox buffer spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume. Figure 9.37b shows the second step in our sketch. First, we superimpose a ladder diagram for Fe on the -axis, using its value of 0.767 V and including the buffer’s range of potentials. Next, we add points representing the pH at 10% of the equivalence point volume (a potential of 0.708 V at 5.0 mL) and at 90% of the equivalence point volume (a potential of 0.826 V at 45.0 mL). We used a similar approach when sketching the acid–base titration curve for the titration of acetic acid with NaOH. The third step in sketching our titration curve is to add two points after the equivalence point. Here the potential is controlled by a redox buffer of Ce and Ce . The redox buffer is at its lower limit of E = – 0.05916 when the titrant reaches 110% of the equivalence point volume and the potential is when the volume of Ce is 2× . Figure 9.37c shows the third step in our sketch. First, we add a ladder diagram for Ce , including its buffer range, using its value of 1.70 V. Next, we add points representing the potential at 110% of (a value of 1.66 V at 55.0 mL) and at 200% of (a value of 1.70 V at 100.0 mL). We used a similar approach when sketching the complexation titration curve for the titration of Mg with EDTA. Next, we draw a straight line through each pair of points, extending the line through the vertical line representing the equivalence point’s volume (Figure 9.37d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.37e). A comparison of our sketch to the exact titration curve (Figure 9.37f) shows that they are in close agreement. Exercise \(\Page {2}\) Sketch the titration curve for the titration of 50.0 mL of 0.0500 M Sn with 0.100 M Tl . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is \[\textrm{Sn}^{2+}(aq)+\textrm{Tl}^{3+}(aq)\rightarrow\textrm{Sn}^{4+}(aq)+\textrm{Tl}^+(aq)\] Compare your sketch to your calculated titration curve from . to review your answer to this exercise. A redox titration’s equivalence point occurs when we react stoichiometrically equivalent amounts of titrand and titrant. As is the case with acid–base and complexation titrations, we estimate the equivalence point of a complexation titration using an experimental end point. A variety of methods are available for locating the end point, including indicators and sensors that respond to a change in the solution conditions. For an acid–base titration or a complexometric titration the equivalence point is almost identical to the inflection point on the steeping rising part of the titration curve. If you look back at Figure 9.7 and Figure 9.28, you will see that the inflection point is in the middle of this steep rise in the titration curve, which makes it relatively easy to find the equivalence point when you sketch these titration curves. We call this a . If the stoichiometry of a redox titration is symmetric—one mole of titrant reacts with each mole of titrand—then the equivalence point is symmetric. If the titration reaction’s stoichiometry is not 1:1, then the equivalence point is closer to the top or to bottom of the titration curve’s sharp rise. In this case we have an . Example \(\Page {1}\) Derive a general equation for the equivalence point’s potential when titrating Fe with MnO . \[5\textrm{Fe}^{2+}(aq)+\textrm{MnO}_4^-(aq)+8\textrm H^+(aq)\rightarrow 5\textrm{Fe}^{3+}(aq)+\textrm{Mn}^{2+}(aq)+\mathrm{4H_2O}\] (We often use H instead of H O when writing a redox reaction.) The half-reactions for Fe and MnO are \[\textrm{Fe}^{2+}(aq)\rightarrow\textrm{Fe}^{3+}(aq)+e^-\] \[\textrm{MnO}_4^-(aq)+8\textrm H^+(aq)+5e^-\rightarrow \textrm{Mn}^{2+}(aq)+4\mathrm{H_2O}(l)\] for which the Nernst equations are \[E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}-0.05916\log\dfrac{[\textrm{Fe}^{2+}]}{[\textrm{Fe}^{3+}]}\] \[E=E^o_\mathrm{\large MnO_4^-/Mn^{2+}}-\dfrac{0.05916}{5}\log\dfrac{[\textrm{Mn}^{2+}]}{\ce{[MnO_4^- ,H^+]^8}}\] Before adding these two equations together we must multiply the second equation by 5 so that we can combine the log terms; thus \[6E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}-0.05916\log\mathrm{\dfrac{[Fe^{2+},Mn^{2+}]}{[Fe^{3+},\ce{MnO_4^-},H^+]^8}}\] At the equivalence point we know that \[[\textrm{Fe}^{2+}]=5\times[\textrm{MnO}_4^-]\] \[[\textrm{Fe}^{3+}]=5\times[\textrm{Mn}^{2+}]\] Substituting these equalities into the previous equation and rearranging gives us a general equation for the potential at the equivalence point. \[6E_\textrm{eq}=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}-0.05916\log\mathrm{\dfrac{5[\ce{MnO_4^-},Mn^{2+}]}{5[Mn^{2+},\ce{MnO_4^-},H^+]^8}}\] \[E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} + 5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}}{6}-\dfrac{0.05916}{6}\log\dfrac{1}{[\textrm H^+]^8}\] \[E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}}{6}+\dfrac{0.05916\times8}{6}\log[\textrm H^+]\] \[E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}}{6}-0.07888\textrm{pH}\] Our equation for the equivalence point has two terms. The first term is a weighted average of the titrand’s and the titrant’s standard state potentials, in which the weighting factors are the number of electrons in their respective half-reactions. (Instead of standard state potentials, you can use formal potentials.) The second term shows that for this titration is pH-dependent. At a pH of 1 (in H SO ), for example, the equivalence point has a potential of \[E_\textrm{eq}=\dfrac{0.768+5\times1.51}{6}-0.07888\times1=1.31\textrm{ V}\] Figure 9.38 shows a typical titration curve for titration of Fe with MnO . Note that the titration’s equivalence point is asymmetrical. Exercise \(\Page {3}\) Derive a general equation for the equivalence point’s potential for the titration of U with Ce . The unbalanced reaction is \[\textrm{Ce}^{4+}(aq)+\textrm U^{4+}(aq)\rightarrow \textrm{UO}_2^{2+}(aq)+\textrm{Ce}^{3+}(aq)\] What is the equivalence point’s potential if the pH is 1? to review your answer to this exercise. Three types of indicators are used to signal a redox titration’s end point. The oxidized and reduced forms of some titrants, such as MnO , have different colors. A solution of MnO is intensely purple. In an acidic solution, however, permanganate’s reduced form, Mn , is nearly colorless. When using MnO as a titrant, the titrand’s solution remains colorless until the equivalence point. The first drop of excess MnO produces a permanent tinge of purple, signaling the end point. Some indicators form a colored compound with a specific oxidized or reduced form of the titrant or the titrand. Starch, for example, forms a dark blue complex with I . We can use this distinct color to signal the presence of excess I as a titrant—a change in color from colorless to blue—or the completion of a reaction consuming I as the titrand—a change in color from blue to colorless. Another example of a specific indicator is thiocyanate, SCN , which forms a soluble red-colored complex of Fe(SCN) with Fe . The most important class of indicators are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. When we add a to the titrand, the indicator imparts a color that depends on the solution’s potential. As the solution’s potential changes with the addition of titrant, the indicator changes oxidation state and changes color, signaling the end point. To understand the relationship between potential and an indicator’s color, consider its reduction half-reaction \[\mathrm{In_{ox}}+ne^-\rightleftharpoons \mathrm{In_{red}}\] where In and In are, respectively, the indicator’s oxidized and reduced forms. For simplicity, In and In are shown without specific charges. Because there is a change in oxidation state, In and In cannot both be neutral. The Nernst equation for this half-reaction is \[E=E^o_\mathrm{In_{\large ox}/In_{\large red}}-\dfrac{0.05916}{n}\log\mathrm{\dfrac{[In_{red}]}{[In_{ox}]}}\] As shown in Figure 9.39, if we assume that the indicator’s color changes from that of In to that of In when the ratio [In ]/[In ] changes from 0.1 to 10, then the end point occurs when the solution’s potential is within the range \[E=E^o_\mathrm{In_{\large ox}/In_{\large red}}\pm\dfrac{0.05916}{n}\] This is the same approach we took in considering acid–base indicators and complexation indicators. Diagram showing the relationship between and an indicator’s color. The ladder diagram defines potentials where In and In are the predominate species. The indicator changes color when is within the range = ± 0.05916/n A partial list of redox indicators is shown in Table 9.16. Examples of appropriate and inappropriate indicators for the titration of Fe with Ce are shown in Figure 9.40. Another method for locating a redox titration’s end point is a potentiometric titration in which we monitor the change in potential while adding the titrant to the titrand. The end point is found by visually examining the titration curve. The simplest experimental design for a potentiometric titration consists of a Pt indicator electrode whose potential is governed by the titrand’s or titrant’s redox half-reaction, and a reference electrode that has a fixed potential. A further discussion of potentiometry is found in Chapter 11. Other methods for locating the titration’s end point include thermometric titrations and spectrophotometric titrations. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical redox titrimetric method. Although each method is unique, the following description of the determination of the total chlorine residual in water provides an instructive example of a typical procedure. The description here is based on Method 4500-Cl B as published in , 20th Ed., American Public Health Association: Washington, D. C., 1998. Representative Method 9.3: The chlorination of public water supplies produces several chlorine-containing species, the combined concentration of which is called the total chlorine residual. Chlorine may be present in a variety of states, including the free residual chlorine, consisting of Cl , HOCl and OCl , and the combined chlorine residual, consisting of NH Cl, NHCl , and NCl . The total chlorine residual is determined by using the oxidizing power of chlorine to convert I to I . The amount of I formed is then determined by titrating with Na S O using starch as an indicator. Regardless of its form, the total chlorine residual is reported as if Cl is the only source of chlorine, and is reported as mg Cl/L. Select a volume of sample requiring less than 20 mL of Na S O to reach the end point. Using glacial acetic acid, acidify the sample to a pH of 3–4, and add about 1 gram of KI. Titrate with Na S O until the yellow color of I begins to disappear. Add 1 mL of a starch indicator solution and continue titrating until the blue color of the starch–I complex disappears (Figure 9.41). Use a blank titration to correct the volume of titrant needed to reach the end point for reagent impurities. 1. Is this an example of a direct or an indirect analysis? This is an indirect analysis because the chlorine-containing species do not react with the titrant. Instead, the total chlorine residual oxidizes I to I , and the amount of I is determined by titrating with Na S O . 2. Why does the procedure rely on an indirect analysis instead of directly titrating the chlorine-containing species using KI as a titrant? Because the total chlorine residual consists of six different species, a titration with I does not have a single, well-defined equivalence point. By converting the chlorine residual to an equivalent amount of I , the indirect titration with Na S O has a single, useful equivalence point. Even if the total chlorine residual is from a single species, such as HOCl, a direct titration with KI is impractical. Because the product of the titration, I , imparts a yellow color, the titrand’s color would change with each addition of titrant, making it difficult to find a suitable indicator. 3. Both oxidizing and reducing agents can interfere with this analysis. Explain the effect of each type of interferent has on the total chlorine residual. An interferent that is an oxidizing agent converts additional I to I . Because this extra I requires an additional volume of Na S O to reach the end point, we overestimate the total chlorine residual. If the interferent is a reducing agent, it reduces back to I some of the I produced by the reaction between the total chlorine residual and iodide. As a result. we underestimate the total chlorine residual. Endpoint for the determination of the total chlorine residual. (a) Acidifying the sample and adding KI forms a brown solution of I . (b) Titrating with Na S O converts I to I with the solution fading to a pale yellow color as we approach the end point. (c) Adding starch forms the deep purple starch–I complex. (d) As the titration continues, the end point is a sharp transition from a purple to a colorless solution. The change in color from (c) to (d) typically takes 1–2 drops of titrant. Although many quantitative applications of redox titrimetry have been replaced by other analytical methods, a few important applications continue to be relevant. In this section we review the general application of redox titrimetry with an emphasis on environmental, pharmaceutical, and industrial applications. We begin, however, with a brief discussion of selecting and characterizing redox titrants, and methods for controlling the titrand’s oxidation state. If a redox titration is to be used in a quantitative analysis, the titrand must initially be present in a single oxidation state. For example, iron can be determined by a redox titration in which Ce oxidizes Fe to Fe . Depending on the sample and the method of sample preparation, iron may initially be present in both the +2 and +3 oxidation states. Before titrating, we must reduce any Fe to Fe . This type of pretreatment can be accomplished using an auxiliary reducing agent or oxidizing agent. A metal that is easy to oxidize—such as Zn, Al, and Ag—can serve as an . The metal, as a coiled wire or powder, is added to the sample where it reduces the titrand. Because any unreacted auxiliary reducing agent will react with the titrant, it must be removed before beginning the titration. This can be accomplished by simply removing the coiled wire, or by filtering. An alternative method for using an auxiliary reducing agent is to immobilize it in a column. To prepare a reduction column an aqueous slurry of the finally divided metal is packed in a glass tube equipped with a porous plug at the bottom. The sample is placed at the top of the column and moves through the column under the influence of gravity or vacuum suction. The length of the reduction column and the flow rate are selected to ensure the analyte’s complete reduction. Two common reduction columns are used. In the the column is filled with amalgamated zinc, Zn(Hg), prepared by briefly placing Zn granules in a solution of HgCl . Oxidation of zinc \[\textrm{Zn(Hg)}(s)\rightarrow \textrm{Zn}^{2+}(aq)+\textrm{Hg}(l)+2e^-\] provides the electrons for reducing the titrand. In the the column is filled with granular Ag metal. The solution containing the titrand is acidified with HCl and passed through the column where the oxidation of silver \[\textrm{Ag}(s)+\textrm{Cl}^-(aq)\rightarrow \textrm{AgCl}(s)+e^-\] provides the necessary electrons for reducing the titrand. Table 9.17 provides a summary of several applications of reduction columns. Several reagents are commonly used as , including ammonium peroxydisulfate, (NH ) S O , and hydrogen peroxide, H O . Peroxydisulfate is a powerful oxidizing agent \[\mathrm{S_2O_8^{2-}}(aq)+2e^-\rightarrow\mathrm{2SO_4^{2-}}(aq)\] capable of oxidizing Mn to MnO , Cr to Cr O , and Ce to Ce . Excess peroxydisulfate is easily destroyed by briefly boiling the solution. The reduction of hydrogen peroxide in acidic solution \[\mathrm{H_2O_2}(aq)+\mathrm{2H^+}(aq)+2e^-\rightarrow\mathrm{2H_2O}(l)\] provides another method for oxidizing a titrand. Excess H O is destroyed by briefly boiling the solution. If it is to be used quantitatively, the titrant’s concentration must remain stable during the analysis. Because a titrant in a reduced state is susceptible to air oxidation, most redox titrations use an oxidizing agent as the titrant. There are several common oxidizing titrants, including MnO , Ce , Cr O , and I . Which titrant is used often depends on how easy it is to oxidize the titrand. A titrand that is a weak reducing agent needs a strong oxidizing titrant if the titration reaction is to have a suitable end point. The two strongest oxidizing titrants are MnO and Ce , for which the reduction half-reactions are \[\ce{MnO_4^-}(aq)+\mathrm{8H^+}(aq)+5e^-\rightleftharpoons \mathrm{Mn^{2+}}(aq)+\mathrm{4H_2O}(l)\] \[\textrm{Ce}^{4+}(aq)+e^-\rightleftharpoons \textrm{Ce}^{3+}(aq)\] Solutions of Ce4+ usually are prepared from the primary standard cerium ammonium nitrate, Ce(NO3)4•2NH4NO3, in 1 M H2SO4. When prepared using a reagent grade material, such as Ce(OH)4, the solution is standardized against a primary standard reducing agent such as Na2C2O4 or Fe2+ (prepared using iron wire) using ferroin as an indicator. Despite its availability as a primary standard and its ease of preparation, Ce4+ is not as frequently used as MnO4– because it is more expensive. Note The standardization reactions are \[\mathrm{Ce^{4+}}(aq)+\mathrm{Fe^{2+}}(aq)\rightarrow \mathrm{Ce^{3+}}(aq)+\mathrm{Fe^{3+}}(aq)\] \[\mathrm{2Ce^{4+}}(aq)+\mathrm{H_2C_2O_4}(aq)\rightarrow \mathrm{2Ce^{3+}}(aq)+\mathrm{2CO_2}(g)+\mathrm{2H^+}(aq)\] Solutions of MnO are prepared from KMnO , which is not available as a primary standard. Aqueous solutions of permanganate are thermodynamically unstable due to its ability to oxidize water. \[\ce{4MnO_4^-}(aq)+\mathrm{2H_2O}(l)\rightleftharpoons\mathrm{4MnO_2}(s)+\mathrm{3O_2}(g)+\mathrm{4OH^-}(aq)\] This reaction is catalyzed by the presence of MnO , Mn , heat, light, and the presence of acids and bases. A moderately stable solution of permanganate can be prepared by boiling it for an hour and filtering through a sintered glass filter to remove any solid MnO that precipitates. Standardization is accomplished against a primary standard reducing agent such as Na C O or Fe (prepared using iron wire), with the pink color of excess MnO signaling the end point. A solution of MnO prepared in this fashion is stable for 1–2 weeks, although the standardization should be rechecked periodically. Note The standardization reactions are \[\ce{MnO_4^-}(aq)+\mathrm{5Fe^{2+}}(aq)+\mathrm{8H^+}(aq)\rightarrow \mathrm{Mn^{2+}}(aq)+\mathrm{5Fe^{3+}}(aq)+\mathrm{4H_2O}(l)\] \[\ce{2MnO_4^-}(aq)+\mathrm{5H_2C_2O_4}(aq)+\mathrm{6H^+}(aq)\rightarrow\mathrm{2Mn^{2+}}(aq)+\mathrm{10CO_2}(g)+\mathrm{8H_2O}(l)\] Potassium dichromate is a relatively strong oxidizing agent whose principal advantages are its availability as a primary standard and the long term stability of its solutions. It is not, however, as strong an oxidizing agent as MnO or Ce , which makes it less useful when the titrand is a weak reducing agent. Its reduction half-reaction is \[\mathrm{Cr_2O_7^{2-}}(aq)+\mathrm{14H^+}(aq)+6e^-\rightleftharpoons \mathrm{2Cr^{3+}}(aq)+\mathrm{7H_2O}(l)\] Although a solution of Cr O is orange and a solution of Cr is green, neither color is intense enough to serve as a useful indicator. Diphenylamine sulfonic acid, whose oxidized form is red-violet and reduced form is colorless, gives a very distinct end point signal with Cr O . Iodine is another important oxidizing titrant. Because it is a weaker oxidizing agent than MnO , Ce , and Cr O , it is useful only when the titrand is a stronger reducing agent. This apparent limitation, however, makes I a more selective titrant for the analysis of a strong reducing agent in the presence of a weaker reducing agent. The reduction half-reaction for I is \[\textrm I_2(aq) + 2e^-\rightleftharpoons 2\textrm I^-(aq)\] Because iodine is not very soluble in water, solutions are prepared by adding an excess of I . The complexation reaction \[\textrm I_2(aq)+\textrm I^-(aq)\rightleftharpoons\textrm I_3^-(aq)\] increases the solubility of I by forming the more soluble triiodide ion, I . Even though iodine is present as I instead of I , the number of electrons in the reduction half-reaction is unaffected. \[\textrm I_3^-(aq)+2e^-\rightleftharpoons 3\textrm I^-(aq)\] Solutions of I are normally standardized against Na S O using starch as a specific indicator for I . Note The standardization reaction is \[\mathrm I_3^-(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow 3\textrm I^-(aq)+\mathrm{2S_4O_6^{2-}}(aq)\] An oxidizing titrant such as MnO , Ce , Cr O , and I , is used when the titrand is in a reduced state. If the titrand is in an oxidized state, we can first reduce it with an auxiliary reducing agent and then complete the titration using an oxidizing titrant. Alternatively, we can titrate it using a reducing titrant. Iodide is a relatively strong reducing agent that could serve as a reducing titrant except that a solution of I– is susceptible to the air-oxidation of I to I . \[3\textrm I^-(aq)\rightleftharpoons \mathrm I_3^-(aq)+2e^-\] Note A freshly prepared solution of KI is clear, but after a few days it may show a faint yellow coloring due to the presence of I . Instead, adding an excess of KI reduces the titrand, releasing a stoichiometric amount of I . The amount of I produced is then determined by a back titration using thiosulfate, S O , as a reducing titrant. \[\mathrm{2S_2O_3^{2-}}(aq)\rightleftharpoons\mathrm{2S_4O_6^{2-}}(aq)+2e^-\] Solutions of S O are prepared using Na S O •5H O, and must be standardized before use. Standardization is accomplished by dissolving a carefully weighed portion of the primary standard KIO in an acidic solution containing an excess of KI. The reaction between IO and I \[\textrm{IO}_3^-(aq)+8\textrm I^-(aq)+6\textrm H^+(aq)\rightarrow \ce{3I_3^-}(aq)+\mathrm{3H_2O}(l)\] liberates a stoichiometric amount of I . By titrating this I with thiosulfate, using starch as a visual indicator, we can determine the concentration of S O in the titrant. Note The standardization titration is \[\mathrm I_3^-(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow 3\textrm I^-(aq)+\mathrm{2S_4O_6^{2-}}(aq)\] which is the same reaction used to standardize solutions of I . This approach to standardizing solutions of S O . is similar to the determination of the total chlorine residual outlined in . Although thiosulfate is one of the few reducing titrants that is not readily oxidized by contact with air, it is subject to a slow decomposition to bisulfite and elemental sulfur. If used over a period of several weeks, a solution of thiosulfate should be restandardized periodically. Several forms of bacteria are able to metabolize thiosulfate, which also can lead to a change in its concentration. This problem can be minimized by adding a preservative such as HgI to the solution. Another useful reducing titrant is ferrous ammonium sulfate, Fe(NH ) (SO ) •6H O, in which iron is present in the +2 oxidation state. A solution of Fe is susceptible to air-oxidation, but when prepared in 0.5 M H SO it remains stable for as long as a month. Periodic restandardization with K Cr O is advisable. The titrant can be used to directly titrate the titrand by oxidizing Fe to Fe . Alternatively, ferrous ammonium sulfate is added to the titrand in excess and the quantity of Fe produced determined by back titrating with a standard solution of Ce or Cr O . One of the most important applications of redox titrimetry is evaluating the chlorination of public water supplies. , for example, describes an approach for determining the total chlorine residual by using the oxidizing power of chlorine to oxidize I to I . The amount of I is determined by back titrating with S O . The efficiency of chlorination depends on the form of the chlorinating species. There are two contributions to the total chlorine residual—the free chlorine residual and the combined chlorine residual. The free chlorine residual includes forms of chlorine that are available for disinfecting the water supply. Examples of species contributing to the free chlorine residual include Cl , HOCl and OCl . The combined chlorine residual includes those species in which chlorine is in its reduced form and, therefore, no longer capable of providing disinfection. Species contributing to the combined chlorine residual are NH Cl, NHCl and NCl . When a sample of iodide-free chlorinated water is mixed with an excess of the indicator , -diethyl- -phenylenediamine (DPD), the free chlorine oxidizes a stoichiometric portion of DPD to its red-colored form. The oxidized DPD is then back titrated to its colorless form using ferrous ammonium sulfate as the titrant. The volume of titrant is proportional to the free residual chlorine. Having determined the free chlorine residual in the water sample, a small amount of KI is added, catalyzing the reduction monochloramine, NH Cl, and oxidizing a portion of the DPD back to its red-colored form. Titrating the oxidized DPD with ferrous ammonium sulfate yields the amount of NH Cl in the sample. The amount of dichloramine and trichloramine are determined in a similar fashion. The methods described above for determining the total, free, or combined chlorine residual also are used to establish a water supply’s chlorine demand. Chlorine demand is defined as the quantity of chlorine needed to completely react with any substance that can be oxidized by chlorine, while also maintaining the desired chlorine residual. It is determined by adding progressively greater amounts of chlorine to a set of samples drawn from the water supply and determining the total, free, or combined chlorine residual. Another important example of redox titrimetry, which finds applications in both public health and environmental analyses is the determination of dissolved oxygen. In natural waters, such as lakes and rivers, the level of dissolved O is important for two reasons: it is the most readily available oxidant for the biological oxidation of inorganic and organic pollutants; and it is necessary for the support of aquatic life. In a wastewater treatment plant dissolved O is essential for the aerobic oxidation of waste materials. If the concentration of dissolved O falls below a critical value, aerobic bacteria are replaced by anaerobic bacteria, and the oxidation of organic waste produces undesirable gases, such as CH and H S. One standard method for determining the dissolved O content of natural waters and wastewaters is the Winkler method. A sample of water is collected without exposing it to the atmosphere, which might change the concentration of dissolved O . The sample is first treated with a solution of MnSO , and then with a solution of NaOH and KI. Under these alkaline conditions the dissolved oxygen oxidizes Mn to MnO . \[\mathrm{2Mn^{2+}}(aq)+\mathrm{4OH^-}(aq)+\mathrm O_2(g)\rightarrow \mathrm{2MnO_2}(s)+\mathrm{2H_2O}(l)\] After the reaction is complete, the solution is acidified with H SO . Under the now acidic conditions I is oxidized to I by MnO . \[\mathrm{MnO_2}(s)+\mathrm{3I^-}(aq)+\mathrm{4H^+}(aq)\rightarrow \mathrm{Mn^{2+}}+\ce{I_3^-}(aq)+\mathrm{2H_2O}(l)\] The amount of I formed is determined by titrating with S O using starch as an indicator. The Winkler method is subject to a variety of interferences, and several modifications to the original procedure have been proposed. For example, NO interferes because it can reduce I to I under acidic conditions. This interference is eliminated by adding sodium azide, NaN , reducing NO to N . Other reducing agents, such as Fe , are eliminated by pretreating the sample with KMnO , and destroying the excess permanganate with K C O . Another important example of redox titrimetry is the determination of water in nonaqueous solvents. The titrant for this analysis is known as the Karl Fischer reagent and consists of a mixture of iodine, sulfur dioxide, pyridine, and methanol. Because the concentration of pyridine is sufficiently large, I and SO react with pyridine (py) to form the complexes py•I and py•SO . When added to a sample containing water, I is reduced to I and SO is oxidized to SO . \[\textrm{py}\bullet\textrm I_2+\textrm{py}\bullet\mathrm{SO_2}+\textrm{py}+\mathrm{H_2O}\rightarrow 2\textrm{py}\bullet\textrm{HI}+\textrm{py}\bullet\mathrm{SO_3}\] Methanol is included to prevent the further reaction of py•SO with water. The titration’s end point is signaled when the solution changes from the product’s yellow color to the brown color of the Karl Fischer reagent. Redox titrimetry also is used for the analysis of organic analytes. One important example is the determination of the chemical oxygen demand (COD) of natural waters and wastewaters. The COD provides a measure of the quantity of oxygen necessary to completely oxidize all the organic matter in a sample to CO and H O. Because no attempt is made to correct for organic matter that can not be decomposed biologically, or for slow decomposition kinetics, the COD always overestimates a sample’s true oxygen demand. The determination of COD is particularly important in managing industrial wastewater treatment facilities where it is used to monitor the release of organic-rich wastes into municipal sewer systems or the environment. A sample’s COD is determined by refluxing it in the presence of excess K Cr O , which serves as the oxidizing agent. The solution is acidified with H SO using Ag SO to catalyze the oxidation of low molecular weight fatty acids. Mercuric sulfate, HgSO , is added to complex any chloride that is present, preventing the precipitation of the Ag catalyst as AgCl. Under these conditions, the efficiency for oxidizing organic matter is 95–100%. After refluxing for two hours, the solution is cooled to room temperature and the excess Cr O is determined by back titrating using ferrous ammonium sulfate as the titrant and ferroin as the indicator. Because it is difficult to completely remove all traces of organic matter from the reagents, a blank titration must be performed. The difference in the amount of ferrous ammonium sulfate needed to titrate the sample and the blank is proportional to the COD. Iodine has been used as an oxidizing titrant for a number of compounds of pharmaceutical interest. Earlier we noted that the reaction of S O with I produces the tetrathionate ion, S O . The tetrathionate ion is actually a dimer consisting of two thiosulfate ions connected through a disulfide (–S–S–) linkage. In the same fashion, I can be used to titrate mercaptans of the general formula RSH, forming the dimer RSSR as a product. The amino acid cysteine also can be titrated with I . The product of this titration is cystine, which is a dimer of cysteine. Triiodide also can be used for the analysis of ascorbic acid (vitamin C) by oxidizing the enediol functional group to an alpha diketone and for the analysis of reducing sugars, such as glucose, by oxidizing the aldehyde functional group to a carboxylate ion in a basic solution. An organic compound containing a hydroxyl, a carbonyl, or an amine functional group adjacent to an hydoxyl or a carbonyl group can be oxidized using metaperiodate, IO , as an oxidizing titrant. \[\ce{IO_4^-}(aq)+\mathrm{H_2O}(l)+2e^-\rightleftharpoons \ce{IO_3^-}(aq)+\mathrm{2OH^-}(aq)\] A two-electron oxidation cleaves the C–C bond between the two functional groups, with hydroxyl groups being oxidized to aldehydes or ketones, carbonyl functional groups being oxidized to carboxylic acids, and amines being oxidized to an aldehyde and an amine (ammonia if a primary amine). The analysis is conducted by adding a known excess of IO to the solution containing the analyte, and allowing the oxidation to take place for approximately one hour at room temperature. When the oxidation is complete, an excess of KI is added, which converts any unreacted IO to IO and I . \[\ce{IO_4^-}(aq)+3\mathrm I^-(aq)+\mathrm{H_2O}(l)\rightarrow \ce{IO_3^-}(aq)+\textrm I_3^-(aq)+\mathrm{2OH^-}(aq)\] The I is then determined by titrating with S O using starch as an indicator. The quantitative relationship between the titrand and the titrant is determined by the stoichiometry of the titration reaction. If you are unsure of the balanced reaction, you can deduce the stoichiometry by remembering that the electrons in a redox reaction must be conserved. Example 9.11 The amount of Fe in a 0.4891-g sample of an ore was determined by titrating with K Cr O . After dissolving the sample in HCl, the iron was brought into the +2 oxidation state using a Jones reductor. Titration to the diphenylamine sulfonic acid end point required 36.92 mL of 0.02153 M K Cr O . Report the ore’s iron content as %w/w Fe O . (Although we can deduce the stoichiometry between the titrant and the titrand without balancing the titration reaction, the balanced reaction \[\mathrm{K_2Cr_2O_7}(aq)+\mathrm{6Fe^{2+}}(aq)+\mathrm{14H^+}(aq)\rightarrow \mathrm{2Cr^{3+}}(aq)+\mathrm{2K^+}(aq)+\mathrm{6Fe^{3+}}(aq)+\mathrm{7H_2O}(l)\] does provide useful information. For example, the presence of H reminds us that the reactionfs feasibility is pH-dependent.) Because we have not been provided with the titration reaction, let’s use a conservation of electrons to deduce the stoichiometry. During the titration the analyte is oxidized from Fe to Fe , and the titrant is reduced from Cr O to Cr . Oxidizing Fe to Fe requires only a single electron. Reducing Cr O , in which each chromium is in the +6 oxidation state, to Cr requires three electrons per chromium, for a total of six electrons. A conservation of electrons for the titration, therefore, requires that each mole of K Cr O reacts with six moles of Fe . The moles of K Cr O used in reaching the end point is \[\mathrm{(0.02153\;M\;K_2Cr_2O_7)\times(0.03692\;L\;K_2Cr_2O_7)=7.949\times10^{-4}\;mol\;K_2Cr_2O_7}\] which means that the sample contains \[\mathrm{7.949\times10^{-4}\;mol\;K_2Cr_2O_7\times\dfrac{6\;mol\;Fe^{2+}}{mol\;K_2Cr_2O_7}=4.769\times10^{-3}\;mol\;Fe^{2+}}\] Thus, the %w/w Fe O in the sample of ore is \[\mathrm{4.769\times10^{-3}\;mol\;Fe^{2+}\times\dfrac{1\;mol\;Fe_2O_3}{2\;mol\;Fe^{2+}}\times\dfrac{159.69\;g\;Fe_2O_3}{mol\;Fe_2O_3}=0.3808\;g\;Fe_2O_3}\] \[\mathrm{\dfrac{0.3808\;g\;Fe_2O_3}{0.4891\;g\;sample}\times100=77.86\%\;w/w\;Fe_2O_3}\] Practice Exercise 9.20 The purity of a sample of sodium oxalate, Na C O , is determined by titrating with a standard solution of KMnO . If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO to reach the titration’s end point, what is the %w/w Na C O in the sample. to review your answer to this exercise. As shown in the following two examples, we can easily extend this approach to an analysis that requires an indirect analysis or a back titration. Example 9.12 A 25.00-mL sample of a liquid bleach was diluted to 1000 mL in a volumetric flask. A 25-mL portion of the diluted sample was transferred by pipet into an Erlenmeyer flask containing an excess of KI, reducing the OCl to Cl , and producing I . The liberated I was determined by titrating with 0.09892 M Na S O , requiring 8.96 mL to reach the starch indicator end point. Report the %w/v NaOCl in the sample of bleach. To determine the stoichiometry between the analyte, NaOCl, and the titrant, Na S O , we need to consider both the reaction between OCl and I , and the titration of I with Na S O . First, in reducing OCl to Cl , the oxidation state of chlorine changes from +1 to –1, requiring two electrons. The oxidation of three I to form I releases two electrons as the oxidation state of each iodine changes from –1 in I to –⅓ in I . A conservation of electrons, therefore, requires that each mole of OCl produces one mole of I . Second, in the titration reaction, I . is reduced to I and S O is oxidized to S O . Reducing I to 3I requires two elections as each iodine changes from an oxidation state of –⅓ to –1. In oxidizing S O to S O , each sulfur changes its oxidation state from +2 to +2.5, releasing one electron for each S O . A conservation of electrons, therefore, requires that each mole of I reacts with two moles of S O . Finally, because each mole of OCl produces one mole of I , and each mole of I reacts with two moles of S O , we know that every mole of NaOCl in the sample ultimately results in the consumption of two moles of Na S O . The balanced reactions for this analysis are: \[\mathrm{OCl^-}(aq)+\mathrm{3I^-}(aq)+\mathrm{2H^+}(aq)\rightarrow \ce{I_3^-}(aq)+\mathrm{Cl^-}(aq)+\mathrm{H_2O}(l)\] \[\mathrm I_3^-(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow \mathrm{S_4O_6^{2-}}(aq)+\mathrm{3I^-}(aq)\] The moles of Na S O used in reaching the titration’s end point is \[\mathrm{(0.09892\;M\;Na_2S_2O_3)\times(0.00896\;L\;Na_2S_2O_3)=8.86\times10^{-4}\;mol\;Na_2S_2O_3}\] which means the sample contains \[\mathrm{8.86\times10^{-4}\;mol\;Na_2S_2O_3\times\dfrac{1\;mol\;NaOCl}{2\;mol\;Na_2S_2O_3}\times\dfrac{74.44\;g\;NaOCl}{mol\;NaOCl}=0.03299\;g\;NaOCl}\] Thus, the %w/v NaOCl in the diluted sample is \[\mathrm{\dfrac{0.03299\;g\;NaOCl}{25.00\;mL}\times100=0.132\%\;w/v\;NaOCl}\] Because the bleach was diluted by a factor of 40 (25 mL to 1000 mL), the concentration of NaOCl in the bleach is 5.28% (w/v). Example 9.13 The amount of ascorbic acid, C H O , in orange juice was determined by oxidizing the ascorbic acid to dehydroascorbic acid, C H O , with a known amount of I , and back titrating the excess I with Na S O . A 5.00-mL sample of filtered orange juice was treated with 50.00 mL of 0.01023 M I . After the oxidation was complete, 13.82 mL of 0.07203 M Na S O was needed to reach the starch indicator end point. Report the concentration ascorbic acid in mg/100 mL. For a back titration we need to determine the stoichiometry between I and the analyte, C H O , and between I and the titrant, Na S O . The later is easy because we know from Example 9.12 that each mole of I reacts with two moles of Na S O . The balanced reactions for this analysis are: \[\mathrm{C_6H_8O_6}(aq)+\ce{I_3^-}(aq)\rightarrow \mathrm{3I^-}(aq)+\mathrm{C_6H_6O_6}(aq)+\mathrm{2H^+}(aq)\] \[\ce{I_3^-}(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow \mathrm{S_4O_6^{2-}}(aq)+\mathrm{3I^-}(aq)\] In oxidizing ascorbic acid to dehydroascorbic acid, the oxidation state of carbon changes from +⅔ in C H O to +1 in C H O . Each carbon releases ⅓ of an electron, or a total of two electrons per ascorbic acid. As we learned in Example 9.12, reducing I requires two electrons; thus, a conservation of electrons requires that each mole of ascorbic acid consumes one mole of I . The total moles of I reacting with C H O and with Na S O is \[\mathrm{(0.01023\;M\;\ce{I_3^-})\times(0.05000\;L\;\ce{I_3^-})=5.115\times10^{-4}\;mol\;\ce{I_3^-}}\] The back titration consumes \[\mathrm{0.01382\;L\;Na_2S_2O_3\times\dfrac{0.07203\;mol\;Na_2S_2O_3}{L\;Na_2S_2O_3}\times\dfrac{1\;mol\;\ce{I_3^-}}{2\;mol\;Na_2S_2O_3}=4.977\times10^{-4}\;mol\;\ce{I_3^-}}\] Subtracting the moles of I reacting with Na S O from the total moles of I gives the moles reacting with ascorbic acid. \[\mathrm{5.115\times10^{-4}\;mol\;\ce{I_3^-} - 4.977\times10^{-4}\;mol\;\ce{I_3^-}=1.38\times10^{-5}\;mol\;\ce{I_3^-}}\] The grams of ascorbic acid in the 5.00-mL sample of orange juice is \[\mathrm{1.38\times10^{-5}\;mol\;\ce{I_3^-}\times\dfrac{1\;mol\;C_6H_8O_6}{mol\;\ce{I_3^-}}\times\dfrac{176.13\;g\;C_6H_8O_6}{mol\;C_6H_8O_6}=2.43\times10^{-3}\;g\;C_6H_8O_6}\] There are 2.43 mg of ascorbic acid in the 5.00-mL sample, or 48.6 mg per 100 mL of orange juice. Practice Exercise 9.21 A quantitative analysis for ethanol, C H O, can be accomplished by a redox back titration. Ethanol is oxidized to acetic acid, C H O , using excess dichromate, Cr O , which is reduced to Cr . The excess dichromate is titrated with Fe , giving Cr and Fe as products. In a typical analysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetric flask. A 10.00-mL sample is taken and the ethanol is removed by distillation and collected in 50.00 mL of an acidified solution of 0.0200 M K Cr O . A back titration of the unreacted Cr O requires 21.48 mL of 0.1014 M Fe . Calculate the %w/v ethanol in the brandy. to review your answer to this exercise. The scale of operations, accuracy, precision, sensitivity, time, and cost of a redox titration are similar to those described earlier in this chapter for acid–base or a complexation titration. As with acid–base titrations, we can extend a redox titration to the analysis of a mixture of analytes if there is a significant difference in their oxidation or reduction potentials. Figure 9.42 shows an example of the titration curve for a mixture of Fe and Sn using Ce as the titrant. A titration of a mixture of analytes is possible if their standard state potentials or formal potentials differ by at least 200 mV. | 52,192 | 679 |
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Many techniques have been used to probe the geometry and electronic structure of the metal-dioxygen moiety in biological systems and in synthetic models. The results form the basis of any understanding of the factors that determine and modulate oxygen affinity. By resonance Raman techniques, the O—O stretch is observed at 744 cm in oxyhemocycanin and at 844 cm in oxyhemerythrin. Dioxygen is therefore coordinated as peroxo species. By use of unsymmetrically labeled dioxygen, O— O, it was established that dioxygen coordinates symmetrically: \(\tag{4.43}\) Carbon monoxide binds to only one of the Cu centers in deoxyhemocyanin, through the C atom,* apparently blocking the second Cu site. Similar behavior is also seen for the nitrosyl adduct. * A report that CO binds to the copper center through the O atom, an unprecedented mode, has been challenged. On the other hand, in an experiment parallel to that just described for hemocyanin, dioxygen was found to bind asymmetrically in oxyhemerythrin: \(\tag{4.45}\) The existence and location of the proton, which cannot be proven in the crystal structure of deoxyhemerythrin or of oxyhemerythrin, are inferred from a model system for the former that contains a hydroxo group bridging two high-spin, weakly antiferromagnetically coupled Fe centers (J = -10 cm ). For oxyhemerythrin (and for one conformation of methydroxohemerythrin), a small change in the position of the symmetric Fe—O—Fe mode is observed when H O is replaced by D O. The strong antiferromagnetic coupling observed for methemerythrin and oxyhemerythrin (-J ~ 100 cm ) is uniquely consistent with a bridging oxo moiety between a pair of Fe centers. Finally, a Bohr effect (release or uptake of protons) is absent in oxygen binding to hemerythrin. These observations are consistent with a \(\mu\)-oxo group slightly perturbed by hydrogen bonding to a coordinated hydroperoxo species. An important role for the protein in hemerythrin is to assemble an asymmetric diiron(II) species. Only a few of the myriad of known \(\mu\)-oxodiiron(III) complexes are asymmetric, and the synthesis of realistic asymmetric models remains a challenge. Deoxyhemocyanin (Cu d ) and deoxyhemerythrin (Fe d ) are colorless. In the oxygenated derivatives there is considerable charge transfer between the coordinated peroxo groups and the metal centers. This phenomenon makes the essentially d-d metal transitions more intense than those for the simple aquated Fe or Cu ions, and permits facile measurement of oxygen-binding curves. The spectral changes accompanying oxygenation are shown in Figures 4.20 and 4.21. Nitric oxide binds to deoxyhemocyanin, to deoxyhemerythrin, and to the mixed valence Fe • • • Fe semimethemerythrin. Carbon monoxide binds to neither form of hemerythrin: apparently the other ligands have insufficiently strong fields to stabilize the low-spin state for which electron density would be available for back donation into the CO \(\pi\)* orbitals. The O—O stretch that is observed by difference infrared techniques at around 1105 cm for oxyhemoglobin and oxymyoglobin clearly categorizes the dioxygen moiety as a superoxo species; that is, the order of the O—O bond is about 1.5. Considerable ink has been spilled about the nature of the Fe—O fragment since Pauling's original suggestion in 1948 that dioxygen binds to iron in an end-on bent fashion: \(\tag{4.46}\) He subsequently reaffirmed this geometry, and proposed that hydrogen bonding between the coordinated dioxygen and the distal imidazole H—N group was important in stabilizing the Fe—O species. In an alternative model Weiss proposed that a low-spin Fe center (S = \(\frac{1}{2}\)) was very strongly antiferromagnetically coupled to a superoxide anion radical (S = \(\frac{1}{2}\)). A triangular peroxo mode has also been advanced. The problem has been how to resolve the observed diamagnetism of oxyhemoglobin with UV-visible, x-ray absorption, and resonance Raman spectroscopic characteristics that are distinctly different from those of Fe systems (such as carbonmonoxyhemoglobin and low-spin six-coordinated hemochromes, such as Fe(Porph)(Py) ) and from unambiguously Fe systems (such as chloromethemoglobin or cyanomethemoglobin). Any adequate theoretical treatment must also explain how iron-porphyrin systems can bind not only O , but also CO, NO, alkyl isocyanides, and alkyl-nitroso moieties. A simple qualitative model presented by Wayland and coworkers conveniently summarizes ligand-binding geometries of cobalt and iron porphyrins. Although a reasonable quantitative theoretical consensus exists for 1:1 cobalt-dioxygen species, the same cannot be said yet for irondioxygen systems. Why does dioxygen bind to iron and cobalt porphyrins in an end-on bentbond fashion as in (4.37) and (4.46)? Why does carbon monoxide bind in a linear manner (Equation 4.40)? Why are six-coordinate dioxygen and carbonmonoxide adducts more stable than five-coordinate ones? A unified picture of ligand binding that addresses these questions is important in understanding properly the specific case of dioxygen binding to hemoglobin and related systems. The splitting of the metal d orbitals for a four-coordinate metalloporphyrin is shown in the center of Figure 4.22. These orbitals contain some porphyrin character and are antibonding with respect to metal-porphyrin bonds. As shown in Figure 4.16, the primary effect of a single \(\sigma\)-donor axial ligand, such as pyridine or 1-methylimidazole, is to elevate the energy of the antibonding d and lower the energy of the d orbital and hence lead to a high-spin species in place of the intermediate-spin four-coordinate one. Thus, for simplicity in highlighting interaction of the metal center with the diatomic \(\sigma\)-donor: \(\pi\)-acid ligands CO, NO, and O , the perturbations wrought by primarily \(\sigma\)-donor ligands, such as 1-methylimidazole, are omitted. For the corresponding cobalt(II) compound, there is an additional electron. The diatomic ligands of interest share a qualitatively similar molecular orbital scheme. The filling of orbitals for CO is shown on the left-hand side. Dioxygen, which is shown on the right-hand side, has two more electrons than CO; these occupy the doubly degenerate \(\pi\)* orbitals. Quantitative calculations show that the energy of the \(\pi\)* orbitals decreases monotonically from CO to NO to O , indicating increasing ease of reduction of the coordinated molecule, a feature that has not been included in the diagram. Only those interactions of molecular orbitals that have appropriate symmetry and energy to interact significantly with the metal d orbitals are shown. Two extremes are shown in Figure 4.22 for the interaction of a diatomic molecule A-B with the metal center: a linear geometry on the left and a bent geometry on the right. A side-on geometry is omitted for the binding of O to a Co or Fe porphyrin, since this would lead to either an M side-on superoxo or an M peroxo species; both these modes of coordination to these metals are currently without precedent. Linear diatomic metal bonding maximizes the metal-d to ligand-p bonding. When a ligand coordinates in a bent manner, axial symmetry is destroyed, and the degeneracy of the ligand p orbitals is lifted. One p orbital is now oriented to combine with the metal d orbital to form a \(\sigma\) bond, and the other is oriented to combine with d and d orbitals to form a \(\pi\) bond. A bent geometry for the diatomic molecule will result when either or both of the metal d or the ligand p orbitals are occupied, since this geometry stabilizes the occupied d orbital in the five-coordinate complex. Thus O binds in a strongly bent manner to Co and Fe porphyrins; NO binds in a strongly bent manner to Co porphyrins; CO binds in a linear fashion to Fe porphyrins. The interaction of NO with Fe porphyrins and CO with Co porphyrins—the resultant species are formally isoelectronic—is more complicated. The degree of bending seen in Fe (TPP)(NO) is midway between the two extremes. For CO the higher-energy p orbitals lead to a greater mismatch in energy between the d and p orbitals, and less effective \(\sigma\) bonding. In EPR experiments the odd electron is found to be localized in a molecular orbital with about 0.87 metal d character for the five-coordinate Co—CO adduct, as expected for a nearly linear geometry. On the other hand, for the Fe—NO adduct the metal d character of the odd electron is about 0.4 to 0.5; a somewhat bent geometry (140°) is observed in the crystal structure of Fe(TPP)(NO). Because the CO ligand is a very weak \(\sigma\) donor, the Co—CO species exists only at low temperatures. Only qualitative deductions can be made from this model about the extent of electron transfer, if any, from the metal onto the diatomic ligand, especially for dioxygen. The higher in energy the metal d orbital is with respect to the dioxygen p orbitals, the closer the superoxo ligand comes to being effectively a coordinated superoxide anion. With an additional electron, the dioxygen ligand in Co—O complexes can acquire greater electron density than it can in Fe—O complexes. From the diagram it may be inferred that a ligand with very strong \(\pi\)-acceptor properties will lower the energy of the d and d orbitals through strong (d , d )-\(\pi\)* interaction. The resultant energy gap between these two orbitals and the other three metal d orbitals may be sufficient to overcome the energy involved in spin-pairing, and hence lead to five-coordinate low-spin species, as happens for complexes containing phosphines and carbon monosulfide. | 9,688 | 680 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 681 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/Pericyclic_Reactions/PR6._Olefin_Metathesis |
Olefin metathesis, or alkene metathesis, is an important process in petroleum refining and in the synthesis of important compounds such as pharmaceuticals. The mechanism of olefin metathesis is related to pericyclic reactions like and [2+2] reactions. In other words, it occurs through the concerted interaction of one molecule with another. In petroleum refining, heating alkenes over metal oxide surfaces results in the formation of longer-chain alkenes. In particular, terminal olefins (with the double bond at the end of the chain) are converted into internal olefins (with the double bond somewhere in the middle of the chain). What does that reaction have to do with addition reactions involving double bonds? Clearly, the alkenes have double bonds. In addition, so do the metal oxides. Metal atoms inside the metal oxides are bridged together by oxygen atoms. The surface of the metal oxides may be covered with a mixture of hydroxyl groups as well as terminal oxides (M=O groups). The terminal oxides on the surface are the important part of the catalyst. When metal alkylidene complexes were developed in the 1970's, it was found that they, too, could catalyze this reaction. In fact, scientists working in petroleum chemistry soon came to believe that metal oxides on the catalyst surface were converted to alkylidenes, which then carried out the work of olefin metathesis. The reaction, it turns out, involves a [2+2] cycloaddition of an alkene to a metal alkylidene (to the metal-carbon double bond). This reaction results in a four-membered ring, called a metallacyclobutane. The [2+2] cycloaddition is quickly followed by the reverse reaction, a retro-[2+2]. The metallacyclobutane pops open to form two new double bonds. This mechanism is called the Chauvin mechanism, after its first proponent, Yves Chauvin of the French Petroleum Institute. Chauvin's proposal of this mechanism shortly after the discovery of metal alkylidenes by Dick Schrock at DuPont earned him a Nobel Prize in 2005. Chauvin and Schrock shared the prize with Bob Grubbs, who made it possible for the reaction to be adapted easily to the synthesis of complex molecules such as pharmaceuticals. Why does olefin metathesis lead to the formation of internal alkenes? The [2+2] addition and retro-[2+2] reactions occur in equilibrium with each other. Each time the metallacyclobutane forms, it can form two different pairs of double bonds through the retro reaction. In the presence of terminal alkenes, one of those pairs of alkenes will eventually include ethene. Since ethene is a gas, it is easily lost from the system, and equilibrium shifts to the right in the equation below. That leaves a longer-chain alkylidene on the metal, ready to be attached to another long chain through subsequent [2+2] addition and reversion reactions. In most cases, a [2+2] addition will not work unless photochemistry is involved, but it does work with metal alkylidenes. The reason for this exception is thought to involve the nature of the metal-carbon double bond. In contrast to an orbital picture for an alkene, an orbital picture for an alkylidene pi bond suggests orbital symmetry that can easily interact with the LUMO on an alkene. That's because a metal-carbon \(pi\) bond likely involves a d orbital on the metal, and the d orbital has lobes alternating in phase like a \(pi\) antibonding orbital. , | 3,395 | 682 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/19%3A_The_First_Law_of_Thermodynamics/19.03%3A_Work_and_Heat_are_not_State_Functions_but_Energy_is_a_State_Function |
Heat (\(q\)) and work (\(w\)) are path functions, not state functions: Functions that depend on the path taken, such as work (\(w\)) and heat (\(q\)), are referred to as . Let's consider a piston that is being compressed at constant temperature (isothermal) to half of its initial volume: The piston will shoot down till the internal and external pressures balance out again and the volume is 1/2 L. Notice that the external pressure was constant at 2 bar during the peg-pulling and that the internal and external pressures were balanced at all time. In a \(P-V\) diagram of an ideal gas, \(P\) is a hyperbolic function of \(V\) under constant temperature (isothermal), but this refers to the pressure of the gas. It is the external one that counts when computing work and they are not necessarily the same. As long as \(P_{external}\) is constant, work is represented by a rectangle. The amount of work being done is equal to the shaded region and in equation: \[w=-\int^{V_2}_{V_1}{PdV}=-P_{ext}(V_2-V_1)=-P\Delta V \nonumber \] This represents the maximum amount of work that can be done for an isothermal compression. Work is being done on the system, so the overall work being done is positive. Let's repeat the experiment, but this time the piston will compress reversibly over infinitesimally small steps where the \(P_{ext}=P_{system}\): For an ideal gas, the amount of work being done along the reversible compression is: \[w=-\int^{V_2}_{V_1}{PdV}=-nRT\int^{V_2}_{V_1}{\frac{1}{V}}=-nRT\ln{\left(\frac{V_2}{V_1}\right)} \nonumber \] The amount of work being done to the two systems are not the same in the two diagrams (see the gray areas). Work is not a state, but a path function, as it depends on the path taken. You may say, what's the big difference. In both cases, the system is compressed from state 1 to state 2. The key is the word . By pegging the position in place for the first compression, we have created a situation where the external pressure is higher than the internal pressure (\(P_{ext}>P\)). Because work is done suddenly by pulling the peg out, the internal pressure is struggling to catch up with the external one. During the second compression, we have \(P_{ext}=P\) at all times. It's a bit like falling off a cliff versus gently sliding down a hill. Path one is called an irreversible path, the second a reversible path. A reversible path is a path that follows a series of states at rest (i.e., the forces are allowed to balance at all times). In an irreversible one the forces only balance at the very end of the process. Notice that less work is being done on the reversible isothermal compression than the one-step irreversible isothermal compression. In fact, the minimum amount of work that can be done during a compression always occurs along the reversible path. Let's consider a piston that is being expanded at constant temperature (isothermal) to twice of its initial volume: The piston will shoot up till the internal and external pressures balance out again and the volume is 2 L. Notice that the external pressure was constant at 1 bar during the peg-pulling and that the internal and external pressures were balanced at all time. The amount of irreversible work being done is again equal to the shaded region and the equation: \[w=-P\Delta V=-P_{ext}(V_2-V_1)=-P\Delta V \nonumber \] This represents the minimum amount of work that can be done for an isothermal expansion. Work is being done on the system, so the overall work being done is negative. Let's repeat the experiment, but this time the piston will compress reversibly over infinitesimally small steps where the \(P_{ext}=P_{system}\): Notice that not only is more work is being done than the one-step irreversible isothermal expansion, but it is the same amount of work being done as the reversible isothermal compression. This is the maximum amount of work that can be done during an expansion. | 3,938 | 683 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/3%3A_Chemical_Bonding_-_The_Classical_Description/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 684 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.03%3A_Binary_Mixture_in_Equilibrium_with_a_Pure_Phase |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) This section considers a binary liquid mixture of components A and B in equilibrium with either pure solid A or pure gaseous A. The aim is to find general relations among changes of temperature, pressure, and mixture composition in the two-phase equilibrium system that can be applied to specific situations in later sections. In this section, \(\mu\A\) is the chemical potential of component A in the mixture and \(\mu\A^*\) is for the pure solid or gaseous phase. We begin by writing the total differential of \(\mu\A/T\) with \(T\), \(p\), and \(x\A\) as the independent variables. These quantities refer to the binary liquid mixture, and we have not yet imposed a condition of equilibrium with another phase. The general expression for the total differential is \begin{equation} \dif(\mu\A/T) = \bPd{(\mu\A/T)}{T}{p,x\A}\!\dif T + \bPd{(\mu\A/T)}{p}{T,x\A}\!\difp + \bPd{(\mu\A/T)}{x\A}{T,p}\!\dx\A \tag{12.3.1} \end{equation} With substitutions from Eqs. 9.2.49 and 12.1.3, this becomes \begin{equation} \dif(\mu\A/T) = -\frac{H\A}{T^2}\dif T + \frac{V\A}{T}\difp + \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A \tag{12.3.2} \end{equation} Next we write the total differential of \(\mu\A^*/T\) for pure solid or gaseous A. The independent variables are \(T\) and \(p\); the expression is like Eq. 12.3.2 with the last term missing: \begin{equation} \dif(\mu\A^*/T) = -\frac{H\A^*}{T^2}\dif T + \frac{V\A^*}{T}\difp \tag{12.3.3} \end{equation} When the two phases are in transfer equilibrium, \(\mu\A\) and \(\mu\A^*\) are equal. If changes occur in \(T\), \(p\), or \(x\A\) while the phases remain in equilibrium, the condition \(\dif(\mu\A/T) = \dif(\mu\A^*/T)\) must be satisfied. Equating the expressions on the right sides of Eqs. 12.3.2 and 12.3.3 and combining terms, we obtain the equation \begin{equation} \frac{H\A-H\A^*}{T^2}\dif T - \frac{V\A-V\A^*}{T}\difp = \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A \tag{12.3.4} \end{equation} which we can rewrite as \begin{gather} \s{ \frac{\Delsub{sol,A}H}{T^2}\dif T - \frac{\Delsub{sol,A}V}{T}\difp = \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A } \tag{12.3.5} \cond{(phases in} \nextcond{equilibrium)} \end{gather} Here \(\Delsub{sol,A}H\) is the molar differential enthalpy of solution of solid or gaseous A in the liquid mixture, and \(\Delsub{sol,A}V\) is the molar differential volume of solution. Equation 12.3.5 is a relation between changes in the variables \(T\), \(p\), and \(x\A\), only two of which are independent in the equilibrium system. Suppose we set \(\difp\) equal to zero in Eq. 12.3.5 and solve for \(\dif T/\dx\A\). This gives us the rate at which \(T\) changes with \(x\A\) at constant \(p\): \begin{gather} \s{ \Pd{T}{x\A}{\!p} = \frac{T^2}{\Delsub{sol,A}H} \bPd{(\mu\A/T)}{x\A}{T,p} } \tag{12.3.6} \cond{(phases in} \nextcond{equilibrium)} \end{gather} We can also set \(\dif T\) equal to zero in Eq. 12.3.5 and find the rate at which \(p\) changes with \(x\A\) at constant \(T\): \begin{gather} \s{ \Pd{p}{x\A}{T} = -\frac{T}{\Delsub{sol,A}V}\bPd{(\mu\A/T)}{x\A}{T,p} } \tag{12.3.7} \cond{(phases in} \nextcond{equilibrium)} \end{gather} Equations 12.3.6 and 12.3.7 will be needed in Secs. 12.4 and 12.5. | 10,511 | 685 |
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There are many ways in which ligand affinity may be perturbed (Figure 4.25). It is convenient to divide these into two groups, referred to as distal and proximal effects. effects are associated with the stereochemistry of the metalloporphyrinato moiety and the coordination of the axial base, and thus their influence on O and CO affinity is indirect. effects pertain to noncovalent interactions of the metal-porphyrinato skeleton and the sixth ligand (O , CO, etc.) with neighboring solvent molecules, with substituents, such as pickets or caps, on the porphyrin, and with the surrounding protein chain. The distal groups that hover over the O -binding site engender the most important distal effects. For convenience, the effects of crystal packing and the protein matrix on porphyrin conformation will be discussed among the proximal effects, although as nonbonded interactions they properly are distal effects. To a first approximation, the effects of substituents on the porphyrin ring, as transmitted through bonds to the metal center, do not perturb the ligand binding properties as much as do distal effects. Thus substituents, such as vinyl and propionic-acid groups on protoporphyrin IX and o-pivalamidophenyl pickets, are ignored; one porphyrin is much like another. At the end of this subsection the various ways ligand affinity may be modulated will be summarized in an augmented version of Figure 4.3. Few molecules have had their conformational properties characterized as exhaustively as have metalloporphyrins. The cyclic aromatic 24-atom porphyrinato skeleton offers a tightly constrained metal-binding site. The conformation of least strain is planar, and the radius of the hole of the dianion is close to 2.00 Å, leading to metal-porphyrinato nitrogen-atom separations, M-N , of 2.00 Å if the metal is centered in the square plane defined by the four porphyrinato nitrogen atoms. Small deviations from planarity are generally observed and attributed to crystal packing effects; large deviations may be induced by bulky substituents on the porphyrin skeleton, especially at the positions, by the crystal matrix, or by the highly anisotropic protein matrix. The 2.00 Å radius hole neatly accommodates low-spin (S = 0) and intermediatespin (S = 1) iron(II), low-spin (S = \(\frac{1}{2}\)) iron(III), and cobalt(II) and cobalt(III) ions. * With few exceptions the metal is centered in or above the central hole for mononuclear porphyrin species; only rarely do M-N bonds show a significant (though still small) scatter about their mean value. * In order to accommodate smaller ions, such as nickel(II), the porphyrin skeleton may contract by ruffling, with little loss of aromaticity; like a pleated skirt the pyrrole rings rotate alternately clockwise and counterclockwise about their respective M-N vectors. This distortion leaves the four porphyrinato nitrogen atoms, N , still coplanar. Alternatively, the porphyrin skeleton may buckle to give a saddle conformation; the N atoms may acquire a small tetrahedral distortion in this process. M-N bonds as short as 1.92 Å have been observed. Metals with one or two electrons in their 3d orbital have a radius larger than 2.00 Å. In order to accommodate them in the plane of the porphyrin, the porphyrin skeleton expands. M-N separations as long as 2.07 Å may occur with the metal still centered in the plane of the N atoms. For five-coordinate complexes the magnitude of the displacement of the metal from the plane of the four nitrogen atoms, M • • • porph, is a consequence of the electronic configuration of ML complexes. Of course, the effect is augmented if the 3d orbital (directed along M-N bonds, Figure 4.16) is occupied. Compare a displacement of 0.14 Å for Co(TPP)(1,2-Me Im) (no 3d occupancy) with 0.43 Å for Fe(PF)(2-MeIm) (3d occupied). For six-coordinate complexes where the two axial ligands, L and L , are different, the M • • • porph displacement usually reflects relative influences. Generally, displacement of the metal from the plane of the porphyrinatonitrogen atoms is within 0.04 Å of the displacement from the 24-atom mean plane of the entire porphyrin skeleton. On occasions this second displacement may be much larger, for example in Fe(TPP)(2-MeIm), where it is 0.15 Å larger than it is for Fe(PF)(2-MeIm). This effect is called , and it is usually attributed to crystal packing forces. Interaction of the porphyrin with protein side chains leads to considerable doming or folding of the heme in vertebrate hemoglobins. The metal-axial ligand separations, M-L (when more than one, L denotes the heterocyclic axial base), are dependent on the nature of the ligand, L. When L and L are different, the M-L separations are sensitive to the relative influences of L and L as well as to steric factors. For example, for Fe(TPP)(1-Melm) , the Fe—N bond length is 2.016(5) Å, whereas for Fe(TPP)(1-Melm)(NO) it is 2.180(4) Å. For sterically active ligands, such as 2-methylimidazole compared to 1-methylimidazole (4.34), the longer Co—N bond occurs for the 2-Melm ligand because of steric clash between the 2-methyl group and the porphyrin. It is possible that combinations of intrinsic bonding and steric factors may give rise to a double minimum and two accessible axial ligand conformations (Figure 4.26). This situation seems to occur in the solid state for Fe(PF)(2-Melm)(O )•EtOH, where a short Fe—N and a long Fe—O bond are observed both from the structure revealed by single-crystal x-ray diffraction methods and by EXAFS data. On the other hand, for solvate-free Fe(PF)(2-MeIm)(O ) and for Fe(PF)(1,2-Me Im)(O ), the EXAFS patterns are interpreted in terms of a short Fe—O and long Fe—Im bond. This parameter is the minimum angle that the plane of the axial base (e.g., pyridine, substituted imidazole, etc.) makes with a plane defined by the N , M, and L atoms (Figure 4.25). If there are two axial ligands, e.g., 1-methylimidazole and O , then, as before, the angle the axial base makes is denoted \(\phi_{1}\) and the other angle \(\phi_{2}\). For a linear CO ligand bound perpendicularly to the porphyrin plane, \(\phi_{2}\) is undefined. Note that the orientation of the second ligand is influenced by distal effects. When \(\phi\) = 0, the axial base eclipses a pair of M-N bonds; contacts with the porphyrin are maximized. When \(\phi\) = 45°, contacts are minimized. Unless the axial base has a 2-substituent, however, the contacts are not excessively close for any value of \(\phi\). With a 2-methyl substituent, the contacts are sufficiently severe that the M-N vector is no longer perpendicular to the porphyrin plane, and the imidazole group is rotated so that the M-N vector no longer approximately bisects the imidazole C—N—C bond angle, as illustrated Figure 4.25. Distal effects arise from noncovalent interactions of the coordinated dioxygen, carbon monoxide, or other ligand with its surroundings. The protein matrix, the pickets, and the caps are functionally equivalent to an anisotropic solvent matrix that contains a variety of solutes. The limits of this simplification are illustrated in the following example. The electronically similar cobalt meso-, deutero-, and protoporphyrin IX complexes bind dioxygen with similar affinities under identical solvent conditions. When they are embedded in globin, larger differences in affinity and changes in cooperativity are observed. These effects are attributed to the slightly different nestling of the porphyrin molecules in the cleft in hemoglobin or, in the generalization introduced, to slightly different solvation effects. Interaction of the coordinated O or CO molecule with solvent molecules or with the protein has a profound influence on kinetics and thermodynamics (see Figure 4.24, and Tables 4.2 and 4.5). As discussed earlier, there is accumulation of negative charge on the dioxygen ligand. The possibility then arises for stabilization of coordination through hydrogen bonding or dipolar interactions with solute molecules, porphyrin substituents (such as amide groups in the picket-fence porphyrins and some species of strapped porphyrins ), or with protein residues* (such as histidine). Destabilization of coordinated ligands and lowered affinity can result if the coordinated ligand is unable, through steric clash, to achieve its optimum stereochemistry or if the closest neighboring groups are electronegative, as are the ether and ester linkages on capped porphyrins. We will describe in detail in the next subsection (III.C) the fascinating variety of means by which ligand binding is modulated by distal amino-acid residues. * For CoMbO no change in EPR parameters occurs on substituting D O for H O. No hydrogen bond between O and a distal group comparable in strength to that in whale CoMbO was inferred. Dissimilar systems may show similar affinities for a ligand as a result of a different mix of the proximal and distal effects enumerated above. These effects are not all of equal magnitude, and an attempt is made here to show the increment in free energy that occurs if the effect is manifest in the deoxy or liganded state of Figure 4.3. Increasing the free energy of the deoxy state while holding that of the liganded state constant leads to an increase in affinity. The reference state is gaseous Fe(TPP)(1-MeIm). The magnitude and sign of these effects are shown in Figure 4.27. For the coordination of alkylisocyanide molecules to hemoglobin, the steric effects of different alkyl groups have been quantified. Lowered affinity occurs with increasing alkyl chain length, with the exception of methyl isocyanide. | 9,668 | 686 |
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If we know \(\mu\) and we have a set of \(N\) data points, the best estimate we can make of the variance is \[\sigma^2=\int^{u_{max}}_{u_{min}}{\left(u-\mu \right)}^2\left(\frac{df}{du}\right)du \approx \sum^N_{i=1}{\left(u_i-\mu \right)}^2\left(\frac{1}{N}\right)\] We have said that if we must use \(\overline{u}\) to approximate the mean, the best estimate of \(\sigma^2\), usually denoted \(s^2\), is \[estimated\ \sigma^2=s^2 =\sum^N_{i=1}{\left(u_i-\overline{u}\right)}^2\left(\frac{1}{N-1}\right)\] The use of \(N-1\), rather than \(N\), in the denominator is distinctly non-intuitive; so much so that this equation often causes great irritation. Let us see how this equation comes about. Suppose that we have a distribution whose mean is \(\mu\) and variance is \(\sigma^2\). Suppose that we draw \(N\) values of the random variable, \(u\), from the distribution. We want to think about the expected value of \({\left(u-\mu \right)}^2\). Let us write \(\left(u-\mu \right)\) as \[\left(u-\mu \right)=\left(u-\overline{u}\right)+\left(\overline{u}-\mu \right).\] Squaring this gives \[{\left(u-\mu \right)}^2={\left(u-\overline{u}\right)}^2+{\left(\overline{u}-\mu \right)}^2+2\left(u-\overline{u}\right)\left(\overline{u}-\mu \right).\] From our definition of expected value, we can write: \[ \begin{array}{l} \text{Expected value of } \left(u-\mu \right)^2= \\ ~~~~ =expected\ value\ of\ \ {\left(u-\overline{u}\right)}^2 \\ \ \ \ \ +expected\ value\ of\ {\left(\overline{u}-\mu \right)}^2 \\ \ \ \ \ +expected\ value\ of\ 2\left(u-\overline{u}\right)\left(\overline{u}-\mu \right) \end{array}\] From our discussion above, we can recognize each of these expected values: Substituting, our expression for the expected value of \({\left(u-\mu \right)}^2\) becomes: \[\sigma^2\approx \sum^N_{i=1} \left(u_i-\overline{u}\right)^2\left(\frac{1}{N}\right)+\frac{\sigma^2}{N}\] so that \[\sigma^2\left(1-\frac{1}{N}\right)=\sigma^2\left(\frac{N-1}{N}\right)\approx \sum^N_{i=1} \frac{\left(u_i-\overline{u}\right)^2}{N}\] and \[\sigma^2 \approx \sum^N_{i=1} \frac{\left(u_i-\overline{u}\right)^2}{N-1}\] That is, as originally stated, when we must use \(\overline{u}\) rather than the true mean, \(\mu\), in the sum of squared differences, the best possible of \(\sigma^2\), usually denoted \(s^2\), is obtained by dividing by \(N-1\), rather than by \(N\) | 2,372 | 689 |
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World production of rhodium (from the Greek rhodon, "rose") is about 10 tons. While the metal itself has few applications, it is an important alloying agent used as a hardener for platinum and palladium. Rhodium was discovered in 1803 by William Hyde Wollaston who named it for the rose-red color of its salts. Rhodium is part of the the Platinum Group Metals (PGM) whic is located in the 5th and 6th rows of the transition metal section of the periodic table and includes , Rhodium, , , , and . Common characteristics include resistance to wear, oxidation, and corrosion, high melting points, and oxidation states of +2 to +4. They are generally non-toxic. It shares the properties of these metals: high corrosion resistance, hardness and ductility. It is the rarest of the group, only occurring to the extent of about 1 part per 200 million in the earth's crust. | 889 | 690 |
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The Beer-Lambert law relates the attenuation of light to the properties of the material through which the light is traveling. This page takes a brief look at the Beer-Lambert Law and explains the use of the terms absorbance and molar absorptivity relating to UV-visible absorption spectrometry. For each wavelength of light passing through the spectrometer, the intensity of the light passing through the reference cell is measured. This is usually referred to as \(I_o\) - that's \(I\) for Intensity. The intensity of the light passing through the sample cell is also measured for that wavelength - given the symbol, \(I\). If \(I\) is less than \(I_o\), then the sample has absorbed some of the light (neglecting reflection of light off the cuvette surface). A simple bit of math is then done in the computer to convert this into something called the absorbance of the sample - given the symbol, \(A\). The absorbance of a transition depends on two external assumptions. relates the absorbance to concentration and can be expressed as \[A \propto c \label{1}\] The absorbance (\(A\)) is defined via the incident intensity \(I_o\) and transmitted intensity \(I\) by \[ A=\log_{10} \left( \dfrac{I_o}{I} \right) \label{2}\] can be expressed as \[A \propto l \label{3}\] Combining Equations \(\ref{1}\) and \(\ref{3}\): \[A \propto cl \label{4}\] This proportionality can be converted into an equality by including a proportionality constant (\(\epsilon\)). \[A = \epsilon c l \label{5}\] This formula is the common form of the , although it can be also written in terms of intensities: \[ A=\log_{10} \left( \dfrac{I_o}{I} \right) = \epsilon l c \label{6} \] The constant \(\epsilon\) is called or and is a measure of the probability of the electronic transition. On most of the diagrams you will come across, the absorbance ranges from 0 to 1, but it can go higher than that. An absorbance of 0 at some wavelength means that no light of that particular wavelength has been absorbed. The intensities of the sample and reference beam are both the same, so the ratio \(I_o/I\) is 1 and the \(\log_{10}\) of 1 is zero. In a sample with an absorbance of 1 at a specific wavelength, what is the relative amount of light that was absorbed by the sample? This question does not need Beer-Lambert Law (Equation \(\ref{5}\)) to solve, but only the definition of absorbance (Equation \(\ref{2}\)) \[ A=\log_{10} \left( \dfrac{I_o}{I} \right)\nonumber\] The relative loss of intensity is \[\dfrac{I-I_o}{I_o} = 1- \dfrac{I}{I_o}\nonumber\] Equation \(\ref{2}\) can be rearranged using the properties of logarithms to solved for the relative loss of intensity: \[ 10^A= \dfrac{I_o}{I}\nonumber\] \[ 10^{-A}= \dfrac{I}{I_o}\nonumber\] \[ 1-10^{-A}= 1- \dfrac{I}{I_o}\nonumber \] Substituting in \(A=1\) \[ 1- \dfrac{I}{I_o}= 1-10^{-1} = 1- \dfrac{1}{10} = 0.9\nonumber\] Hence 90% of the light at that wavelength has been absorbed and that the transmitted intensity is 10% of the incident intensity. To confirm, substituting these values into Equation \(\ref{2}\) to get the absorbance back: \[\dfrac{I_o}{I} = \dfrac{100}{10} =10 \label{7a}\] and \[\log_{10} 10 = 1 \label{7b}\] You will find that various different symbols are given for some of the terms in the equation - particularly for the concentration and the solution length. The Greek letter epsilon in these equations is called the - or sometimes the molar absorption coefficient. The larger the molar absorptivity, the more probable the electronic transition. In uv spectroscopy, the concentration of the sample solution is measured in mol L and the length of the light path in cm. Thus, given that absorbance is unitless, the units of molar absorptivity are L mol cm . However, since the units of molar absorptivity is always the above, it is customarily reported without units. has a maximum absorbance of 275 nm. \(\epsilon_{275} = 8400 M^{-1} cm^{-1}\) and the path length is 1 cm. Using a spectrophotometer, you find the that \(A_{275}= 0.70\). What is the concentration of guanosine? To solve this problem, you must use Beer's Law. \[A = \epsilon lc \] 0.70 = (8400 M cm )(1 cm)(\(c\)) Next, divide both side by [(8400 M cm )(1 cm)] \(c\) = 8.33x10 mol/L There is a substance in a solution (4 g/liter). The length of cuvette is 2 cm and only 50% of the certain light beam is transmitted. What is the extinction coefficient? Using Beer-Lambert Law, we can compute the absorption coefficient. Thus, \(- \log \left(\dfrac{I_t}{I_o} \right) = - \log(\dfrac{0.5}{1.0}) = A = {8} \epsilon\) Then we obtain that \(\epsilon\) = 0.0376 In Example 3 above, what is the molar absorption coefficient if the molecular weight is 100? It can simply obtained by multiplying the absorption coefficient by the molecular weight. Thus, \(\epsilon\) = 0.0376 x 100 = 3.76 L·mol ·cm The proportion of the light absorbed will depend on how many molecules it interacts with. Suppose you have got a strongly colored organic dye. If it is in a reasonably concentrated solution, it will have a very high absorbance because there are lots of molecules to interact with the light. However, in an incredibly dilute solution, it may be very difficult to see that it is colored at all. The absorbance is going to be very low. Suppose then that you wanted to compare this dye with a different compound. Unless you took care to make allowance for the concentration, you couldn't make any sensible comparisons about which one absorbed the most light. In Example \(\Page {3}\) above, how much is the beam of light is transmitted when 8 g/liter ? Since we know \(\epsilon\), we can calculate the transmission using Beer-Lambert Law. Thus, \(log(1) - log(I_t) = 0 - log(I_t)\) = 0.0376 x 8 x 2 = 0.6016 \(log(I_t)\) = -0.6016 Therefore, \(I_t\) = 0.2503 = 25% The absorption coefficient of a glycogen-iodine complex is 0.20 at light of 450 nm. What is the concentration when the transmission is 40 % in a cuvette of 2 cm? It can also be solved using Beer-Lambert Law. Therefore, \[- \log(I_t) = - \log_{10}(0.4) = 0.20 \times c \times 2\] Then \(c\) = 0.9948 Suppose this time that you had a very dilute solution of the dye in a cube-shaped container so that the light traveled 1 cm through it. The absorbance is not likely to be very high. On the other hand, suppose you passed the light through a tube 100 cm long containing the same solution. More light would be absorbed because it interacts with more molecules. Again, if you want to draw sensible comparisons between solutions, you have to allow for the length of the solution the light is passing through. Both concentration and solution length are allowed for in the Beer-Lambert Law. The Beer-Lambert law (Equation \(\ref{5}\)) can be rearranged to obtain an expression for \(\epsilon\) (the molar absorptivity): \[ \epsilon = \dfrac{A}{lc} \label{8}\] Remember that the absorbance of a solution will vary as the concentration or the size of the container varies. Molar absorptivity compensates for this by dividing by both the concentration and the length of the solution that the light passes through. Essentially, it works out a value for what the absorbance would be under a standard set of conditions - the light traveling 1 cm through a solution of 1 mol dm . That means that you can then make comparisons between one compound and another without having to worry about the concentration or solution length. Values for molar absorptivity can vary hugely. For example, ethanal has two absorption peaks in its UV-visible spectrum - both in the ultra-violet. One of these corresponds to an electron being promoted from a lone pair on the oxygen into a pi anti-bonding orbital; the other from a \(\pi\) bonding orbital into a \(\pi\) anti-bonding orbital. Table 1 gives values for the molar absorptivity of a solution of ethanal in hexane. Notice that there are no units given for absorptivity. That's quite common since it assumes the length is in cm and the concentration is mol dm , the units are mol dm cm . The ethanal obviously absorbs much more strongly at 180 nm than it does at 290 nm. (Although, in fact, the 180 nm absorption peak is outside the range of most spectrometers.) You may come across diagrams of absorption spectra plotting absorptivity on the vertical axis rather than absorbance. However, if you look at the figures above and the scales that are going to be involved, you aren't really going to be able to spot the absorption at 290 nm. It will be a tiny little peak compared to the one at 180 nm. To get around this, you may also come across diagrams in which the vertical axis is plotted as log (molar absorptivity). If you take the logs of the two numbers in the table, 15 becomes 1.18, while 10,000 becomes 4. That makes it possible to plot both values easily, but produces strangely squashed-looking spectra! Jim Clark ( ) | 8,857 | 693 |
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More than 100 years ago, Kekule recognized the possible existence of other conjugated cyclic polyalkenes, which at least superficially would be expected to have properties like benzene. The most interesting of these are cyclobutadiene, \(23\), and cyclooctatetraene, \(24\): For each we can write two equivalent planar VB structures, and the qualitative VB method would suggest that both compounds, like benzene, have substantial electron-delocalization energies. However, the planar structures would have abnormal \(\ce{C-C=C}\) angles, and consequently at least some degree of associated with these bond angles ( ). Nonetheless, estimation of the strain energies show that while they are substantial, they are not prohibitive. Should then these molecules be stabilized by resonance in the same sense as benzene is postulated to be? In 1911 a German chemist, R. Willstatter (Nobel Prize 1915), reported an extraordinary thirteen-step synthesis of cyclooctatetraene from a rare alkaloid called pseudopelletierine isolated from the bark of pomegranate trees. The product was reported to be a light-yellow, highly unsaturated compound that absorbed four moles of hydrogen to form cyclooctane. Numerous tries to repeat the Willstatter synthesis were unsuccessful, and in the 1930s the prevailing opinion was that the product had been misidentified. However, during the Second World War, the German chemist W. Reppe found that cyclooctatetraene can be made in reasonable yields by the tetramerization of ethyne under the influence of a nickel cyanide catalyst: The properties of the product substantiated Willstatter's reports and it became clear that cyclooctatetraene is like benzene. Subsequent studies of the geometry of the molecule revealed further that it is , with alternating single and double bonds, \(25a\): This geometry precludes the possibility of two equivalent VB structures, as for benzene, because, as you will see if you try to make a ball-and-stick model, \(25b\) is highly strained and not energetically equivalent to \(25a\) at all. Thus we can conclude that the delocalization energy of cyclooctatetraene is not large enough to overcome the angle strain that would develop if the molecule were to become planar and allow the \(\pi\) electrons to form equivalent \(\pi\) bonds between all of the pairs of adjacent carbons. Cyclobutadiene, \(23\), eluded Kekule, Willstatter, and a host of other investigators for almost 100 years. As more work was done, it became increasingly clear that the molecule, when formed in reactions, was immediately converted to something else. Finally, the will-o'-the-wisp was captured in an essentially rigid matrix of argon at \(8 \: \text{K}\). It was characterized by its spectral properties (not by combustion analysis). On warming to even \(35 \: \text{K}\), it dimerizes to yield \(26\): One possibility for the lack of stability\(^4\) of cyclobutadiene is that the angle strain associated with having four \(sp^2\) carbons in a four-membered ring is much greater than estimated. However, the stable existence of many compounds with four such \(sp^2\) carbons, for example \(27\) and \(28\), make this argument weak, if not invalid: Why, then, is cyclobutadiene so unstable and reactive? On this point, and also with respect to the nonaromatic character of cyclooctatetraene, the simple qualitative VB method that we have outlined is no help whatsoever. There is no way simply to look at the electron-pairing schemes \(23\) and \(24\) and see any difference between them and the corresponding schemes for benzene.\(^5\) It is in this area that qualitative MO procedures have great success because there are general characteristics of the \(\pi\) molecular orbitals of monocyclic, conjugated polyene systems that predict differences in the properties of cyclobutadiene, benzene, cyclooctatetraene, and other similar compounds that are not obvious from the simple VB method. As a rule, for \(N\) parallel atomic \(p\) orbitals overlapping in the \(\pi\) manner in a monocyclic array, there will be just one lowest molecular orbital, with all the atomic orbitals having the same phase. This will be seen for benzene in Figure 21-5. What is harder to understand without going through the calculations is that the higher-energy molecular orbitals for cyclic conjugated polyenes are predicted to come in successive degenerate\(^6\) pairs, as shown in Figure 21-13 for \(N = 3\) to \(9\). The qualitative ordering and, indeed, the numerical values of the energies of the \(\pi\) molecular orbitals for a cyclic system of \(N\) \(p\) orbitals can be derived in a very simple way. It is necessary only to inscribe a regular polygon with \(N\) sides inside a circle of radius \(2 \beta\) with . For example, for \(N = 5\) we get the following: The molecular orbital energies are in units of \(\beta\) at . The nonbonding level corresponds to the horizontal dashed line drawn through the of the circle. The data of Figure 21-13 provide a rationale for the instability of cyclobutadiene and cyclooctatetraene. For cyclobutadiene, we can calculate that four \(\pi\) electrons in the lowest orbitals will lead to a predicted \(\pi\)-electron energy of \(2 \left( \alpha + 2 \beta \right) + 2 \left( \alpha \right) = 4 \alpha + 4 \beta\), which is just the \(\pi\)-electron energy calculated for (see Figure 21-3). The delocalization energy of the \(\pi\) electrons of cyclobutadiene therefore is predicted to be zero! Another feature of the \(\pi\) system of cyclobutadiene is that the four \(\pi\) electrons do not suffice to fill the three lowest orbitals and, if we apply Hund's rule ( ), the best way to arrange the electrons is as in \(29\), with two electrons, which is known as a :\(^7\) With the MO predictions of zero delocalization energy and an electronic configuration with unpaired electrons, we should not be surprised that cyclobutadiene readily dimerizes to give \(26\) even at very low temperatures. The energies of the molecular orbitals calculated for cyclooctatetraene (Figure 21-13) lead to a predicted delocalization energy of \(\left( 8 \alpha + 9.64 \beta \right) - \left( 8 \alpha + 8 \beta \right) = 1.64 \beta\) \(\left( \sim 31 \: \text{kcal} \right)\), which is smaller than that of benzene, even though there are eight atomic orbitals instead of six through which the electrons are delocalized. Furthermore, the lowest electronic configuration for the planar molecule is, like cyclobutadiene, predicted to be a triplet. Experimental evidence indicates that the positions of the double bonds of cyclooctatetraene shift slowly as the result of formation of the molecule in the unstable planar state. The energy input required to flatten the molecule is about \(15 \: \text{kcal mol}^{-1}\): As Huckel formulated, the \(4n + 2\) rule applies only to monocyclic systems. However, as a practical matter it can be used to predict the properties of polycyclic conjugated polyenes, provided the important VB structures involve only the perimeter double bonds, as in the following examples: Application of the \(4n + 2\) rule to other \(\pi\) systems, such as \(30\) and \(31\), is not valid because good VB structures cannot be written that involve changes in the pairing schemes of the perimeter electrons all at once. The hydrogens of the \(\ce{-CH_2}-\) group of 1,3-cyclopentadiene are acidic. In fact, they are considerably more acidic than the ethyne hydrogens of the 1-alkynes ( ). This means that 1,3-cyclopentadiene is at least \(10^{30}\) times more acidic than the ordinary alkanes. The reason is that loss of one of the \(\ce{CH_2}\) protons of cyclopentadiene results in formation of an especially stabilized anion: The structure of the anion may be described as a hybrid of energetically equivalent structures, \(34a\) through (34e). The unshared electron pair therefore is delocalized over five carbon atoms, and the resulting delocalized anion is much more stable than expected for any of the equivalent localized structures: This looks very reasonable, although the simple beauty is seemingly destroyed by the fact that the cyclopentadienyl is not very stable, despite the five structures, \(35a\) through \(35e\), that may be written for it: Extension of these ideas to the other ring sizes of Figure 21-13 suggests that all of the following ions, which have \(\left( 4n + 2 \right)\) \(\pi\) electrons, should be unusually stable: In contrast, the following should be unstable with \(4n\) \(\pi\) electrons and triplet electronic configurations: These predictions indeed are borne out by many experiments, some of which we will discuss later. \(^4\)It should be recognized that the term "stability" is subject to many interpretations. One criterion of stability would be whether an isolated molecule would fragment spontaneously in interstellar space, such as one would expect for a "molecule" consisting of two neon atoms \(1.5 \: \text{Å}\) apart (see Figure 4-6). A different criterion would be whether a molecule could be preserved in the presence of the same or other kinds of molecules at some specified temperature. The classical criterion would be whether the substance could be isolated, put into a bottle and preserved for at least a short time. All of the existing evidence indicates that cyclobutadiene molecules would not spontaneously decompose in interstellar space, but they do react with each other extremely readily, even at low temperatures, and can be preserved only by being held isolated from one another in a rigid matrix of a very inert material, such as solid argon. "Stability" in the sense of "lack of reactivity" has to be carefully defined in terms of experimental conditions. For example, is very in the presence of nucleophiles such as water or methanol, whereas it is quite in "super-acid solutions" where no good nucleophiles are present ( ). \(^5\)A rather simple extension of the VB method by what is called the "orbital-phase continuity principle" does permit the qualitative judgment that cyclobutadiene should be less stable than benzene [see W. A. Goddard III, , 743 (1972), for applications to many processes for which VB theory generally has been regarded as incapable of giving much insight]. \(^6\)Degenerate orbitals have the same energy; see . \(^7\)The name "triplet state" is used because a system with two unpaired electrons has different energy states in a magnetic field. and (1977) | 10,466 | 696 |
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are highly specific catalysts for biochemical reactions, with each enzyme showing a selectivity for a single reactant, or . For example, the enzyme acetylcholinesterase catalyzes the decomposition of the neurotransmitter acetylcholine to choline and acetic acid. Many enzyme–substrate reactions follow a simple mechanism that consists of the initial formation of an enzyme–substrate complex, \(ES\), which subsequently decomposes to form product, releasing the enzyme to react again. This is described within the following multi-step mechanism \[ E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\overset{k_2}{\rightleftharpoons}} E + P \label{13.20}\] where \(k_1\), \(k_{–1}\), \(k_2\), and \(k_{–2}\) are rate constants. The reaction’s rate law for generating the product \([P]\) is \[ rate = \dfrac{d[P]}{dt} = k_2[ES] - k_{-2}[E,P] \label{13.21A}\] However, if we make measurement early in the reaction, the concentration of products is negligible, i.e., \[[P] \approx 0\] and we can ignore the back reaction (second term in right side of Equation \(\ref{13.21A}\)). Then under these conditions, the reaction’s rate is \[ rate = \dfrac{d[P]}{dt} = k_2[ES] \label{13.21}\] To be analytically useful we need to write Equation \(\ref{13.21}\) in terms of the reactants (e.g., the concentrations of enzyme and substrate). To do this we use the , in which we assume that the concentration of \(ES\) remains essentially constant. Following an initial period, during which the enzyme–substrate complex first forms, the rate at which \(ES\) forms \[ \dfrac{d[ES]}{dt} =k_1[E] [S] = k_1([E]_0 − [ES])[S] \label{13.22}\] is equal to the rate at which it disappears \[ − \dfrac{d[ES]}{dt} = k_{−1}[ES] + k_2[ES] \label{13.23}\] where \([E]_0\) is the enzyme’s original concentration. Combining Equations \(\ref{13.22}\) and \(\ref{13.23}\) gives \[k_1([E]_0 − [ES]) [S] = k_{−1}[ES] + k_2[ES]\] which we solve for the concentration of the enzyme–substrate complex \[ [ES] = \dfrac{ [E]_0[S] }{ \dfrac{k_{−1} + k_2}{k_1} + [S] } = \dfrac{[E]_0[S] }{K_m + [S]} \label{Eq13.24}\] where \(K_m\) is the . Substituting Equation \(\ref{Eq13.24}\) into Equation \(\ref{13.21}\) leaves us with our final rate equation. \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \label{Eq13.25} \] A plot of Equation \(\ref{Eq13.25}\), as shown in Figure \(\Page {1}\), is instructive for defining conditions where we can use the rate of an enzymatic reaction for the quantitative analysis of an enzyme or substrate. For high substrate concentrations, where \([S] \gg K_m\), Equation \(\ref{Eq13.25}\) simplifies to \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{ K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{[S]} = k_2[E]_0 = V_{max} \label{Eq13.26}\] where \(V_{max}\) is the maximum rate for the catalyzed reaction. Under these conditions the reaction is in substrate and we can use \(V_{max}\) to calculate the enzyme’s concentration, typically using a variable-time method. At lower substrate concentrations, where \([S] \ll K_m\), Equation \(\ref{Eq13.25}\) becomes \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{ K_m} =\dfrac{V_{max}[S]}{K_m} \label{13.27}\] The reaction is now in substrate, and we can use the rate of the reaction to determine the substrate’s concentration by a fixed-time method. The Michaelis constant \(K_m\) is the substrate concentration at which the reaction rate is at half-maximum, and is an inverse measure of the substrate's affinity for the enzyme—as a small \(K_m\) indicates high affinity, meaning that the rate will approach \(V_{max}\) more quickly. The value of \(K_m\) is dependent on both the enzyme and the substrate, as well as conditions such as temperature and pH. The Michaelis constant \(K_m\) is the substrate concentration at which the reaction rate is at half-maximum. From the last two terms in Equation \(\ref{13.27}\), we can express \(V_{max}\) in terms of a (\(k_{cat}\)): \[ V_{max} = k_{cat}[E]_o\] where \([E]_0\) is the enzyme concentration and \(k_{cat}\) is the turnover number, defined as the maximum number of substrate molecules converted to product per enzyme molecule per second. Hence, the turnover number is defined as the maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration \([E]_o\). Acetylcholinesterase (AChE) may be one of the fastest enzymes. It hydrolyzes acetylcholine to choline and an acetate group. One of the earliest values of the turnover number was \(3 \times 10^7\) (molecules of acetylcholine) per minute per molecule of enzyme. A more recent value at 25°C, pH = 7.0, acetylcholine concentration of \(2.5 \times 10^{−3}\; M\), was found to be \(7.4 \times 10^5\; min^{−1}\) ( ). There may be some 30 active centers per molecule. AChE is a serine hydrolase that reacts with acetylcholine at close to . A diffusion-controlled reaction occurs so quickly that the reaction rate is the rate of transport of the reactants through the solution; a s quickly as the reactants encounter each other, they react. The Michaelis-Menten model is used in a variety of biochemical situations other than enzyme-substrate interaction, including antigen-antibody binding, DNA-DNA hybridization, and protein-protein interaction. It can be used to characterize a generic biochemical reaction, in the same way that the can be used to model generic adsorption of biomolecular species. When an empirical equation of this form is applied to microbial growth. The experimentally determined parameters values vary wildly between enzymes (Table \(\Page {1}\)): While \(K_m\) is equal to the substrate concentration at which the enzyme converts substrates into products at half its maximal rate and hence is related to the affinity of the substrate for the enzyme. The catalytic rate \(k_{cat}\) is the rate of product formation when the enzyme is saturated with substrate and therefore reflects the enzyme's maximum rate. The rate of product formation is dependent on both how well the enzyme binds substrate and how fast the enzyme converts substrate into product once substrate is bound. For a kinetically perfect enzyme, every encounter between enzyme and substrate leads to product and hence the reaction velocity is only limited by the rate the enzyme encounters substrate in solution. From Equation \(\ref{Eq13.24}\), the catalytic efficiency of a protein can be evaluated. \[ \dfrac{k_{cat}}{K_m} = \dfrac{k_2}{K_m} = \dfrac{k_1k_2}{k_{-1} + k_2}\] This \(k_{cat}/K_m\) ratio is called the specificity constant measure of how efficiently an enzyme converts a substrate into product. It has a theoretical upper limit of ; enzymes working close to this, such as fumarase, are termed superefficient (Table \(\Page {1}\)). Determining \(V_m\) and \(K_m\) from experimental data can be difficult and the most common way is to determine initial rates, \(v_0\), from experimental values of \([P]\) or \([S]\) as a function of time. Hyperbolic graphs of \(v_0\) vs. \([S]\) can be fit or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine \(K_d\). These included: \[\dfrac{1}{r} = \dfrac{K_M + \left[ \text{S} \right]}{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]} = \dfrac{K_M}{k_2 \left[ \text{E} \right]_0} \dfrac{1}{\left[ \text{S} \right]} + \dfrac{1}{k_2 \left[ \text{E} \right]_0} \label{Eq28}\] Tthe Lineweaver–Burk plot (or double reciprocal plot) is a graphical representation of the Lineweaver–Burk equation of enzyme kinetics, described by Hans Lineweaver and Dean Burk in 1934 (Figure \(\Page {2}\)). \(1/V_{max}\) via Equation \(\ref{Eq13.26}\). The plot provides a useful graphical method for analysis of the Michaelis–Menten equation: \[V = \dfrac{V_{\max} [S]}{K_m + [S]}\] Taking the reciprocal gives \[\dfrac{1}{V} = \dfrac {K_m + [S]} {V_{max}[S]} = \dfrac{K_m}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}}\] where The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as \(K_m\) and \(V_{max}\), before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of \(V_{max}\); the x-intercept of the graph represents \(−1/K_m\). It also gives a quick, visual impression of the different forms of enzyme inhibition. The reaction between nicotineamide mononucleotide and ATP to form nicotineamide–adenine dinucleotide and pyrophosphate is catalyzed by the enzyme nicotinamide mononucleotide adenylyltransferase. The following table provides typical data obtained at a pH of 4.95. The substrate, S, is nicotinamide mononucleotide and the initial rate, , is the μmol of nicotinamide–adenine dinucleotide formed in a 3-min reaction period. Determine values for and . Figure 13.12 shows the Lineweaver–Burk plot for this data and the resulting regression equation. Using the -intercept, we calculate as = 1 / −intercept = 1 / 1.708 mol = 0.585 mol and using the slope we find that is = slope × = 0.7528 molimM × 0.585 mol = 0.440 mM The following data are for the oxidation of catechol (the substrate) to -quinone by the enzyme -diphenyl oxidase. The reaction is followed by monitoring the change in absorbance at 540 nm. The data in this exercise are adapted from jkimball. 0.3 0.6 1.2 4.8 0.020 0.035 0.048 0.081 Determine values for and . to review your answer to this exercise. The double reciprocal plot distorts the error structure of the data, and it is therefore unreliable for the determination of enzyme kinetic parameters. Although it is still used for representation of kinetic data, non-linear regression or alternative linear forms of the Michaelis–Menten equation such as the or are generally used for the calculation of parameters. The Lineweaver–Burk plot is classically used in older texts, but is prone to error, as the -axis takes the reciprocal of the rate of reaction – in turn increasing any small errors in measurement. Also, most points on the plot are found far to the right of the -axis (due to limiting solubility not allowing for large values of \([S]\) and hence no small values for \(1/[S]\)), calling for a large extrapolation back to obtain - and -intercepts When used for determining the type of enzyme inhibition, the Lineweaver–Burk plot can distinguish competitive, non-competitive and uncompetitive inhibitors. Competitive inhibitors have the same y-intercept as uninhibited enzyme (since \(V_{max}\) is unaffected by competitive inhibitors the inverse of \(V_{max}\) also doesn't change) but there are different slopes and x-intercepts between the two data sets. Non-competitive inhibition produces plots with the same x-intercept as uninhibited enzyme (\(K_m\) is unaffected) but different slopes and y-intercepts. Uncompetitive inhibition causes different intercepts on both the y- and x-axes but the same slope. The Eadie–Hofstee plot is a graphical representation of enzyme kinetics in which reaction rate is plotted as a function of the ratio between rate and substrate concentration and can be derived from the Michaelis–Menten equation (\(\ref{Eq13.25}\)) by inverting and multiplying with \(V_{max}\): \[ \dfrac{V_{max}}{v} = \dfrac{V_{max}(K_m+[S])}{V_{max}[S]} = \dfrac{K_m+[S]}{[S]}\] Rearrange: \[V_{max} = \dfrac{vK_m}{[S]} + \dfrac{v[S]}{[S]} = \dfrac{v K_m}{[S]} + v\] Isolate v: \[v = -K_m \dfrac{v}{[S]} + V_{max}\] A plot of v against \(v/[S]\) will hence yield \(V_{max}\) as the y-intercept, \(V_{max}/K_m\) as the x-intercept, and \(K_m\) as the negative slope (Figure \(\Page {3}\)). Like other techniques that linearize the Michaelis–Menten equation, the Eadie-Hofstee plot was used historically for rapid identification of important kinetic terms like \(K_m\) and \(V_{max}\), but has been superseded by nonlinear regression methods that are significantly more accurate and no longer computationally inaccessible. It is also more robust against error-prone data than the Lineweaver–Burk plot, particularly because it gives equal weight to data points in any range of substrate concentration or reaction rate (the Lineweaver–Burk plot unevenly weights such points). Both Eadie-Hofstee and Lineweaver–Burk plots remain useful as a means to present data graphically. One drawback from the Eadie–Hofstee approach is that neither ordinate nor abscissa represent independent variables: both are dependent on reaction rate. Thus any experimental error will be present in both axes. Also, experimental error or uncertainty will propagate unevenly and become larger over the abscissa thereby giving more weight to smaller values of v/[S]. Therefore, the typical measure of goodness of fit for linear regression, the correlation coefficient R, is not applicable. | 12,887 | 698 |
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The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons. How electrons fill in their positions around an atom is called the Aufbau Process (German: "building-up" process). The Aufbau Process is all about keeping electrons at their lowest possible energy and is the direct result of the . A corollary of Coulomb's law is that the energy of an electron is affected by attractive and repulsive forces. The closer an electron is to the nucleus, the lower its energy. The closer an electron is to another electron, the higher its energy. Of course, a basic principle of thermodynamics is that a system will proceed to the lowest energy possible. That means, if an atom has only one electron, the electron will have quantum numbers that place it at the lowest possible energy. It will be as close as possible to the positive nucleus. If an atom has a second electron, it will also be as close as possible to the nucleus. It could have the same quantum numbers as the first electron, except for spin. There is a trade-off, of course, because those two electrons will be close enough to repel each other. However, if it is a choice between that and taking a position much farther from the nucleus, the second electron will go ahead and pair up. These two electrons are often described as being "in the same orbital" and share their first three quantum numbers, so they are found somewhere in the same region of space. This first orbital, which has no directional restrictions, is called the 1s orbital. There is only room for one orbital at this distance from the nucleus. A third electron has to occupy another orbital farther away, the 2s orbital. Again, this is a spherical orbital: the electron can be found in any direction. The 2 in 2s means the Principal Quantum Number is two (corresponding to the second general energy level). The s is a code for other quantum numbers; it means the electron can be found in any direction, just like the 1s electrons. The second energy level is large enough to accommodate additional orbitals, but they are a little farther from the nucleus. These are called the 2p orbitals. They are regions of space along the x, y and z axes. There are three orbitals of this type, and they are just called p , p and p to remind us that they are orthogonal to each other. Expressed in a different way, an electron with Principal Quantum Number 2 can have four different combinations of its other quantum numbers. These combinations are denoted 2s, 2p , 2p and 2p . The three 2p combinations are a little higher in energy than the 2s orbital. A third electron will go into the 2s orbital. It is the lowest in energy. What about a fourth? Does it go into the 2s or a 2p? Once again the pairing energy is not quite as big as the energy jump up to the 2p orbital. The fourth electron pairs up in the 2s orbital. A fifth electron goes into one of the 2p orbitals. It does not matter which one. We will say it is the p , arbitrarily. A sixth electron again could either pair up in the p , or it could go into the p . But the p level is really the same as the p , just in a different direction. The energy is the same. That means a sixth electron will go into the p rather than pair up in the p , where it would experience extra repulsion. Note that the p orbitals are often drawn a little differently. For example, p orbitals are usually drawn in a way that shows that they have phase. Either the two lobes are colored differently to show that they are out of phase with each other, or they are shown with one lobe shaded and the other left blank. This pattern is generally followed for all of the elements. The tally of how many electrons are found in each orbital is called the electron configuration. For example, hydrogen has only one electron. Its ground state configuration (that means, assuming the electron has not been excited to another orbital via addition of energy) is 1s . On the other hand, an atom with six electrons, such as the element carbon, has the configuration 1s 2s 2p 2p . You may already know that electron configuration is the reason the periodic table works the way it does. Mendelev and others noticed certain elements had very similar properties, and that is because they have very similar electron configurations. Lithium has configuration 1s 2s and its alkali sister, sodium, has configuration 1s 2s 2p 2p 2p 3s . In both cases, the last electron added is an unpaired s electron. The last electron, or last few electrons, added to an atom generally play a strong role in how the atom behaves. This "frontier" electron is the one at the outer limits of the atom. If the atom is to interact with anything, the frontier electron will encounter the thing first. In contrast, the "core" electrons closer to the nucleus are more protected from the outside. There are some shortcuts we take with electron configurations. We tend to abbreviate "filled shells" (meaning all the possible orbitals with a given Principal Quantum Number are filled with electrons) and filled "sub-shells" (like 2s or 2p). First of all, in the case of p orbitals, if all the p orbitals are filled, we might just write 2p instead of 2p 2p 2p , because there is only one way all the orbitals could be filled. However, we would not necessarily write 2p instead of 2p 2p , because we may wish to make clear that the configuration does not involve two electrons in one p orbital at that point, as in 2p 2p . Also, we dispense with orbital labels entirely to ignore core electrons in a filled shell. For example, instead of writing 1s 2s for lithium, we can write [He]2s . Instead of writing 1s 2s 2p 2p 2p 3s for sodium, we write [Ne]3s . The [He] means all the electrons found in a helium atom, which is a noble gas. The [Ne] means all the electrons found in a neon atom. A noble gas is an unreactive element with a filled shell: helium, neon, argon, krypton, xenon or radon. The electrons beyond the noble gas shell are called the electrons. Principal Quantum Number 3 actually allows a third set of orbitals. These are called the d orbitals that are a little like p orbitals, but they are two-dimensional rather than one-dimensional. A d electron, for example, might extend along the x axis and the y axis, but not in between the axes. The d orbitals have five allowed orientations with: The other two orbitals extend between the axes instead, d rotated 45 degrees away from one of the other d orbitals and extends between the x and y axes. In the same way, you could imagine an orbital between the x and z and between the y and z axes, but that would make six different orientations. Quantum mechanical rules do not allow that. As a result, two of the possible combinations collapse into a mathematical sum, making just one orbital. We call this one the d orbital. However, just as the 2p orbitals are higher in energy than the 2s orbitals, the 3d orbitals are higher in energy than the 3p orbitals. The 3d level is very similar in energy to the 4s level. For that reason, calcium's last electrons go into a 4s orbital, not a 3d orbital. Calcium behaves much like magnesium as a result. The order of electrons in an atom, from lowest to highest energy, is: The atomic orbital energies do not properly follow this trend as shown pictorially below. For example, the relative energies of the 3d and 4s orbitals switch between calcium and scandium. Write electron configurations for the following elements. Write abbreviated electron configurations for the following elements. Write abbreviated electron configurations for the following elements. , | 7,893 | 700 |
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Rhenium is a dense, silvery white metal that takes its name from the Latin, Rhenus, for the Rhine river. It was discovered in 1925 by Ida and Walter Noddack along with Otto Berg. (L. : Rhine) Discovery of rhenium is generally attributed to Noddack, Tacke, and Berg, who announced in 1925 they had detected the element in platinum ore and columbite. They also found the element in gadolinite and molybdenite. By working up 660 kg of molybdenite in 1928 they were able to extract 1 g of rhenium. Rhenium was another of the "missing" elements proposed by Mendeleev. The first sample was concentrated 100,000 fold from a gadolinium ore sample. Just enough was obtained for a spectroscopic study in which previously unseen lines were observed. The metal is acid resistant and has one of the highest melting points. But its scarcity (and therefore expense) makes practical use limited. | 900 | 703 |
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The synthesis of element 105 was reported by Soviet scientists at the research station in Dubna as early as 1967 and by scientists at U.C. Berkeley in 1970. The Soviet scientists reported synthesis by bombarding Am-243 with Ne-22. At Berkeley, Cf-249 was bombarded with N-15. At the purported time of discovery the Soviets suggested the name Nielsbohrium (Ns) for the element while the American group suggested Hahnium (Ha). But the ensuing controversy over the discovery of this and other 6d-elements led the IUPAC to adopt its numerical system for naming elements beyond 103 until some agreements could be reached (hence Unp or Unnilpentium for 105). The longest lived isotope has a half-life of 34 seconds. Dubnium is the name approved by the IUPAC in August 1997. | 787 | 704 |
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Discovered in a museum in 1801, niobium was named after Niobe, the mythological daughter of Tantalus (the metal is chemically related to Tantalum). Occurring in the earth's crust in proportions of about 20 parts-per-million, niobium is used as an alloying agent with iron and nickel. It has commercial uses in atomic reactors and has superconductive properties when alloyed with tin or aluminum. In many places niobium is known as columbium since it was originally "discovered" in a mineral named columbite. However, careful subsequent analysis indicated that the "new" element was actually Tantalum. The rediscovery of the element we know now as niobium was not even associated with columbium until after the official statistics were in. Pure niobium looks much like steel but resists corrosion better due to a thin coating of oxide that forms on all exposed surfaces. The only acid that attacks Nb at room temperature is HF. Above 200° the metal becomes more reactive. | 990 | 705 |
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More abundant than better known metals such as silver and gold, hafnium (from the Latin Hafnia, a name for Copenhagen) was not discovered until 1923 by Coster and de Hevesy. The reason is the similarity of hafnium to zirconium. Mendeleev had predicted the existence of element 72 but had wrongly suggested it might be found along with titanium ores. Instead it lay hidden with "pure" samples of zirconium. Later Niels Bohr predicted the arrangement of outer electrons for element 72 and it was using X-ray techniques to study this very thing that led to the identification of hafnium as a separate element (thus the connection to Copenhagen in the name for the metal: Bohr's hometown). Hafnium metal is used in the manufacture of control rods for nuclear reactors because of its ability to absorb neutrons. | 826 | 707 |
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Like its immediate predecessors, element 108 (Hs), is of purely theoretical interest and has an extremely short half-life, decaying by spontaneous fission. It is so unstable, in fact, that it was not discovered until after element 109 had been positively confirmed. The search for elements beyond uranium has occupied the attention of various groups of scientists around the world since the late 1940's. The principals involved hail from the laboratories at Berkeley, California, what was previously the Soviet Joint Institute for Nuclear Research at Dubna, and the Heavy-Ion Research Laboratory in Darmstadt, Germany. Originally a committee of the IUPAC recommended that element 108 be named Hahnium, after a German scientist, Otto Hahn. That name met with some opposition and the name approved in August 1997 is Hassium, named for the German state of Hesse. | 879 | 709 |
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Nickel had been in use centuries before its actual discovery and isolation. As far back as 3500 BC Syrian bronzes contained a small amount of the element. In 235 BC, coins in China were minted from nickel. However there was no real documentation of the element until thousands of years later. In the 17 century, German miners discovered a red Colored ore they believed to contain copper. They discovered upon analysis that there was no copper but that a useless, smelly material was actually present. Thinking the ore was evil they dubbed it "Kupfernickel" or Old Nick's Copper, which meant false or bad copper. Swedish scientist Baron Axel Frederich Cronstedt in 1751 finally isolated nickel from an ore closely resembling kupfernickel. Hence, he named this new element after the traditional mineral. At the time of its discovery nickel was thought to be useless but as its valuable properties came to light the demand for the metal increased dramatically. The usefulness of nickel as a material in alloys was eventually discovered as the strength, corrosion resistance and hardness it adds to other metals came to be appreciated. In the 1800s, the technique of silver plating was developed with a nickel-copper-zinc alloy being utilized in the process. Today, stainless steel, another nickel containing alloy, is one of the most valuable materials of the 20 century. The U.S. five-cent piece is 25% nickel. The element is also an important alloying agent for stainless steels and in very powerful magnets. Nickel is found in the earth's crust to the extent of about 70 parts per million, about the same as copper and zinc. There is a good chance that a high proportion of the core of the earth is molten nickel. Unlike its near neighbor copper on the periodic table, nickel is only a fair electrical conductor. But like its other neighbor, cobalt, it is very useful in making strong permanent magnets. It is also highly resistant to attack by alkalis and is used to store and transport concentrated sodium and potassium hydroxide. Nickel reacts with most acids to produce hydrogen gas and the green Ni2+ ion. Nickel is the earth's 22 most abundant element and the 7 most abundant transition metal. It is a silver white crystalline metal that occurs in meteors or combined with other elements in ores. Two important groups of ores are: Canada is the world's leading nickel producer and the Sudbury Basin of Ontario contains one of the largest nickel deposits in the world. In 1899 Ludwig Mond developed a process for extracting and purifying nickel. The so-called "Mond Process" involves the conversion of nickel oxides to pure nickel metal. The oxide is obtained from nickel ores by a series of treatments including concentration, roasting and smelting of the minerals. In the first step of the process, nickel oxide is reacted with water gas, a mixture of H and CO, at atmospheric pressure and a temperature of 50 °C. The oxide is thus reduced to impure nickel. Reaction of this impure material with residual carbon monoxide gives the toxic and volatile compound, nickel tetracarbonyl, Ni(CO) . This compound decomposes on heating to about 230 °C to give pure nickel metal and CO, which can then be recycled. The actual temperatures and pressures used in this process may very slightly from one processing plant to the next. However the basic process as outlined is common to all. The process can be summarized as follows: \[ \ce{Ni} + \ce{4CO} \ce{->[ 50°C]} \underbrace{\ce{Ni(CO)_4}}_{impure} \ce{->[ 230°C]} \underbrace{\ce{Ni}}_{pure} + \ce{4CO} \] Nickel is a hard silver white metal, which occurs as cubic crystals. It is malleable, ductile and has superior strength and corrosion resistance. The metal is a fair conductor of heat and electricity and exhibits magnetic properties below 345°C. Five isotopes of nickel are known. In its metallic form nickel is chemically unreactive. It is insoluble in cold and hot water and ammonia and is unaffected by concentrated nitric acid and alkalis. It is however soluble in dilute nitric acid and sparingly soluble in dilute hydrochloric and sulfuric acids. Nickel is known primarily for its divalent compounds since the most important oxidation state of the element is +2. There do exist however certain compounds in which the oxidation state of the metal is between -1 to +4. Blue and green are the characteristic colors of nickel compounds and they are often hydrated. Nickel hydroxide usually occurs as green crystals that can be precipitated when aqueous alkali is added to a solution of a nickel (II) salt. It is insoluble in water but dissolves readily in acids and ammonium hydroxide. Nickel oxide is a powdery green solid that becomes yellow on heating. It is difficult to prepare this compound by simply heating nickel in oxygen and it is more conveniently obtained by heating nickel hydroxide, carbonate or nitrate. Nickel oxide is readily soluble in acids but insoluble in hot and cold water. Thermal decomposition of Ni(OH) , NiCO , or NiNO generates NiO. Nickel sulfides consist of NiS , which has a pyrite structure, and Ni S , which has a spinel structure. All the nickel dihalides are known to exist. These compounds are usually yellow to dark brown in Color. Preparation directly from the elements is possible for all except NiF , which is best prepared from reaction of F on NiCl at 350°C. Most are soluble in water and crystallization of the hexahydrate containing the [Ni(H O) ] ion can be achieved. NiF however is only slightly soluble in water from which the trihydrate crystallizes. The only nickel trihalide known to exist is an impure specimen of NiF . Nickel carbonate usually occurs as a light green crystalline solid or a brown powder. It dissolves in ammonia and dilute acids but is insoluble in hot water. It exhibits vigorous reaction with iodine, hydrogen sulphide or a mixture of barium oxide and air. It decomposes on heating before melting occurs. Ni + F 55°C /slow → NiF
Ni + Cl EtOH/ 20°C → NiCl
Ni + Br red heat → NiBr
NiCl + 2NaI → NiI + 2NaCl Nickel carbonyl is a colorless, volatile, liquid. It is soluble in alcohol, benzene, and nitric acid but only slightly soluble in water, and insoluble in dilute acids and alkalis. Upon heating or in contact with acid or acid fumes, nickel carbonyl emits toxic carbon monoxide gas, a property exploited in preparation of nickel metal. When exposed to heat or flame the compound explodes and it can react violently with air, oxygen and bromine. Identification of nickel compounds can be achieved by employing the use of an organic reagent dimethylglyoxine. This compound forms a red flocculent precipitate on addition to a solution of a nickel compound. The Nickel (II) ion forms many stable complexes as predicted by the Irving Williams series. Whilst there are no other important oxidation states to consider, the Ni(II) ion can exist in a wide variety of CN's which complicates its coordination chemistry. For example, for CN=4 both tetrahedral and square planar complexes can be found. For CN=5 both square pyramid and trigonal bipyramid complexes are formed. The phrase has been used to describe this behavior and the fact that equilibria often exist between these forms. Some examples include: (a) substituted acacs react with Ni to give green dihydrates (6 coordinate). On heating, the two coordinated water groups are generally removed to give tetrahedral species. The unsubstituted acac complex, Ni(acac) normally exists as a trimer, see below. Lifschitz salts containing substituted 1,2-diaminoethanes can be isolated as either 4 or 6 coordinate species depending on the presence of coordinated solvent. (b) Ni(acac) is only found to be monomeric at temperatures around 200°C in non-coordinating solvents such as n-decane. 6-coordinate monomeric species are formed at room temperature in solvents such as pyridine, but in the solid state Ni(acac) is a trimer, where each Ni atom is 6-coordinate. Note that Co(acac) actually exists as a tetramer. [Ni(acac) ] [Co(acac) ] (c) Complexes of the type NiL X , where L are phosphines, can give rise to either tetrahedral or square planar complexes. It has been found that: for L= mixed aryl and alkyl phosphines, both stereochemistries can occur in the same crystalline substance. The energy of activation for conversion of one form to the other has been found to be around 50kJ mol . Similar changes have been observed with variation of the X group: where Φ is shorthand for C H . Ni reacts with CN- to give Ni(CN) .nH O (blue-green) which on heating at 180-200°C is dehydrated to yield Ni(CN) . Reaction with excess KCN gives K Ni(CN) .H O (orange crystals) which can be dehydrated at 100°C. Addition of strong concentrations of KCN produces red solutions of Ni(CN) . The crystal structure of the double salt prepared by addition of Cr(en) to Ni(CN) showed that two types of Ni stereochemistry were present in the crystals in approximately equal proportions. 50% as square pyramid and 50% as trigonal bipyramid. The primary use of nickel is in the preparation of alloys such as stainless steel, which accounts for approximately 67% of all nickel used in manufacture. The greatest application of stainless steel is in the manufacturing of kitchen sinks but it has numerous other uses as well. Other nickel alloys also have important applications. An alloy of nickel and copper for example is a component of the tubing used in the desalination of sea water. Nickel steel is used in the manufacture of armor plates and burglar proof vaults. Nickel alloys are especially valued for their strength, resistance to corrosion and in the case of stainless steel for example, esthetic value. Electroplating is another major use of the metal. Nickel plating is used in protective coating of other metals. In wire form, nickel is used in pins, staples, jewelry and surgical wire. Finely divided nickel catalyzes the hydrogenation of vegetable oils. Nickel is also used in the coloring of glass to which it gives a green hue. Other applications of nickel include: Nickel compounds also have useful applications. Ceramics, paints and dyes, electroplating and preparation of other nickel compounds are all applications of these compounds. Nickel oxide for example is used in porcelain painting and in electrodes for fuel cells. Nickel acetate is used as a mordant in the textiles industry. Nickel carbonate finds use in ceramic colors and glazes. For an account of the limited information known about the role of nickel in human health see The first crystallization of an enzyme was reported in the 1920's. The enzyme was urease which converts urea to ammonia and bicarbonate. One source of the enzyme is the bacterium . The release of ammonia is beneficial to the bacterium since it partially neutralizes the very acidic environment of the stomach (whose function in part helps kill bacteria). In the initial study it was claimed that there were no metals in the enzyme. Fifty years later this was corrected when it was discovered that nickel ions were present and an integral part of the system. The Nobel Prize in Physiology or Medicine for 2005 was awarded to Barry J. Marshall and J. Robin Warren "for their discovery of the bacterium and its role in gastritis and peptic ulcer disease". | 11,338 | 710 |
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Algal blooms like the ones in the photos below may be harmful when the algae are toxic, or if they reduce the oxygen concentration enough to emperil other organisms. . Blooms are visible because algae concentrations may reach millions of cells per milliliter. Blooms often result when one is supplied to the environment, either naturally, or through Human activities. A limiting reagent is one of several reactants that is necessary for a reaction to occur, but which is present in low concentration, so no reaction occurs even though there is an excess of all other reactants. Coccolithophore algal bloom in the Bering Sea in 1998 A "red tide" which may poison seafood and cause human illness or death, caused by a dinoflagellate species. The limiting reagent that prevents uncontrolled algae growth is often phosphorus, and it may be in low concentrations because phosphate mineral sources ("phosphate rock", like apatite) are insoluble. Phosphate often limits growth of foodcrops, so producing soluble phosphate is a significant sector of the fertilizer industry. Runoff from agricultural fields is often the cause of algal blooms. Phosphate rock is solubilized for fertilizer by treatment with sulfuric acid, giving phosphoric acid and, as a byproduct, gypsum (CaSO · 2 H O used in Plaster of Paris and "drywall"). The reaction is: Phosphoric Acid When 100.0 g of chloroapetite rock is reacted with 100.0 g of sulfuric acid to form phosphoric acid and gypsum, which is the limiting reagent? --- The balanced equation tells us that according to the atomic theory, 1 mol Ca (PO ) Cl is required for every 5 moles of H SO . That is, the stoichiometric ratio S(Ca (PO ) Cl /H SO ) = 1 mol Ca (PO ) Cl/ 5 mol H SO . Let us see how many moles of each we actually have \(\begin{align}
& n_{\text{Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}=\text{100}\text{.0 g}\times \frac{\text{1 mol}}{\text{520}\text{.8 g}}=\text{0}\text{.192 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl} \\
& n_{\text{H}_{\text{2}}\text{SO}_{\text{4}}}=\text{100}\text{.0 g}\times \frac{\text{1 mol}_{\text{2}}}{\text{98}\text{.1 g}}=\text{1}\text{.02 mol H}_{\text{2}}\text{SO}_{\text{4}} \\
\end{align}\) If all the H SO were to react, it would require 1.02 mol H SO x (1 mol Ca (PO ) Cl / 5 mol H SO ) = 0.204 mol Ca (PO ) Cl, but only 0.192 mol is present. So Ca (PO ) Cl is the , an all of the H SO cannot react. When the reaction ends, 0.960 mol H SO will have reacted with 0.192 mol Ca (PO ) Cl and there will be (1.02 – 0.960) mol H SO = 0.06 mol H SO left over. Ca (PO ) Cl is therefore the limiting reagent. These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. Calculations are shown for each possible case, assuming that one reactant is completely consumed and determining if enough of the other reactants is present to consume it. If not, that scenario is discarded. From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts \(\frac{n_{\text{Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}\text{(initial)}}}{n_{\text{H}_{\text{2}}\text{SO}_{\text{4}}}\text{(initial)}}=\frac{\text{0.192 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{1.02 mol H}_{\text{2}}\text{SO}_{\text{4}}}= \frac{\text{0.188 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{mol H}_{\text{2}}\text{SO}_{\text{4}} }\) was less than the stoichiometric ratio \(\text{S}\left( \frac{\text{Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{H}_{\text{2}}\text{SO}_{\text{4}}} \right)=\frac{\text{1 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{5 mol H}_{\text{2}}\text{SO}_{\text{4}}}= \frac{\text{0.200 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{mol H}_{\text{2}}\text{SO}_{\text{4}} }\) This indicated that there was not enough Ca (PO ) Cl to react with all the H SO and Ca (PO ) Cl was the limiting reagent. The corresponding general rule, for any reagents X and Y, is \(\begin{align}
& \text{If}\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \\
& \\
& \text{If}\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \\
\end{align}\) (Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example. The phosphoric acid may be applied as a fertilizer solution, or converted to trisodium phosphate (TSP), a solid: TSP is used as a cleaning and degreasing agent, and was used extensively in household detergents in the US until its deleterious effects on the environment were appreciated in the 1970s. Since release of TSP and phosphoric acid may not be controlled in all countries, and in all activities, they often provide a source for phosphorus which is a limiting reagent in nature because of the low solubility of mineral phosphates. As we explained when we discussed the significance of formulas, ecological stoichiometry examines the stoichiometric relationship between the nutritional demands of a species and the food available to the species. . Proponents say that "Ecological stoichiometry recognizes that organisms themselves are outcomes of chemical reactions and thus their growth and reproduction can be constrained by supplies of key chemical elements [especially carbon (C), nitrogen (N) and phosphorus (P)]". For example, by writing an approximate chemical equation for photosynthesis in oceanic algae, we can predict which nutrients (nitrogen as potassium nitrate, KNO , phosphorus as phosphoric acid, H PO , etc.) are required for algae growth, and what products result from algal respiration. 106 CO ( ) + 16 KNO ( ) + H PO ( ) + 122 H O( ) + 16 H ( ) ↔ C H O N P ( ) + 138 O ( ) + 16 K (1) The "formula" for algae, (C H O N P )(M = 3553.259 g/mol) does not represent a single molecule, but just the overall composition of the algae (one might call it an "average" molecular formula) . Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. In the environment, inexpensive reagents like atmospheric O are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the . In a small scale experiment to model fertilizer runoff, water containing 10 g of H PO contaminates a small pond which already contains 300 g of KNO . If the pond contains stable algae which forms according to the equation above, and plenty of CO and other reactants are available, (a) what is the limiting reagent, and (b) how much algae can form as a result of the runoff? a) The stoichiometric ratio connecting KNO and H PO is \(\text{S}\left( \frac{\text{KNO}_{\text{3}}}{\text{H}_{\text{3}}\text{PO}_{\text{4}}} \right)=\frac{\text{16 mol KNO}_{\text{3}}}{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}} \) The initial amounts of KNO and H PO are calculated using appropriate molar masses \(\begin{align}
& \text{ }n_{\text{KNO}_{\text{3}}}\text{(initial)}=\text{300 g}\times \frac{\text{1 mol KNO}_{\text{3}}}{\text{101}\text{.1 g}}=\text{2.96}\text{mol KNO}_{\text{3}} \\
& \\
& n_{\text{H}_{\text{3}}\text{PO}_{\text{4}}}\text{(initial)}=\text{10}\text{.0 g}\times \frac{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}{\text{98.0 g}}=\text{0}\text{.102}\text{mol H}_{\text{3}}\text{PO}_{\text{4}} \\
\end{align}\) Their ratio is \(\frac{n_{\text{KNO}_{\text{3}}}\text{(initial)}}{n_{\text{H}_{\text{3}}\text{PO}_{\text{4}}}\text{(initial)}}=\frac{\text{2}\text{.96}\text{mol KNO}_{\text{3}}}{\text{0}\text{.102}\text{mol H}_{\text{3}}\text{PO}_{\text{4}}}=\frac{\text{29}\text{.0 mol KNO}_{\text{3}}}{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}\) Since this ratio is larger than the stoichiometric ratio, you have more than enough KNO to react with all the H PO . H PO is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was . Some of the excess reactant KNO will be left over, but all the initial amount of H PO will be consumed. Therefore we use (initial) to calculate how much algae can be obtained \(n_{\text{H}_{\text{3}}\text{PO}_{\text{4}}}\text{ }\xrightarrow{S\text{(Algae/H}_{\text{3}}\text{PO}_{\text{4}}\text{)}}\text{ }n_{\text{Algae}}\xrightarrow{M_{\text{Algae}}}\text{ }m_{\text{Algae}}\) \(m_{\text{Algae}}=\text{0.102}\text{ mol H}_{\text{3}}\text{PO}_{\text{4}}\text{ }\times \text{ }\frac{\text{1 mol Algae}}{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}\text{ }\times \text{ }\frac{\text{3553 g}}{\text{mol Algae}}=\text{362 g Algae}\) Note that only 10 g of H PO allowed the consumption of 456 g of carbon dioxide to make 362 g of Algae! These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. The H and K have been omitted. 450 As you can see from the example, in a case where there is a limiting reagent, . Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed. The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. | 11,140 | 711 |
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Copper occupies the same family of the periodic table as silver and gold, since they each have one s-orbital electron on top of a filled electron shell which forms metallic bonds. This similarity in electron structure makes them similar in many characteristics. All have very high thermal and electrical conductivity, and all are malleable metals. Among pure metals at room temperature, copper has the second highest electrical and thermal conductivity, after silver. The use of Copper dates back far into history. Copper beads have been found in what is now modern Iraq, dating back to 9000 BC. The metal is relatively easy to mine and refine, contributing to its early and widespread use. Being soft, however, it is unsuitable for making reliable tools and weapons. Early metalsmiths as far back as 3000 BC learned to combine copper with other metals to produce more durable alloys. Brass (copper and zinc) and bronze (copper and tin) are two examples. The symbol and name for copper are from the Latin , which literally means "from the island of Cyprus", an early source of copper ore. Before 1982 U.S. pennies were pure copper. Now they are mostly zinc with a thin shell of copper. Most copper that is mined today is refined and drawn into wire for use in the electrical industries. A significant portion is also used in manufacturing water pipe. Copper, of course, has a characteristic color which most people recognize. It is one of the best electrical conductors and resists corrosion from most acids (except nitric and hot concentrated sulfuric). When exposed to the elements for a period of time it develops a greenish coating or patina which is copper(II) carbonate, a protective coating that prevents further wear. Copper occurs both in combined state and free state and also in many ores. The important ores of copper are copper pyrites (\(CuFeS_2\)), cuprite and copper glance. The copper ores are mostly found in the north of India. The extraction of copper also involves many steps. The ore used for extraction is copper pyrites, which is crushed, concentrated and then heated in the presence of air. During heating the moisture gets expelled and the copper pyrites gets converted to ferrous sulfide and cuprous sulfide. \[ 2CuFeS_2 + O_2 \rightarrow Cu_2S + 2FeS + SO_2 \] Blast furnace is used to heat the mixture of roasted ore, powdered coke and sand. In the blast furnace oxidation reactions takes place. Ferrous sulfide forms ferrous oxide which combines with silica and forms the slag (\(FeSiO_2\)). \[ 2FeS + 3O_2 \rightarrow 2FeO + 2SO_2\] \[ FeO + SiO_2 \rightarrow FeSiO_3\] Cuprous sulfide forms cuprous oxide which is partially converted to cuprous sulfide. \[ Cu_2S + 3 O_2 \rightarrow 2Cu_2O + 2SO_2\] \[ Cu_2O + FeS \rightarrow Cu_2S + FeO\] This cuprous sulfide contains some amount of ferrous sulfide and this is called matte. Matte is removed from the base outlet of blast furnace. The removed matte is shifted into Bessemer converter which is lined inside with magnesium oxide. This converter has pipes through which hot air and \(SiO_2\) are sent in. In this converter \(Cu_2S\) converts to \(Cu_2O\) and \(FeS\) converts to \(FeO\). The ferrous oxide forms slag with \(SiO_2\). The cuprous oxide formed reacts with Cu2S and forms copper. \[ 2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2\] The copper thus formed is to be purified by . The simplest ion that copper forms in solution is the typical blue hexaaquacopper(II) ion - [Cu(H O) ] . Hydroxide ions (from, say, sodium hydroxide solution) remove hydrogen ions from the water ligands attached to the copper ion. Once a hydrogen ion has been removed from two of the water molecules, you are left with a complex with no charge - a neutral complex. This is insoluble in water and a precipitate is formed. The color coding is to show that this is not a ligand exchange reaction. The oxygens which were originally attached to the copper are still attached in the neutral complex. In the test-tube, the color change is: The ammonia acts as both a base and a ligand. With a small amount of ammonia, hydrogen ions are pulled off the hexaaqua ion exactly as in the hydroxide ion case to give the same neutral complex. \[ [Cu(H_2O)_6]^{2+} + 2NH_3 \rightarrow [Cu(H_2O)_4(OH)_2] + 2NH_4^+\] That precipitate dissolves if you add an excess of ammonia. The ammonia replaces water as a ligand to give tetraamminediaquacopper(II) ions. Notice that only 4 of the 6 water molecules are replaced. \[ [Cu(H_2O)_6]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + H_2O\] You might wonder why this second equation is given starting from the original hexaaqua ion rather than the neutral complex. Explaining why the precipitate redissolves is quite complicated. You will find the explanation in full on the page about the . The color changes are: You simply get a precipitate of what you can think of as copper(II) carbonate. \[ Cu^{2+} + CO_3^{2-} \rightarrow CuCO_3(s)\] If you add concentrated hydrochloric acid to a solution containing hexaaquacopper(II) ions, the six water molecules are replaced by four chloride ions. The reaction taking place is reversible. \[ Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\] Because the reaction is reversible, you get a mixture of colors due to both of the complex ions. The color of the tetrachlorocuprate(II) ion may also be described as olive-green or yellow. If you add water to the green solution, it returns to the blue color. Copper(II) ions oxidize iodide ions to molecular iodine, and in the process are themselves reduced to copper(I) iodide. The initial mucky brown mixture separates into an off-white precipitate of copper(I) iodide under an iodine solution. If you pipette a known volume of a solution containing copper(II) ions into a flask, and then add an excess of potassium iodide solution, you get the reaction we have just described. \[2Cu^{2+} + 4I^- \rightarrow 2CuI(s) +I_2 (aq)\] You can find the amount of iodine liberated by titration with sodium thiosulphate solution. \[ 2S_2O_3^{2-} (aq) + I_2 (aq) \rightarrow S_4O_6^{2-} (aq) + 2I^- (aq)\] As the sodium thiosulfate solution is run in from a burette, the color of the iodine fades. When it is almost all gone, you add some starch solution. This reacts reversibly with iodine to give a deep blue starch-iodine complex which is much easier to see. You add the last few drops of the sodium thiosulfate solution slowly until the blue color disappears. If you trace the reacting proportions through the two equations, you will find that for every 2 moles of copper(II) ions you had to start with, you need 2 moles of sodium thiosulfate solution. If you know the concentration of the sodium thiosulfate solution, it is easy to calculate the concentration of the copper(II) ions. Copper(I) chemistry is limited by a reaction which occurs involving simple copper(I) ions in solution. This is a good example of disproportionation - a reaction in which something oxidises and reduces itself. Copper(I) ions in solution disproportionate to give copper(II) ions and a precipitate of copper. The reaction is: Any attempt to produce a simple copper(I) compound in solution results in this happening. For example, if you react copper(I) oxide with hot dilute sulfuric acid, you might expect to get a solution of copper(I) sulfate and water produced. In fact you get a brown precipitate of copper and a blue solution of copper(II) sulfate because of the disproportionation reaction. \[ Cu_2O + H_2SO_4 \rightarrow Cu + CuSO_4 + H_2O\] We've already seen that copper(I) iodide is produced as an off-white precipitate if you add potassium iodide solution to a solution containing copper(II) ions. The copper(I) iodide is virtually insoluble in water, and so the disproportionation reaction does not happen. Similarly copper(I) chloride can be produced as a white precipitate (reaction described below). Provided this is separated from the solution and dried as quickly as possible, it remains white. In contact with water, though, it slowly turns blue as copper(II) ions are formed. The disproportionation reaction only occurs with simple copper(I) ions in solution. Forming copper(I) complexes (other than the one with water as a ligand) also stabilizes the copper(I) oxidation state. For example, both [Cu(NH ) ] and [CuCl ] are copper(I) complexes which do not . The chlorine-containing complex is formed if copper(I) oxide is dissolved in concentrated hydrochloric acid. You can think of this happening in two stages. First, you get copper(I) chloride formed: \[Cu_2O_{(s)} + 2HCl_{(aq)} \rightarrow 2CuCl_{(s)} + H_2O_{(l)}\] But in the presence of excess chloride ions from the HCl, this reacts to give a stable, soluble copper(I) complex. \[ CuCl_{(s)} + Cl^-_{(aq)} \rightarrow [CuCl_2]^-_{(aq)}\] You can get the white precipitate of copper(I) chloride (mentioned above) by adding water to this solution. This reverses the last reaction by stripping off the extra chloride ion. Jim Clark ( ) | 9,048 | 712 |
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The molecular properties of alkadienes depend on the relationship between the double bonds, that is, whether they are , , or : \[\ce{CH_3CH=C=CHCH_3}\] \[\ce{CH_2=CH-CH=CHCH_3}\] \[\ce{CH_2=CH-CH_2-CH=CH_2}\] The properties of a compound with isolated double bonds, such as 1,4-pentadiene, generally are similar to those of simple alkenes because the double bonds are essentially isolated from one another by the intervening \(\ce{CH_2}\) group. However, with a conjugated alkadiene, such as 1,3-pentadiene, or a cumulated alkadiene, such as 2,3-pentadiene, the properties are sufficiently different from those of simple alkenes (and from each other) to warrant separate discussion. Some aspects of the effects of conjugation already have been mentioneed, such as the influence on spectroscopic properties ( ). The emphasis here will be on the effects of conjugation on chemical properties. The reactions of greatest interest are addition reactions, and this chapter will include various types of addition reactions: electrophilic, radical, cycloaddition, and polymerization. and (1977) | 1,108 | 713 |
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The solutions to Schrödinger's equation for atomic orbitals can be expressed in terms of spherical coordinates: \(r\), \(\theta\), and \(\phi\). For a point \((r, \theta, \phi)\), the variable \(r\) represents the distance from the center of the nucleus, \(\theta\) represents the angle to the positive -axis, and \(\phi\) represents the angle to the positive -axis in the -plane. Because the atomic orbitals are described with a time-independent potential , Schrödinger’s equation can be solved using the technique of , so that any wavefunction has the form: \(\Psi(r,\theta,\phi) = R(r) Y(\theta,\phi)\) where \(R(r)\) is the and \(Y(\theta,\phi)\) is the : \(Y(\theta,\phi) = \Theta(\theta) \; \; \Phi(\phi)\) Each set of quantum numbers, (\(n\), \(l\), \(m_l\)), describes a different wave function. The radial wave function is only dependent on \(n\) and \(l\), while the angular wavefunction is only dependent on \(l\) and \(m_l\). So a particular orbital solution can be written as: \(\Psi_{n,l,m_l}(r,\theta,\phi) = {R}_{n,l}(r) Y_{l,m_l}(\theta,\phi)\) Where \(n = 1, 2, 3, …\) \(l = 0, 1, …, n-1\) \(m_l = -l, … , -2, -1, 0, +1, +2, …, l\) A wave function occurs at points where the wave function is zero and changes signs. The electron has zero probability of being located at a node. Because of the separation of variables for an electron orbital, the wave function will be zero when any one of its component functions is zero. When \(R(r)\) is zero, the node consists of a sphere. When \(\Theta(\theta)\) is zero, the node consists of a cone with the z-axis as its axis and apex at the origin. In the special case \(\Theta(\pi/2)\) = 0, the cone is flattened to be the x-y plane. When \(\Phi(\phi)\) is zero, the node consists of a plane through the z-axis. The shape and extent of an orbital only depends on the square of the magnitude of the wave function. However, when considering how bonding between atoms might take place, the signs of the wave functions are important. As a general rule a bond is stronger, i.e. it has lower energy, when the orbitals of the shared electrons have their wavefunctions match positive to positive and negative to negative. Another way of expressing this is that the bond is stronger when the wave functions with each other. When the orbitals overlap so that the wave functions match positive to negative, the bond will be weaker or may not form at all. The radial wavefunctions are of the general form: \(R(r) = N \; p(r) \; e^{-kr}\) Where The exponential factor is always positive, so the nodes and sign of \(R(r)\) depends on the behavior of \(p(r)\). Because the exponential factor has a negative sign in the exponent, \(R(r)\) will approach 0 as \(r\) goes to infinity. the probability of the electron being at a particular point. The probability distribution, \(P(r)\) is the probability that the electron will be at any point that is \(r\) distance from the nucleus. For any type of orbital, since is separable into radial and angular components that are each appropriately normalized, and a sphere of radius has area proportional to \(r^2\), we have: \(P(r) = r^2R^2(r)\) The angular wave function \(Y(\theta,\phi)\)does much to give an orbital its distinctive shape. \(Y(\theta,\phi)\) is typically normalized so the the integral of \(Y^2(\theta,\phi)\) over the unit sphere is equal to one. In this case, \(Y^2(\theta,\phi)\) serves as a probability function. The probability function can be interpreted as the probability that the electron will be found on the ray emitting from the origin that is at angles \((\theta,\phi)\) from the axes. The probability function can also be interpreted as the probability distribution of the electron being at position \((\theta,\phi)\) on a sphere of radius , given that it is distance from the nucleus. The angular wave functions for a hydrogen atom, \(Y_{l,m_l}(\theta,\phi)\) are also the wavefunction solutions to Schrödinger’s equation for a rigid rotor consisting of two bodies, for example a diatomic molecule. The simplest case to consider is the hydrogen atom, with one positively charged proton in the nucleus and just one negatively charged electron orbiting around the nucleus. It is important to understand the orbitals of hydrogen, not only because hydrogen is an important element, but also because they serve as building blocks for understanding the orbitals of other atoms. The hydrogen s orbitals correspond to \(l=0\) and only allow \(m_l = 0\). In this case, the solution for the angular wavefunction, \(Y_{0,0}(\theta,\phi)\) is a constant. As a result, the \(\Psi_{n,0,0}(r,\theta,\phi)\) wavefunctions only depend on \(r\) and the s orbitals are all spherical in shape. Because \(\Psi_{n,0,0}\) depends only on , the probability distribution function of the electron: \(\Psi^2_{n,0,0}(r,\theta,\phi) = \dfrac{1}{4\pi}R^2_{n,0}(r)\) Graphs of the three functions, \(R(r)\) in green, \(R^2(r)\) in purple and \(P(r)\) in orange are given below for = 1, 2, and 3. The graph of the functions have been variously scaled along the vertical axis to allow an easy comparison of their shapes and where they are zero, positive and negative. The vertical scales for different functions, either within or between diagrams, are not necessarily the same. In addition, a cross-section contour diagram is given for each of the three orbitals. These contour diagrams indicate the physical shape and size of the orbitals and where the probabilities are concentrated. An electron will be in the most-likely-10% (purple) regions 10% of the time, and it will be in the most-likely-50% regions (including the most-likely-10% regions, dark blue and purple) 50% of the time. Nodes are shown in orange in the contour diagrams. In all of these contour diagrams, the x-axis is horizontal, the z-axis is vertical, and the y-axis comes out of the diagram. The actual 3-dimensional orbital shape is obtained by rotating the 2-dimensional cross-section about the axis of symmetry, which is shown as a blue dashed line. The contour diagrams also indicate for regions that are separated by nodes, whether the wave function is positive (+) or negative (-) in that region. In order for the wave function to change sign, one must cross a node. From these diagrams, we see that the 1s orbital does not have any nodes, the 2s orbital has one node, and the 3s orbital has 2 nodes. Because for the s orbitals, \(\Psi^2 = R^2(r)\), it is interesting to compare the \(R^2(r)\) graphs and the \(P(r)\) graphs. By comparing maximum values, in the 1s orbital, the \(R^2(r)\) graph shows that the most likely place for the electron is at the nucleus, but the \(P(r)\) graph shows that the most likely radius for the electron is at \(a_0\), the Bohr radius. Similarly, for the other s orbitals, the one place the electron is most likely to be is at the nucleus, but the most likely radius for the electron to be at is outside the outermost node. Something that is not readily apparent from these diagrams is that the average radius for the 1s, 2s, and 3s orbitals is 1.5 a , 6 a , and 13.5 a , forming ratios of 1:4:9. In other words, the average radius is proportional to \(n^2\). The hydrogen p orbitals correspond to = 1 when ? 2 and allow = -1, 0, or +1. The diagrams below describe the wave function for = 0. The angular wave function \(Y_{1,0}(\theta,\phi) = cos\;\theta\) only depends on \(\theta\). Below, the angular wavefunction shown with a node at \(\theta = \pi/2\). The radial wavefunctions and orbital contour diagrams for the p orbitals with = 2 and 3 are: As in the case of the s orbitals, the actual 3-dimensional p orbital shape is obtained by rotating the 2-dimensional cross-sections about the axis of symmetry, which is shown as a blue dashed line. The p orbitals display their distinctive dumbbell shape. The angular wave function creates a nodal plane (the horizontal line in the cross-section diagram) in the x-y plane. In addition, the 3p radial wavefunction creates a spherical node (the circular node in the cross-section diagram) at r = 6 a . For \(m_l = 0\), the axis of symmetry is along the z axis. The wavefunctions for = +1 and -1 can be represented in different ways. For ease of computation, they are often represented as real-valued functions. In this case, the orbitals have the same shape and size as \(m_l = 0\), except that they are oriented in a different direction: the axis of symmetry is along the x axis with the nodal plane in the y-z plane or the axis of symmetry is along the y-axis with the nodal plane in the x-z plane. These correspond to wavefunctions that are the sum and the difference of the two = +1 and -1 wavefunctions \[\psi_{x}= \psi_{m_j=+1} + \psi_{m_j=-1} \notag \] \[\psi_{y}= \psi_{m_j=+1} - \psi_{m_j=-1} \notag \] the \(\psi_{z}\) wavefunction has a magnetic quantum number of =0, but the \(\psi_{x}\) and \(\psi_{y}\) are mixtures of the wavefunctions corresponding to = +1 and -1 and do not have unique magnetic quantum numbers. The hydrogen d orbitals correspond to = 2 when = 3 and allow = -2, -1, 0, +1, or +2. There are two basic shapes of d orbitals, depending on the form of the angular wave function. The first shape of a d orbital corresponds to = 0. In this case, \(Y_{2,0}(\theta,\phi)\) only depends on \(\theta\). The graphs of the angular wavefunction, and for \(n = 3\), the radial wave function and orbital contour diagram are as follows: As in the case of the s and p orbitals, the actual 3-dimensional d orbital shape is obtained by rotating the 2-dimensional cross-section about the axis of symmetry, which is shown as a blue dashed line. This first d orbital shape displays a dumbbell shape along the z axis, but it is surrounded in the middle by a doughnut (corresponding to the regions where the wavefunction is negative). The angular wave function creates nodes which are cones that open at about 54.7 degrees to the z-axis. At n=3, the radial wave function does not have any nodes. The second d orbital shape is illustrated for = +1 and = 3. In this case, \(Y_{2,1}(\theta,\phi)\) depends on both \(\theta\) and \(\phi\), and can be shown as a surface curving over and under a rectangular domain. As a result, separate diagrams are shown for \(Y_{2,1}(\theta,\phi)\) on the left and \(Y^2_{2,1}(\theta,\phi)\) on the right. Unlike previous orbital diagrams, this contour diagram indicates more than one axis of symmetry. Each axis of symmetry is at 45 degrees to the x- and z-axis. Each axis of symmetry only applies to the region surrounding it and bounded by nodes. Each of the four arms of the contour is rotated about its axis of symmetry to produce the 3-dimensional shape. However, the rotation is a non-standard rotation, producing only radial symmetry about the axis, not circular symmetry as was the case with other orbitals. This produces a double dumbell shape, with nodes in the x-y plane and the y-z plane. Similar to the p orbitals, the wavefunctions for =+2, -1, and -2 can be represented as real-valued functions that have the same shape as for =+1, just oriented in different directions. In two cases, the shape is re-oriented so the the axes of symmetry are in the x-y plane or in the z-y plane. In both of those cases, the axes of symmetry are at 45 degrees to their respective coordinate axes, just as with =+1. For the third and final case, the orbital shape is re-oriented so the axes of symmetry are in the x-y plane, but also laying along the x and y axes. It is often the case that the orbitals in the d subshell corresponding to the magnetic quantum numbers = ±1 and = ±2 are, as for the \(\psi_{x}\) and \(\psi_{y}\) orbitals, represented as sums and differences of the wavefunctions corresponding to = ±1 and = ±2. This, as for the p orbitals, better represents the spatial orientation of bonds formed with these orbitals. The orbitals d and d are sums and differences of the two orbitals with = ±1 and lie in the xz and yz planes. = ±2 similarly corresponds to d and d ; both lie in the xy plane. = 0 is the d orbital, which is oriented along the -axis. Hydrogenic atoms are atoms that only have one electron orbiting around the nucleus, even though the nucleus may have more than one proton and one or more neutrons. In this case, the electron has the same orbitals as the hydrogen atom, except that they are scaled by a factor of 1/Z. Z is the atomic number of the atom, the number of protons in the nucleus. The increased number of positively charged protons shrinks the size of the orbitals. Thus, the same graphs for hydrogen above apply to hydrogenic atoms, except that instead of expressing the radius in units of a , the radius is expressed in units of a /Z. Correspondingly, the values have to be renormalized by a factor of (Z/a ) . So a He atom has orbitals that are the same shape but half the size of the corresponding hydrogen orbitals and a Li atom has orbitals that are the same shape but one third the size of the corresponding hydrogen orbitals. | 13,054 | 714 |
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Discovered in 1803 by Smithson Tennant (most famous for his determination that diamond is just a form of carbon), osmium is a very dense, blue-white hard metal. Its name is taken from the Greek, osme, for "odor". The oxides of osmium emit highly toxic gases and form readily when the metal is exposed to air. Thus there are few commercial applications for osmium except as a minor alloying agent where it reduces frictional wear ("osmiroid" ball point pen tips, for example). Most osmium is recovered as a by-product of the refining of platinum and nickel ores. Osmium tetroxide, \(\ce{OsO4}\), is very poisonous, has a pungent smell and can cause blindness; it is a powerful oxidizing agent, and is used to stain biological materials. Palladium is part of the the Platinum Group Metals (PGM) whic is located in the 5th and 6th rows of the transition metal section of the periodic table and includes , , , Osmium, , and . Common characteristics include resistance to wear, oxidation, and corrosion, high melting points, and oxidation states of +2 to +4. They are generally non-toxic. \(OsO_4\) (Osmium Tetroxide) is a valuable catalyst in organic chemistry reactions. | 1,193 | 715 |
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In chemistry and physics, selection rules define the transition probability from one eigenstate to another eigenstate. In this topic, we are going to discuss the transition moment, which is the key to understanding the intrinsic transition probabilities. Selection rules have been divided into the electronic selection rules, vibrational selection rules (including Franck-Condon principle and vibronic coupling), and rotational selection rules. The transition probability is defined as the probability of particular spectroscopic transition to take place. When an atom or molecule absorbs a photon, the probability of an atom or molecule to transit from one energy level to another depends on two things: the nature of initial and final state wavefunctions and how strongly photons interact with an eigenstate. Transition strengths are used to describe transition probability. Selection rules are utilized to determine whether a transition is allowed or not. Electronic dipole transitions are by far the most important for the topics covered in this module. In an atom or molecule, an electromagnetic wave (for example, visible light) can induce an oscillating electric or magnetic moment. If the frequency of the induced electric or magnetic moment is the same as the energy difference between one eigenstate Ψ and another eigenstate Ψ , the interaction between an atom or molecule and the electromagnetic field is resonant (which means these two have the same frequency). Typically, the amplitude of this (electric or magnetic) moment is called the transition moment. In quantum mechanics, the transition probability of one molecule from one eigenstate Ψ to another eigenstate Ψ is given by |\(\vec{M}_{21}\)| , and \(\vec{M}_{21}\) is called the transition dipole moment, or transition moment, from Ψ to Ψ . In mathematical form it can be written as \[\vec{M}_{21}=\int \Psi_2\vec{\mu}\Psi_1d\tau\] The Ψ and Ψ are two different eigenstates in one molecule, \(\vec{M}_{21}\) is the electric dipole moment operator. If we have a system with n molecules and each has charge Q , and the dipole moment operator is can be written as \[ \vec{\mu}=\sum_{n}Q_n\vec{x}_n\] the \(\vec{x}_{n}\) is the position vector operator. Based on the Born-Oppenheimer approximation, the fast electronic motion can be separated from the much slower motion of the nuclei. As a result, the total wavefunction can be separated into electronic, vibrational, and rotational parts: \(\Psi (r,R) = \psi _e (r,R_e )\psi _v (R)\psi _r (R)\) The Born-Oppenheimer approximation assumes that the electronic wavefunction, \(\psi _e\), is approximated in all electronic coordinates at the equilibrium nuclear coordinates (R ). Since mass of electrons is much smaller than nuclear mass, the rotational wavefunction, \(\psi _r\), only depends on nuclear coordinates. The rotational wavefunction could provide important information for rotational selection rules, but we will not consider the rotational wavefuntion any further for simplicity because most of the spectra are not rotationally resolved. With the rotational part removed, the transition moment integral can be expressed as \(\ M = \iint {\psi _e '(r,R_e )}\cdot \psi _v '(R)(\mu _e + \mu _n )\psi _e ''(r,R_e ) \cdot \psi _v ''(R)drdR\) where the prime and double prime represent the upper and lower states respectively. Both the nuclear and electronic parts contribute to the dipole moment operator. The above equation can be integrated by two parts, with \(\mu _n\) and \(\mu _e\) respectively. A product of two integral is obtained: \(\ M = \int {\psi _e '(r,R_e )} \cdot \mu _e \cdot \psi _e ''(r,R_e )dr\int {\psi _v '(R)} \cdot \psi _v ''(R)dR + \int {\psi _v '(R)} \cdot \mu _n \cdot \psi _v ''(R)dR\int {\psi _e '(r,R_e )} \psi _e ''(r,R_e )dr\) Because different electronic wavefunctions must be orthogonal to each other, hence \(\int {\psi _e '(r,R_e )} \psi _e ''(r,R_e )dr\) is zero, the second part of the integral should be zero. The transition moment integral can be simplified as \(\ M = \int {\psi _e '(r,R_e )} \cdot \mu _e \cdot \psi _e ''(r,R_e )dr\int {\psi _v '(R)} \cdot \psi _v ''(R)dR\) The above equation is of great importance because the first integral defines the electronic selection rules, while the second integral is the basis of vibrational selection rules. Atoms are described by the primary quantum number n, angular momentum quantum number L, spin quantum number S, and total angular momentum quantum number J. Based on approximation of electron coupling, the atomic term symbol can be represented as L . The electronic-state configurations for molecules can be described by the primary quantum number n, the angular momentum quantum number Λ, the spin quantum number S, which remains a good quantum number, the quantum number Σ (S, S-1, ..., -S), and the projection of the total angular momentum quantum number onto the molecular symmetry axis Ω, which can be derived as Ω=Λ+Σ. The for the electronic states can be represented as \[^{2S+1} \Lambda_{\Omega,(g/u)}^{(+/-)}\] Group theory makes great contributions to the prediction of the electronic selection rules for many molecules. An example is used to illustrate the possibility of electronic transitions via group theory. 1. The total spin cannot change, ΔS=0; the rule ΔΣ=0 holds for multiplets; If the spin-orbit coupling is not large, the electronic spin wavefunction can be separated from the electronic wavefunctions. Since the electron spin is a magnetic effect, electronic dipole transitions will not alter the electron spin. As a result, the spin multiplicity should not change during the electronic dipole transition. 2. The total orbital angular momentum change should be ΔΛ=0, \(\pm\)1; For heteronuclear diatomic molecules with \(\ C_{\infty v} \) symmetry, a Σ \(\leftrightarrow\) \(\Pi\) transition is allowed according to ΔΛ selection rule. In order to prove the allowance of this transition, the direct product of \(\Sigma ^ + \otimes \Pi\) yields \(\Pi\) irreducible representation from the direct product table. Based on the \(\ C_{\infty v} \) character table below, the operator in the x and y direction have doubly degenerate \(\Pi\) symmetry. Therefore, the transition between Σ \(\leftrightarrow\) \(\Pi\) must be allowed since the multiplication of any irreducible representation with itself will provide the totally symmetric representation. The electronic transition moment integral can be nonzero. We can use the same kind of argument to illustrate that \(\Sigma ^ - \leftrightarrow \Phi\) transition is forbidden. \(C_{\infty v}\) character table 3. Parity conditions are related to the symmetry of the molecular wavefunction reflecting against its symmetry axis. For homonuclear molecules, the g \(\leftrightarrow\) u transition is allowed. For heteronuclear molecules, + \(\leftrightarrow\) + and - \(\leftrightarrow\) - transitions apply; The geometry of vibrational wavefunctions plays an important role in vibrational selection rules. For diatomic molecules, the vibrational wavefunction is symmetric with respect to all the electronic states. Therefore, the Franck-Condon integral is always totally symmetric for diatomic molecules. The vibrational selection rule does not exist for diatomic molecules. For polyatomic molecules, the nonlinear molecules possess 3N-6 normal vibrational modes, while linear molecules possess 3N-5 vibrational modes. Based on the harmonic oscillator model, the product of 3N-6 normal mode wavefunctions contribute to the total vibrational wavefunction, i.e. \[\psi _{vib} = \prod\limits_{3N - 6} {\psi _1 \psi _2 } \psi _3 ...\psi _{3N - 6}\] where each is represented by the wavefunction \(\psi_i\). Comparing to the Franck-Condon factor for diatomic molecules with single vibrational overlap integral, a product of 3N-6 (3N-5 for linear molecules) overlap integrals needs to be evaluated. Based on the symmetry of each normal vibrational mode, polyatomic vibrational wavefunctions can be totally symmetric or non-totally symmetric. If a normal mode is totally symmetric, the vibrational wavefunction is totally symmetric with respect to all the vibrational quantum number v. If a normal mode is non-totally symmetric, the vibrational wavefunction alternates between symmetric and non-symmetric wavefunctions as v alternates between even and odd number. If a particular normal mode in both the upper and lower electronic state is totally symmetric, the vibrational wavefunction for the upper and lower electronic state will be symmetric, resulting in the totally symmetric integrand in the Franck-Condon integral. If the vibrational wavefunction of either the lower or upper electronic state is non-totally symmetric, the Franck-Condon integrand will be non-totally symmetric. We will use CO as an example to specify the vibrational selection rule. CO has four vibrational modes as a linear molecule. The are illustrated in the figure below: The vibrational wavefunction for the totally symmetric C-O stretch, v , is totally symmetric with respect to all the vibrational quantum numbers. However, the vibrational wavefunctions for the doubly degenerate bending modes, v , and the antisymmetric C-O stretch, v , are non-totally symmetric. Therefore, the vibrational wavefunctions are totally symmetric for even vibrational quantum numbers (v=0, 2, 4...), while the wavefunctions remain non-totally symmetric for v odd (v=1, 3, 5...). Therefore, any value of Δv is possible between the upper and lower electronic state for mode v . On the other hand, modes v and v include non-totally symmetric vibrational wavefunctions, so the vibrational quantum number can only change evenly, such as Δv=\(\pm\) 2, \(\pm\) 4, etc.. Franck-Condon principle was proposed by German physicist James Franck (1882-1964) and U.S. physicist Edward U. Condon (1902-1974) in 1926. This principle states that when an electronic transition takes place, the time scale of this transition is so fast compared to nucleus motion that we can consider the nucleus to be static, and the vibrational transition from one vibrational state to another state is more likely to happen if these states have a large overlap. It successfully explains the reason why certain peaks in a spectrum are strong while others are weak (or even not observed) in absorption spectroscopy. \(\ M = \int {\psi _e '(r,R_e )} \cdot \mu _e \cdot \psi _e ''(r,R_e )dr\int {\psi _v '(R)} \cdot \psi _v ''(R)dR\) The second integral in the above equation is the vibrational overlap integral between one eigenstate and another eigenstate. In addition, the square of this integral is called the Franck-Condon factor: \(\displaystyle \ Franck-Condon \; Factor = \mid \int {\psi _v '}\psi _v ''dR\mid^2\) It governs the vibrational transition contribution to the transition probability and shows that in order to have a large vibrational contribution, the vibrational ground state and excited state must have a strong overlap. The above figure shows the Franck-Condon principle energy diagram, since electronic transition time scale is small compared to nuclear motion, the vibrational transitions are favored when the vibrational transition have the smallest change of nuclear coordinate, which is a vertical transition in the figure above. The electronic eigenstates favors the vibrational transition v'=0 in the ground electronic state to v"=2 in the excited electronic state, while peak intensity of v'=0 to v"=0 transition is expected to be low because the overlap between the v'=0 wavefunction and v''=0 wavefunction is very low. The figure above is the photoelectron spectrum of the ionization of hydrogen molecule (H ), it is also a beautiful example to formulate the Franck-Condon principle. It shows the appropriate energy curves and the vibrational energy levels, and with the help of the Franck-Condon principle, the transition between ground vibrational state v " and excited vibrational state v ' is expected to be the most intense peak in the spectrum. Why can some electronic-forbidden transitions be observed as weak bands in spectrum? It can be explained by the interaction between the electronic and vibrational transitions. The word "vibronic" is the combination of the words "vibrational" and "electronic." Because the energy required for one electronic state to another electronic state (electronic transition, usually in the UV-Vis region) is larger than one vibrational state to another vibrational state (vibrational transition, usually in the IR region), sometimes energy (a photon) can excite a molecule to an excited electronic and vibrational state. The bottom figure shows the pure electronic transition (no vibronic coupling) and the electronic transition couples with the vibrational transition (vibronic coupling). We now can go back to the original question: Why can some electronic-forbidden transitions be observed as weak bands in spectrum? For example, the d-d transitions in the octahedral transition metal complexes are Laporte forbidden (same symmetry, parity forbidden), but they can be observed in the spectrum and this phenomenon can be explained by vibronic coupling. Now we consider the Fe(OH ) complex which has low spin d as ground state (see figure above). Let's examine the one-electron excitation from t molecular orbital to e molecular orbital. The octahedral complex has O symmetry and therefore the ground state has A symmetry from the character table. The symmetry of the excited state, which is the direct product of all singly occupied molecular orbitals (e and t in this case): \(\displaystyle \Gamma_{t_{2g}}\otimes\Gamma_{e_g}= T_{1g}+T_{2g}\) The transition moment integral for the electronic transition can be written as \(\vec{M}=\int \psi^{'*}\vec{\mu}\psi d\tau\) where ψ is the electronic ground state and ψ' is the electronic excited state. The condition for the electronic transition to be allowed is to make the transition moment integral nonzero. The \(\hat{\mu}\) is the transition moment operator, which is the symmetry of the \(\hat{x}\), \(\hat{y}\), \(\hat{z}\) operators from the character table. For octahedral symmetry, these operators are degenerate and have T symmetry. Therefore, we can see that both T ← A and T ← A are electronically forbidden by the direct product (do not contain the totally symmetric A ): \(\displaystyle T_{1g}\times\ T_{1u}\times\ A_{1g}= A_{1u}+E_{u}+T_{1u}+T_{2u}\) \(\displaystyle T_{2g}\times\ T_{1u}\times\ A_{1g}= A_{2u}+E_{u}+T_{1u}+T_{2u}\) For octahedral complex, there are 15 . From the O character table we can get these irreducible representations: \(\displaystyle \Gamma_{vib}= a_{1g}+e_{g}+2t_{1u}+t_{2g}+t_{2u}\) When we let the vibrational transition to couple with the electronic transition, the transition moment integral has the form: \(\displaystyle \vec{M}=\int \psi_{e'}^{*} \psi_{v'}^{*} \vec{\mu} \psi_e \psi_v d\tau\) For the vibronic coupling to be allowed, the transition moment integral has to be nonzero. Use these vibrational symmetries of the octahedral complex to couple with the electronic transition: \(\displaystyle A_{1g}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= A_{1u}+E_{u}+T_{1u}+T_{2u}\) \(\displaystyle E_{g}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= E_{g}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1u}+A_{2u}+2E_{u}+2T_{1u}+2T_{2u}\) \(\displaystyle T_{1u}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= T_{1u}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1g}+ \ldots\) \(\displaystyle T_{2g}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= T_{2g}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1u}+A_{2u}+2E_{u}+3T_{1u}+4T_{2u}\) \(\displaystyle T_{2u}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= T_{2u}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1g}+ \ldots\) Since the t and t representation can generate the totally symmetric representation in the integrand. Therefore, we can see that t and t can couple with the electronic transition to form the allowed vibronic transition. Therefore, the d-d transition band for Fe(OH ) complex can be observed through vibronic coupling. 1. Transitions with ΔJ=\(\pm\)1 are allowed; Photons do not have any mass, but they have angular momentum. The conservation of angular momentum is the fundamental criteria for spectroscopic transitions. As a result, the total angular momentum has to be conserved after a molecule absorbs or emits a photon. The rotational selection rule relies on the fact that photon has one unit of quantized angular momentum. During the photon emission and absorption process, the angular moment J cannot change by more than one unit. Let's consider a single photon transition process for a diatomic molecule. The rotational selection rule requires that transitions with ΔJ=\(\pm\)1 are allowed. Transitions with ΔJ=1 are defined as R branch transitions, while those with ΔJ=-1 are defined as P branch transitions. Rotational transitions are conventional labeled as P or R with the rotational quantum number J of the lower electronic state in the parentheses. For example, R(2) specifies the rotational transition from J=2 in the lower electronic state to J=3 in the upper electronic state. 2. ΔJ=0 transitions are allowed when two different electronic or vibrational states are involved: (X'', J''=m) \(\to\) (X', J'=m). The Q branch transitions will only take place when there is a net orbital angular momentum in one of the electronic states. Therefore, Q branch does not exist for \(\ {}^1\Sigma \leftrightarrow {}^1\Sigma\) electronic transitions because \(\Sigma\) electronic state does not possess any net orbital angular momentum. On the other hand, the Q branch will exist if one of the electronic states has angular momentum. In this situation, the angular momentum of the photon will cancel out with the angular momentum of the electronic state, so the transition will take place without any change in the rotational state. The schematic of P, Q, and R branch transitions are shown below: With regard to closed-shell non-linear polyatomic molecules, the selection rules are more complicated than diatomic case. The rotational quantum number J remains a good quantum number as the total angular momentum if we don't consider the nuclear spin. Under the effect of single photon transition, the change of J is still limited to a maximum of \(\pm\) 1 based on the conservation of angular momentum. However, the possibility of Q branch is greatly enhanced irrelevant to the symmetry of the lower and upper electronic states. The rotational quantum number K is introduced along the inertial axis. For polyatomic molecules with symmetric top geometry, the transition moment is polarized along inertial axis. The selection rule becomes ΔK=0. then we will find out it is impossible for a high spin d5 octahedral complex since it violates Pauli exclusion principle. Therefore, the d-d transition for this is and also spin forbidden, resulting in a pale color. 4. The symmetry of the excited state will have the symmetry: \[\Gamma_{a_{1g}}\otimes\Gamma_{b_{1g}}= B_{1g}\] In the point group of D , the dipole moment operator transforms as e (x,y) and a (z). Then by evaluating the transition moment integral for the electronic transition, we can easily find out that there is no totally symmetric irreducible representation in the integrand.Therefore, the electronic transition is not allowed: \[B_{1g}\times\dbinom{E_u}{A_{2u}}\times\ A_{1g}= \dbinom{E_u}{B_{2u}}\] Next step we examine the vibrational irreducible representation a D complex has: \[ \Gamma_{vib}= a_{1g}+b_{1g}+b_{2g}+a_{2u}+b_{2u}+2e_{u}\] we can see that e and b can potentially serve as the promoting mode for vibronic transition, that is, generating the totally symmetric A in the integrand. Therefore, this electronic transition for D complex is forbidden, however, the transition can be allowed through vibronic coupling. 5. If the vibrational mode is totally symmetric, any change in the vibrational quantum number v is possible between the lower electronic state and upper electronic state. If the vibrational mode is non-totally symmetric, the vibrational wavefunction for the lower and upper electronic states are non-totally symmetric. The vibrational quantum number can only change evenly, such as Δv=\(\pm\) 2, \(\pm\) 4, etc.. 6.The Q branch transitions will only take place when there is a net orbital angular momentum in one of the electronic states. | 20,415 | 716 |
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The name "polyprotic" literally means many protons. Therefore, in this section we will be observing some specific acids and bases which either lose or accept proton. Then, we will be talking about the equations used in finding the degree of dissociation. Finally, with given examples, we will be able to approach problems dealing with polyprotic acids and bases. are specific acids that are capable of more than a single proton per molecule in acid-base reactions. (In other words, acids that have more than one ionizable H atom per molecule). Protons are lost through several stages (one at each stage), with the first proton being the fastest and most easily lost. Contrast with in section . From the table above, we see that sulfuric acid is the strongest. It is important to know that K >K >K , where K stands for the acidity constant or (first, second, and third, respectively). These constants are used to measure the degree of dissociation of hydrogens in the acid. For a more in depth discussion on this, go to . To find K of Hydrosulfuric acid (H S), you must first write the reaction: \[H_2S \rightleftharpoons H^+ + HS^- \nonumber \] Dividing the products by the reactants, we then have: \[K_{a1} = \dfrac{[H^+] [HS^-]}{ [HS-]} \nonumber \] To find K , we start with the reaction: \[HS^- \rightleftharpoons H^+ + S_2^- \nonumber \] Then, like when finding \(K_{a1}\), write the products over the reactants: \[K_{a2} = \dfrac{[H^+] [S_2^-]}{[HS^-]} \nonumber \] From these reactions we can observe that it takes two steps to fully remove the H ion. This also means that this reaction will produce two or . The equivalence point, by definition, is the point during an acid-base titration in which there has been equal amounts of acid and base reacted. If we were to graph this, we would be able to see exactly just what two equivalence points looks like. Let's check it out: Note the multiple equivalence points and notice that they are almost straight lines at that point, indicating equal added quantities of acid and base. In strong acid + strong base titrations, the pH changes slowly at first, rapidly through the equivalence point of pH=7, and then slows down again. If it is being titrated in a strong acid, the pH will go up as the base is added to it. Conversely, if it is in a strong base, the pH will fall down as acid is added. Next, let's take a look at sulfuric acid. This unique polyprotic acid is the only one to be completely deprotonated after the first step: \[H_2SO_{4(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + HSO^-_{4(aq)} \nonumber \] Now let's try something a little harder. The ionization of phosphoric acid (three dissociation reactions this time) can be written like this: Start with H PO : \[K_{a1}: H_3PO_{4(aq)} \rightleftharpoons H^+_{(aq)} + H_2PO^-_{4(aq)} \nonumber \] \[K_{a2} : H_2PO^-_{4(aq)} \rightleftharpoons HPO_{4(aq)} + H^+_{(aq)} \nonumber \] \[K_{a3} : HPO^-_{4(aq)} \rightleftharpoons H^+_{(aq)} + PO^{3-}_{4(aq)} \nonumber \] So from above reactions we can see that it takes three steps to fully remove the H ion. This also means that this reaction will produce three equivalence points. are bases that can at least one H ion, or proton, in acid-base reactions. First, start with the reaction A + H O ? HA + OH K = [OH ,HA ]/[A ]=K /K Then, we plug in the products over the reactants: HA + H O ? H A + OH K = [OH ,H A ]/[HA ]=K /K Finally, we are left with the third dissociation, or K : H A + H O ? H A + OH K = [OH ,H A]/[H A ]=K /K | 3,551 | 717 |
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Because the operators \(x\) and \(p\) are not compatible, \([\hat{X},\hat{P}]\neq 0\), there is measurement that can precisely determine both \(x\) and \(p\) simultaneously. Hence, there must be an uncertainty relation between them that specifies how uncertain we are about one quantity given a definite precision in the measurement of the other. Presumably, if one can be determined with infinite precision, then there will be an infinite uncertainty in the other. Recall that we had defined the uncertainty in a quantity by \[\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle ^2} \tag{1} \] Thus, for \(x\) and \(p\), we have \[ \Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2} \tag{2a} \] \[ \Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle ^2} \tag{2b} \] These quantities can be expressed explicitly in terms of the wave function \(\Psi (x, t)\) using the fact that \[\langle x \rangle = \langle \Psi(t)\vert x\vert\Psi(t)\rangle = \int dx \langle \Psi (t) \vert x \rangle \langle \vert x\vert X\vert\Psi(t)\rangle =\int dx \Psi^*(x,t) x \Psi(x,t) \tag{3} \] and \[\langle x^2 \rangle = \langle \Psi(t)\vert x^2\vert\Psi(t)\rangle = \int \Psi^*(x,t) x^2 \Psi(x,t) \tag{4} \] Similarly, \[\langle p \rangle = \langle \Psi(t)\vert p \vert\Psi(t)\rangle = \int dx \langle \Psi (t) \vert x \rangle \langle \vert p \vert \Psi (t) \rangle = \int dx \Psi^*(x,t){\hbar \over i}{\partial \over \partial x}\Psi(x,t) \tag{5} \] and \[\langle p^2 \rangle = \langle \Psi(t)\vert p^2\vert\Psi(t)\rangle = \int dx \Psi ^* (x, t)\left(-\hbar^2{\partial^2 \over \partial x^2}\right)\Psi(x,t) \tag{6} \] Then, the Heisenberg uncertainty principle states that \[\Delta x \Delta p \stackrel{>}{\sim} \hbar \tag{7} \] which essentially states that the greater certainty with which a measurement of \(x\) or \(p\) can be made, the greater will be the in the other. | 1,889 | 718 |
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We already have alluded to the difficulties encountered in the interpretation of the structure of benzene in and . Our task here is to see what new insight the VB and MO treatments can give us about benzene, but first we will indicate those properties of benzene that are difficult to explain on the basis of simple structure theory. From x-ray diffraction and spectroscopic measurements, benzene is known to be a planar molecule with six carbons \(1.390\) Å apart in a hexagonal ring, \(5\). Six hydrogen atoms, one associated with each carbon, are located \(1.09\) Å from those carbons. All \(\ce{H-C-C}\) and \(\ce{C-C-C}\) bond angles are \(120^\text{o}\): The 1,3,5-cyclohexatriene structure, \(6\), proposed for benzene in 1866 by Kekule, has alternating single and double bonds around the ring, which would be predicted to have bond lengths of \(1.48\) Å and \(1.34\) Å, respectively (see Table 2-1): The knowledge that the bond lengths are equal in the ring in benzene is a point against the Kekule formulation, but a more convincing argument is available from a comparison of the chemistry of benzene with that of 1,3,5-hexatriene, \(7\): Benzene also is more stable by about \(36\)-\(38 \: \text{kcal mol}^{-1}\) than anticipated for the 1,3,5-cyclohexatriene structure. You will recall from earlier discussions that the heat of combustion of one mole of benzene is \(38 \: \text{kcal}\) than calculated for cyclohexatriene (see ). Also, the heat of hydrogenation of benzene is only \(49.8 \: \text{kcal mol}^{1}\), which is \(36 \: \text{kcal}\) than expected for 1,3,5-cyclohexatriene; this estimate is based on the assumption that the heat of hydrogenation of 1,3,5-cyclohexatriene (with three double bonds) would be three times that of cyclohexane (\(28.5 \: \text{kcal mol}^{-1}\), for one double bond), or \(3 \times 28.2 = 85.5 \: \text{kcal mol}^{-1}\). The extra stability of benzene relative to the hypothetical 1,3,5-cyclohexatriene can be called its . Most (but not all) of this stabilization may be ascribed to resonance or electron delocalization. In an atomic model of benzene was discussed in some detail. Each carbon in the ring was considered to form three coplanar \(sp^2\)-hybrid \(\sigma\) bonds at \(120^\text{o}\) angles. These carbon-carbon and carbon-hydrogen \(\sigma\) bonds use three of the four valence electrons of each carbon. The remaining six carbon electrons are in parallel \(p\) orbitals, one on each of the six carbons. Each of the \(\pi\) electrons can be regarded as being paired with its immediate neighbors all around the ring, as shown by \(8\): As mentioned in , delocalization of the electrons over all six centers in benzene should give a more stable electron distribution than any structure in which the electrons are localized in pairs between adjacent carbons (as in the classical 1,3,5-cyclohexatriene structure). The simple MO and VB treatments of benzene begin with the atomic-orbital model and each treats benzene as a six-electron \(\pi\)-bonding problem. The assumption is that the \(\sigma\) bonds of benzene should not be very much different from those of ethene and may be regarded as independent of the \(\pi\) system. Extension of the ideas of for the MO treatment of an electron-pair bond between two nuclei to the \(\pi\) bonding in benzene is fairly straightforward. What is very important to understand is that there must be more than one molecular orbital for the \(\pi\) electrons because there are six \(\pi\) electrons, and the Pauli principle does not allow more than two paired electrons to occupy a given orbital. In fact, combination (or mixing) os the six \(2p\) orbitals of benzene, shown in \(8\), gives \(\pi\) molecular orbitals. Without exception, . The details of the mathematics of the mixing process to give an optimum set of molecular orbitals will not be described here,\(^1\) but the results are shown in Figure 21-5. Of the six predicted molecular orbitals, three are bonding and three are antibonding. The six \(\pi\) electrons are assigned to the three bonding orbitals in pairs and are calculated to have a total \(\pi\)-electron energy of \(6 \alpha + 8 \beta\). The calculation that leads to the results shown in Figure 21-5 is not very sophisticated. It is based on the assumption that the \(\pi\) bonding between each carbon and its immediate neighbors is equal all around the ring and that bonding involving carbons more than \(2\) Å apart is unimportant. What happens if we use the MO method to calculate the \(\pi\)-electron energy of classical 1,3,5-cyclohexatriene? The procedure is exactly as for benzene, except that we decree that each carbon \(p\) orbital forms a \(\pi\) bond with only of its neighboring \(p\) orbitals. The results are shown in Figure 21-6. The \(\pi\)-electron energy turns out to be three times that of ethene, or \(6 \alpha + 6 \beta\) compared to \(6 \alpha + 8 \beta\) for benzene. The calculated for benzene is the difference between these quantities, or \(\left( 6 \alpha + 8 \beta \right) - \left( 6 \alpha + 6 \beta \right) = 2 \beta\). That is to say, the calculated delocalization energy is the difference between the energy of benzene with full \(\pi\) bonding and the energy of 1,3,5-cyclohexatriene with alternating single and double bonds. If the electron delocalization energy \(\left( 2 \beta \right)\) is equal to the stabilization energy \(\left( 38 \: \text{kcal mol}^{-1} \right)\), then \(\beta = 19 \: \text{kcal mol}^{-1}\). Whether this is a valid method for determining \(\beta\) has been a matter of dispute for many years. Irrespective of this, the results of the calculations do account for the fact that benzene is more stable than would be expected for 1,3,5-cyclohexatriene. However, do the results also account for the low reactivity toward the various reagents in Figure 214, such as those that donate \(\ce{Br}^\oplus\) to double bonds (see )? To settle this question, we have calculated the changes in \(\pi\)-electron energy that occur in each of the following reactions: This means calculating the \(\pi\)-electron energies of all four entities and assuming that the differences in the \(\sigma\)-bond energies cancel between the two reactions. The result of this rather simple calculation is that attack of \(\ce{Br}^\oplus\) on benzene is thermodynamically favorable than on 1,3,5-cyclohexatriene by about \(\beta\). If \(\beta\) is \(19 \: \text{kcal mol}^{-1}\), this is clearly a sizable energy difference, and we can conclude that the simple MO method does indeed account for the fact that benzene is attacked by \(\ce{Br}^\oplus\) far less readily than is 1,3,5-cyclohexatriene. Extension of the basic ideas of the VB treatment described in to the atomic-orbital model of benzene is straightforward. We can write VB structures that represent pairing schemes of electrons in the atomic orbitals as shown in \(9\) through \(13\): Pairing schemes \(9\) and \(10\) correspond to Kekule's structures, whereas \(11\), \(12\), and \(13\) are called "Dewar structures" because J. Dewar suggested, in 1869, that benzene might have a structure such as \(14\): The electrons are paired in the configurations represented by \(11\), \(12\), and \(13\), but these pairing schemes are not as energetically favorable as \(9\) and \(10\). The reason is that the two electrons paired according to the dashed lines in \(11\), \(12\), and \(13\) are on nuclei separated by \(2.8\) Å, which is too far apart for effective bonding. The dashed lines between the distant carbons in \(11\), \(12\), and \(13\) are significant in that they define a pairing scheme. Such lines sometimes are said to represent "formal bonds". We hope that it is clear from what we have said here and previously that the ; indeed, the energy of the actual molecule is than any one of the contributing structures. The double-headed arrow between the structures is used to indicate that they represent different electron-pairing schemes for a molecule and different forms of the molecule in equilibrium with one another. When we use the resonance method in a qualitative way, we consider that the contribution of each of the several structures is to be weighted in some way that accords with the degree of bonding each would have, it were to represent an actual molecule with the specified geometry. Thus the Kekule-type electron-pairing schemes, \(9\) and \(10\), are to be taken as contributing and to the hybrid structure of benzene - equally because they are energetically equivalent, and predominantly because they can contribute much more to the overall bonding than \(11\), \(12\), and \(13\). In using the resonance method, we assume that all the resonance structures contributing to a given resonance hybrid have the same spatial arrangements of the nuclei but different pairing schemes for the electrons. Therefore \(11\), \(12\), and \(13\) are not to be confused with bicyclo[2.2.0]-2,5-hexadiene, \(15\), because \(15\) is a known (albeit not very stable) molecule with different atom positions and therefore vastly different bond angles and bond lengths from benzene: The electron-pairing schemes \(9\) and \(10\) represent the electron pairing that \(15\) would have if it were grossly distorted, with each carbon at the corner of a regular hexagon and a formal bond in place of a carbon-carbon single bond. Thus \(9\) and \(10\) would contribute in a significant way to the resonance hybrid of \(15\). Clearly, it is inconvenient and tedious to write the structures of the contributing forms to show the structure of a resonance hybrid. A shorthand notation is therefore desirable. Frequently, dashed rather than full lines are used where the bonding electrons are expected to be delocalized over several atoms. For benzene, \(16a\) or \(16b\) is quite appropriate: Unfortunately, although these are clear and explicit renderings, they are tedious to draw. As a result, many authors use (as we will most often) a single Kekule structure to represent benzene, with the understanding that all the \(\ce{C-C}\) bonds are equivalent. Other authors choose to represent benzene as a hexagon with an inscribed circle: This is a simple notation for benzene, but is quite uninformative and even can be actively misleading with some aromatic ring systems, and thus should be used with this limitation in mind. In calculations of the resonance energy of benzene, the five electronic configurations (valence-bond structures \(9\) through \(13\)) are combined mathematically to give five hybrid states, and of these the lowest-energy state is assumed to correspond to the normal state of the molecule. Thus benzene is considered by this approach to be a resonance hybrid of the valence-bond structures \(9\) through \(13\). In this simple treatment, \(9\) and \(10\) are calculated to contribute about \(80\%\) and \(11\), \(12\), and \(13\) about \(20\%\) to the hybrid. The actual numerical VB calculations, which are much more difficult to carry through than the corresponding MO calculations, give an energy of \(Q + 2.61 J\) for benzene and \(Q + 1.50 J\) for classical 1,3,5-cyclohexatriene.\(^3\) The resonance or delocalization energy then is \(\left( Q + 2.61 J \right) - \left( Q + 1.50 J \right) = 1.11 J\), which makes \(J \sim 35 \: \text{kcal mol}^{-1}\) if the resonance energy is taken to be equal to the \(38 \: \text{kcal}\) value obtained for the stabilization energy. If one carries through a simple VB calculation of the energy change associated with attack of \(\ce{Br}^\oplus\) on benzene as compared to 1,3,5-cyclohexatriene, the value obtained is \(0.63 J\), which corresponds to \(22 \: \text{kcal}\). This is in excellent agreement with the \(19 \: \text{kcal}\) value obtained by the MO method ( ). \(^1\)There are many excellent books that cover this subject in great detail; however, the simplest introductory work is J. D. Roberts, , W. A. Benjamin, Inc., Menlo Park, Calif., 1961. \(^2\) Note that in \(9\) and also in \(10\), we show a particular way of pairing the electrons. However, just as \(1 \leftrightarrow 2\), and \(4a \leftrightarrow 4b\), we also must consider other sets that represent exchanges of electrons across the dashed lines of \(9\) and also of \(10\). \(^3\)\(Q\) and \(J\) are negative VB energy parameters that correspond roughly to the MO parameters \(\alpha\) and \(\beta\). and (1977) | 12,441 | 719 |
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Nuclear reactions that transform atomic nuclei alter their identity and spontaneously emit radiation via processes of radioactive decay. In 1889, Ernest Rutherford recognized and named two modes of radioactive decay, showing the occurrence of both processes in a decaying sample of natural uranium and its daughters. Rutherford named these types of radiation based on their penetrating power: heavier alpha and lighter beta radiation. Gamma rays, a third type of radiation, were discovered by P. Villard in 1900 but weren't recognized as electromagnetic radiation until 1914. Since gamma radiation is only the discharge of a high-energy photon from an over-excited nucleus, it does not change the identity of the atom from which it originates and therefore will not be discussed in depth here. Because nuclear reactions involve the breaking of very powerful intra nuclear bonds, massive amounts of energy can be released. At such high energy levels, the matter can be converted directly to energy according to Einstein's famous Mass-Energy relationship . The sum of mass and energy are conserved in nuclear decay. The of any spontaneous reaction must be negative according to thermodynamics (ΔG < 0), and is essentially equal to the energy change of nuclear reactions because ΔE is so massive. Therefore, a nuclear reaction will occur spontaneously when: \[ΔE = Δmc^2 < 0\] \(ΔE < 0\) or \(Δm < 0\) When the mass of the products of a nuclear reaction weigh less than the reactants, the difference in mass has been converted to energy. There are three types of nuclear reactions that are classified as beta decay processes. Beta decay processes have been observed in 97% of all known unstable nuclides and are thus the most common mechanism for radioactive decay by far. The first type (here referred to as ) is also called Negatron Emission because a negatively charged beta particle is emitted, whereas the second type ( ) emits a positively charged beta particle. In , an orbital electron is captured by the nucleus and absorbed in the reaction. All these modes of decay represent changes of one in the atomic number Z of the parent nucleus but no change in the mass number A. Alpha decay is different because both the atomic and mass number of the parent nucleus decrease. In this article, the term beta decay will refer to the first process described in which a true beta particle is the product of the nuclear reaction. Nuclides can be radioactive and undergo nuclear decay for many reasons. Beta decay can occur in nuclei that are rich in neutrons - that is - the nuclide contains more neutrons than stable isotopes of the same element. These "proton deficient" nuclides can sometimes be identified simply by noticing that their mass number A (the sum of neutrons and protons in the nucleus) is significantly more than twice that of the atomic number Z (number of protons in nucleus). In order to regain some stability, such a nucleus can decay by converting one of its extra neutrons into a proton, emitting an electron and an antineutrino( ). The high energy electron emitted in this reaction is called a and is represented by \( _{-1}^{0}\textrm{e}^{-} \) in nuclear equations. Lighter atoms (Z < 60) are the most likely to undergo beta decay. The decay of a neutron to a proton, a beta particle, and an antineutrino (\(\bar{\nu}\)) is \[ \ce{_{0}^{1}n^0 \rightarrow _{0}^{1}p^+ + _{-1}^{0}e^-}+ \bar{\nu} \] Some examples of are \[ \ce{_{2}^{6}He \rightarrow _{3}^{6}Li + _{-1}^{0}e^-} +\bar{\nu} \] \[ \ce{_{11}^{24}Na \rightarrow _{12}^{24}Mg + _{-1}^{0}e^-} + \bar{\nu} \] In order for beta decay to occur spontaneously according to Δm < 0, the mass of the parent (not atom) must have a mass greater than the sum of the masses of the daughter nucleus and the beta particle: [ Z] > [ (Z+1)] + [ e ]
(Parent nucleus) > (Daughter nucleus) + (electron) The mass of the antineutrino is almost zero and can therefore be neglected. The equation above can be reached easily from any beta decay reaction, however, it is not useful because mass spectrometers measure the mass of rather than just their nuclei. To make the equation useful, we must make these nuclei into neutral atoms by adding the mass of electrons to each side of the equation. The parent nucleus then becomes the neutral atom [ Z] plus the mass of one electron, while the daughter nucleus and the beta particle on the right side of the equation become the neutral atom [ (Z+1)] plus the mass of the beta particle. The extra electron on the left cancels the mass of the beta particle on the right, leaving the inequality [ Z] > [ (Z+1)]
(Parent atom) > (Daughter atom) The change in mass then equals Δm = [ (Z+1)] - [ Z] The energy released in this reaction is carried away as kinetic energy by the beta particle and antineutrino, with an insignificant of energy causing recoil in the daughter nucleus. The beta particle can carry anywhere from all to none of this energy, therefore the maximum kinetic energy of a beta particle in any instance of beta decay is . Nuclides that are imbalanced in their ratio of protons to neutrons undergo decay to correct the imbalance. Nuclei that are rich in protons relative to their number of neutrons can decay by conversion of a proton to a neutron, emitting a (\(^0_1e^+\)) and a neutrino ( . Positrons are the antiparticles of electrons, therefore a positron has the same mass as an electron but with the opposite (positive) charge. In positron emission, the atomic number Z by 1 while the mass number A remains the same. Some examples of are \[\ce{^8_5B \rightarrow ^8_4Be + ^0_1e^{+}} + \nu_e\] \[\ce{^{50}_{25}Mg \rightarrow ^{50}_{24}Cr + ^0_1e^+} + \nu_e\] Positron emission is only one of the two types of decay that tends to happen in "neutron deficient" nuclides, therefore it is very important to establish the correct mass change criterion. Positron emission occurs spontaneously when [ Z] > [ (Z-1)] + [ e ]
(Parent nucleus) > (Daughter nucleus) + (positron) In order to rewrite this inequality in terms of the masses of neutral atoms, we add the mass of electrons to both sides of the equation, giving the mass of a neutral [ Z] atom on the left and the mass of a neutral [ (Z-1)] atom, plus an extra electron, (since only Z-1 electrons are needed to make the neutral atom), and a positron on the right. Because positrons and electrons have equal mass, the inequality can be written as [ Z] > [ (Z-1)] + 2 [ e ]
(Parent atom) > (Daughter atom) + (2 electrons) The change in mass for positron emission decay is Δm = [ (Z)] - [ Z] - 2 [ e ] As with beta decay, the kinetic energy is split between the emitted particles - in this case the positron and neutrino. As mentioned before, there are two ways in which neutron-deficient / proton-rich nuclei can decay. When the mass change \(Δm < 0\) yet is insufficient to cause spontaneous positron emission, a neutron can form by an alternate process known as electron capture. An outside electron is pulled inside the nucleus and combined with a proton to make a neutron, emitting only a neutrino. \[ \ce{^1_1p + ^0_{-1}e^{-} → ^1_0n + \nu }\] Some examples of are \[\ce{^{231}_{92}U + ^0_{-1}e^{-} → ^{231}_{91}Pa + \nu }\] \[\ce{ ^{81}{36}Kr + ^0_{-1}e^- → ^{81}_{35}Br + \nu }\] Electron capture happens most often in the heavier neutron-deficient elements where the mass change is smallest and positron emission isn't always possible. For \(Δm < 0\), the following inequality applies: [ Z] + [ e ] > [ (Z-1)]
(Parent nucleus) + (electron) > (Daughter nucleus) Adding \(Z\) electrons to each side of the inequality changes it to its useful form in which the captured electron on the left cancels out the extra electron on the right [ Z] > [ (Z-1)]
(Parent atom) > (Daughter atom) The change in mass then equals Δm = [ (Z-1)] - [ Z] When the loss of mass in a nuclear reaction is greater than zero, but less than 2 [ e ], the process cannot occur by positron emission and is spontaneous for electron capture. The other three processes of nuclear decay involve the formation of a neutron or a proton inside the nucleus to correct an existing imbalance. In alpha decay, unstable, heavy nuclei (typically \(Z > 83\)) reduce their mass number \(A\) by 4 and their atomic number \(Z\) by 2 with the emission of a helium nuclei (\(\ce{^4_2He^{2+}}\)), known as an . Some examples of are \[ \ce{^{222}_{88}Ra} \rightarrow \ce{^{218}_{86}Rn + ^4_2He^{2+}}\] \[ \ce{^{233}_{92}U} \rightarrow \ce{^{229}_{90}Th + ^4_2He^{2+}}\] As with beta decay and electron capture, Δm must only be less than zero for spontaneous alpha decay to occur. Since the number of total protons on each side of the reaction does not change, equal numbers of electrons are added to each side to make neutral atoms. Therefore, the mass of the parent atom must simply be greater than the sum of the masses of its daughter atom and the helium atom. [ Z] > [ (Z-2)] + [ He ] The change in mass then equals Δm = [ (Z)] - [ (Z-2)] - [ He ] The energy released in an alpha decay reaction is mostly carried away by the lighter helium, with a small amount of energy manifesting itself in the recoil of the much heavier daughter nucleus. Alpha decay is a form of spontaneous fission, a reaction in which a massive nuclei can lower its mass and atomic number by splitting. Other heavy unstable elements undergo fission reactions in which they split into nuclei of about equal size. Proton-deficient or neutron-deficient nuclei undergo nuclear decay reactions that serve to correct unbalanced neutron/proton ratios. Proton-deficient nuclei undergo - emitting a beta particle (electron) and an antineutrino to convert a neutron to a proton - thus raising the elements atomic number Z by one. Neutron-deficient nuclei can undergo or (depending on the mass change), either of which synthesizes a neutron - emitting a positron and a neutrino or absorbing an electron and emitting a neutrino respectively - thus lowering Z by one. Nuclei with Z > 83 which are unstable and too massive will correct by , emitting an alpha particle (helium nucleus) and decreasing both mass and atomic number. Very proton-deficient or neutron-deficient nuclei can also simply eject an excess particle directly from the nucleus. These types of decay are called and . These processes are summarized in the table below. | 10,405 | 720 |
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Quantum mechanics can account for the periodic structure of the elements, by any measure a major conceptual accomplishment for any theory. Although accurate computations become increasingly more challenging as the number of electrons increases, the general patterns of atomic behavior can be predicted with remarkable accuracy. According to the orbital approximation, which was introduced in the last Chapter, an -electron atom contains spinorbitals spinorbitals spinorbital wavefunction \[ \Psi (1,2 \ldots N ) = \frac{1}{\sqrt{N !}} \begin{vmatrix} \phi_a(1) & \phi_b(1) & \ldots & \phi_n(1) \\ \phi_a(2) & \phi_b(2) & \ldots & \phi_n(2) \\ & & \vdots & \\ \phi_a(N) & \phi_b(N) & \ldots & \phi_n(N) \\ \end{vmatrix} \label{1} \] Since interchanging any two rows (or columns) of a determinant multiplies it by \(-1\), the antisymmetry property (8.15) is fulfilled, for pair of electrons. The Hamiltonian for an atom with electrons around a nucleus of charge can be written \[ \hat{H} = \sum_{i=1}^N \left\{-\frac{1}{2}\bigtriangledown^2_i - \frac{Z}{r_i} \right\} + \sum_{i<j}^N \frac{1}{ r_{ij}}\label{2} \] The sum over electron repulsions is written so that each pair { , } is counted just once. The energy of the state represented by a Slater determinant (Equation \(\ref{1}\)) can be obtained after a lengthy derivation. We give just the final result \[ \tilde{E} = \sum_{a} I_a+\frac{1}{2}\sum_{a,b} \left( J_{ab}-K_{ab} \right) \label{3} \] where the sums run over all occupied spinorbitals. The one-electron, Coulomb and exchange integrals have the same form as those defined for helium atom in Eqs (8.22-24). The only difference is that an exchange integral equals zero unless the spins of orbitals and are both \(\alpha\) or both \(\beta\). The factor / corrects for the double counting of pairs of spinorbitals in the second sum. The contributions with = can be omitted since = . This effectively removes the Coulomb interaction of an orbital with itself, which is spurious. The Hartree-Fock or self-consistent field (SCF) method is a procedure for optimizing the orbital functions in the Slater determinant (1), so as to minimize the energy (Equation \(\ref{3}\)). SCF computations have been carried out for all the atoms of the periodic table, with predictions of total energies and ionization energies generally accurate in the \(1-2\%\) range. means "building-up." Aufbau principles determine the order in which atomic orbitals are filled as the atomic number is increased. For the hydrogen atom, the order of increasing orbital energy is given by 1 < 2 = 2 < 3 = 3 = 3 , etc. The dependence of energy on alone leads to extensive degeneracy, which is however removed for orbitals in many-electron atoms. Thus 2 lies below 2 , as already observed in helium. Similarly, 3 , 3 and 3 increase energy in that order, and so on. The 4 is lowered sufficiently that it becomes comparable to 3 . The general ordering of atomic orbitals is summarized in the following scheme: \[ 1s < 2s < 2p < 3s < 3p < 4s \sim 3d < 4p < 5s \sim 4d\\< 5p < 6s \sim 5d \sim 4f < 6p < 7s \sim 6d \sim 5f \label{4}\] and illustrated in Figure 1. This provides enough orbitals to fill the ground states of all the atoms in the periodic table. For orbitals designated as comparable in energy, e.g., 4 \(\sim\) 3 , the actual order depends which other orbitals are occupied. The sequence of orbitals pictured above increases in the order \(n+\frac{1}{2}\) , except that = 4 (rather than 3) is used for an -orbital. The tabulation below shows the ground-state electron configuration and term symbol for selected elements in the first part of the periodic table. From the term symbol, one can read off the total orbital angular momentum and the total spin angular momentum . The code for the total orbital angular momentum mirrors the one-electron notation, but using upper-case letters, as follows: = 0 1 2 3 4
S P D F G The total spin is designated, somewhat indirectly, by the spin multiplicity 2 + 1 written as a superscript the S, P, D. . . symbol. For example S (singlet S) , P (singlet P). . . mean = 0; S (doublet S) , P (doublet P). . . mean = 1/2; S (triplet S) , P (triplet P). . . mean = 1, and so on. Please do not confuse the spin quantum number with the orbital designation S. The vector sum of the orbital and spin angular momentum is designated \[ \bf{J} = \bf{L} + \bf{S} \label{5} \] The possible values of the total angular momentum quantum number runs in integer steps between | - and + . The value is appended as a subscript on the term symbol, eg, S , P , P . The energy differences between states is a result of , a magnetic interaction between the circulating charges associated with orbital and spin angular momenta. For atoms of low atomic number, the spin-orbit coupling is a relatively small correction to the energy, but it can become increasingly significant for heavier atoms. We will next consider in some detail the Aufbau 2 S 1 S 2 P To build the carbon atom, we add a second 2 electron. Since there are three degenerate 2 orbitals, the second electron can go into either the already-occupied 2 orbital or one of the unoccupied 2 orbitals. Clearly, two electrons in different 2 orbitals will have less repulsive energy than two electrons crowded into the same 2 orbital. In terms of the Coulomb integrals, we would expect, for example \[ J(2px, 2py) < J(2px, 2px) \label{6} \] For nitrogen atom, with three 2 electrons, we expect, by the same line of reasoning, that the third electron will go into the remaining unoccupied 2 orbital. The half-filled 2 subshell has an interesting property. If the three occupied orbitals are 2 2 and 2 , then their total electron density is given by \[ \rho_{2p} = \psi^{2}_{2p_{x}} + \psi^{2}_{2p_{y}} + \psi^{2}_{2p_{z}} = \left(x^2 + y^2 + z^2\right) \times \text{function of r} = \text{function of r} \label{7} \] noting that \( x^2 + y^2 + z^2 = r^2 \). But spherical symmetry implies zero angular momentum, like an -orbital. In fact, any half filled subshell, such as , , , will contribute zero angular momentum. The same is, of course true as well for subshells, such as , , . These are all S terms. Another way to understand this vector cancelation of angular momentum is to consider the alternative representation of the degenerate 2 -orbitals: 2 ; 2 and 2 . Obviously, the -components of angular momentum now add to zero, and since only this one component is observable, the total angular momentum must also be zero. Returning to our unfinished consideration of carbon, the 2 subshell can be regarded, in concept, as a half-filled 2 subshell plus an electron "hole." The advantage of this picture is that the total orbital angular momentum must be equal to that of the hole, namely = 1. This is shown below: Thus the term symbol for the carbon ground state is P. It remains to determine the total spins of these subshells. Recall that exchange integrals are non-zero only if the orbitals and have the same spin. Since exchange integrals enter the energy formula (3) with negative signs, the more nonvanishing integrals, the lower the energy. This is achieved by having the maximum possible number of electrons with spins. We conclude that = 1 for carbon and = 3/2 for nitrogen, so that the complete term symbols are P and S, respectively. The allocation electrons among degenerate orbitals can be formalized by : For an atom in its ground state, the term with the highest multiplicity has the lowest energy. Resuming Aufbau of the periodic table, oxygen with four 2 electrons must have one of the 2 -orbitals doubly occupied. But the remaining two electrons will choose unoccupied orbitals with parallel spins. Thus oxygen has, like carbon, a P ground state. Fluorine can be regarded as a complete shell with an electron hole, thus a P ground state. Neon completes the 2 2 shells, thus term symbol S. The chemical stability and high ionization energy of all the noble-gas atoms can be attributed to their electronic structure of complete shells. The third row of the periodic table is filled in complete analogy with the second row. The similarity of the outermost electron shells accounts for the periodicity of chemical properties. Thus, the alkali metals Na and K belong in the same family as Li, the halogens Cl and Br are chemically similar to F, and so forth. The transition elements, atomic numbers 21 to 30, present further challenges to our understanding of electronic structure. A complicating factor is that the energies of the 4 and 3 orbitals are very close, so that interactions among occupied orbitals often determines the electronic state. Ground-state electron configurations can be deduced from spectroscopic and chemical evidence, and confirmed by accurate self-consisent field computations. The 4 orbital is the first to be filled in K and Ca. Then come 3 electrons in Sc, Ti and V. A discontinuity occurs at Cr. The groundstate configuration is found to be 4 3 , instead of the extrapolated 4 3 . This can be attributed to the enhanced stability of a half-filled 3 -shell. All six electrons in the valence shells have parallel spins, maximizing the number of stabilizing exchange integrals and giving the observed S term. An analogous discontinuity occurs for copper, in which the 4 subshell is again raided to complete the 3 subshell. The order in which orbitals are filled is not necessarily consistent with the order in which electrons are removed. Thus, in all the positive ions of transition metals, the two 4 -electrons are removed first. The inadequacy of any simple generalizations about orbital energies is demonstrated by comparing the three ground-state electron configurations: Ni 4 3 , Pd 5 4 and Pt 6 5 . The periodic structure of the elements is evident for many physical and chemical properties, including chemical valence, atomic radius, electronegativity, melting point, density, and hardness. The classic prototype for periodic behavior is the variation of the first ionization energy with atomic number, which is plotted in in Figure 2. (Professor Emeritus of Chemistry and Physics at the , ) | 10,243 | 721 |
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In addition to having synthetic applications, catalytic hydrogenation is useful for analytical and thermochemical purposes. The analysis of a substance for the number of carbon-carbon double bonds it contains is carried out by measuring the uptake of hydrogen for a known amount of sample. Measurement of the heat evolved in the hydrogenation of alkenes gives information as to the relative stabilities of alkenes, provided that the differences in \(\Delta S^0\) values are small. The experimental values of \(\Delta H^0\) for hydrogenation of a number of alkenes and alkynes are listed in Table 11-2. The \(\Delta H^0\) calculated from average bond energies is \(-30 \: \text{kcal/mol}\) for a double bond and \(-69 \: \text{kcal/mol}\) for a triple bond. The divergences from these values reflect the influence of structure on the strengths of multiple bonds. Some important generalizations can be made: 3. Conjugated dienes are more stable than isolated dienes (compare 1,3- and 1,4-pentadiene). 4. Cumulated dienes appear to be less stable than conjugated or isolated dienes (see 1,2-propadiene). and (1977) | 1,132 | 722 |
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For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. This is mathematically: \[pV = constant\] That means that, for example, if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times. Is this consistent with pV = nRT ? That means that everything on the right-hand side of pV = nRT is constant, and so pV is constant - which is what we have just said is a result of Boyle's Law. This is easiest to see if you think about the effect of decreasing the volume of a fixed mass of gas at constant temperature. Pressure is caused by gas molecules hitting the walls of the container. With a smaller volume, the gas molecules will hit the walls more frequently, and so the pressure increases. You might argue that this isn't actually what Boyle's Law says - it wants you to increase the pressure first and see what effect that has on the volume. But, in fact, it amounts to the same thing. If you want to increase the pressure of a fixed mass of gas without changing the temperature, the only way you can do it is to squeeze it into a smaller volume. That causes the molecules to hit the walls more often, and so the pressure increases. But everything in the nR/p part of this is constant. That means that V = constant x T, which is . Jim Clark ( ) | 1,372 | 723 |
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This page describes the manufacture of ethanol by the direct hydration of ethene, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture and the rate of the reaction. Ethanol is manufactured by reacting ethene with steam. The reaction is reversible, and the formation of the ethanol is exothermic. Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this: The equation shows that the ethene and steam react 1 : 1. In order to get this ratio, you would have to use equal volumes of the two gases. Because water is cheap, it would seem sensible to use an excess of steam in order to move the position of equilibrium to the right according to Le Chatelier's Principle. In practice, an excess of ethene is used. This is very surprising at first sight. Even if the reaction was one-way, you couldn't possibly convert all the ethene into ethanol. There isn't enough steam to react with it. The reason for this oddity lies with the nature of the catalyst. The catalyst is phosphoric(V) acid coated onto a solid silicon dioxide support. If you use too much steam, it dilutes the catalyst and can even wash it off the support, making it useless. You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of ethanol in the equilibrium mixture. The forward reaction (the production of ethanol) is exothermic. According to Le Chatelier's Principle, this will be favoured if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. In order to get as much ethanol as possible in the equilibrium mixture, you need as low a temperature as possible. However, 300°C isn't particularly low. The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much ethanol as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of ethanol if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor. 300°C is a compromise temperature producing an acceptable proportion of ethanol in the equilibrium mixture, but in a very short time. Under these conditions, about 5% of the ethene reacts to give ethanol at each pass over the catalyst. Notice that there are 2 molecules on the left-hand side of the equation, but only 1 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much ethanol as possible in the equilibrium mixture, you need as high a pressure as possible. High pressures also increase the rate of the reaction. However, the pressure used isn't all that high. There are two quite separate problems in this case: The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of ethanol in the equilibrium mixture. Its only function is to speed up the reaction. In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor. Jim Clark ( ) | 3,846 | 726 |
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Cobalt (Co) lies with the transition metals on the periodic table. The atomic number of Cobalt is 27 with an atomic mass of 58.933195. Cobalt was first discovered in 1735 by George Brandt in Stockholm Sweden. It is used in many places today, such as, magnets materials, paint pigments, glasses, and even cancer therapy. The word cobalt is from the German word kobold, which means "goblin" or "evil spirit" this term was used by miners that was really difficult to mine and harmful to the miners health. Cobalt is a sturdy, gray metal which resembles iron and nickel. Although cobalt is ductile it is also somewhat malleable. Ductile means the metals ability to be drawn into thin wires. Malleable means is the ability of being hammered into thin sheets. Next to nickel and , cobalt is one of the three naturally occurring magnetic metals. When cobalt is combined with another metal to make an alloy its magnetic properties are even more apparent than the individual metal of cobalt. Cobalt's melting point is 1495 degrees C with a boiling point of 2870 degrees C. The density is 8.9 grams per cubic centimeter. Cobalt has one naturally occurring isotope which is cobalt 59. Isotopes are two or more forms a element. Isotopes are different because of their mass number. The number written on the right of an element i.e 59 is the mass number. The mass number is the number of protons and neutrons contained in the nucleus of the atom.The number of neutrons in a atom is how elements vary, this variation is called an isotope. There are also 10 radioactive isotopes of cobalt that are currently known. A radioactive isotope which splits apart and emits radiation. Cobalt-60 is one of the most commonly used radioactive isotopes and is used in medicine to find a treat certain diseases including the which determines if a patients body is making and using effectively. Co-57,58 are also used for the same purpose. Cobalt-60 is also used to treat cancer, because the radiation it gives off kills cancer cells. Cobalt's most important application is the production of super alloys. Super alloys consist of iron, , , , tungsten, , and . Super alloys do not rust and retain their physical and chemical properties at high temperatures. Superalloys are used where metals are placed under a great deal of stress and high temperatures.Super alloys are commonly used in airplane parts such as jet engine components and gas turbines. Superalloys are used in gas turbines because is it is used to make electricity, which operates at a really high temperature. Cobalt is also used to make magnetic alloys. Cobalt magnetic alloys are used in devices that need to hold a magnetic field. An example of this is electric motors and generators. Cobalt can also be used for the production of cemented carbides. Cementation is the process where one metal is used to cover the coating of a second metal. Through the process of cementation a extremely hard, strong alloy such as those used in drilling tools is produced. Cobalt is relatively abundant in nature at around 10 to 30 parts per million. The common ores of cobalt include: cobaltite, smaltite, chloranthite, and linnaeite. The largest suppliers of cobalt are Zambia, Canada, Russia, Australia, Zaire, and Cuba. Cobalt it not currently mined in the United States. Cobalt compounds are commonly used to make colored glass, glazes, paints, rubber, inks, cosmetics, and pottery. These compounds compounds include: cobalt oxide, cobalt potassium nitrite, cobalt aluminate, and cobalt ammonium phosphate. Cobalt compounds can also be used as catalyst. A catalyst is a substance that is used to speed up or lower a chemical reaction, by increasing or decreasing its activation energy. For all the cobalt used in the United States about 25% of them are cobalt compounds. Cobalt is a trace mineral that the human body needs in only small amounts. When trace metals are absent in a diet this leads to health problems. Animals use trace minerals to make essential enzymes which function as catalysts. These enzymes speed up changes that occur in the human body. Enzymes are needed for living cells to function properly For example cobalt is needed for the natural production of B-12 vitamins. B-12 vitamins ensure that enough red blood cells are produced in the human body. Cobalt also affects other animals because of its lack of presence in the soil. For example sheep in Australia become infected with a disease called Coast disease, caused by cobalt deficiency. Excess cobalt can also lead to health problems. People who work around cobalt can inhale its dust which causes vomiting, diarrhea, or breathing problems. If cobalt is presence on the skin in can lead to an irritation and rash. Cobalt is somewhat of a reactive element. It combines with oxygen in the air, but does not catch on fire and burn unless it is in powder form. Cobalt has the ability to react with most acids to produce hydrogen gas. However cobalt does not react with water that is at room temperature. The simplest ion that cobalt forms in solution is the pink hexaaquacobalt(II) ion - [Co(H O) ] . Cobalt(II) chloride is often used in humidity indicators because in dry form it is blue and when hydrated it is pink. The radioactive isotope of cobalt, Co-60, is used in portable "x-ray" devices. It actually produces gamma rays which can be used to detect internal flaws in metal in much the same way as x-rays by producing a shadowy image on film. Co-60 is also used in cancer treatment. Hydroxide ions remove hydrogen ions from the water ligands attached to the cobalt ion. Once a hydrogen ion has been removed from two of the water molecules, you are left with a complex with no charge - a neutral complex. This is insoluble in water and a precipitate is formed. \[ \ce{ [Co(H2O)6]^{2+} + 2OH^{-} \rightarrow [Co(H2O)4(OH)2] + 2H2O} \nonumber\] In the test-tube, the color changes are: The ammonia acts as both a base and a ligand. With a small amount of ammonia, hydrogen ions are pulled off the hexaaqua ion exactly as in the hydroxide ion case to give the same neutral complex. \[ [Co(H_2O)_6]^{2+} + 2NH_3 \rightarrow [Co(H_2O)_4(OH)_2] + 2NH_4^+\] That precipitate dissolves if you add an excess of ammonia. The ammonia replaces water as a ligand to give hexaamminecobalt(II) ions. \[ [Co(H_2O)_6]^{2+} + 6NH_3 \rightarrow [Co(NH_3)_6]^{2+} + 6H_2O\] The color changes are: The hexaamminecobalt(II) complex is very easily oxidized to the corresponding cobalt(III) complex. In the test-tube this is seen as a rapid darkening to a deep red-brown solution. In fact the hexaamminecobalt(III) ion is yellow! What you see is a mixture of this ion and various other cobalt(III) ions involving ligand exchange reactions with both water molecules and negative ions present in the solution. Rather than relying on oxidation by the air, you can add an such as hydrogen peroxide. You can do this after the addition of ammonia as in the last case, or you can do it following addition of sodium hydroxide solution. The reaction with ammonia solution followed by hydrogen peroxide produces the same dark reddish-brown solution as before - only faster. The equation for the oxidation of the a hexammine complex is: \[ 2Co(NH_3)_6^{2+} + H_2O_2 \rightarrow 2Co(NH_3)_6^{3+} + 2OH^-\] You get the variably colored precipitate of the cobalt(II) hydroxide complex when you add the sodium hydroxide solution. Addition of hydrogen peroxide produces lots of bubbles of oxygen and a dark chocolate brown precipitate. The final precipitate contains cobalt in the +3 oxidation state. The nature of the precipate of thie reaction is under. One resource describes it as a "hydrous \(Co_2O_3\), i.e., cobalt(III) oxide with closely associated water. Another suggests a the priciptates is \(CoO(OH)\). The oxygen seen in the reaction is produced from the decomposition of the hydrogen peroxide in a side reaction. Many things catalyze this decomposition - presumably, in this case, one or more of the various cobalt compounds present. You simply get a precipitate of what you can think of as cobalt(II) carbonate. \[ Co^{2+} (aq) + CO_3^{2-} (aq) \rightarrow CoCO_3 (s)\] The the precipitate is better described as a with a formula of the type \(xCoCO_3,yCo(OH)_2,zH_2O\). A greater discussion of general is discussed elsewhere. If you add concentrated hydrochloric acid to a solution containing hexaaquacobalt(II) ions, the solution turns from its original pink color to a rich blue. The six water molecules are replaced by four chloride ions. The reaction taking place is reversible. \[ [Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\] If you add water to the blue solution, it returns to the pink color. This reaction is explored in more detail on the module addressing . Jim Clark ( ) | 8,819 | 728 |
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The particle-in-a box model is used to approximate the Hamiltonian operator for the \(\pi\) electrons because the full Hamiltonian is quite complex. The full Hamiltonian operator for each electron consists of the kinetic energy term \(\dfrac {-\hbar ^2}{2m} \dfrac {d^2}{dx^2}\) and the sum of the Coulomb potential energy terms \(\dfrac {q_1q_2}{4\pi \epsilon _0 r_{12}}\) for the interaction of each electron with all the other electrons and with the nuclei (\(q\) is the charge on each particle and \(r\) is the distance between them). Considering these interactions, the Hamiltonian for electron i given below. \[ \hat {H} _i = \dfrac {- \hbar ^2}{2m} \dfrac {d^2}{dx^2} + \underset{\text{sum over electrons}}{ \sum _{j } \dfrac {e^2}{4 \pi \epsilon _0 r_{i, j}}} - \underset{\text{sum over nuclei}}{ \sum _{n} \dfrac {e^2 Z_n}{4 \pi \epsilon _0 r_{i,n}} } \label {4-1}\] The Schrödinger equation obtained with this Hamiltonian cannot be solved analytically by anyone because of the electron-electron interaction terms. Some approximations for the potential energy must be made. We want a model for the dye molecules that has a particularly simple potential energy function because we want to be able to solve the corresponding Schrödinger equation easily. The particle-in-a-box model has the necessary simple form. It also permits us to get directly at understanding the most interesting feature of these molecules, their absorption spectra. As mentioned in the previous section, we assume that the π-electron motion is restricted to left and right along the chain in one dimension. The average potential energy due to the interaction with the other electrons and with the nuclei is taken to be a constant except at the ends of the molecule. At the ends, the potential energy increases abruptly to a large value; this increase in the potential energy keeps the electrons bound within the conjugated part of the molecule. Figure \(\Page {1}\) shows the classical particle-in-a-box potential function and the more realistic potential energy function. We have defined the constant potential energy for the electrons within the molecule as the zero of energy. One end of the molecule is set at \(x = 0\), the other at \(x = L\), and the potential energy is goes to infinity at these points. For one electron located within the box, i.e. between \(x = 0\) and \(x = L\), the Hamiltonian is \[\hat {H} = \dfrac {-\hbar ^2}{2m} \dfrac {d^2}{dx^2}\] because \(V =0\), and the (time-independent) Schrödinger equation that needs to be solved is then \[\dfrac {- \hbar ^2}{2m} \dfrac {d^2}{dx^2} \psi (x) = E \psi (x) \label {4-2}\] We need to solve this differential equation to find the wavefunction and the energy. In general, differential equations have multiple solutions (solutions that are families of functions), so actually by solving this equation, we will find all the wavefunctions and all the energies for the particle-in-a-box. There are many ways of solving differential equations, and you will see some of them illustrated here and in subsequent chapters. One way is to recognize functions that might satisfy the equation. This equation says that differentiating the function twice produces the function times a constant. What kinds of functions have you seen that regenerate the function after differentiating twice? Exponential functions and sine and cosine functions come to mind. Use \(\sin(kx)\), \(\cos(kx)\), and \(e^{ikx}\) for the possible wavefunctions in Equation \(\ref{4-2}\) and differentiate twice to demonstrate that each of these functions satisfies the Schrödinger equation for the particle-in-a-box. Exercise \(\Page {1}\) leads you to the following three equations. \[\dfrac {\hbar ^2 k^2}{2m} \sin (kx) = E \sin (kx) \label {4-3}\] \[\dfrac {\hbar ^2 k^2}{2m} \cos (kx) = E \cos (kx) \label {4-4}\] \[\dfrac {\hbar ^2 k^2}{2m} e^{ikx} = E e^{ikx} \label {4-5}\] For the equalities expressed by these equations to hold, \(E\) must be given by \[E = \dfrac {\hbar ^2 k^2}{2m} \label {4-6}\] Kinetic energy is the momentum squared divided by twice the mass \(p^2/2m\), so we conclude from Equation \(\ref{4-6}\) that \(ħ^2k^2 = p^2\). Solutions to differential equations that describe the real world also must satisfy conditions imposed by the physical situation. These conditions are called boundary conditions. For the particle-in-a-box, the particle is restricted to the region of space occupied by the conjugated portion of the molecule, between \(x = 0\) and \(x = L\). If we make the large potential energy at the ends of the molecule infinite, then the wavefunctions must be zero at \(x = 0\) and \(x = L\) because the probability of finding a particle with an infinite energy should be zero. Otherwise, the world would not have an energy resource problem. This boundary condition therefore requires that \(ψ(0) = ψ(L) = 0\). Which of the functions \(sin(kx)\), \(cos(kx)\), or \(e^{ikx}\) is 0 when x = 0? As you discovered in Exercise \(\Page {2}\) for these three functions, only \(sin(kx) = 0\) when \(x = 0\). Consequently only \(sin(kx)\) is a physically acceptable solution to the Schrödinger equation. The boundary condition described above also requires us to set \(ψ(L) = 0\). \[ψ(L) = \sin(kL) = 0 \label {4.7}\] The sine function will be zero if \(kL = nπ\) with \(n = 1,2,3, \cdots\). In other words, \[ k = \dfrac {n \pi}{L} \label {4-8}\] with \(n = 1, 2, 3 \cdots\) Note that \(n = 0\) is here because this makes the wave vector zero \(k = 0\), so \(\sin(kx) = 0\), and thus \(ψ(x)\) is zero everywhere. If the wavefunction were zero everywhere, it means that the probability of finding the electron is zero. This clearly is not acceptable because it means there is no electron. Show that \(\sin(kx) = 0\) at \(x = L\) if \(k = nπ/L\) and \(n\) is an integer. It appears that a negative integer also would work for \(n\) because \[\sin \left ( \dfrac {-n \pi}{L} x \right ) = - \sin \left ( \dfrac {n \pi}{L} x \right ) \label {4.9}\] which also satisfies the boundary condition at \(x = L\). The reason negative integers are not used is a bit subtle. Changing \(n\) to \(–n\) just changes the sign (also called the phase) of the wavefunction from + to -, and does not produce a function describing a new state of the particle. Note that the probability density for the particle is the absolute square of the function, and the energies are the same for \(n\) and \(–n\). Also, since the wave vector k is associated with the momentum (p = ħk), n > 0 means k > 0 corresponding to momentum in the positive direction, and \(n < 0\) means \(k < 0\) corresponding to momentum in the negative direction. By using Euler’s formula one can show that the sine function incorporates both \(k\) and \(–k\) since \[ \sin (kx) = \dfrac {1}{2i} ( e^{ikx} - e^{-ikx} ) \label {4-10}\] so changing \(n\) to \(–n\) and \(k\) to \(–k\) does not produce a function describing new state, because both momentum states already are included in the sine function. The set of wavefunctions that satisfies both boundary conditions is given by \[ \psi _n (x) = N \sin \left ( \dfrac {n \pi}{L} x \right ) \text {with } n = 1, 2, 3, \cdots \label {4-11}\] The normalization constant N is introduced and evaluated to satisfy the normalization requirement. \[ \int \limits _0^L \psi ^* (x) \psi (x) dx = 1 \label {4- 12}\] \[N^2 \int \limits _0^L \sin ^2 \left ( \dfrac {n \pi x}{L} \right ) dx = 1 \label {4-13}\] \[N = \sqrt{ \dfrac{1}{\int \limits _0^L \sin ^2 \dfrac {n\pi x}{L} dx} } \label {4-14}\] \[ N = \sqrt{ \dfrac {2}{L}} \label {4-15}\] Finally we write the wavefunction: \[ \psi _n (x) = \sqrt{ \dfrac {2}{L} } \sin \left ( \dfrac {n \pi}{L} x \right ) \label {4-16}\] Evaluate the integral in Equation \(\ref{4-13}\) and show that \(N = (2/L)^{1/2}\). By finding the solutions to the Schrödinger equation and imposing boundary conditions, we have found a whole set of wavefunctions and corresponding energies for the particle-in-a box. The wavefunctions and energies depend upon the number n, which is called a quantum number. In fact there are an infinite number of wavefunctions and energy levels, corresponding to the infinite number of values for n \(n = 1 \rightarrow \infty\). The wavefunctions are given by Equation \(\ref{4-16}\) and the energies by Equation \(\ref{4-6}\). If we substitute the expression for k from Equation \(\ref{4-8}\) into Equation \(\ref{4-6}\), we obtain the equation for the energies \(E_n\) \[ E_n = \dfrac {n^2 \pi ^2 \hbar ^2}{2mL^2} = n^2 \left (\dfrac {h^2}{8mL^2} \right ) \label {4-17}\] Substitute the wavefunction, Equation \(\ref{4-16}\), into Equation \(\ref{4.2}\) and differentiate twice to obtain the expression for the energy given by Equation \(\ref{4-17}\). From Equation \(\ref{4-17}\) we see that the energy is quantized in units of \(\dfrac {h^2}{8mL^2}\); i.e. only certain values for the energy of the particle are possible. This quantization, the dependence of the energy on integer values for n, results from the boundary conditions requiring that the wavefunction be zero at certain points. We will see in other chapters that quantization generally is produced by boundary conditions and the presence of Planck’s constant in the equations. The lowest-energy state of a system is called the ground state. Note that the ground state (\(n = 1\)) energy of the particle-in-a-box is not zero. This energy is called the zero-point energy. Here is a neat way to deduce or remember the expression for the particle-in-a-box energies. The momentum of a particle has been shown to be equal to \(ħk\). Show that this momentum, with \(k\) constrained to be equal to \(nπ/L\), combined with the classical expression for the kinetic energy in terms of the momentum \((p^2/2m)\) produces Equation \(\ref{4.17}\). Determine the units for \(\dfrac {h^2}{8mL^2}\) from the units for \(h\), \(m\), and \(L\). Why must the wavefunction for the particle-in-a-box be normalized? Show that φ(x) in Equation \(\ref{4-16}\) is normalized. Use a spreadsheet program, Mathcad, or other suitable software to construct an accurate energy level diagram and to plot the wavefunctions and probability densities for a particle-in-a-box with \(n = 1\) to \(6\). You can make your graphs universal, i.e. apply to any particle in any box, by using the quantity \((h^2/8mL^2)\) as your unit of energy and \(L\) as your unit of length. To make these universal graphs, plot \(n^2\) on the y-axis of the energy-level diagram, and plot \(x/L\) from \(0\) to \(1\) on the x-axis of your wavefunction and probability density graphs. How does the energy of the electron depend on the size of the box and the quantum number n? What is the significance of these variations with respect to the spectra of cyanine dye molecules with different numbers of carbon atoms and pi electrons? Plot \(E(n_2)\), \(E(L_2)\), and \(E(n)\) on the same figure and comment on the shape of each curve. The quantum number serves as an index to specify the energy and wavefunction or state. Note that \(E_n\) for the particle-in-a-box varies as \(n^2\) and as \(1/L^2\), which means that as \(n\) increases the energies of the states get further apart, and as \(L\) increases the energies get closer together. How the energy varies with increasing quantum number depends on the nature of the particular system being studied; be sure to take note of the relationship for each case that is discussed in subsequent chapters. ") | 11,465 | 730 |
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The following headers are meant as a guide to the sections that you need in your module. Here, before the table of contents, give a two-four sentence overview of the Module. Be straight to the point. No examples and no figures in this section. In this section, give a short introduction to your topic to put it in context. The module should be easy to read; please reduce excess space, crop figures to remove white spaces, full justify all text, with the exception of equations, which should use the equation command for construction (see FAQ). This is where you put the core text of your module. Add any number of headings necessary for your topic. Try to reduce unnecessary discussion and get to the point in a terse, yet informative, manner possible. You can delete the header for this section and place your own related to the topic. Remember to hyperlink your module to other modules via the link button on the editor toolbar. Make up some practice problems for the future readers. Five original with varying difficulty questions (and answers) are ideal. Tags below. If no tags exist, then add two new ones: "Vet1" and the level of the module content (e.g. "Fundamental"). See FAQ for more details. | 1,223 | 731 |
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We pointed out in the importance of using systematic names for compounds such that the name uniquely describes the structure. It is equally important to be able to unambiguously describe the configuration of a compound. The convention that is used to designate the configurations of chiral carbons of naturally occurring compounds is called the \(D, L\) system. To use it, we view the molecule of interest according to the following rules: 1. The main carbon chain is oriented vertically with the numbered carbon at the . The numbering used for this purpose must follow the IUPAC rules: 2. Next, the structure must be arranged at the particular chiral carbon whose configuration is to be assigned so the horizontal bonds to that carbon extend toward you and the vertical bonds extend away from you. This arrangement will be seen to be precisely the same as the convention of projection formulas such as \(5c\) and \(6c\) ( ). 3. Now the relative positions of the substituents on the horizontal bonds at the chiral centers are examined. If the main substituent is the of the main chain, the \(L\) configuration is assigned; if this substituent is on the , the \(D\) configuration is assigned. For example, the two configurations of the amino acid, alanine, would be represented in perspective or projection as \(15\) and \(16\). The carboxyl carbon is \(C1\) and is placed at the top. The substituents at the chiral carbon connected to the horizontal bonds are amino (\(-NH_2\)) and hydrogen. The amino substituent is taken to be the main substituent; when this is on the the acid has the \(L\) configuration, and when it is on the , the \(D\) configuration. All of the amino acids that occur in natural proteins have been shown to have the \(L\) configuration. Glyceraldehyde, \(CH_2OHCHOHCHO\), which has one chiral carbon bonded to an aldehyde function, hydrogen, hydroxyl, and hydroxymethyl (\(CH_2OH\)), is of special interest as the simplest chiral prototype of sugars (carbohydrates). Perspective views and Fischer projections of the \(D\) and \(L\) forms correspond to \(17\) and \(18\), respectively, where the carbon of the aldehyde function (\(-CH=O\)) is \(C1\): The \(D,L\) system of designating configuration only can be applied when there a main chain, and when we can make an unambiguous choice of the main substituent groups. Try, for instance, to assign \(D\) and \(L\) configurations to enantiomers of bromochlorofluoromethane. An excellent set of rules has been worked out for such cases that leads to unambiguous configurational assignments by what is called the . We discuss the \(R,S\) system in detail in Chapter 19 and, if you wish, you can turn to it now. However, for the next several chapters, assigning configurations is much less important to us than knowing what kinds of stereoisomers are possible. and (1977) | 2,872 | 732 |
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It is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. The collision model explains why chemical reactions often occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. Before delving into the relationship between temperature and reaction rate, we must discuss three microscopic factors that influence the observed macroscopic reaction rates. Central to collision model is that a chemical reaction can occur only when the reactant molecules, atoms, or ions collide. Hence, the observed rate is influence by the frequency of collisions between the reactants. The is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. While deriving the collisional frequency (\(Z_{AB}\)) between two species in a gas is , it is beyond the scope of this text and the equation for collisional frequency of \(A\) and \(B\) is the following: \[Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \dfrac{8\pi k_{B}T}{\mu_{AB}}} \label{freq} \] with The specifics of Equation \ref{freq} are not important for this conversation, but it is important to identify that \(Z_{AB}\) increases with increasing density (i.e., increasing \(N_A\) and \(N_B\)), with increasing reactant size ( . A Discussing Collision Theory of Kinetics: Previously, we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time. The collision model of chemical kinetics explains this behavior by introducing the concept of (\(E_a\)). We will define this concept using the reaction of \(\ce{NO}\) with ozone, which plays an important role in the depletion of ozone in the ozone layer: \[\ce{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)} \nonumber \] Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate. Experimental rate law for this reaction is \[\text{rate} = k [\ce{NO},\ce{O3}] \nonumber \] and is used to identify how the reaction rate (not the rate constant) vares with concentration. The rate constant, however, does vary with temperature. gure \(\Page {1}\) shows a plot of the rate constant of the reaction of \(\ce{NO}\) with \(\ce{O3}\) at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (e.g, the ). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier. In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the , was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the or the of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily. Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence. We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. shows a plot for the NO–O system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction (\(ΔE\)) is negative, which means that the reaction releases energy. (In this case, \(ΔE\) is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction (\(E_a\) is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur. re \(\Page {3a}\) illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, F re \(\Page {3b}\) illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and \(Δ > 0\). Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive; there is a class of reactions called barrierless reactions, but those are discussed elsewhere. For similar reactions under comparable conditions, the one with the smallest will occur most rapidly. Whereas \(ΔE\) is related to the tendency of a reaction to occur spontaneously, \(E_a\) gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. For two similar reactions under comparable conditions, the reaction with the smallest \(E_a\) will occur more rapidly. Figure \(\Page {4}\) shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than ; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than . Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier. Discussing Transition State Theory: Even when the energy of collisions between two reactant species is greater than \(E_a\), most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For \(\ce{NO}\) and \(\ce{O3}\) to produce \(\ce{NO2}\) and \(\ce{O2}\), a terminal oxygen atom of \(\ce{O3}\) must collide with the nitrogen atom of \(\ce{NO}\) at an angle that allows \(\ce{O3}\) to transfer an oxygen atom to \(\ce{NO}\) to produce \(\ce{NO2}\) ( re \(\Page {4}\)). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of \(\ce{NO}\) and \(\ce{O3}\) result in a reaction at kinetic energies greater than \(E_a\), most collisions of \(\ce{NO}\) and \(\ce{O3}\) are unproductive. The fraction of orientations that result in a reaction is called the (\(\rho\)) and its value can range from \(\rho=0\) (no orientations of molecules result in reaction) to \(\rho=1\) (all orientations result in reaction). The collision model explains why most collisions between molecules do not result in a chemical reaction. For example, nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 10 times per second. If every collision produced two molecules of \(\ce{NO}\), the atmosphere would have been converted to \(\ce{NO}\) and then \(\ce{NO2}\) a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide. For an \(A + B\) elementary reaction, all three microscopic factors discussed above that affect the reaction rate can be summarized in a single relationship: \[\text{rate} = (\text{collision frequency}) \times (\text{steric factor}) \times (\text{fraction of collisions with } E > E_a ) \nonumber \] where \[\text{rate} = k[A,B] \label{14.5.2} \] Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, \(A\), called the : \[k=Ae^{-E_{\Large a}/RT} \label{14.5.3} \] The frequency factor is used to convert concentrations to collisions per second (scaled by the steric factor). Because the frequency of collisions depends on the temperature, \(A\) is actually not constant (Equation \ref{freq}). Instead, \(A\) increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time. is known as the and summarizes the collision model of chemical kinetics, where \(T\) is the absolute temperature (in K) and is the ideal gas constant [8.314 J/(K·mol)]. \(E_a\) indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large \(E_a\) increases rapidly with increasing temperature, whereas the reaction rate with a smaller \(E_a\) increases much more slowly with increasing temperature. If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of , \[\begin{align} \ln k &=\ln A+\left(-\dfrac{E_{\textrm a}}{RT}\right) \\[4pt] &=\ln A+\left[\left(-\dfrac{E_{\textrm a}}{R}\right)\left(\dfrac{1}{T}\right)\right] \label{14.5.4} \end{align} \] is the equation of a straight line, \[y = mx + b \nonumber \] where \(y = \ln k\) and \(x = 1/T\). This means that a plot of \(\ln k\) versus \(1/T\) is a straight line with a slope of \(−E_a/R\) and an intercept of \(\ln A\). In fact, we need to measure the reaction rate at only two temperatures to estimate \(E_a\). Knowing the \(E_a\) at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining \(E_a\) from reaction rates measured at several temperatures is illustrated in Example \(\Page {1}\). A Discussing The Arrhenius Equation: Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping (\(f\)) as a function of temperature (\(T\)). Use the data in the following table, along with the graph of ln[chirping rate] versus \(1/T\) to calculate \(E_a\) for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F). chirping rate at various temperatures activation energy and chirping rate at specified temperature If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of \(\ln f\) versus \(1/T\) should give a straight line ( ). Also, the slope of the plot of \(\ln f\) versus \(1/T\) should be equal to \(−E_a/R\). We can use the two endpoints in to estimate the slope: \[\begin{align*}\textrm{slope}&=\dfrac{\Delta\ln f}{\Delta(1/T)} \\[4pt] &=\dfrac{5.30-4.37}{3.34\times10^{-3}\textrm{ K}^{-1}-3.48\times10^{-3}\textrm{ K}^{-1}} \\[4pt] &=\dfrac{0.93}{-0.14\times10^{-3}\textrm{ K}^{-1}} \\[4pt] &=-6.6\times10^3\textrm{ K}\end{align*} \nonumber \] A computer best-fit line through all the points has a slope of −6.67 × 10 K, so our estimate is very close. We now use it to solve for the activation energy: \[\begin{align*} E_{\textrm a} &=-(\textrm{slope})(R) \\[4pt] &=-(-6.6\times10^3\textrm{ K})\left(\dfrac{8.314 \textrm{ J}}{\mathrm{K\cdot mol}}\right)\left(\dfrac{\textrm{1 KJ}}{\textrm{1000 J}}\right) \\[4pt] &=\dfrac{\textrm{55 kJ}}{\textrm{mol}} \end{align*} \nonumber \] If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use to express the known rate constant (\(k_1\)) at the first temperature (\(T_1\)) as follows: \[\ln k_1=\ln A-\dfrac{E_{\textrm a}}{RT_1} \nonumber \] Similarly, we can express the unknown rate constant (\(k_2\)) at the second temperature (\(T_2\)) as follows: \[\ln k_2=\ln A-\dfrac{E_{\textrm a}}{RT_2} \nonumber \] These two equations contain four known quantities ( , , , and ) and two unknowns ( and ). We can eliminate by subtracting the first equation from the second: \[\begin{align*} \ln k_2-\ln k_1 &=\left(\ln A-\dfrac{E_{\textrm a}}{RT_2}\right)-\left(\ln A-\dfrac{E_{\textrm a}}{RT_1}\right) \\[4pt] &=-\dfrac{E_{\textrm a}}{RT_2}+\dfrac{E_{\textrm a}}{RT_1} \end{align*} \nonumber \] Then \[\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \nonumber \] To obtain the best prediction of chirping rate at 308 K ( ), we try to choose for and the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for = 296 K, where = 158, and using the \(E_a\) calculated previously, \[\begin{align*} \ln\dfrac{k_{T_2}}{k_{T_1}} &=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \\[4pt] &=\dfrac{55\textrm{ kJ/mol}}{8.314\textrm{ J}/(\mathrm{K\cdot mol})}\left(\dfrac{1000\textrm{ J}}{\textrm{1 kJ}}\right)\left(\dfrac{1}{296 \textrm{ K}}-\dfrac{1}{\textrm{308 K}}\right) \\[4pt] &=0.87 \end{align*} \nonumber \] Thus / = 2.4 and = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute. The equation for the decomposition of \(\ce{NO2}\) to \(\ce{NO}\) and \(\ce{O2}\) is second order in \(\ce{NO2}\): \[\ce{2NO2(g) → 2NO(g) + O2(g)} \nonumber \] Data for the reaction rate as a function of temperature are listed in the following table. Calculate \(E_a\) for the reaction and the rate constant at 700 K. \(E_a\) = 114 kJ/mol; = 18,600 M ·s = 1.86 × 10 M ·s . What \(E_a\) results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C? about 51 kJ/mol A Discussing Graphing Using the Arrhenius Equation: For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation. The Arrhenius equation is \(k=Ae^{-E_{\Large a}/RT}\). A minimum energy (activation energy,v\(E_a\)) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the , the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: \(k=Ae^{-E_{\Large a}/RT}\). A plot of the natural logarithm of versus 1/ is a straight line with a slope of − / . | 17,336 | 733 |
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The polynomial coefficient, \(C\), is a function of the factorials of large numbers. Since \(N!\) quickly becomes very large as \(N\) increases, it is often impractical to evaluate \(N!\) from the definition, \[N!=\left(N\right)\left(N-1\right)\left(N-2\right)\dots \left(3\right)\left(2\right)\left(1\right)\] Fortunately, an approximation, known as or is available. Stirling’s approximation is a product of factors. Depending on the application and the required accuracy, one or two of these factors can often be taken as unity. Stirling’s approximation is \[N!\approx N^N \left(2\pi N\right)^{1/2}\mathrm{exp}\left(-N\right)\mathrm{exp}\left(\frac{1}{12N}\right)\approx N^N\left(2\pi N\right)^{1/2}\mathrm{exp}\left(-N\right)\approx N^N\mathrm{exp}\left(-N\right)\] In many statistical thermodynamic arguments, the important quantity is the natural logarithm of \(N!\) or its derivative, \({d ~ { \ln N!\ }}/{dN}\). In such cases, the last version of Stirling’s approximation is usually adequate, even though it affords a rather poor approximation for \(N!\) itself. | 1,085 | 734 |
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We previously saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination. To determine whether a system has reached equilibrium, chemists use a quantity called the Reaction Quotient (\(Q\)). The expression for the Reaction Quotient has precisely the same form as the equilibrium constant expression from the , except that \(Q\) may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction: the Reaction Quotient is defined as follows: \[Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Eq1}\] To understand how information is obtained using a Reaction Quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide, \[\ce{N2O4(g) \rightleftharpoons 2NO2(g)}. \label{Rx}\] For this reaction, \(K = 4.65 \times 10^{−3}\) at 298 K. We can write \(Q\) for this reaction as follows: The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of \(Q\) were calculated for each. Each experiment begins with different proportions of product and reactant: As these calculations demonstrate, \(Q\) can have any numerical value between 0 and infinity (undefined); that is, \(Q\) can be greater than, less than, or equal to K. Comparing the magnitudes of \(Q\) and K enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes \(Q\) approach K. If \(Q = K\), for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed. If \(Q < K\), then the ratio of the concentrations of products to the concentrations of reactants is less than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants. Conversely, if \(Q > K\), then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products. These points are illustrated graphically in Figure \(\Page {1}\). If \(Q < K\), the reaction will proceed to the right as written. If \(Q > K\), the reaction will proceed to the left as written. If \(Q = K\), then the system is at equilibrium. At elevated temperatures, methane (\(CH_4\)) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction: \[\ce{CH4(g) + H2O (g) \rightleftharpoons CO (g) + 3H2(g)}
\nonumber \] \(K = 2.4 \times 10^{−4}\) at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If \(1.2 × 10^{−2}\) mol of \(CH_4\), \(8.0 × 10^{−3}\) mol of \(H_2O\), \(1.6 \times 10^{−2}\) mol of \(CO\), and \(6.0 \times 10^{−3}\) mol of \(H_2\) are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce \(CO\) and \(H_2\) or to the left to form \(CH_4\) and \(H_2O\)? : balanced chemical equation, \(K\), amounts of reactants and products, and volume : direction of reaction : : We must first find the initial concentrations of the substances present. For example, we have \(1.2 \times 10^{−2} mol\) of \(CH_4\) in a 2.0 L container, so \[\begin{align*} [CH_4]&=\dfrac{1.2\times 10^{−2} mol}{2.0\; L} \\[4pt] &=6.0 \times 10^{−3} M \end{align*}\] We can calculate the other concentrations in a similar way: We now compute \(Q\) and compare it with \(K\): \[\begin{align*} Q&=\dfrac{[CO,H_2]^3}{[CH_4,H_2O]} \\[4pt] &=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})} \\[4pt] &=9.0 \times 10^{−6} \end{align*}\] Because \(K = 2.4 × 10^{−4}\), we see that \(Q < K\). Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming \(\ce{H2}\) and \(\ce{CO}\) at the expense of \(H_2O\) and \(CH_4\). In the water–gas shift reaction introduced in Example \(\Page {1}\), carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen: \[\ce{ CO (g) + H2O (g) \rightleftharpoons CO2(g) +H2(g)}
\nonumber\] \(K = 0.64\) at 900 K. If 0.010 mol of both \(\ce{CO}\) and \(\ce{H_2O}\), 0.0080 mol of \(\ce{CO_2}\), and 0.012 mol of \(\ce{H_2}\) are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written? \(Q = 0.96\) (Q > K), so the reaction will proceed to the left, and \(CO\) and \(H_2O\) will form. By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which \(Q = K\). Such a graph allows us to predict what will happen to a reaction when conditions change so that \(Q\) no longer equals \(K\), such as when a reactant concentration or a product concentration is increased or decreased. Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation: \[\ce{PbCO3(s) \rightleftharpoons PbO(s) + CO2(g)} \label{Eq3}\] Because \(PbCO_3\) and \(PbO\) are solids, the equilibrium constant is simply \(K = [CO_2]\). At a given temperature, therefore, any system that contains solid \(PbCO_3\) and solid \(PbO\) will have exactly the same concentration of \(CO_2\) at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in Figure \(\Page {2}\), which shows a plot of \([CO_2]\) versus the amount of \(PbCO_3\) added. Initially, the added \(PbCO_3\) decomposes completely to \(CO_2\) because the amount of \(PbCO_3\) is not sufficient to give a \(CO_2\) concentration equal to \(K\). Thus the left portion of the graph represents a system that is not at equilibrium because it contains only CO2(g) and PbO(s). In contrast, when just enough \(PbCO_3\) has been added to give \([CO_2] = K\), the system has reached equilibrium, and adding more \(PbCO_3\) has no effect on the \(CO_2\) concentration: the graph is a horizontal line. Thus any \(CO_2\) concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough \(PbCO_3\) and \(PbO\) are present. For example, the point labeled A in Figure \(\Page {2}\) lies above the horizontal line, so it corresponds to a \([CO_2]\) that is greater than the equilibrium concentration of \(CO_2\) (Q > K). To reach equilibrium, the system must decrease \([CO_2]\), which it can do only by reacting \(CO_2\) with solid \(PbO\) to form solid \(PbCO_3\). Thus the reaction in Equation \(\ref{Eq3}\) will proceed to the left as written, until \([CO_2] = K\). Conversely, the point labeled B in Figure \(\Page {2}\) lies below the horizontal line, so it corresponds to a \([CO_2]\) that is less than the equilibrium concentration of \(CO_2\) (Q < K). To reach equilibrium, the system must increase \([CO_2]\), which it can do only by decomposing solid \(PbCO_3\) to form \(CO_2\) and solid \(PbO\). The reaction in Equation \(\ref{Eq3}\) will therefore proceed to the right as written, until \([CO_2] = K\). In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor: \[\ce{CdO (s) + H2(g) \rightleftharpoons Cd (s) + H2O(g)} \label{Eq4}\] and the equilibrium constant K is \([H_2O]/[H_2]\). If \([H_2O]\) is doubled at equilibrium, then [H2] must also be doubled for the system to remain at equilibrium. A plot of \([H_2O]\) versus \([H_2]\) at equilibrium is a straight line with a slope of K (Figure \(\Page {3}\)). Again, only those pairs of concentrations of \(H_2O\) and \(H_2\) that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in Equation \(\ref{Eq4}\) will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point A in Figure \(\Page {3}\) lies below the line, indicating that the \([H_2O]/[H_2]\) ratio is less than the ratio of an equilibrium mixture (Q < K). Thus the reaction in Equation \(\ref{Eq4}\) will proceed to the right as written, consuming \(H_2\) and producing \(H_2O\), which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point B in Figure \(\Page {3}\) lies above the line, indicating that the \([H_2O]/[H_2]\) ratio is greater than the ratio of an equilibrium mixture (Q > K). Thus the reaction in Equation \(\ref{Eq4}\) will proceed to the left as written, consuming \(H_2O\) and producing \(H_2\), which causes the concentration ratio to move down and to the right toward the equilibrium line. In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures: \[ NH_4I_{(s)} \rightleftharpoons NH_{3(g)}+HI_{(g)} \label{Eq5}\] For this system, \(K\) is equal to the product of the concentrations of the two products: \([NH_3,HI]\). If we double the concentration of NH3, the concentration of HI must decrease by approximately a factor of 2 to maintain equilibrium, as shown in Figure \(\Page {4}\). As a result, for a given concentration of either \(HI\) or \(NH_3\), only a single equilibrium composition that contains equal concentrations of both \(NH_3\) and HI is possible, for which \([NH_3] = [HI] = K^{1/2}\). Any point that lies below and to the left of the equilibrium curve (such as point A in Figure \(\Page {4}\)) corresponds to \(Q < K\), and the reaction in Equation \(\ref{Eq5}\) will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in Figure \(\Page {4}\)) corresponds to \(Q > K\), and the reaction in Equation \(\ref{Eq5}\) will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium. Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for \(K_c\)) or partial pressure (for \(K_p\)). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium. As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure. Consider what happens when we increase the pressure on a system in which \(\ce{NO}\), \(\ce{O_2}\), and \(\ce{NO_2}\) are at equilibrium: \[\ce{2NO}_{(g)}+\ce{O}_{2(g)} \rightleftharpoons \ce{2NO}_{2(g)} \label{Eq8}\] The formation of additional amounts of \(\ce{NO2}\) decreases the total number of molecules in the system because each time two molecules of \(\ce{NO_2}\) form, a total of three molecules of \(\ce{NO}\) and \(\ce{O_2}\) are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of \(\ce{NO_2}\) into \(\ce{NO}\) and \(\ce{O_2}\), which tends to restore the pressure. Now consider this reaction: \[\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g) \label{Eq9}\] Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide. Changing concentration or pressure perturbs an equilibrium because the Reaction Quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle. When hydrogen reacts with gaseous iodine, heat is evolved. \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI (g)} \;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{Eq10}\] Because this reaction is exothermic, we can write it with heat as a product. \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI (g) + heat} \label{Eq11}\] Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of \(\ce{H2}\) and \(\ce{I2}\) and a reduction in the concentration of \(\ce{HI}\). Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide. When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the \(\ce{HI}\) system increases the equilibrium constant: At the new equilibrium the concentration of \(\ce{HI}\) has increased and the concentrations of \(\ce{H2}\) and \(\ce{I2}\) decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C. Temperature affects the equilibrium between \(\ce{NO_2}\) and \(\ce{N_2O_4}\) in this reaction \[\ce{N_2O}_{4(g)} \rightleftharpoons \ce{2NO}_{2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{Eq12}\] The positive \(ΔH\) value tells us that the reaction is endothermic and could be written \[\ce{heat}+\ce{N_2O}_{4(g)} \rightleftharpoons \ce{2NO}_{2(g)} \label{Eq13}\] At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown \(\ce{NO_2}\) molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless \(\ce{N_2O_4}\) increases, and the concentration of brown \(\ce{NO_2}\) decreases, causing the brown color to fade. It is not uncommon that textbooks and instructors to consider heat as a independent "species" in a reaction. While this is rigorously incorrect because one cannot "add or remove heat" to a reaction as with species, it serves as a convenient mechanism to predict the shift of reactions with changing temperature. For example, if heat is a "reactant" (\(\Delta{H} > 0 \)), then the reaction favors the formation of products at elevated temperature. Similarly, if heat is a "product" (\(\Delta{H} < 0 \)), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below. The commercial production of hydrogen is carried out by treating natural gas with steam at high temperatures and in the presence of a catalyst (“steam reforming of methane”): \[\ce{CH_4 + H_2O <=> CH_3OH + H_2} \nonumber\] Given the following boiling points: CH (methane) = –161°C, H O = 100°C, CH OH = 65°, H = –253°C, predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C. To identify the influence of changing pressure on the three reaction conditions, we need to identify the correct reaction including the phase of each reactant and product. Calculate the change in the moles of gas for each process: What will happen to the equilibrium when the volume of the system is decreased? \[\ce{2SO2 (g) + O2 (g) \rightleftharpoons 2SO3 (g)} \nonumber\] Decreasing the volume leads to an increase in pressure which will cause the equilibrium to shift towards the side with fewer moles. In this reaciton, there are three moles on the reactant side and two moles on the product side, so the new equilibrium will shift towards the products (to the right). The Reaction Quotient (Q) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. Reaction Quotient: \[Q =\dfrac{[C]^c[D]^d}{[A]^a[B]^b}
\nonumber\] The Reaction Quotient (\(Q\) or \(Q_p\)) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, \(Q = K\). Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side. ). | 19,556 | 735 |
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are highly specific catalysts for biochemical reactions, with each enzyme showing a selectivity for a single reactant, or . For example, the enzyme acetylcholinesterase catalyzes the decomposition of the neurotransmitter acetylcholine to choline and acetic acid. Many enzyme–substrate reactions follow a simple mechanism that consists of the initial formation of an enzyme–substrate complex, \(ES\), which subsequently decomposes to form product, releasing the enzyme to react again. This is described within the following multi-step mechanism \[ E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\overset{k_2}{\rightleftharpoons}} E + P \label{13.20}\] where \(k_1\), \(k_{–1}\), \(k_2\), and \(k_{–2}\) are rate constants. The reaction’s rate law for generating the product \([P]\) is \[ rate = \dfrac{d[P]}{dt} = k_2[ES] - k_{-2}[E,P] \label{13.21A}\] However, if we make measurement early in the reaction, the concentration of products is negligible, i.e., \[[P] \approx 0\] and we can ignore the back reaction (second term in right side of Equation \(\ref{13.21A}\)). Then under these conditions, the reaction’s rate is \[ rate = \dfrac{d[P]}{dt} = k_2[ES] \label{13.21}\] To be analytically useful we need to write Equation \(\ref{13.21}\) in terms of the reactants (e.g., the concentrations of enzyme and substrate). To do this we use the , in which we assume that the concentration of \(ES\) remains essentially constant. Following an initial period, during which the enzyme–substrate complex first forms, the rate at which \(ES\) forms \[ \dfrac{d[ES]}{dt} =k_1[E] [S] = k_1([E]_0 − [ES])[S] \label{13.22}\] is equal to the rate at which it disappears \[ − \dfrac{d[ES]}{dt} = k_{−1}[ES] + k_2[ES] \label{13.23}\] where \([E]_0\) is the enzyme’s original concentration. Combining Equations \(\ref{13.22}\) and \(\ref{13.23}\) gives \[k_1([E]_0 − [ES]) [S] = k_{−1}[ES] + k_2[ES]\] which we solve for the concentration of the enzyme–substrate complex \[ [ES] = \dfrac{ [E]_0[S] }{ \dfrac{k_{−1} + k_2}{k_1} + [S] } = \dfrac{[E]_0[S] }{K_m + [S]} \label{Eq13.24}\] where \(K_m\) is the . Substituting Equation \(\ref{Eq13.24}\) into Equation \(\ref{13.21}\) leaves us with our final rate equation. \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \label{Eq13.25} \] A plot of Equation \(\ref{Eq13.25}\), as shown in Figure \(\Page {1}\), is instructive for defining conditions where we can use the rate of an enzymatic reaction for the quantitative analysis of an enzyme or substrate. For high substrate concentrations, where \([S] \gg K_m\), Equation \(\ref{Eq13.25}\) simplifies to \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{ K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{[S]} = k_2[E]_0 = V_{max} \label{Eq13.26}\] where \(V_{max}\) is the maximum rate for the catalyzed reaction. Under these conditions the reaction is in substrate and we can use \(V_{max}\) to calculate the enzyme’s concentration, typically using a variable-time method. At lower substrate concentrations, where \([S] \ll K_m\), Equation \(\ref{Eq13.25}\) becomes \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{ K_m} =\dfrac{V_{max}[S]}{K_m} \label{13.27}\] The reaction is now in substrate, and we can use the rate of the reaction to determine the substrate’s concentration by a fixed-time method. The Michaelis constant \(K_m\) is the substrate concentration at which the reaction rate is at half-maximum, and is an inverse measure of the substrate's affinity for the enzyme—as a small \(K_m\) indicates high affinity, meaning that the rate will approach \(V_{max}\) more quickly. The value of \(K_m\) is dependent on both the enzyme and the substrate, as well as conditions such as temperature and pH. The Michaelis constant \(K_m\) is the substrate concentration at which the reaction rate is at half-maximum. From the last two terms in Equation \(\ref{13.27}\), we can express \(V_{max}\) in terms of a (\(k_{cat}\)): \[ V_{max} = k_{cat}[E]_o\] where \([E]_0\) is the enzyme concentration and \(k_{cat}\) is the turnover number, defined as the maximum number of substrate molecules converted to product per enzyme molecule per second. Hence, the turnover number is defined as the maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration \([E]_o\). Acetylcholinesterase (AChE) may be one of the fastest enzymes. It hydrolyzes acetylcholine to choline and an acetate group. One of the earliest values of the turnover number was \(3 \times 10^7\) (molecules of acetylcholine) per minute per molecule of enzyme. A more recent value at 25°C, pH = 7.0, acetylcholine concentration of \(2.5 \times 10^{−3}\; M\), was found to be \(7.4 \times 10^5\; min^{−1}\) ( ). There may be some 30 active centers per molecule. AChE is a serine hydrolase that reacts with acetylcholine at close to . A diffusion-controlled reaction occurs so quickly that the reaction rate is the rate of transport of the reactants through the solution; a s quickly as the reactants encounter each other, they react. The Michaelis-Menten model is used in a variety of biochemical situations other than enzyme-substrate interaction, including antigen-antibody binding, DNA-DNA hybridization, and protein-protein interaction. It can be used to characterize a generic biochemical reaction, in the same way that the can be used to model generic adsorption of biomolecular species. When an empirical equation of this form is applied to microbial growth. The experimentally determined parameters values vary wildly between enzymes (Table \(\Page {1}\)): While \(K_m\) is equal to the substrate concentration at which the enzyme converts substrates into products at half its maximal rate and hence is related to the affinity of the substrate for the enzyme. The catalytic rate \(k_{cat}\) is the rate of product formation when the enzyme is saturated with substrate and therefore reflects the enzyme's maximum rate. The rate of product formation is dependent on both how well the enzyme binds substrate and how fast the enzyme converts substrate into product once substrate is bound. For a kinetically perfect enzyme, every encounter between enzyme and substrate leads to product and hence the reaction velocity is only limited by the rate the enzyme encounters substrate in solution. From Equation \(\ref{Eq13.24}\), the catalytic efficiency of a protein can be evaluated. \[ \dfrac{k_{cat}}{K_m} = \dfrac{k_2}{K_m} = \dfrac{k_1k_2}{k_{-1} + k_2}\] This \(k_{cat}/K_m\) ratio is called the specificity constant measure of how efficiently an enzyme converts a substrate into product. It has a theoretical upper limit of ; enzymes working close to this, such as fumarase, are termed superefficient (Table \(\Page {1}\)). Determining \(V_m\) and \(K_m\) from experimental data can be difficult and the most common way is to determine initial rates, \(v_0\), from experimental values of \([P]\) or \([S]\) as a function of time. Hyperbolic graphs of \(v_0\) vs. \([S]\) can be fit or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine \(K_d\). These included: \[\dfrac{1}{r} = \dfrac{K_M + \left[ \text{S} \right]}{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]} = \dfrac{K_M}{k_2 \left[ \text{E} \right]_0} \dfrac{1}{\left[ \text{S} \right]} + \dfrac{1}{k_2 \left[ \text{E} \right]_0} \label{Eq28}\] Tthe Lineweaver–Burk plot (or double reciprocal plot) is a graphical representation of the Lineweaver–Burk equation of enzyme kinetics, described by Hans Lineweaver and Dean Burk in 1934 (Figure \(\Page {2}\)). \(1/V_{max}\) via Equation \(\ref{Eq13.26}\). The plot provides a useful graphical method for analysis of the Michaelis–Menten equation: \[V = \dfrac{V_{\max} [S]}{K_m + [S]}\] Taking the reciprocal gives \[\dfrac{1}{V} = \dfrac {K_m + [S]} {V_{max}[S]} = \dfrac{K_m}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}}\] where The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as \(K_m\) and \(V_{max}\), before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of \(V_{max}\); the x-intercept of the graph represents \(−1/K_m\). It also gives a quick, visual impression of the different forms of enzyme inhibition. The reaction between nicotineamide mononucleotide and ATP to form nicotineamide–adenine dinucleotide and pyrophosphate is catalyzed by the enzyme nicotinamide mononucleotide adenylyltransferase. The following table provides typical data obtained at a pH of 4.95. The substrate, S, is nicotinamide mononucleotide and the initial rate, , is the μmol of nicotinamide–adenine dinucleotide formed in a 3-min reaction period. Determine values for and . Figure 13.12 shows the Lineweaver–Burk plot for this data and the resulting regression equation. Using the -intercept, we calculate as = 1 / −intercept = 1 / 1.708 mol = 0.585 mol and using the slope we find that is = slope × = 0.7528 molimM × 0.585 mol = 0.440 mM The following data are for the oxidation of catechol (the substrate) to -quinone by the enzyme -diphenyl oxidase. The reaction is followed by monitoring the change in absorbance at 540 nm. The data in this exercise are adapted from jkimball. 0.3 0.6 1.2 4.8 0.020 0.035 0.048 0.081 Determine values for and . to review your answer to this exercise. The double reciprocal plot distorts the error structure of the data, and it is therefore unreliable for the determination of enzyme kinetic parameters. Although it is still used for representation of kinetic data, non-linear regression or alternative linear forms of the Michaelis–Menten equation such as the or are generally used for the calculation of parameters. The Lineweaver–Burk plot is classically used in older texts, but is prone to error, as the -axis takes the reciprocal of the rate of reaction – in turn increasing any small errors in measurement. Also, most points on the plot are found far to the right of the -axis (due to limiting solubility not allowing for large values of \([S]\) and hence no small values for \(1/[S]\)), calling for a large extrapolation back to obtain - and -intercepts When used for determining the type of enzyme inhibition, the Lineweaver–Burk plot can distinguish competitive, non-competitive and uncompetitive inhibitors. Competitive inhibitors have the same y-intercept as uninhibited enzyme (since \(V_{max}\) is unaffected by competitive inhibitors the inverse of \(V_{max}\) also doesn't change) but there are different slopes and x-intercepts between the two data sets. Non-competitive inhibition produces plots with the same x-intercept as uninhibited enzyme (\(K_m\) is unaffected) but different slopes and y-intercepts. Uncompetitive inhibition causes different intercepts on both the y- and x-axes but the same slope. The Eadie–Hofstee plot is a graphical representation of enzyme kinetics in which reaction rate is plotted as a function of the ratio between rate and substrate concentration and can be derived from the Michaelis–Menten equation (\(\ref{Eq13.25}\)) by inverting and multiplying with \(V_{max}\): \[ \dfrac{V_{max}}{v} = \dfrac{V_{max}(K_m+[S])}{V_{max}[S]} = \dfrac{K_m+[S]}{[S]}\] Rearrange: \[V_{max} = \dfrac{vK_m}{[S]} + \dfrac{v[S]}{[S]} = \dfrac{v K_m}{[S]} + v\] Isolate v: \[v = -K_m \dfrac{v}{[S]} + V_{max}\] A plot of v against \(v/[S]\) will hence yield \(V_{max}\) as the y-intercept, \(V_{max}/K_m\) as the x-intercept, and \(K_m\) as the negative slope (Figure \(\Page {3}\)). Like other techniques that linearize the Michaelis–Menten equation, the Eadie-Hofstee plot was used historically for rapid identification of important kinetic terms like \(K_m\) and \(V_{max}\), but has been superseded by nonlinear regression methods that are significantly more accurate and no longer computationally inaccessible. It is also more robust against error-prone data than the Lineweaver–Burk plot, particularly because it gives equal weight to data points in any range of substrate concentration or reaction rate (the Lineweaver–Burk plot unevenly weights such points). Both Eadie-Hofstee and Lineweaver–Burk plots remain useful as a means to present data graphically. One drawback from the Eadie–Hofstee approach is that neither ordinate nor abscissa represent independent variables: both are dependent on reaction rate. Thus any experimental error will be present in both axes. Also, experimental error or uncertainty will propagate unevenly and become larger over the abscissa thereby giving more weight to smaller values of v/[S]. Therefore, the typical measure of goodness of fit for linear regression, the correlation coefficient R, is not applicable. | 12,887 | 737 |
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Electron Spin or Spin Quantum Number is the fourth quantum number for electrons in atoms and molecules. Denoted as \(m_s\), the electron spin is constituted by either upward (\(m_s=+1/2\)) or downward (\(m_s=-1/2\)) arrows. In 1920, Otto Stern and Walter Gerlach designed an experiment, which unintentionally led to the discovery that electrons have their own individual, continuous spin even as they move along their orbital of an atom. Today, this electron spin is indicated by the fourth quantum number, also known as the and denoted by . In 1925, Samuel Goudsmit and George Uhlenbeck made the claim that features of the hydrogen spectrum that were unexamined might by explained by assuming electrons act as if it has a spin. This spin can be denoted by an arrow pointing up, which is +1/2, or an arrow pointing down, which is -1/2. The experiment mentioned above by Otto Stern and Walter Gerlach was done with silver which was put in an oven and vaporized. The result was that silver atoms formed a beam that passed through a magnetic field in which it split in two. An explanation of this is that an electron has a magnetic field due to its spin. When electrons that have opposite spins are put together, there is no net magnetic field because the positive and negative spins cancel each other out. The silver atom used in the experiment has a total of 47 electrons, 23 of one spin type, and 24 of the opposite. Because electrons of the same spin cancel each other out, the one unpaired electron in the atom will determine the spin. There is a high likelihood for either spin due to the large number of electrons, so when it went through the magnetic field it split into two beams. A total of four quantum numbers were developed to better understand the movement and pathway of electrons in its designated orbital within an atom. The lines represent how many orientations each orbital has, (e.g. the s-orbital has one orientation, a p-orbital has three orientations, etc.) and each line can hold up to two electrons, represented by up and down arrows. An electron with an up arrow means it has an electron spin of +\(\frac{1}{2}\), and an electron with a down arrow means it has an electron spin of -\(\frac{1}{2}\). Significance: determines if an atom will or will not generate a magnetic field (For more information, scroll down to ). Although the electron spin is limited to or , certain rules apply when assigning electrons of different spins to fill a subshell (orientations of an orbital) . For more information, scroll down to . An atom with electrons are termed as An atom with electrons are termed as Applying concepts of magnetism with liquid nitrogen and liquid oxygen: The magnetic spin of an electron follows in the direction of the magnetic field lines as shown below. {{media("www.youtube.com/watch?v=uj0DFDfQajw\) An effective visual on how to assign spin directions can be represented by the orbital diagram (shown previously and below.) Restrictions apply when assigning spin directions to electrons, so the following Pauli Exclusion Principle and Hund's Rule help explain this. When one is filling an orbital, such as the p orbital, you must fill all orbitals possible with one electron spin before assigning the opposite spin. For example, when filling the fluorine, which will have a total of two electrons in the s orbital and a total of five electrons in the p orbital, one will start with the s orbital which will contain two electrons. So, the first electron one assigns will be spin up and the next spin down. Moving on to the three p orbitals that one will start by assigning a spin up electron in each of the three orbitals. That takes up three of the five electrons, so with the remaining two electrons, one returns to the first and second p orbital and assigns the spin down electron. This means there will be one unpaired electron in fluorine so it will be paramagnetic. The declares that there can only be a maximum of two electrons for every one orientation, and the two electrons must be opposite in spin direction; meaning one electron has \(m_s = +\frac{1}{2}\) and the other electron has \(m_s = -\frac{1}{2}\). declares that the electrons in the orbital are filled up first by the +\(\frac{1}{2}\) spin. Once all the orbitals are filled with unpaired +\(\frac{1}{2}\) spins, the orbitals are then filled with the -\(\frac{1}{2}\) spin. ( Sulfur - S (16 electrons) Electronic Configuration: When given a principal quantum number, n, with either the s, p, d or f-orbital, identify all the possibilities of L, m and m . Given 5f, identify all the possibilities of the four quantum numbers. In this problem, the principal quantum number is n = 5 (the subshell number placed in front of the orbital, the f-orbital in this case). Since we are looking at the f-obital, therefore L = 3. (Look under "Subshells" in the module for more information) Knowing L = 3, we can interpret that m = 0, \(\pm\) 1, \(\pm\) 2, \(\pm\) 3 since m = -L,...,-1, 0, 1,...+L. As for m , since it isn't specified in the problem as to whether it is -\(\frac{1}{2}\) or +\(\frac{1}{2}\), therefore for this problem, it could be both; meaning that the electron spin quantum number is \(\pm\)\(\frac{1}{2}\). Given 6s and m = +1, identify all the possibilities of the four quantum numbers. The principal quantum number is n = 6. Looking at the s-orbital, we know that L = 0. Knowing that m = -L,...,-1, 0, 1,...+L, therefore m = +1 is since in this problem, the interval of m can only equal to 0 according to the angular momentum quantum number, L. Given 4d and m = +\(\frac{1}{2}\), identify all the possibilities of the four quantum numbers. The principal quantum number is n = 4. Given that it is a d-orbital, we know that L = 2. Therefore, m = 0, \(\pm\) 1, \(\pm\) 2 since m = -L,...,-1, 0, 1,...+L. For m , this problem specifically said m = +\(\frac{1}{2}\); meaning that the electron spin quantum number is +\(\frac{1}{2}\). First, draw a table labeled n, L, m and m , as shown below: Then, depending on what the question is asking for, fill in the boxes accordingly. Finally, determine the number of electrons for the given quantum number, n, with regards to L, m and m . How many electrons can have n = 5 and L = 1? This problem includes both -\(\frac{1}{2}\), +\(\frac{1}{2}\), therefore the answer is electrons based on the m . How many electrons can have n = 5 and m = -\(\frac{1}{2}\)? This problem only wants the Spin Quantum Number to be -\(\frac{1}{2}\), the answer is electrons based on the m . How many electrons can have n = 3, L = 2 and m = 3? How many electrons can have n = 3, and m = +\(\frac{1}{2}\)? This problem only wants the Spin Quantum Number to be +\(\frac{1}{2}\), the answer is electrons based on the m . Sodium (Na) --> Electronic Configuration [Ne] 3s Spin direction for the valence electron or m = +\(\frac{1}{2}\) Sodium (Na) with a neutral charge of zero is paramagnetic, meaning that the electronic configuration for Na consists of one or more unpaired electrons. Chlorine (Cl) --> Electronic Configuration [Ne] 3s 3p Spin direction for the valence electron or m = +\(\frac{1}{2}\) Chlorine (Cl) with a neutral charge of zero is paramagnetic. Calcium (Ca) --> Electronic Configuration [He] 4s Spin direction for the valence electron or m = \(\pm\)\(\frac{1}{2}\) Whereas for Calcium (Ca) with a neutral charge of zero, it is diamagnetic; meaning that ALL the electrons are paired as shown in the image above. Given 5p and m = -\(\frac{1}{2}\), identify all the possibilities of the four quantum numbers. The principal quantum number is n = 5. Given that it is a p-orbital, we know that L = 1. And based on L, m = 0, \(\pm\) 1 since m = -L,...,-1, 0, 1,...+L. As for m , this problem specifically says m = -\(\frac{1}{2}\), meaning that the spin direction is -\(\frac{1}{2}\), pointing downwards ("down" spin). Given 6f, identify all the possibilities of the four quantum numbers. The principal quantum number is n = 6. Given that it is a f-orbital, we know that L = 3. Based on L, m = 0, \(\pm\) 1, \(\pm\) 2, \(\pm\) 3 since m = -L,...,-1, 0, 1,...+L. As for m , since it isn't specified in the problem as to whether it is -\(\frac{1}{2}\) or +\(\frac{1}{2}\), therefore for this problem, it could be both; meaning that the electron spin quantum number is \(\pm\)\(\frac{1}{2}\). How many electrons can have n = 4 and L = 1? This problem includes both -\(\frac{1}{2}\), +\(\frac{1}{2}\), therefore the answer is electrons based on the m . How many electrons can have n = 4, L = 1, m = -2 and m = +\(\frac{1}{2}\)? Since , m is not possible because L = 1, so it is impossible for m to be equal to 2 when m MUST be with the interval of -L and +L. So, there is electron. How many electrons can have n = 5, L = 3, and m = +\(\frac{1}{2}\)? This problem only wants the Spin Quantum Number to be +\(\frac{1}{2}\) and m = \(\pm\) 2, therefore electrons can have n = 5, L = 3, m = \(\pm\) 2 and m = +\(\frac{1}{2}\). How many electrons can have n = 5, L = 4 and ? This problem includes both -\(\frac{1}{2}\) and +\(\frac{1}{2}\) and given that m = +3, therefore the answer is electrons. How many electrons can have n = 4, L = 2 and ? This problem includes both -\(\frac{1}{2}\) and +\(\frac{1}{2}\) and given that m = \(\pm\) 1, therefore the answer is electrons. How many electrons can have n = 3, L = 3, m = +2 and m = -\(\frac{1}{2}\)? Since there is electron, not possible because in this problem, n = L = 3. | 9,583 | 738 |
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Aufbau comes from the German word "Aufbauen" which means "to build". In essence when writing electron configurations we are building up electron orbitals as we proceed from atom to atom. As we write the electron configuration for an atom, we will fill the orbitals in order of increasing atomic number. The Aufbau principle originates from the which says that no two (e.g., electrons) in an atom can have the same set of , hence they have to "pile up" or "build up" into higher energy levels. How the electrons build up is a topic of . If we follow the pattern across a period from B (Z=5) to Ne (Z=10) the number of electrons increase and the subshells are filled. Here we are focusing on the p subshell in which as we move towards Ne, the p subshell becomes filled. | 790 | 739 |
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At the turn of the nineteenth century Sir William Herschel discovered invisible radiation beyond the red end of the visible region of the electromagnetic spectrum. This radiation appropriately is called , meaning “beneath the red,” and it encompasses the wavelength region from \(10^3 \: \text{nm}\) to \(10^6 \: \text{nm}\). You probably are familiar with the common applications of infrared to radiant heating and photography. In addition to these uses, infrared spectroscopy has become the most widely used spectroscopic technique for investigating organic structures. Infrared spectroscopy was the province of physicists and physical chemists until about 1940. At that time, the potential of infrared spectroscopy as an analytical tool began to be recognized by organic chemists. The change was due largely to the production of small, quite rugged infrared spectrophotometers and instruments of this kind now are virtually indispensable for chemical analysis. A brief description of the principles and practice of this spectroscopic method is the topic of this section. Absorption of infrared radiation causes transitions between energy states of a molecule. A simple diatomic molecule, such as \(\ce{H-Cl}\), has only one vibrational mode available to it, a stretching vibration somewhat like balls on the ends of a spring: Molecules with three or more atoms can vibrate by stretching and also by bending of the chemical bonds, as indicated below for carbon dioxide: The absorption frequencies in the infrared spectra of molecules correspond to changes in the stretching or bending vibrations or both. In general, a polyatomic molecule with \(n\) atoms will have \(3n - 6\) modes of vibration of which \(n -1\) are stretching vibrations and \(2n - 5\) are bending vibrations. There are circumstances, however, where fewer vibrational modes are possible. If the molecule is linear, like \(CO_2\), then there are \(3n-5\) possible vibrations, and some of these vibrations may be equivalent (degenerate vibrations in the language of spectroscopists). For example, \(CO_2\) should have \(3n - 5\) or \(4\) vibrational modes, two of which are stretching and two of which are bending modes. However, the two bending modes are equivalent because the in which the molecule bends is immaterial; in-plane or out-of-plane bending are the same: Diatomic molecules such as \(HCl\) have one vibrational mode, but it is important to note that , such as \(O_2\), \(N_2\), \(Cl_2\), \(F_2\), and \(H_2\), . This is because absorption cannot occur if the vibration is electrically symmetrical. Fortunately, then, the infrared spectra can be recorded in air because the main components of air, \(N_2\) and \(O_2\), do not interfere. In practice, infrared spectra can be obtained with gaseous, liquid, or solid samples. The sample containers (cells) and the optical parts of the instrument are made of rock salt (\(NaCl\)) or similar material that transmits infrared radiation (glass is opaque). Typical infrared spectra are shown in Figure 9-9 for 2-propanone (acetone), \(\ce{CH_3-CO-CH_3}\), and 2-butanone (methyl ethyl ketone), \(\ce{CH_3-CO-CH_2-CH_3}\). In accord with current practice, the position of absorption (horizontal scale) is recorded in units of wave numbers (\(\bar{\nu}\), \(\text{cm}^{-1}\); see ). The vertical scale measures the intensity of radiation transmitted through the sample. Zero transmission means complete absorption of radiation by the sample as at \(1740 \: \text{cm}^{-1}\) in Figure 9-9. The other absorption bands in Figure 9-9 that correspond to excitation of stretching or bending vibrations are not as intense as the absorption at \(1740 \: \text{cm}^{-1}\). What information can we derive about molecular structure from the vibrational bands of infrared spectra? Absorption of radiation in the range of \(5000\)-\(1250 \: \text{cm}^{-1}\) is characteristic of the types of bonds present in the molecule, and corresponds for the most part to stretching vibrations. For example, we know that the \(\ce{C-H}\) bonds of alkanes and alkyl groups have characteristic absorption bands around \(2900 \: \text{cm}^{-1}\); an unidentified compound that shows absorption in this region will very likely have alkane-type \(\ce{C-H}\) bonds. More explicitly, the band observed for 2-propanone (Figure 9-9a) at \(3050 \: \text{cm}^{-1}\) arises from absorption of infrared radiation, which causes transitions between the ground vibrational state (or lowest vibrational energy level) of a \(\ce{C-H}\) bond and the first excited vibrational energy level for stretching of that \(\ce{C-H}\) bond. The band at \(1740 \: \text{cm}^{-1}\) corresponds to the infrared absorption that causes transitions between vibrational energy levels of the \(\ce{C=O}\) bond. The reason that these are transitions from the vibrational ground state is because, at room temperature, by far the largest portion of the molecules are in this state (cf. Exercise 9-9). Stretching frequencies characteristic of the most important types of bonds found in organic molecules are given in Table 9-2. You will notice that the absorption band for each bond type is described by its position within a more or less broad frequency range and by its shape (broad, sharp) and intensity (strong, medium, weak). A qualitative discussion of the factors that determine infrared band position and band intensities follows. To a first approximation, a chemical bond resembles a mechanical spring that vibrates with a stretching frequency \(\bar{\nu}\) (\(\text{cm}^{-1}\)), \[ \bar{\nu} = \dfrac{1}{2\pi c} \sqrt{\dfrac{k}{\dfrac{m_1m_2}{m_1+m_2}}} \tag{9.3}\] in which \(k\) is the force constant, and \(m_1\) and \(m_2\) are the masses of the individual atoms at each end of the bond. The force constant \(k\) is a measure of the stiffness of the bond and usually is related to the bond strength. From Equation 9-3, we can see that the heavier the bonded atoms, the smaller will be the vibrational frequency of the bond provided \(k\) remains essentially constant.\(^6\) Thus if we increase \(m_2\) while holding \(k\) and \(m_1\) constant we expect the frequency to decrease. This is just what occurs when we change the \(C-H\) bond to a \(C-D\) bond. We also see that the frequency decreases in the order \(\ce{C-H}\) > \(\ce{C-C}\) > \(\ce{C-N}\) > \(\ce{C-O}\), which also is in the order of increasing \(m_2\), but here matters are more complicated because \(k\) also changes. Other things being equal, it requires more energy to stretch a bond than to bend it. Therefore the infrared bands arising from changes in the stretching vibrations are found at higher frequencies than are those arising from changes in the bending vibrations. Another consequence of Equation 9-3 is that if \(m_1\) and \(m_2\) remain the same, the larger the value of \(k\), the higher will be the vibrational frequency. Because \(k\) is expected to run more or less parallel to the bond strength, and because multiple bonds are stronger than single bonds, the absorption frequencies of multiple bonds are higher than for single bonds. Examples are the absorption of \(C=C\) at \(2100 \: \text{cm}^{-1}\), \(C=C\) at \(1650 \: \text{cm}^{-1}\), and \(C-C\) at \(1000 \: \text{cm}^{-1}\). Other effects besides mass and bond strength also affect infrared absorption frequencies. The structural environment of a bond is particularly important. Thus the absorption frequency of a \(\ce{C-H}\) bond depends on whether it is an alkyl, alkenyl, alkynyl, or aryl \(\ce{C-H}\) bond (see Table 9-2). The intensity of an infrared absorption band arising from changes in the vibrational energy is related to the electrical symmetry of the bond. More symmetrical, less polarized bonds give weaker absorptions. In fact, if the bond is completely symmetrical, there is no infrared absorption. In contrast, unsymmetrical molecules in which the bonds are quite polarized, such as \(C=O\) bonds, show strong infrared absorptions. Notice in Figure 9-9 that infrared spectra of organic molecules do not show very sharp absorption lines. This is because changes in rotational energies can occur together with the vibrational changes. The reason can be seen more clearly in Figure 9-10, in which each vibrational level, such as \(E_1\) and \(E_2\), of a molecule has associated with it closely spaced rotational levels. Transitions between \(E_1\) and \(E_2\) also may involve changes in rotational levels. This gives a “band” of closely spaced lines for any given vibrational change. For complex molecules, particularly in the liquid state, the “rotational fine structure” of a given vibrational band usually cannot be resolved. Some Characteristic Infrared Absorption Frequencies Infrared absorption bands between \(1250 \: \text{cm}^{-1}\) and \(675 \: \text{cm}^{-1}\) generally are associated with complex vibrational and rotational energy changes of the molecule as a whole and are quite characteristic of particular molecules. This part of the spectrum is often called the "fingerprint" region and is extremely useful for determining whether the samples are chemically identical. The spectra of 2-propanone and 2-butanone are seen to be very similar in the region \(4000 \: \text{cm}^{-1}\) to \(1250 \: \text{cm}^{-1}\) but quite different from \(1250 \: \text{cm}^{-1}\) to \(675 \: \text{cm}^{-1}\). The fingerprint region of the spectrum is individual enough so that if the infrared spectra of two samples are indistinguishable in the range of frequencies from \(3600 \: \text{cm}^{-1}\) to \(675 \: \text{cm}^{-1}\), it is highly probable that the two samples are of the same compound (or the same mixture of compounds). Characteristic stretching and bending frequencies occur in the fingerprint region, but they are less useful for identifying functional groups, because they frequently overlap with other bands. This region is sufficiently complex that a complete analysis of the spectrum is seldom possible. The infrared spectra of the alkanes show clearly absorptions corresponding to the \(C-H\) stretching frequencies at \(2850 \: \text{cm}^{-1}\) to \(3000 \: \text{cm}^{-1}\). The \(C-C\) stretching absorptions have variable frequencies and are usually weak. Methyl (\(CH_3-\)) and methylene (\(-CH_2-\)) groups normally have characteristic \(C-H\) bending vibrations at \(1400 \: \text{cm}^{-1}\) to \(1470 \: \text{cm}^{-1}\). Methyl groups also show a weaker band near \(1380 \: \text{cm}^{-1}\). Two sample infrared spectra that illustrate these features are given in Figure 9-11. The infrared spectra of the cycloalkanes are similar to those of the alkanes, except that when there are no alkyl substituents the characteristic bending frequencies of methyl groups at \(1380 \: \text{cm}^{-1}\) are absent. A moderately strong \(CH_2\) "scissoring" frequency is observed between \(1440 \: \text{cm}^{-1}\) and \(1470 \: \text{cm}^{-1}\), the position depending somewhat on the size of the ring. These features of the infrared spectra of cycloalkanes are illustrated in Figure 9-12 using cyclooctane and methylcyclohexane as examples. Infrared spectra are very useful both for identification of specific organic compounds, and for determining types of compounds. For example, Figure 9-13 shows the infrared spectrum of a substance, \(C_4H_6O_2\), for which we wish to determine the compound type and, if possible, the specific structure. The most informative infrared absorptions for determining the compound type are between \(1500 \: \text{cm}^{-1}\) and \(3600 \: \text{cm}^{-1}\). Two groups of bands in this region can be seen at about \(1700 \: \text{cm}^{-1} \left( s \right)\) and \(3000 \: \text{cm}^{-1} \left( s \right)\), where \(\left( s \right)\) means strong; if we used \(\left( m \right)\) it would mean medium, and \(\left( w \right)\) would mean weak. From Table 9-2 we can see that these bands are indicative of \(C=O\) (\(1700 \: \text{cm}^{-1}\)) and hydrogen-bonded \(OH\) of carboxylic acids (\(3000 \: \text{cm}^{-1}\)). The presumption is that there is a \(-CO_2H\) group in the molecule, and we can derive some reassurance from the fact that the molecular formula \(C_4H_6O_2\) has enough oxygens to allow for this possibility. Table 9-2 also shows that a \(-CO_2H\) group should have a \(C-O\) absorption band between \(1350 \: \text{cm}^{-1}\) and \(1400 \: \text{cm}^{-1}\) and \(O-H\) absorption (bending frequency) between \(1000 \: \text{cm}^{-1}\) and \(1410 \: \text{cm}^{-1}\), and there is indeed a band of medium intensity at \(1350 \: \text{cm}^{-1}\) and a strong band at \(1240 \: \text{cm}^{-1}\). These absorptions, being in the fingerprint region, do not that the compound is a carboxylic acid; but if there were no absorptions in the \(1000 \: \text{cm}^{-1}\) to \(1400 \: \text{cm}^{-1}\) range, the presence of a \(-CO_2H\) group would be highly questionable. Tentatively, then, we may write a partial structure for \(C_4H_6O_2\) as Final identification may be possible by comparison with an authentic spectrum of cyclopropanecarboxylic acid, if it is available in one of the several standard compendia of infrared spectra. A total of about 150,000 infrared spectra are available for comparison purposes. You should check with the reference section of your library to see what atlases of spectral data are available to you. The foregoing example illustrates the way structures can be determined from infrared spectral data. For many purposes, the infrared frequencies given in Table 9-2 are both approximate and incomplete. However, you could be easily frustrated in interpreting spectral data by being burdened with a very detailed table in which the unimportant is mixed with the important. The ability to use extensive tables effectively comes with experience. You should remember that tabulated infrared frequencies indicate only the in which a given vibrational transition will fall. The exact value for a particular compound usually is meaningless because it will change depending on whether the spectrum is taken of the solid, liquid, or gaseous states, the solvent used, the concentration, and the temperature. \(^6\)Remember that lower frequency means longer wavelengths and lower energy. and (1977) | 14,244 | 740 |
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According to the suggested approach to planning a synthesis, the primary consideration is how to construct the target carbon skeleton starting with smaller molecules (or, alternatively, to reconstruct an existing skeleton). Construction of a skeleton from smaller molecules almost always will involve formation of carbon-carbon bonds. Up to this point we have discussed only a few reactions in which carbon-carbon bonds are formed and these are summarized in Table 13-4. Other important reactions that can be used to enlarge a carbon framework will be discussed in later chapters. The most logical approach to planning the synthesis of a particular carbon framework requires that one work by mentally fragmenting the molecule into smaller pieces that can be “rejoined” by known \(\ce{C-C}\) bond-forming reactions. The first set of pieces in turn is broken into smaller pieces, and the mental fragmentation procedure is repeated until the pieces correspond to the carbon skeletons of readily available compounds. There almost always will be several different backward routes, and each is examined for its potential to put the desired functional groups at their proper locations. In almost all cases it is important to use reactions that will lead to pure compounds without having to separate substances with similar physical properties. A typical synthesis problem would be to devise a preparation of -2-octene, given the restrictions that the starting materials have fewer than eight carbons, and that we use the \(\ce{C-C}\) bond-forming reactions we have discussed up to now. The reasoning involved in devising an appropriate synthesis with the given restrictions will be outlined for this example in detail. First, we can see that the carbon skeleton of the desired product can be divided to give the following combinations of fragments: Next, we have to decide what reaction or reactions would be useful to put these fragments together to reform the \(\ce{C_8}\) chain. If we look at the list of available reactions in Table 13-4\(^4\) for \(\ce{C-C}\) bond formation, we can rule out 1, 2, 4, 8, and 9: 1 and 4 because the reactions are unsuitable for making unbranched chains; 8 and 9 because they make rings, not chains; and 2 because it does not work well in the absence of activating groups. Reaction 3 could be used to combine \(\ce{C_1}\) and \(\ce{C_7}\) units to give \(\ce{C_8}\), as by the radical addition of \(\ce{CBrCl_3}\) to 1-heptene: Reactions 6 and 7 could also be used to make \(\ce{C_8}\), 6 by linking \(\ce{C_7}\) to \(\ce{C_1}\) and 7 by putting together two \(\ce{C_4}\) units.\(^5\) Reaction 5 could be useful for all of the possible ways of dividing \(\ce{C_8}\). Some of the possible combinations are: This does not exhaust the possibilities because, as 5b and 5e show, Reaction 5 can be used to make the same \(\ce{C_8}\) compound from different sets of starting materials. We now have to consider how to convert the \(\ce{C_8}\) materials that we might make into -2-octene. The possibilities are: Of these, d is the obvious choice for first consideration because it has its functionality, a single triple bond, between the same two carbons we wish to have joined by a cis double bond in the product. Now, we have to ask if there are reactions that will convert \(\ce{-C \equiv C}-\) to - . Two possibilities were mentioned previously - hydrogenation of a triple bond with the Lindlar catalyst ( ) and hydroboration followed by treatment with propanoic acid ( ): Either of these two reactions provides a simple and straightforward way of converting 2-octyne to -2-octene, so a satisfactory answer to the original problem is You can see that even with having available only seven \(\ce{C-C}\) bond-forming reactions and two ways of converting \(\ce{-C \equiv C}- \rightarrow\) , considerable amount of logical screening is required to eliminate unsuitable possibilities. The skilled practitioner makes this kind of diagnosis quickly in his head; at the outset you will find it useful to write the steps in your screening in the same way as we have done for this example. \(^4\)You may wish to review the sections cited for each reaction to be sure you understand the judgments we make here as to the suitability of particular reactions for the purpose at hand. \(^5\)Reaction 7 could also be used to make \(\ce{C_8}\) by other combinations, such as of \(\ce{C_5}\) and \(\ce{C_3}\), but these would give undesirable mixtures of products. Thus, \(\ce{C-C-C-C \equiv CH} + \ce{HC \equiv C-C} \rightarrow \ce{C-C-C-C \equiv C-C \equiv C-C} + \ce{C-C \equiv.C-C \equiv C-C} + \ce{C-C-C-C \equiv C-C \equiv C-C-C-C}\) and (1977) | 4,682 | 741 |
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According to the suggested approach to planning a synthesis, the primary consideration is how to construct the target carbon skeleton starting with smaller molecules (or, alternatively, to reconstruct an existing skeleton). Construction of a skeleton from smaller molecules almost always will involve formation of carbon-carbon bonds. Up to this point we have discussed only a few reactions in which carbon-carbon bonds are formed and these are summarized in Table 13-4. Other important reactions that can be used to enlarge a carbon framework will be discussed in later chapters. The most logical approach to planning the synthesis of a particular carbon framework requires that one work by mentally fragmenting the molecule into smaller pieces that can be “rejoined” by known \(\ce{C-C}\) bond-forming reactions. The first set of pieces in turn is broken into smaller pieces, and the mental fragmentation procedure is repeated until the pieces correspond to the carbon skeletons of readily available compounds. There almost always will be several different backward routes, and each is examined for its potential to put the desired functional groups at their proper locations. In almost all cases it is important to use reactions that will lead to pure compounds without having to separate substances with similar physical properties. A typical synthesis problem would be to devise a preparation of -2-octene, given the restrictions that the starting materials have fewer than eight carbons, and that we use the \(\ce{C-C}\) bond-forming reactions we have discussed up to now. The reasoning involved in devising an appropriate synthesis with the given restrictions will be outlined for this example in detail. First, we can see that the carbon skeleton of the desired product can be divided to give the following combinations of fragments: Next, we have to decide what reaction or reactions would be useful to put these fragments together to reform the \(\ce{C_8}\) chain. If we look at the list of available reactions in Table 13-4\(^4\) for \(\ce{C-C}\) bond formation, we can rule out 1, 2, 4, 8, and 9: 1 and 4 because the reactions are unsuitable for making unbranched chains; 8 and 9 because they make rings, not chains; and 2 because it does not work well in the absence of activating groups. Reaction 3 could be used to combine \(\ce{C_1}\) and \(\ce{C_7}\) units to give \(\ce{C_8}\), as by the radical addition of \(\ce{CBrCl_3}\) to 1-heptene: Reactions 6 and 7 could also be used to make \(\ce{C_8}\), 6 by linking \(\ce{C_7}\) to \(\ce{C_1}\) and 7 by putting together two \(\ce{C_4}\) units.\(^5\) Reaction 5 could be useful for all of the possible ways of dividing \(\ce{C_8}\). Some of the possible combinations are: This does not exhaust the possibilities because, as 5b and 5e show, Reaction 5 can be used to make the same \(\ce{C_8}\) compound from different sets of starting materials. We now have to consider how to convert the \(\ce{C_8}\) materials that we might make into -2-octene. The possibilities are: Of these, d is the obvious choice for first consideration because it has its functionality, a single triple bond, between the same two carbons we wish to have joined by a cis double bond in the product. Now, we have to ask if there are reactions that will convert \(\ce{-C \equiv C}-\) to - . Two possibilities were mentioned previously - hydrogenation of a triple bond with the Lindlar catalyst ( ) and hydroboration followed by treatment with propanoic acid ( ): Either of these two reactions provides a simple and straightforward way of converting 2-octyne to -2-octene, so a satisfactory answer to the original problem is You can see that even with having available only seven \(\ce{C-C}\) bond-forming reactions and two ways of converting \(\ce{-C \equiv C}- \rightarrow\) , considerable amount of logical screening is required to eliminate unsuitable possibilities. The skilled practitioner makes this kind of diagnosis quickly in his head; at the outset you will find it useful to write the steps in your screening in the same way as we have done for this example. \(^4\)You may wish to review the sections cited for each reaction to be sure you understand the judgments we make here as to the suitability of particular reactions for the purpose at hand. \(^5\)Reaction 7 could also be used to make \(\ce{C_8}\) by other combinations, such as of \(\ce{C_5}\) and \(\ce{C_3}\), but these would give undesirable mixtures of products. Thus, \(\ce{C-C-C-C \equiv CH} + \ce{HC \equiv C-C} \rightarrow \ce{C-C-C-C \equiv C-C \equiv C-C} + \ce{C-C \equiv.C-C \equiv C-C} + \ce{C-C-C-C \equiv C-C \equiv C-C-C-C}\) and (1977) | 4,682 | 742 |
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A coordinate bond (also called a dative covalent bond) is a covalent bond (a shared pair of electrons) in which electrons come from the same atom. A covalent bond is formed by two atoms sharing a pair of electrons. The atoms are held together because the electron pair is attracted by both of the nuclei. In the formation of a simple covalent bond, each atom supplies one electron to the bond - but that does not have to be the case. If these colorless gases are allowed to mix, a thick white smoke of solid ammonium chloride is formed. The reaction is \[ \ce{NH3 (g) + HCl (g) \rightarrow NH4Cl (s)} \nonumber \] Ammonium ions, NH , are formed by the transfer of a hydrogen ion (a proton) from the hydrogen chloride molecule to the lone pair of electrons on the ammonia molecule. When the ammonium ion, NH , is formed, the fourth hydrogen is attached by a dative covalent bond, because only the hydrogen's nucleus is transferred from the chlorine to the nitrogen. The hydrogen's electron is left behind on the chlorine to form a negative chloride ion. Once the ammonium ion has been formed it is impossible to tell any difference between the dative covalent and the ordinary covalent bonds. Although the electrons are shown differently in the diagram, there is no difference between them in reality. In simple diagrams, a coordinate bond is shown by an arrow. The arrow points from the atom donating the lone pair to the atom accepting it. Something similar happens. A hydrogen ion (H ) is transferred from the chlorine to one of the lone pairs on the oxygen atom. \[ \ce{H_2O + HCl \rightarrow H_3O^{+} + Cl^{-}} \nonumber] The H O ion is variously called the hydroxonium ion, the hydronium ion or the oxonium ion. In an introductory chemistry course, whenever you have talked about hydrogen ions (for example in acids), you have actually been talking about the hydroxonium ion. A raw hydrogen ion is simply a proton, and is far too reactive to exist on its own in a test tube. If you write the hydrogen ion as \(\ce{H^{+}(aq)}\), the "\(\ce{(aq)}\)" represents the water molecule that the hydrogen ion is attached to. When it reacts with something (an alkali, for example), the hydrogen ion simply becomes detached from the water molecule again. Note that once the coordinate bond has been set up, all the hydrogens attached to the oxygen are exactly equivalent. When a hydrogen ion breaks away again, it could be any of the three. Boron trifluoride is a compound that does not have a noble gas structure around the boron atom (a notorious ). The boron only has three pairs of electrons in its bonding level, whereas there would be room for four pairs. \(BF_3\) is described as being electron deficient. The lone pair on the nitrogen of an ammonia molecule can be used to overcome that deficiency, and a compound is formed involving a coordinate bond. Using lines to represent the bonds, this could be drawn more simply as: The second diagram shows another way that you might find coordinate bonds drawn. The nitrogen end of the bond has become positive because the electron pair has moved away from the nitrogen towards the boron - which has therefore become negative. We shall not use this method again - it's more confusing than just using an arrow. Aluminum chloride sublimes (phase transition from solid to gas) at about 180°C. If it simply contained ions it would have a very high melting and boiling point because of the strong attractions between the positive and negative ions. The implication is that it when it sublimes at this relatively low temperature, it must be covalent. The dots-and-crosses diagram shows only the outer electrons. AlCl , like BF , is electron deficient. There is likely to be a similarity, because aluminum and boron are in the same group of the Periodic Table, as are fluorine and chlorine. Measurements of the relative formula mass of aluminum chloride show that its formula in the vapor at the sublimation temperature is not AlCl , but Al Cl . It exists as a dimer (two molecules joined together). The bonding between the two molecules is coordinate, using lone pairs on the chlorine atoms. Each chlorine atom has 3 lone pairs, but only the two important ones are shown in the line diagram. The uninteresting electrons on the chlorines have been faded in color to make the coordinate bonds show up better. There's nothing special about those two particular lone pairs - they just happen to be the ones pointing in the right direction. Energy is released when the two coordinate bonds are formed, and so the dimer is more energetically stable than two separate AlCl molecules. Water molecules are strongly attracted to ions in solution - the water molecules clustering around the positive or negative ions. In many cases, the attractions are so great that formal bonds are made, and this is true of almost all positive metal ions. Ions with water molecules attached are described as hydrated ions. Although aluminum chloride is a covalent compound, when it dissolves in water, ions are produced. Six water molecules bond to the aluminum to give an ion with the formula Al(H O) . It's called the hexaaquaaluminum complex ion with as six ("hexa") water molecules ("aqua") wrapped around an aluminum ion. The bonding in this (and the similar ions formed by the great majority of other metals) is coordinate (dative covalent) using lone pairs on the water molecules. The electron configuration of aluminum is 1s 2s 2p 3s 3p . When it forms an Al ion it loses the 3-level electrons to leave 1s 2s 2p . That means that all the 3-level orbitals are now empty. The aluminum reorganizes (hybridizes) six of these (the 3s, three 3p, and two 3d) to produce six new orbitals all with the same energy. These six hybrid orbitals accept lone pairs from six water molecules. You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules it is possible to fit around an aluminum ion (and most other metal ions). By making the maximum number of bonds, it releases most energy and is the most energetically stable. Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminum and so is not involved in the bonding. The resulting ion looks like this: Because of the movement of electrons towards the center of the ion, the 3+ charge is no longer located entirely on the aluminum, but is now spread over the whole of the ion. Dotted arrows represent lone pairs coming from water molecules behind the plane of the screen or paper. Wedge shaped arrows represent bonds from water molecules in front of the plane of the screen or paper. Carbon monoxide, CO, can be thought of as having two ordinary covalent bonds between the carbon and the oxygen plus a coordinate bond using a lone pair on the oxygen atom. In nitric acid, HNO , one of the oxygen atoms can be thought of as attaching to the nitrogen via a coordinate bond using the lone pair on the nitrogen atom. In fact this structure is misleading because it suggests that the two oxygen atoms on the right-hand side of the diagram are joined to the nitrogen in different ways. Both bonds are actually identical in length and strength, and so the arrangement of the electrons must be identical. There is no way of showing this using a dots-and-crosses picture. The bonding involves delocalization. Jim Clark ( ) | 7,418 | 743 |
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When a molten metal is mixed with another substance, there are two mechanisms that can cause an alloy to form: (1) or (2) . The relative size of each element in the mix plays a primary role in determining which mechanism will occur. When the atoms are relatively similar in size, the atom exchange method usually happens, where some of the atoms composing the metallic crystals are substituted with atoms of the other constituent. This is called a . Examples of substitutional alloys include bronze and brass, in which some of the copper atoms are substituted with either tin or zinc atoms. The bonding between two metals is best described as a combination of metallic electron "sharing" and covalent bonding, one can't occur without the other and the proportion of one to the other changes depending on the constituents involved. Metals share there electrons throughout there structure, this flow of electrons is the reason behind many of the characteristics associated with metals, including their ability to act as conductors. The different amount and strength of covalent bonds can change depending on the different specific metals involved and how they are mixed. The covalent bonding is what is responsible for the crystal structure as well as the melting point and various other physical properties. As the similarities between the electron structure of the metals involved in the alloy increase, the metallic characteristics of the alloy decrease. Pure metals are useful but their applications are often limited to each individual metal's properties. Alloys allow metal mixtures that have increased resistance to oxidation, increased strength, conductivity, and melting point; Essentially any property can be manipulated by adjusting alloy concentrations. An example could be Brass Door fixtures, they are strong and resist corrosion better then pure zinc or copper, the two major metals that constitute a brass alloy. The combination also has a low melting point allowing it to be easily cast into many different shapes and sizes.(1) There are many other aspects of substitutional alloys that could be explored in depth, but the basic concept is the idea that each individual metal in an alloy give the final product its chemical and physical properties. Substitutional alloys played an important role in the development of human society and culture as we know it today. The Bronze age itself is named after the Substitutional alloy consisting of tin in a metallic solution of copper. Ancient bronzes are very impure, or even mislabeled, containing large amounts of zinc and arsenic as well as lots of impurities. These many substitutional alloys allowed for stronger tools and weapons, they allowed for increased productivity in the workshop as well as on the battlefield. The need for raw materials like tin and copper for the production of bronze also spurred an increase in trade, since their ores are rarely found together. The current chemical understanding of substitutional alloys would not be so in depth if it weren't for the usefulness of the alloys to humans. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compo | 3,603 | 744 |
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The energies of the hydrogenlike orbitals of various atoms were mentioned in Chapter 6 and, in particular, we showed a diagram of the most stable state \(\left( 1s \right)^2 \left( 2s \right)^2 \left( 2p \right)^2\) of a carbon atom (Figure 6-4). Transfer of one of the \(2p\) electrons to the \(3s\) orbital requires excitation of the atom to a higher energy state and this can be achieved by absorption of electromagnetic radiation of the proper wavelength. The usual way that such excitation occurs is by absorption of a single of radiant energy, and we can say that the absorption of this amount of energy \(\Delta{E_{12}}\), corresponds to excitation of the atom from the with energy \(E_1\) to an of configuration \(\left( 1s \right)^2 \left( 2s \right)^2 \left( 2p \right)^1 \left( 3s \right)^1\) and energy \(E_2\): The difference in energy, \(\Delta{E_{12}}\), is related directly to the frequency (\(\nu\), \(\text{sec}^{-1}\)) or wavelength (\(\lambda\), \(\text{nm}\))\(^4\) of the absorbed quantum of radiation by the equation \[ \Delta{E_{12}} = h \nu = \dfrac{hc}{\lambda} \tag{9.1}\] in which \(h\) is Planck's constant and \(c\) is the velocity of light. The relationship \(\Delta{E} = h\nu\) often is called the . For chemical reactions, we usually express energy changes in \(\text{kcal mol}^{-1}\). For absorption of one quantum of radiation by each atom (or each molecule) in one mole, the energy change is related to \(\lambda\) by \[ \Delta{E_{12}} = \dfrac{28,600}{\lambda(nm)} \; kcal/mol \tag{9.2}\] As defined, \(\Delta{E_{12}}\) corresponds to one of radiation. What we have developed here is the idea of a spectroscopic change being related to a change in energy associated with the absorption of a quantum of energy. are the result of searches for such absorptions over a range of wavelengths (or frequencies). If one determines and plots the degree of absorption by a monoatomic gas such as sodium vapor as a function of wavelength, a series of very sharp absorption bands or lines are observed, hence the name . The lines are sharp because they correspond to specific changes in electronic configuration without complication from other possible energy changes. \(^4\)See for discussion of the units of frequency and wavelength. and (1977) | 2,302 | 746 |
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Two simple mechanisms can be written for the reaction of chloromethane with hydroxide ion in aqueous solution that differ in the of bond breaking relative to bond making. In the first mechanism, \(A\), the overall reaction is the result of two steps, the first of which involves a dissociation of chloromethane to solvated methyl carbocation\(^4\) and solvated chloride ion. The second step involves a reaction between the carbocation and hydroxide ion (or water) to yield methanol. \(A\): or In the second mechanism, \(B\), the reaction proceeds in a single step. Attack of hydroxide ion at carbon occurs simultaneously with the loss of chloride ion; that is, the carbon-oxygen bond is formed as the carbon-chlorine bond is broken: \(B\): Both of these mechanisms are important in the displacement reactions of alkyl compounds, although chloromethane appears to react by Mechanism \(B\). Now we will discuss the criteria for distinguishing between the concerted and stepwise mechanisms. Of the two mechanisms, \(A\) requires that the reaction rate be determined solely by the rate of the first step (cf. earlier discussion in Section 4-4C). This means that the rate at which methanol is formed (measured in moles per unit volume per unit time) will depend on the chloromethane concentration, but on the hydroxide ion concentration, because hydroxide ion is not utilized except in a reaction. In contrast, Mechanism \(B\) requires the rate to depend on the concentrations of both reagents because the slow step involves collisions between hydroxide ions and chloromethane molecules. \[ v = k_A [CH_3Cl] \label{8-1}\] \[ v = k_B [CH_3Cl,OH^-] \label{8-2}\] More precisely, the reaction rate (\(\nu\)) may be expressed in terms of Equation \(\ref{8-1}\) for Mechanism \(A\) and Equation \(\ref{8-2}\) for Mechanism \(B\): Customarily, \(\nu\) is expressed in moles of product formed per liter of solution per unit of time (most frequently in seconds). The concentration terms \(\left[ CH_3Cl \right]\) and \(\left[ OH^\ominus \right]\) are then in units of moles per liter, and the proportionality constant \(k\) (called the specific rate constant) has the units of \(\text{sec}^{-1}\) for Mechanism \(A\) and \(\text{mol}^{-1} \times \text{L} \times \text{sec}^{-1}\) for Mechanism \(B\). It is important to recognize the difference between and the . The order of a reaction with respect to a reactant is the power to which the concentration of must be raised to have direct proportionality between concentration and reaction rate. According to Equation \(\ref{8-2}\) the rate of the chloromethane-hydroxide ion reaction is with respect to chloromethane and with respect to hydroxide ion. In Equation \(\ref{8-1}\) the rate is with respect to chloromethane and with respect to hydroxide ion because \(\left[ OH^\ominus \right]^0 = 1\). The of reaction is the of the orders of the respective reactants. Thus Equations \(\ref{8-1}\) and \(\ref{8-2}\) express the rates of overall and reactions, respectively. We can use the overall reaction order to distinguish between the two possible mechanisms, \(A\) and \(B\). Experimentally, the rate of formation of methanol is found to be proportional to the concentrations of chloromethane and of hydroxide ion. Therefore the reaction rate is second order overall and is expressed correctly by Equation \(\ref{8-2}\). This means that the mechanism of the reaction is the single-step process \(B\). Such reactions generally are classified as , often designated \(\text{S}_\text{N}2\), \(\text{S}\) for substitution, \(\text{N}\) for nucleophilic, and \(2\) for bimolecular, because there are reactant molecules in the transition state. To summarize: For an \(\text{S}_\text{N}2\) reaction, \[ v = k[RX,Y] \label{8-3}\] The stepwise Mechanism \(A\) is a and accordingly is designated \(\text{S}_\text{N}1\). The numeral \(1\) (or \(2\)) used in these designations does refer to the kinetic order of the reaction, but refers to the number of molecules (not including solvent molecules) that make up the transition state. Thus for \(\text{S}_\text{N}1\), \[ v = k[RX] \label{8-4}\] or Many \(\text{S}_\text{N}\) reactions are carried out using the solvent as the nucleophilic agent. They are called solvolysis reactions and involve solvents such as water, ethanol, ethanoic acid, and methanoic acid. Two examples are In these examples, solvolysis is necessarily a first-order reaction, because normally the solvent is in such great excess that its concentration does not change appreciably during reaction, and hence its contribution to the rate does not change. However, that the rate is first order does not mean the reaction necessarily proceeds by an \(\text{S}_\text{N}1\) mechanism, particularly in solvents such as water, alcohols, or amines, which are reasonably good nucleophilic agents. The solvent can act as the displacing agent in an \(\text{S}_\text{N}2\) reaction. To distinguish between \(\text{S}_\text{N}1\) and \(\text{S}_\text{N}2\) mechanisms requires other criteria, notably stereochemistry (Sections 8-5 and 8-6), and the effect of added nucleophiles on the rate and nature of the reaction products. For example, it often is possible to distinguish between \(\text{S}_\text{N}1\) and \(\text{S}_\text{N}2\) solvolysis by adding to the reaction mixture a relatively small concentration of a substance that is expected to be a more powerful nucleophile than the solvent. If the reaction is strictly \(\text{S}_\text{N}1\), the rate at which \(RX\) disappears should remain essentially unchanged because it reacts only as fast as \(R^\oplus\) forms, and the rate of this step is not changed by addition of the nucleophile, even if the nucleophile reacts with \(R^\oplus\). However, if the reaction is \(\text{S}_\text{N}2\), the rate of disappearance of \(RX\) should because \(RX\) reacts with the nucleophile in an \(\text{S}_\text{N}2\) reaction and now the rate depends on both the nature and the concentration of the nucleophile. \(^4\)Many organic chemists, and indeed the previous versions of this book, use the term "carbonium ion" for species of this kind. However, there is well-established usage of the suffix, for ammonium, oxonium, chloronium, and so on, to denote positively charged atoms with valence shells. In the interest of greater uniformity of nomenclature we shall use "carbocation" for carbon positive ions that have unfilled valence shells (6 electrons). and (1977) | 6,520 | 747 |
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This page takes a simple look at solids, liquids and gases, and changes of state such as melting and boiling, in terms of the behavior of the particles present. In a gas, the particles are entirely free to move. At ordinary pressures, the distance between individual particles is of the order of ten times the diameter of the particles. At that distance, any attractions between the particles are fairly negligible at ordinary temperatures and pressures. The average energy of the particles in a liquid is governed by the temperature. The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average. Some of the more energetic particles on the surface of the liquid can be moving fast enough to escape from the attractive forces holding the liquid together. They evaporate. Notice that evaporation only takes place on the surface of the liquid. That's quite different from boiling which happens when there is enough energy to disrupt the attractive forces throughout the liquid. That's why, if you look at boiling water, you see bubbles of gas being formed all the way through the liquid. If you look at water which is just evaporating in the sun, you don't see any bubbles. Water molecules are simply breaking away from the surface layer. Eventually, the water will all evaporate in this way. The energy which is lost as the particles evaporate is replaced from the surroundings. As the molecules in the water jostle with each other, new molecules will gain enough energy to escape from the surface. Now imagine what happens if the liquid is in a closed container. Common sense tells you that water in a sealed bottle doesn't seem to evaporate - or at least, it doesn't disappear over time. But there is constant evaporation from the surface. Particles continue to break away from the surface of the liquid - but this time they are trapped in the space above the liquid. As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there. There will rapidly be an equilibrium set up in which the number of particles leaving the surface is exactly balanced by the number rejoining it. In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid. When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapor pressure (also known as saturation vapor pressure) of the liquid. Solids can also lose particles from their surface to form a vapor, except that in this case we call the effect rather than evaporation. Sublimation is the direct change from solid to vapor (or vice versa) without going through the liquid stage. In most cases, at ordinary temperatures, the saturated vapor pressures of solids range from low to very, very, very low. The forces of attraction in many solids are too high to allow much loss of particles from the surface. However, there are some which do easily form vapors. For example, naphthalene (used in old-fashioned "moth balls" to deter clothes moths) has quite a strong smell. Molecules must be breaking away from the surface as a vapor, because otherwise you wouldn't be able to smell it. Another fairly common example (discussed in detail elsewhere on the site) is solid carbon dioxide - "dry ice". This never forms a liquid at atmospheric pressure and always converts directly from solid to vapor. That's why it is known as dry ice. Jim Clark ( ) | 3,567 | 748 |
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In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously. The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot measure multiple quantum variables simultaneously. Until the dawn of quantum mechanics, it was held as a fact that all variables of an object could be known to exact precision simultaneously for a given moment. Newtonian physics placed no limits on how better procedures and techniques could reduce measurement uncertainty so that it was conceivable that with proper care and accuracy all information could be defined. Heisenberg made the bold proposition that there is a lower limit to this precision making our knowledge of a particle inherently uncertain. Matter and photons are waves, implying they are spread out over some distance. What is the position of a particle, such as an electron? Is it at the center of the wave? The answer lies in how you measure the position of an electron. Experiments show that you will find the electron at some definite location, unlike a wave. But if you set up exactly the same situation and measure it again, you will find the electron in a different location, often far outside any experimental uncertainty in your measurement. Repeated measurements will display a statistical distribution of locations that appears wavelike (Figure 1.9.1
). After de Broglie proposed the wave nature of matter, many physicists, including Schrödinger and Heisenberg, explored the consequences. The idea quickly emerged that, . However, each particle goes to a definite place (Figure 1.9.1
). After compiling enough data, you get a distribution related to the particle’s wavelength and diffraction pattern. There is a certain of finding the particle at a given location, and the overall pattern is called a . Those who developed quantum mechanics devised equations that predicted the probability distribution in various circumstances. It is somewhat disquieting to think that you cannot predict exactly where an individual particle will go, or even follow it to its destination. Let us explore what happens if we try to follow a particle. Consider the double-slit patterns obtained for electrons and photons in Figure 1.9.2
. The interferrence patterns build up statistically as individual particles fall on the detector. This can be observed for photons or electrons—for now, let us concentrate on electrons. You might imagine that the electrons are interfering with one another as any waves do. To test this, you can lower the intensity until there is never more than one electron between the slits and the screen. The same interference pattern builds up! This implies that a particle’s probability distribution spans both slits, and the particles actually interfere with themselves. Does this also mean that the electron goes through both slits? An electron is a basic unit of matter that is not divisible. But it is a fair question, and so we should look to see if the electron traverses one slit or the other, or both. One possibility is to have coils around the slits that detect charges moving through them. What is observed is that an electron always goes through one slit or the other; it does not split to go through both. But there is a catch. If you determine that the electron went through one of the slits, you no longer get a double slit pattern—instead, you get single slit interference. There is no escape by using another method of determining which slit the electron went through. Knowing the particle went through one slit forces a single-slit pattern. If you do not observe which slit the electron goes through, you obtain a double-slit pattern. How does knowing which slit the electron passed through change the pattern? The answer is fundamentally important— . Information can be lost, and in some cases it is impossible to measure two physical quantities simultaneously to exact precision. For example, you can measure the position of a moving electron by scattering light or other electrons from it. Those probes have momentum themselves, and by scattering from the electron, they change its momentum . There is a limit to absolute knowledge, even in principle. It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles. First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio \(h/λ\), where h represents Planck’s constant and \(\lambda\) represents the wavelength of the photon. The position of a photon of light is simply its wavelength (\(\lambda\)). To represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore, \[\Delta{p}=\dfrac{h}{\lambda} \label{1.9.1} \] \[\Delta{x}= \lambda \label{1.9.2} \] By substituting \(\Delta{x}\) for \(\lambda\) into Equation \(\ref{1.9.1}\), we derive \[\Delta{p}=\dfrac{h}{\Delta{x}} \label{1.9.3} \] or, \[\underset{\text{early form of uncertainty principle }}{\Delta{p}\Delta{x}=h} \label{1.9.4} \] Equation \(\ref{1.9.4}\) can be derived by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let \(\Delta p=mv\), and \(Δx=h/(m v)\) (from De Broglie’s expression for the wavelength of a particle). Substituting in \(Δp\) for \(mv\) in the second equation leads to Equation \(\ref{1.9.4}\). Equation \ref{1.9.4} was further refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as \[\Delta{p_x}\Delta{x} \ge \dfrac{h}{4\pi} = \dfrac{\hbar}{2} \label{1.9.5} \] with \(\hbar = \dfrac{h}{2\pi}= 1.0545718 \times 10^{-34}\; m^2 \cdot kg / s\). Equation \(\ref{1.9.5}\) reveals that the more accurately a particle’s position is known (the smaller \(Δx\) is), the less accurately the momentum of the particle in the x direction (\(Δp_x\)) is known. Mathematically, this occurs because the smaller \(Δx\) becomes, the larger \(Δp_x\) must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known (Figure 1.9.2
). Equation \(\ref{1.9.5}\) relates the uncertainty of momentum and position. An immediate questions that arise is if \(\Delta x\) represents the full range of possible \(x\) values or if it is half (e.g., \(\langle x \rangle \pm \Delta x\)). \(\Delta x\) is the standard deviation and is a statistic measure of the spread of \(x\) values. The use of half the possible range is more accurate estimate of \(\Delta x\). As we will demonstrated later, once we construct a wavefunction to describe the system, then both \(x\) and \(\Delta x\) can be explicitly derived. However for now, Equation \ref{1.9.5} will work. For example: If a problem argues a particle is trapped in a box of length, \(L\), then the uncertainly of it position is \(\pm L/2\). So the value of \(\Delta x\) used in Equation \(\ref{1.9.5}\) should be \(L/2\), not \(L\). An electron is confined to the size of a magnesium atom with a 150 pm radius. What is the uncertainty in its velocity? The uncertainty principle (Equation \(\ref{1.9.5}\)): \[\Delta{p}\Delta{x} \ge \dfrac{\hbar}{2} \nonumber \] can be written \[\Delta{p} \ge \dfrac{\hbar}{2 \Delta{x}} \nonumber \] and substituting \(\Delta p=m \Delta v \) since the mass is not uncertain. \[\Delta{v} \ge \dfrac{\hbar}{2\; m\; \Delta{x}} \nonumber \] the relevant parameters are \[ \begin{align*} \Delta{v} &\ge \dfrac{1.0545718 \times 10^{-34} \cancel{kg} m^{\cancel{2}} / s}{(2)\;( 9.109383 \times 10^{-31} \; \cancel{kg}) \; (150 \times 10^{-12} \; \cancel{m}) } \\[4pt] &= 3.9 \times 10^5\; m/s \end{align*} \nonumber \] What is the maximum uncertainty of velocity the electron described in Example 1.9.1
? Infinity. There is no limit in the maximum uncertainty, just the minimum uncertainty. It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously. Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light. The speed of a 1.0 g projectile is known to within \(10^{-6}\;m/s\). From Equation \(\ref{1.9.5}\), the \(\Delta{p_x} = m \Delta v_x\) with \(m=1.0\;g\). Solving for \(\Delta{x}\) to get \[ \begin{align*} \Delta{x} &= \dfrac{\hbar}{2m\Delta v} \\[4pt] &= \dfrac{1.0545718 \times 10^{-34} \; m^2 \cdot kg / s}{(2)(0.001 \; kg)(10^{-6} \;m/s)} \\[4pt] &= 5.3 \times 10^{-26} \,m \end{align*} \nonumber \] This negligible for all intents and purpose as expected for any macroscopic object. Unlimited (or the size of the universe). The Heisenberg uncertainty principles does not quantify the maximum uncertainty. Estimate the minimum uncertainty in the speed of an electron confined to a hydrogen atom within a diameter of \(1 \times 10^{-10} m\)? We need to quantify the uncertainty of the electron in position. We can estimate that as \(\pm 5 \times 10^{-10} m\). Hence, substituting the relavant numbers into Equation \ref{1.9.5} and solving for \(\Delta v\) we get \[\Delta v= 1.15 \times 10^6\, km/s \nonumber \] Notice that the uncertainty is significantly greater for the electron in a hydrogen atom than in the magnesium atom (Example 1.9.1
) as expected since the magnesium atom is appreciably bigger. Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom. | 12,324 | 749 |
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Many separation methods are based on , that is, separation of the components of a mixture by differences in the way they become distributed (or ) between two different phases. To illustrate with an extreme example, suppose we have a mixture of gaseous methane and ammonia and contact this mixture with water. Ammonia, being very soluble in water (~\(90 \: \text{g}\) per \(100 \: \text{g}\) water at \(1 \: \text{atm}\) pressure), will mostly go into the , whereas the methane, being almost insoluble (~\(0.003 \: \text{g}\) per \(100 \: \text{g}\) of water) will essentially remain entirely in the . Such a separation of methane and ammonia would be a one-stage partitioning between gas and liquid phases and, clearly, could be made much more efficient by contacting the gas layer repeatedly with fresh water. Carried through many separate operations, this partitioning procedure is, at best, a tedious process, especially if the compounds to be separated are similar in their distributions between the phases. However, partitioning can be achieved nearly automatically by using , which permit a to be contacted by a . To illustrate, suppose a sample of a gaseous mixture of ammonia and methane is injected into a long tube (column) filled with glass beads moistened with water (the stationary phase), and a slow stream of an inert , such as nitrogen or helium, is passed in to push the other gases through. A multistage partitioning would occur as the ammonia dissolves in the water and the resulting gas stream encounters fresh water as it moves along the column. Carrier gas enriched with methane would emerge first and effluent gas containing ammonia would come out later. This is a crude description of the method of (abbreviated often as glc, GC, or called vapor-phase chromatography, vpc). This technique has become so efficient as to revolutionize the analysis and separation of almost any organic substance that has even a slight degree of volatility at some reasonably attainable temperature. The most modern glc equipment runs wholly under computer control, with preprogrammed temperatures and digital integration of the detector output. A wide variety of schemes is available for measuring the concentration of materials in the effluent carrier gas, and some of these are of such extraordinary sensitivity that only very small samples are necessary (\(10^{-9} \: \text{g}\), or less). In the usual glc procedure, a few microliters of an organic liquid to be analyzed are injected into a vaporizer and carried with a stream of gas (usually helium) into a long heated column that is packed with a porous solid (such as crushed firebrick) impregnated with a nonvolatile liquid. Gas-liquid partitioning occurs, and small differences between partitioning of the components can be magnified by the large number of repetitive partitions possible in a long column. Detection often is achieved simply by measuring changes in thermal conductivity of the effluent gases. A schematic diagram of the apparatus and a typical separation pattern are shown in Figures 9-1 and 9-2. The method is extraordinarily useful for detection of minute amounts of impurities provided these are separated from the main peak. Glc also can be used effectively to purify materials as well as to detect impurities. To do this, the sample size and the size of the apparatus may be increased, or an automatic system may be used wherein the products from many small-scale runs are combined. Liquid-solid chromatography originally was developed for the separation of colored substances, hence the name chromatography, which stems from the Greek word meaning color. In a typical examination, a colored substance suspected of containing colored impurities is dissolved in a suitable solvent and the solution allowed to percolate down through a column packed with a solid adsorbent, such as alumina or silica, as shown in Figure 9-3. The "chromatogram" then is "developed" by passing through a suitable solvent that washes the down through the column. What one hopes for, but may not always find, is that the components of the mixture will be adsorbed by the solid phase so distinct bands or zones of color appear. The bands at the top of the column contain the most strongly adsorbed components and the bands at the bottom the least strongly held components. The zones may be separated mechanically, or sufficient solvent can be added to wash, or , the zones of adsorbed materials sequentially from the column for further analysis. Liquid-solid chromatography in the form just described was developed first by the Russian biochemist M. S. Tswett, about 1906. In recent years, many variations have been developed that provide greater convenience, better separating power, and wider applicability. In , which is especially useful for rapid analyses, a solid adsorbent containing a suitable binder is spread evenly on a glass plate, a drop of solution to be analyzed is placed near one edge and the plate is placed in a container with the edge of the plate below the spot, dipping into an eluting solvent. The solvent ascends the plate and the materials in the spot move upward at different rates, as on a Tswett column. Various detecting means are used - simple visual observation for colored compounds, differential fluorescence under ultraviolet light, and spraying of the plate with substances that will give colored materials with the compounds present. In favorable cases, this form of liquid-solid chromatography can be carried out with submicrogram quantities of materials. An extremely important improvement on the Tswett procedure is . Increasing the input pressure on the system to \(20\)-\(70 \: \text{atm}\) improves the speed of separations by permitting the use of much smaller solid particles (with more surface area) than would be practical for gravity-flow Tswett columns. Automatic monitoring of the column effluent by ultraviolet spectroscopy ( ) or by changes in the refractive index usually provides an effective means of determining how the separation is proceeding. With such techniques chromatograms similar to Figure 9-2 are obtained. High-pressure liquid chromatography (hplc) has great advantages for analysis and separation of high-molecular-weight heat-sensitive compounds that are unsuitable for glc. An ingenious variation of solid-liquid chromatography is to use a solid support to which material is attached that has a specific affinity for a particular substance to be separated. The technique is especially useful for separating enzymes, and the immobile phase can be constructed from compounds known to react with, or be complexed by, the enzyme. Some other forms of chromatography are discussed in and . Observation of a single peak in a given chromatographic procedure is evidence, albeit not definitive evidence, for purity. Contaminants with nearly the same properties may be very difficult to separate and, if knowing the degree of purity is highly important, one can run chromatograms with a variety of different adsorbents to see if each gives the same result. If they do, the presumption of purity improves, although it is desirable to determine whether the spectroscopic techniques to be described in the following section permit the same conclusion. and (1977) | 7,323 | 750 |
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We begin our discussion of solubility and —those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibriums, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression. When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
\[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \tag{17.1.1}\] The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product ( ) of the salt. Because the concentration of a pure solid such as Ca (PO ) is a constant, it does not appear explicitly in the equilibrium constant expression. (For more information on the equilibrium constant expression, see .) The equilibrium constant expression for the dissolution of calcium phosphate is therefore
\[K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \tag{17.1.2a}\] \[[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \tag{17.1.2b}\] At 25°C and pH 7.00, for calcium phosphate is 2.07 × 10 , indicating that the concentrations of Ca and PO ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of for some common salts are listed in ; they show that the magnitude of varies dramatically for different compounds. Although is not a function of pH in , changes in pH can affect the solubility of a compound, as you will discover in . As with , the concentration of a pure solid does not appear explicitly in . Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of of solute per 100 mL of solvent, , like , is defined in terms of the concentrations of the component ions. Kidney stones form from sparingly soluble calcium salts and are largely composed of Ca(O CCO )·H O and Ca (PO ) . © Thinkstock Calcium oxalate monohydrate [Ca(O CCO )·H O, also written as CaC O ·H O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca (PO ) ]. Its solubility in water at 25°C is 7.36 × 10 g/100 mL. Calculate its . solubility in g/100 mL Write the balanced dissolution equilibrium and the corresponding solubility product expression. Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate . We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox ) are as follows: \(\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}},\mathrm{ox^{2-}}]\) Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant. Next we need to determine [Ca ] and [ox ] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows: The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows: Because of the stoichiometry of the reaction, the concentration of Ca and ox ions are both 5.04 × 10 M. Inserting these values into the solubility product expression, \[K_{sp} = [Ca^{2+},ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \notag \] In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value. Exercise One crystalline form of calcium carbonate (CaCO ) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its . 4.5 × 10 The reaction of weakly basic anions with H O tends to make the actual solubility of many salts higher than predicted. When a transparent crystal of calcite is placed over a page, we see images of the letters. Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding. Tabulated values of can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example 1. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations as we did in , remembering that the concentration of the pure solid is essentially constant. We saw that the for Ca (PO ) is 2.07 × 10 at 25°C. Calculate the aqueous solubility of Ca (PO ) in terms of the following: molar concentration and mass of salt that dissolves in 100 mL of water Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C. Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent. Although the of solid Ca (PO ) changes as some of it dissolves, its does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression ( ): This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2 and 3 , which means that
[PO ] = 2.28 × 10 and [Ca ] = 3.42 × 10 . Exercise The solubility product of silver carbonate (Ag CO ) is 8.46 × 10 at 25°C. Calculate the following: The ion product ( ) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient ( ) discussed for gaseous equilibriums in . Whereas describes equilibrium concentrations, the ion product describes concentrations that are necessarily equilibrium concentrations. The ion product is analogous to the reaction quotient for gaseous equilibriums. As summarized in , there are three possible conditions for an aqueous solution of an ionic solid: The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed. We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na SO is added to 100 mL of 3.2 × 10 M BaCl ? Recall that NaCl is highly soluble in water. and volumes and concentrations of reactants whether precipitate will form Write the balanced equilibrium equation for the precipitation reaction and the expression for . Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product ( ). Compare the values of and to decide whether a precipitate will form. The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO because NaCl is highly soluble. The equation for the precipitation of BaSO is as follows: \[BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO^{2−}_{4(aq)} \notag \] The solubility product expression is as follows: To solve this problem, we must first calculate the ion product— = [Ba ,SO ]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba when the solutions are mixed is the total number of moles of Ba in the original 100 mL of BaCl solution divided by the final volume (100 mL + 10.0 mL = 110 mL): Similarly, the concentration of SO after mixing is the total number of moles of SO in the original 10.0 mL of Na SO solution divided by the final volume (110 mL): We can now calculate : We now compare with the . If > , then BaSO will precipitate, but if < , it will not. Because > , we predict that BaSO will precipitate when the two solutions are mixed. In fact, BaSO will continue to precipitate until the system reaches equilibrium, which occurs when [Ba ,SO ] = = 1.08 × 10 . Exercise The solubility product of calcium fluoride (CaF ) is 3.45 × 10 . If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10 M solution of Ca(NO ) , will CaF precipitate? yes ( = 4.7 × 10 > ) The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect discussed in : adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, . Consider, for example, the effect of adding a soluble salt, such as CaCl , to a saturated solution of calcium phosphate [Ca (PO ) ]. We have seen that the solubility of Ca (PO ) in water at 25°C is 1.14 × 10 M ( = 2.07 × 10 ). Thus a saturated solution of Ca (PO ) in water contains 3 × (1.14 × 10 M) = 3.42 × 10 M Ca and 2 × (1.14 × 10 M) = 2.28 × 10 M PO , according to the stoichiometry shown in (neglecting hydrolysis to form HPO as described in ). If CaCl is added to a saturated solution of Ca (PO ) , the Ca ion concentration will increase such that [Ca ] > 3.42 × 10 M, making > . The only way the system can return to equilibrium is for the reaction in to proceed to the left, resulting in precipitation of Ca (PO ) . This will decrease the concentration of both Ca and PO until = . The common ion effect usually decreases the solubility of a sparingly soluble salt. Calculate the solubility of calcium phosphate [Ca (PO ) ] in 0.20 M CaCl . concentration of CaCl solution solubility of Ca (PO ) in CaCl solution Write the balanced equilibrium equation for the dissolution of Ca (PO ) . Tabulate the concentrations of all species produced in solution. Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca (PO ) . The balanced equilibrium equation is given in the following table. If we let equal the solubility of Ca (PO ) in moles per liter, then the change in [Ca ] is once again +3 , and the change in [PO ] is +2 . We can insert these values into the table. The expression is as follows: Because Ca (PO ) is a sparingly soluble salt, we can reasonably expect that << 0.20. Thus (0.20 + 3 ) M is approximately 0.20 M, which simplifies the expression as follows: This value is the solubility of Ca (PO ) in 0.20 M CaCl at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example 2—here the initial [Ca ] was 0.20 M rather than 0. Exercise Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10 at 25°C. 2.9 × 10 M (versus 1.3 × 10 M in pure water) The equilibrium constant for a dissolution reaction, called the , is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of of solute per 100 mL of solvent, is defined in terms of the concentrations of the component ions. In contrast, the describes concentrations that are not necessarily equilibrium concentrations. Comparing and enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion. Write an expression for for each salt. Some species are not represented in a solubility product expression. Why? Describe the differences between and . How can an ion product be used to determine whether a solution is saturated? When using to directly compare the solubilities of compounds, why is it important to compare only the values of salts that have the same stoichiometry? Describe the effect of a common ion on the solubility of a salt. Is this effect similar to the common ion effect found in buffers? Explain your answer. Explain why the presence of MgCl decreases the molar solubility of the sparingly soluble salt MgCO . For a 1:1 salt, the molar solubility is simply \(\sqrt{K_{\textrm{sp}}}\) ; for a 2:1 salt, the molar solubility is \(\sqrt[3]{K_{\textrm{sp}}/4}\) Consequently, the magnitudes of can be correlated with molar solubility if the salts have the same stoichiometry. Because of the common ion effect. Adding a soluble Mg salt increases [Mg ] in solution, and Le Chatelier’s principle predicts that this will shift the solubility equilibrium of MgCO to the left, decreasing its solubility. Predict the molar solubility of each compound using the values given in . Predict the molar solubility of each compound using the values given. A student prepared 750 mL of a saturated solution of silver sulfate (Ag SO ). How many grams of Ag SO does the solution contain? = 1.20 × 10 . Given the values in and , predict the molar concentration of each species in a saturated aqueous solution. Given the values in and , predict the molar concentration of each species in a saturated aqueous solution. Silicon dioxide, the most common binary compound of silicon and oxygen, constitutes approximately 60% of Earth’s crust. Under certain conditions, this compound can react with water to form silicic acid, which can be written as either H SiO or Si(OH) . Write a balanced chemical equation for the dissolution of SiO in basic solution. Write an equilibrium constant expression for the reaction. The of Mg(OH) is 5.61 × 10 . If you tried to dissolve 24.0 mg of Mg(OH) in 250 mL of water and then filtered the solution and dried the remaining solid, what would you predict to be the mass of the undissolved solid? You discover that only 1.0 mg remains undissolved. Explain the difference between your expected value and the actual value. The of lithium carbonate is 8.15 × 10 . If 2.34 g of the salt is stirred with 500 mL of water and any undissolved solid is filtered from the solution and dried, what do you predict to be the mass of the solid? You discover that all of your sample dissolves. Explain the difference between your predicted value and the actual value. You have calculated that 24.6 mg of BaSO will dissolve in 1.0 L of water at 25°C. After adding your calculated amount to 1.0 L of water and stirring for several hours, you notice that the solution contains undissolved solid. After carefully filtering the solution and drying the solid, you find that 22.1 mg did not dissolve. According to your measurements, what is the of barium sulfate? In a saturated silver chromate solution, the molar solubility of chromate is 6.54 × 10 . What is the ? A saturated lead(II) chloride solution has a chloride concentration of 3.24 × 10 mol/L. What is the ? From the solubility data given, calculate for each compound. From the solubility data given, calculate for each compound. Given the following solubilities, calculate for each compound. Given the following solubilities, calculate for each compound. The of the phosphate fertilizer CaHPO ·2H O is 2.7 × 10 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 3.0 L of water at this temperature? The of zinc carbonate monohydrate is 5.5 × 10 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 2.0 L of water at this temperature? Silver nitrate eye drops were formerly administered to newborn infants to guard against eye infections contracted during birth. Although silver nitrate is highly water soluble, silver sulfate has a of 1.20 × 10 at 25°C. If you add 25.0 mL of 0.015 M AgNO to 150 mL of 2.8 × 10 M Na SO , will you get a precipitate? If so, what will its mass be? Use the data in to predict whether precipitation will occur when each pair of solutions is mixed. What is the maximum volume of 0.048 M Pb(NO ) that can be added to 250 mL of 0.10 M NaSCN before precipitation occurs? = 2.0 × 10 for Pb(SCN) . Given 300 mL of a solution that is 0.056 M in lithium nitrate, what mass of solid sodium carbonate can be added before precipitation occurs (assuming that the volume of solution does not change after adding the solid)? = 8.15 × 10 for Li CO . Given the information in the following table, calculate the molar solubility of each sparingly soluble salt in 0.95 M MgCl . 3.37 g 22.4 mg; a secondary reaction occurs, where OH from the dissociation of the salt reacts with H from the dissociation of water. This reaction causes further dissociation of the salt (Le Chatelier’s principle). 1.2 × 10 1.70 × 10 7.4 × 10 M; 2.1 mg Precipitation will occur in all cases. 8.27 g | 19,942 | 751 |
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We have presented a microscopic view of the chemical reaction between oxygen and hydrogen. The equation \[\ce{O2 (g) + 2 H2 (g) -> 2H2O (l)} \nonumber \] represents the same event in terms of chemical symbols and formulas. We also saw that plants form glucose from CO and H O: \[\ce{6CO2 (g) + 6 H2O (l) + Light Energy -> C6H12O6 (s) + 6O2(g)} \nonumber \] But how does a practicing chemist what is occurring on the microscopic scale? When a reaction is observed for the first time, little is known about the microscopic nature of the products. It is therefore necessary to determine the composition and formula of a newly synthesized substance. The first step in such a procedure is usually to separate and purify the products of a reaction. The products above are easy to separate, because they are liquids or solids, while the reactants are gaseous. But how could you determine that the formula should be H O, and not H O ? And plants produce mixtures of carbohydrates during photosynthesis, which must be separated by chromatography or other techniques before they can be identified. Glucose is perhaps the most common simple sugar, and it exists in two molecular forms, with the chemical formula C H O : the open chain form of glucose the cyclic form of glucose Plants may produce other simple sugars with similar structures, like D-ribulose (C H O ), D-fructose (C H O ) and di- or polysaccharides (with two or more simple sugars linked together) like sucrose (C H O ) or (C H O ). How can the actual product be identified? One answer involves —the determination of the percentage by mass of each element in the compound. Such data are usually reported as the . When 10.0 g oxygen reacts with sufficient hydrogen, 11.26 g of a pure compound is formed. Calculate the percent composition from these experimental data. The percentage of oxygen is the mass of oxygen divided by the total mass of compound times 100 percent: \[ \text{%O} = \tfrac {\text{m}_{\text{O}}} {\text{m}_{\text{compound}}} \cdot \text{100%} = \tfrac{\text{10.0 g}} {\text{11.26 g}} \cdot \text{100%} = \text{88.81%} \nonumber \] The remainder of the compound (11.26 g – 10 g = 1.26 g) is hydrogen: \[\text{%H} = \tfrac{\text{m}_{\text{H}}} {\text{m}_{\text{compound}}} \cdot \text{100%} = \tfrac{\text{1.26 g}} {\text{11.26 g}} \cdot \text{100%} = \text{11.19%} \nonumber \] As a check, verify that the percentages add to 100: 88.81% + 11.19% = 100% To obtain the formula from percent-composition data, we must find how many hydrogen atoms there are per oxygen atom. On a macroscopic scale this corresponds to the ratio of the amount of hydrogen to the amount of oxygen. If the formula is H O, it not only indicates that there are two hydrogen per oxygen , it also says that there are 2 of hydrogen atoms for each 1 of oxygen atoms. That is, the of hydrogen is twice the of oxygen. The numbers in the ratio of the amount of bromine to the amount of mercury (2:1) are the subscripts of hydrogen and oxygen in the formula. Determine the formula for the compound whose percent composition was calculated in the previous example. For convenience, assume that we have 100 g of the compound. Of this, 88.81 g (88.81%) is oxygen and 11.19 g is hydrogen. Each mass can be converted to an amount of substance \[\begin{align} & n_{\text{O}}=\text{88}\text{.81 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{5}\text{.550 mol O} \\ & n_{\text{H}}=\text{11}\text{.19 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{11}\text{.101 mol H} \\ \end{align} \nonumber \] So the formula is H O , but we know there are no fractions of atoms, so we have to put this in a standard form. To do this, we divide the larger amount by the smaller: \[\frac{n_{\text{H} }}{n_ { \text{O} }}=\frac{\text{11.19 mol H}}{\text{5.550 mol O}}=\frac{\text{1.998 mol H}}{\text{1 mol O}} \nonumber \] To the formula is H O , but again, there seems to be a problem. The ratio 1.998 mol H to 1 mol O also implies that there are 1.998 H atoms per 1 O atom. If the atomic theory is correct, there is no such thing as 0.998 atom; but remembering that our measurements are only good to three significant figures, we round 1.998 to 2.00 and write the formula as H O. An oxide of hydrogen has the composition 94.07% O and 5.93% H. Find its formula. Again assume a 100-g sample and calculate the amount of each element: \begin{align} & n_{\text{O}}=\text{94}\text{.07 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{5}\text{.88 mol O} \\ & n_{\text{H}}=\text{5}\text{.93 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{5}\text{.88 mol H} \\ \end{align} The ratio is \[\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{5.88 mol H}}{\text{5.88 mol O}}=\frac{\text{1 mol H}}{\text{1 mol O}} \nonumber \] So the formula is H O , which is a 1:1 ratio within experimental precision. We would therefore assign the formula HO. The formula obtained in Example 3 does not correspond to either of the two hydrogen oxides we have already discussed. Is it a third one? The answer is no because our method can only determine the of H to O. The ratio 1:1 is the same as 2:2, and so our method will give the same result for HO or H O (or H O , for that matter, should it exist). The formula determined by this method is called the or . Occasionally, as in the case of hydrogen peroxde, the empirical formula differs from the actual molecular composition, or the . Experimental determination of the molecular weight in addition to percent composition permits calculation of the molecular formula. A compound whose molecular weight is 28 contains 85.6% C and 14.4% H. Determine its empirical and molecular formulas. \[\begin{align} & n_{\text{C}}=\text{85}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{7}\text{.13 mol C} \\ & n_{\text{H}}=\text{14}\text{.4 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{14}\text{.3 mol H} \\ \end{align} \nonumber \] So the formula is C H , but to get integral subscripts we divide each by the smaller: \[\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{14.3 mol H}}{\text{7.13 mol C}}=\frac{\text{2.01 mol H}}{\text{1 mol C}} \nonumber \] The empirical formula is therefore CH . The molecular weight corresponding to the empirical formula is 12.01 + 2 × 1.008 = 14.03 D-Xylose contains 40.0% C, 6.71% H, and 53.29% O. What is its empirical formula? \[\begin{align} & n_{\text{H}}=\text{6}\text{.71 g}\times \tfrac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{6}\text{.66 mol H} \\ & n_{\text{C}}=\text{40}\text{.00 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{3}\text{.33 mol C} \\ & n_{\text{O}}=\text{53}\text{.29 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{3}\text{.33 mol O} \\ \end{align} \nonumber \] So the formula is C H O , and dividing all three by the smallest amount of substance we obtain CH2O. \[\begin{align} & \frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{3}\text{.33 mol C}}{\text{3}\text{.33 mol O}}=\frac{\text{1}\text{. mol H}}{\text{1 mol O}} \\ & \frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{4}\text{.44 mol H}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.00 mol H}}{\text{1 mol O}} \\ \end{align} \nonumber \] Aspirin contains 60.0% C, 4.48% H, and 35.5% O. What is its empirical formula? \begin{align} & n_{\text{H}}=\text{14}\text{.4 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{14}\text{.3 mol H} \\ & n_{\text{C}}=\text{85}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{7}\text{.13 mol C} \\ & n_{\text{O}}=\text{35}\text{.5 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{2}\text{.22 mol O} \\ \end{align} \begin{align} & \frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{5}\text{.00 mol C}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.25 mol H}}{\text{1 mol O}} \\ & \frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{4}\text{.44 mol H}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.00 mol H}}{\text{1 mol O}} \\ \end{align} \[\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{2.25 mol C}}{\text{1 mol O}}=\frac{\text{9 mol C}}{\text{4 mol O}} \nonumber \] We can also write \[\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{2 mol H}}{\text{1 mol O}}=\frac{\text{8 mol H}}{\text{4 mol O}} \nonumber \] Thus the empirical formula is C H O . Once someone has determined a formula–empirical or molecular—it is possible for someone else to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows. As we saw above, all carbohydrates have the general formula C (H O) . All simple sugars have the general formula C (H O) , so they all have the same percentage of C, H, and O. a. C (H O) contains 6 mol C, 12 mol H, and 6 mol O. The molar mass is thus = (6 x 12.011) + (12 x 1.008) + (6 x 16) = 180 g mol A 1-mol sample would weigh 180.56 g. The mass of 6 mol C it contains is \[\text{m }=\text{6 mol C}\cdot\frac{\text{12.011 g}}{\text{1 mol C}}=\text{72.066 g} \nonumber \] \[\text{%C}=\frac{\text{m }}{\text{m }}\cdot\text{ 100%}=\frac{\text{72.066 g}}{\text{180.56 g}}\cdot\text{100%}=\text{40.00%} \nonumber \] \begin{align} & m_{\text{H}}=\text{12 mol H}\cdot=\frac{\text{1.008 g}}{\text{1 mol H}}=\text{12.096 g}\\&\text{%H}= \frac{\text{12.096 g}}{\text{180.56 g}}\cdot\text{100%}=\text{6.70%}\\& m_{\text{O}}=\text{6 mol O}\cdot=\frac{\text{16.00 g}}{\text{1 mol O}}=\text{96.00 g}\\& \text{%O}= \frac{\text{96.00 g}}{\text{180.56 g}}\cdot\text{100%}=\text{53.17%}\\\end{align} Note that for any simple sugar, \[\text{%C}=\tfrac{\textit{n}\text{(12.011)}}{\textit{n}\text{(12.011)} +\textit{n}\text{(18)}} = \text{40%} \nonumber \] regardless of how many carbons it contains (simple sugars are trioses, C H O , tetroses, C H O , pentoses, and hexoses. The same goes for the percent H and O. b. The molar mass of C (H O) is 168.15; %C = [(8 x 12.011)/168.15] x 100% = 57.14%, not the 40% characteristic of simple sugars. Similarly, %H = [(8 x 1.008) / 168.15] x 100% = 4.80% and the percent O is 38.06%. | 10,060 | 752 |
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Hydrocarbons are organic compounds that contain carbon and hydrogen. The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors. The classifications for hydrocarbons, defined by nomenclature of organic chemistry are as follows: Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds. The inherent ability of hydrocarbons to bond to themselves is known as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that the bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30). | 1,262 | 753 |
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You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO H). The major families of organic compounds are characterized by their functional groups. Figure \(\Page {1}\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group. The first family listed in Figure \(\Page {1}\) is the hydrocarbons. These include alkanes, with the general molecular formula C H where n is an integer; alkenes, represented by C H ; alkynes, represented by C H ; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO group. The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH ) C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde. Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH group in dimethyl benzene is indicated with a 1, but the second CH group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(\Page {2}\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene. We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached. Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom. | 4,548 | 754 |
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Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light. In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1): (1) These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1. Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light. When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows: (2) Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum. The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows: (3) If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See for more detail) As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2. Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed The peak positions, intensities, widths, and shapes all provide useful information Fast and easy technique Sensitive technique (Micrograms of materials can be detected routinely) Inexpensive Atoms or monatomic ions do not have infrared spectra Homonuclear diatomic molecules do not posses infrared spectra Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy Table 2. The Advantage and Disadvantage of Infrared Spectroscopy The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light. (4) Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M M )/(M +M ) where M and M are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively. The reduced masses of C- H and C- D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm . Table 4. An Example of an electronic Effect When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm . These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm . The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because \( \frac{\partial \mu}{\partial x}\) is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm , and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration. Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law, (5) The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on ( The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration. In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light. Figure 1. Hydrogen Bonding in water molecules In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed. An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm . This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample. Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum The region of the infrared spectrum from 1200 to 700 cm is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule. One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules. Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending. In alkanes, which have very few bands, each band in the spectrum can be assigned: In alkenes compounds, each band in the spectrum can be assigned: In alkynes, each band in the spectrum can be assigned: In aromatic compounds, each band in the spectrum can be assigned: Figure 6. shows the spectrum of toluene. Figure 6. Infrared Spectrum of Toluene Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations. The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears: If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches. The carbonyl stretch C=O of esters appears: The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm . The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding. Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine. For more Infrared spectra is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks. Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum. Main infrared bands of some common inorganic ions: Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen) Characteristic infrared bands(cm ) of triatomic inorganic molecules: 1388, 1286 3311 2053 714, 784 327 667 712 486, 471 380 249 2349 2049 748 2219 842 Bent Molecules H O O SnCl 3675 1135 354 1595 716 120 3756 1089 334 There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation: | 11,021 | 755 |
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The most important synthetic use of Grignard reagents and organolithium reagents is to form new carbon-carbon bonds by addition to multiple bonds, particularly carbonyl bonds. An example is the addition of methyl-magnesium iodide to methanal: The yields of addition products in reactions of this kind are generally high. The adducts have metal-oxygen bonds that can be broken readily by acid hydrolysis to give the organic product. Grignard reagents seldom add to carbon-carbon multiple bonds (however, see ). With suitable variations of the carbonyl compound, a wide range of compounds can be built up from substances containing fewer carbon atoms per molecule. The products formed when several types of carbonyl compounds react with Grignard reagents are listed in Table 14-4. The sequence of reactions starting with an organic halide, \(\ce{RX}\), amounts to the addition of \(\ce{R-H}\) across a carbonyl bond.\(^2\) can be prepared by the addition of \(\ce{RMgX}\) or \(\ce{RLi}\) to methanal, \(\ce{CH_2=O}\), Alcohols of formula \(\ce{RCH_2CH_2OH}\) can be prepared by addition of \(\ce{RMgX}\) to oxacyclopropane (oxirane): are obtained from aldehydes, whereas ketones give : Hydrolysis of the intermediate \(\ce{R-OMgX}\) compound is achieved best with aqueous ammonium chloride solution. Addition of water gives an unpleasant mess of \(\ce{Mg(OH)_2}\) whereas addition of strong acids such as \(\ce{HCl}\) or \(\ce{H_2SO_4}\) can lead to side reactions of dehydration and so on, especially with tertiary alcohols ( ): also are obtained from both acyl halides, \(\ce{RCOCl}\), and esters, \(\ce{RCO_2R}\), by the addition of moles of Grignard reagent. The first mole of \(\ce{RMgX}\) adds to the carbonyl bond to give the adducts \(13\) or \(14\): However, these first-formed adducts are unstable and decompose to a ketone, \(\ce{CH_3COR}\), and magnesium salts, \(\ce{MgXCl}\) or \(\ce{MgXOC_2H_5}\). The ketone usually cannot be isolated, but reacts rapidly with more \(\ce{RMgX}\) ultimately to give a tertiary alcohol: behave very much like Grignard reagents, but with increased reactivity. They offer advantages over the magnesium compounds when the \(\ce{R}\) group or the carbonyl compound is highly branched. For instance, isopropyllithium adds in good yield to 2,4-dimethyl-3-pentanone, whereas isopropylmagnesium bromide fails completely to give the normal addition product: Failure of Grignard reagents to add in the normal way generally is because reactions by alternative paths occur more rapidly. If the Grignard reagent has a hydrogen on the carbon adjacent to the point of attachment of \(\ce{-MgX}\) (i.e., a \(\beta\) hydrogen), then reduction can occur, with the effect of adding \(\ce{H_2}\) to the carbonyl group. Furthermore, if the carbonyl compound has a hydrogen located on the carbon next to the carbonyl group, the Grignard reagent can behave as a base and remove this hydrogen as a proton. The result is that the compound becomes an enolate salt and \(\ce{RMgX}\) becomes \(\ce{RH}\). Apparently, the complicating side reactions observed with \(\ce{RMgX}\) are not nearly as important with \(\ce{RLi}\). The reasons for this difference are not well understood. The reaction of carbon dioxide with Grignard reagents initially gives a magnesium salt of a carboxylic acid, \(\ce{RCO_2MgX}\): This salt, which has a carbonyl group, in principle could add a second \(\ce{RMgX}\). However, further addition is usually slow, and for most practical purposes the reaction stops at this stage. If the reaction can go further, the way to run it is by bubbling \(\ce{CO_2}\) into the Grignard solution. This exposes the first-formed \(\ce{RCO_2MgX}\) to excess \(\ce{RMgX}\) and may lead to further addition reactions. The easy way to avoid this problem is to pour the \(\ce{RMgX}\) solution onto powdered dry ice (solid \(\ce{CO_2}\)). Hydrolysis of the product (here a stronger acid than \(\ce{NH_4Cl}\) is required) generates the carboxylic acid, \(\ce{RCO_2H}\): Although organomagnesium compounds are not sufficiently reactive to add to carboxylate anions, alkyllithium compounds add quite well. A useful synthesis of methyl ketones involves the addition of methyllithium to the lithium salt of a carboxylic acid: Other methods begin with acid chlorides or esters and attempt to add only mole of \(\ce{RMgX}\): The disadvantage of using Grignard reagents for this purpose is that they add very rapidly to the ketone as it is formed. There are two ways in which this disadvantage can be minimized. First, one can add the Grignard solution to an excess of acid chloride solution (the so-called “inverse addition” procedure) to keep the concentration of \(\ce{RMgX}\) in the reaction mixture low, and hope that the reaction will stop at the ketone stage. However, this device seldom works very well with acid chlorides. Better results can be obtained with \(\ce{RMgX}\) and \(\ce{R'CON(CH_3)_2}\). The second method is to use a less reactive organometallic compound - one that will react with \(\ce{RCOCl}\) but not with \(\ce{R_2C=O}\). One easy way to do this is to add cadmium chloride to the Grignard solution, whereby an organocadmium compound, \(\ce{R_2Cd}\), is formed (cf. , Method 3). In the presence of magnesium halides, \(\ce{R_2Cd}\) reacts moderately rapidly with acid chlorides, but only slowly with ketones. The addition therefore can be arrested at the ketone stage: \[2 \ce{RMgCl} + \ce{CdCl_2} \rightarrow \ce{R_2Cd} + 2 \ce{MgCl_2}\] Alkylcopper compounds, \(\ce{R-Cu}\), also are selective reagents that react with acid chlorides to give ketones, but do not add to esters, acids, aldehydes, or ketones. The \(\ce{R-Cu}\) compounds can be prepared from\(\ce{CuI}\) and the alkyllithium. With an excess of the alkyllithium, the alkylcopper is converted to \(\ce{R_2CuLi}\): A conjugated alkenone, , can react with an organometallic reagent by a normal 1,2 addition across the carbonyl group, or by 1,4 addition to the conjugated system. On hydrolysis, the 1,4 adduct first yields the corresponding enol, but this is normally unstable and rearranges rapidly to the ketone. The final product therefore corresponds to addition of \(\ce{R-H}\) across the carbon-carbon double bond: Organomagnesium and organolithium compounds can add both 1,2 and 1,4 to alkenones, and the relative importance of each mode of addition depends on the structure of the reactants. This sort of dual behavior can be a nuisance in synthetic work because it leads to separation problems and low yields. Organocopper compounds are a great help in this situation because they show a very high selectivity for 1,4 addition and add to unsaturated ketones in excellent yield: Grignard reagents react with oxygen, sulfur, and halogens to form substances containing \(\ce{C-O}\), \(\ce{C-S}\), and \(\ce{C-X}\) bonds, respectively: These reactions are not often important for synthesis because the products, \(\ce{ROH}\), \(\ce{RSH}\), and \(\ce{RX}\), can be obtained more conveniently and directly from alkyl halides by \(S_\text{N}1\) and \(S_\text{N}2\) displacement reactions, as described in Chapter 8. However, when both \(S_\text{N}1\) and \(S_\text{N}2\) reactions are slow or otherwise impractical, as for neopentyl derivatives, the Grignard reactions can be very useful: Also, oxygenation of a Grignard reagent at temperatures provides an excellent method for the synthesis of hydroperoxides: To prevent formation of excessive amounts of the alcohol, is desirable (i.e., a solution of Grignard reagent is added to ether through which oxygen is bubbled rather than bubbling oxygen through a solution of the Grignard reagent). \(^2\)It is not possible to add \(\ce{RH}\) to directly because \(\Delta G^0\) generally is somewhat unfavorable [\(+5 \: \text{kcal}\) for \(\ce{CH_4} + \ce{(CH_3)_2C=O} \rightarrow \ce{(CH_3)_3COH}\)]. How we get around this unfavorable equilibrium in practice provides an interesting example of how energy can be (and is) squandered to achieve some particular desired result; for example, the reaction \(\ce{CH_3CH_3} + \ce{CH_3CHO} \rightarrow \ce{CH_3CH_3CH(CH_3)OH}\) has \(\Delta H^0 = -12 \: \text{kcal}\) but \(\Delta G^0 = + 0.5 \: \text{kcal}\). A possible sequence is The overall result is the expenditure of \(10 + 76 + 71 = 157 \: \text{kcal}\) to achieve a reaction that itself has \(\Delta H^0 = -12 \: \text{kcal}\), but an unfavorable \(\Delta G^0\). (\(\ce{Li}\) is used in this example rather than \(\ce{Mg}\) because the heat of formation of \(\ce{C_2H_5MgBr}\) is not available.) and (1977) | 8,600 | 756 |
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This page describes what an infra-red spectrum is and how it arises from bond vibrations within organic molecules. You probably know that visible light is made up of a continuous range of different electromagnetic frequencies - each frequency can be seen as a different color. Infra-red radiation also consists of a continuous range of frequencies - it so happens that our eyes can't detect them. If you shine a range of infra-red frequencies one at a time through a sample of an organic compound, you find that some frequencies get absorbed by the compound. A detector on the other side of the compound would show that some frequencies pass through the compound with almost no loss, but other frequencies are strongly absorbed. How much of a particular frequency gets through the compound is measured as percentage transmittance. A percentage transmittance of 100 would mean that all of that frequency passed straight through the compound without any being absorbed. In practice, that never happens - there is always some small loss, giving a transmittance of perhaps 95% as the best you can achieve. A transmittance of only 5% would mean that nearly all of that particular frequency is absorbed by the compound. A very high absorption of this sort tells you important things about the bonds in the compound. A graph is produced showing how the percentage transmittance varies with the frequency of the infra-red radiation. Notice that an unusual measure of frequency is used on the horizontal axis. Wavenumber is defined like this: Similarly, don't worry about the change of scale half-way across the horizontal axis. You will find infra-red spectra where the scale is consistent all the way across, infra-red spectra where the scale changes at around 2000 cm , and very occasionally where the scale changes again at around 1000 cm . As you will see when we look at how to interpret infra-red spectra, this does not cause any problems - you simply need to be careful reading the horizontal scale. Each frequency of light (including infra-red) has a certain energy. If a particular frequency is being absorbed as it passes through the compound being investigated, it must mean that its energy is being transferred to the compound. Energies in infra-red radiation correspond to the energies involved in bond vibrations. In covalent bonds, atoms aren't joined by rigid links - the two atoms are held together because both nuclei are attracted to the same pair of electrons. The two nuclei can vibrate backwards and forwards - towards and away from each other - around an average position. The diagram shows the stretching that happens in a carbon-oxygen single bond. There will, of course, be other atoms attached to both the carbon and the oxygen. For example, it could be the carbon-oxygen bond in methanol, CH OH. The energy involved in this vibration depends on things like the length of the bond and the mass of the atoms at either end. That means that each different bond will vibrate in a different way, involving different amounts of energy. Bonds are vibrating all the time, but if you shine exactly the right amount of energy on a bond, you can kick it into a higher state of vibration. The amount of energy it needs to do this will vary from bond to bond, and so each different bond will absorb a different frequency (and hence energy) of infra-red radiation. As well as stretching, bonds can also bend. The diagram shows the bending of the bonds in a water molecule. The effect of this, of course, is that the bond angle between the two hydrogen-oxygen bonds fluctuates slightly around its average value. Imagine a lab model of a water molecule where the atoms are joined together with springs. These bending vibrations are what you would see if you shook the model gently. Again, bonds will be vibrating like this all the time and, again, if you shine exactly the right amount of energy on the bond, you can kick it into a higher state of vibration. Since the energies involved with the bending will be different for each kind of bond, each different bond will absorb a different frequency of infra-red radiation in order to make this jump from one state to a higher one. Look again at the infra-red spectrum of propan-1-ol, CH CH CH OH: In the diagram, three sample absorptions are picked out to show you the bond vibrations which produced them. Notice that bond stretching and bending produce different troughs in the spectrum. Jim Clark ( ) | 4,467 | 758 |
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Dietary intake of vegetable oils and hydrogenated vegetable oils has significant health effects. Not only do they have about twice as many calories per gram as sugars and proteins, but they have long term effects on circulatory system health. Crisco , with saturated oils, may not be as healthy as olive oil, with more unsaturated oils. Crisco contains hydrogenated vegetable oil olive oil is 55-83% oleic acid To understand these effects, we need to look at the structure of . The triglyceride [1] is an important part of the blood test done with an annual physical exam. Vegetable oils are all triglycerides, which contain a glycerol ( ) three carbon "backbone" with 3 long chain attached through ester linkages, as in the figure below. The long chain fatty acids may be with hydrogen atoms, in which case they have all single bonds like the top fatty acid in the Figure (which is palmitic acid). If they have fewer hydrogen atoms, they are and have double bonds like the middle fatty acid in the Figure (which is oleic acid). The bottom fatty acid is , with multiple double bonds (it is linolenic acid). Various cooking oils have |known concentrations of saturated and unsaturated fatty acids. Generally, triglycerides with more unsaturated fatty acid substituents are more healthful, but food companies them to make them solid saturated fats (like margarine or Crisco), and to reduce the tendency to spoil. Unsaturated fats have kinks in their molecular structures that reduce the tendency for them to cause arthereosclerosis (clogged arteries), pretty much for the same reason that kinks reduce the tendency to pack efficiently and form solids. Saturated fats have more linear fatty acid chains that pack well and solidify easily. Compare the Jmol models of saturated palmitic acid and unsaturated oleic acid below. Partial hydrogenation of polyunsaturated fats also produces -fatty acids, which have structures like saturated fats and consequently are as unhealty (see Elaidic Acid Below): - fatty acids have the hydrogens on opposite sides of the C=C double bond, like this while -fatty acids have the hydrogen atoms on the same side, like this . As of 2010 Crisco is made of soybean oil, fully hydrogenated cottonseed oil, and partially hydrogenated soybean and cottonseed oils. According to the product information label, one 12 g serving of Crisco contains 3 g of saturated fat, 0g of trans fat, 6 g of polyunsaturated fat, and 2.5 g of monounsaturated fat. Notice that the fat masses don't add up because the weights of glycerol are not included in the separately listed components. fatty acids are now recognized as a major dietary risk factor for cardiovascular diseases, and the US FDA has revised food labeling requirements to include trans fats. The composition of the soybean oil in Crisco is shown below. Quite often a mixture of two or more products is formed in a chemical reaction. For example, when a vegetable oil like palm oil is hydrogenated, we might want to make just mono-unsaturated products. But the many triglycerides it contains with varied fatty acid chains. No single process could work for all of them. Suppose we start with just one possible palm oil molecule, a glycerol with 2 linolenic acid, and 1 linoleic acid substituents (we'll abbreviate it GLLL). The desired product might be the oil with three oleic acid substituents (we'll abbreviate it GOOO, which also might be a good description of it) so the equation is: (C H O ) (C H O ) -(C H O ) + 5 H → (C H O ) (C H O ) -(C H O ) "GLLL" + 5 H → "GOOO" A large excess of hydrogen is usually present under pressure, with a palladium or "Raney Nickel" catalyst . A large number of products is obtained, including completely saturated fats like Stearin (glyceryl tristearate), and trans fats. The products are usually analyzed by converting the oils to simpler (methyl) esters and running a gas chromatogam. The effectiveness of the reaction is usually evaluated in terms of of the desired product. A is calculated by assuming that all the limiting reagent is converted to product. The experimentally determined mass of product is then compared to the theoretical yield and expressed as a percentage: \(\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent}\) Suppose a |hydrogenation of 100.0 g of (C H O ) (C H O ) -(C H O ), abbreviated "GLLL" (M = 875.4 g/mol) is carried out with 2.000 g H , sealed in a high pressure steel reaction vessel with a catalyst at 55°C. The products include 90.96 g(C H O ) (C H O ) -(C H O ), abbreviated "GOOO" (M = 885.5 g/mol). Calculate the percent yield. We must calculate the theoretical yield of (C H O ) (C H O ) -(C H O ), and to do this, we must first discover whether (C H O ) (C H O ) -(C H O ) or H is the limiting reagent. For the balanced equation above, The stoichiometric ratio of the reactants is \[\text{S}\left( \frac{\text{GLLL}}{\text{H}_{\text{2}}} \right)=\frac{\text{1 mol GLLL}}{\text{5 mol H}_{\text{2}}} \nonumber \] Now, the initial amounts of the two reagents are and \[\begin{align*} & n_{\text{GLLL}}\text{(initial)}=\text{100}\text{.0 g GLLL}\times \frac{\text{1 mol stearin}}{\text{875}\text{.4 g GLLL}}=\text{0}\text{.1142 mol GLLL} \\ & \\ & n_{\text{H}_2}\text{(initial)}=\text{2}\text{.000 g H}_2\times \frac{\text{1 mol H}_2}{\text{2}\text{.016 g H}_2}=\text{0}\text{.9921 mol H}_2 \\ \end{align*} \] The ratio of initial amounts is thus \[\frac{n_{\text{GLLL}}\text{(initial)}}{n_{\text{H}_2}\text{(initial}} ~=~ \frac{\text{0}\text{.1142 mol stearin}}{\text{0}\text{.9921 mol H}_2} ~=~\frac{\text{0}\text{0.1151 mol stearin}}{\text{1 mol H}_2} \nonumber \] Since this ratio is less than \(\text{S}\left( \frac{\text{GLLL}}{\text{H}_2} \right)~=~0.20\), there is an excess of H . GLLL is the limiting reagent. Accordingly we must use 0.1142 mol GLLL and 0.5712 mol H (rather than 0.9921 mol H ) to calculate the theoretical yield of (C H O ) (C H O ) -(C H O ), or "GOOO". We then have \[n_{\text{GOOO}}\text{(theoretical)}=\text{0}\text{.1142 mol GLLL}\times \frac{\text{1 mol GOOO}}{\text{1 mol GLLL}}=\text{0}\text{.1142 mol GOOO} \nonumber \] so that \[\text{m}_{\text{GOOO}}\text{(theoretical)}=\text{0}\text{.1142 mol GOOO}\times \frac{\text{885}\text{.5 g GOOO}}{\text{1 mol GOOO}}=\text{101}\text{.2 g GOOO} \nonumber \] We can organize these calculations in a table: The percent yield is then \[\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent }=\frac{\text{90}\text{.96 g}}{\text{101}\text{.2 g}}\times \text{100 percent}=\text{89}\text{.9 percent} \nonumber \] | 6,666 | 760 |
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High performance liquid chromatography consists of the injection of a liquid sample into a liquid mobile phase which is passed through a column of solid or supported liquid stationary phase. Separation is the result of partition, adsorption, size exclusion or ion exchange between the stationary and mobile phases and the separated constituents of the sample are usually detected by an ultra-violet absorption detector. A flow chart of the typical instrumentation used is illustrated in figure D10.1. Coupling with chemiluminescence detection adds the sensitivity of this technique to selectivity of a powerful separation method. It requires measurement of the emitted light due to a post-column reaction between the analytes in the column eluents and the chemiluminescence reagents, which are delivered by additional pumps with the incorporation of devices for rapid mixing. Measurement of the chemiluminescence intensity at its maximum requires optimization of the transit time (dependent on length of tubing and flow rate) between the mixing point and the detector. The most important problem in designing the coupling instrumentation is ensuring compatibility between the conditions necessary for efficient chromatographic separation and those needed for intense chemiluminescence. Separation depends heavily on mobile phase composition, whereas chemiluminescence emission is known to be affected by solvent, pH, reaction temperature and the presence of enhancers and/or catalysts . Interfaces between chromatography columns and chemiluminescence can become very complex. For example, peroxy-oxalate chemiluminescence is frequently coupled with HPLC. As it detects only fluorescent analytes (see chapter B5), successful detection depends on derivatization of the analytes eluted from the column before the addition of the chemiluminescence reagents. Figure D10.2 shows part of the post-column arrangements used for the determination of catecholamines by peroxy-oxalate chemiluminescence after reaction with ethylene diamine, which produces fluorescent derivatives . For simplicity, the arrangements for different temperatures in different parts of the system have not been shown. | 2,201 | 761 |
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Nuclear waste is radioactive waste, meaning that it spontaneously emits radiation. It usually originates from the by-products of nuclear reactions in applications such as medicine and research. Radioactive waste degrades with time, releasing alpha, beta, and gamma radiation that pose many health risks to the environment and most organisms, including humans. Due to the harmful nature of nuclear waste, there is strict government regulation on the safe disposal of it. There are several types of waste, and as such, many different ways of discarding it. The use of nuclear energy provides cheaper and more powerful energy through fission reactions. Fission is the process by which a heavy nucleus splits into two smaller, separate nuclei. For example, the splitting of a large atom such as Uranium creates neutrons and a substantial amount of energy. This energy is created due to the difference in mass between the two nuclei products and the large nucleus product. However, the remains of the nuclear reactor becomes nuclear waste, which is extremely lethal due to its radioactivity. As a response, researchers have found several ways to shield and isolate radioactive waste till it degrades completely. The controversy behind nuclear technology is due to the radioactive waste it creates. Some elements used in nuclear reactors have extremely long half-lives and must be shielded from humans and the environment for thousands of years. For example, plutonium-239, an isotope used in the production of nuclear weapons, has a half-life of 24,200 years while uranium-235 Hiroshima has a half-life of 700 million years. These elements emit large quantities of radioactivity that is extremely dangerous. Too much exposure can be followed by Acute Radiation Syndrome (ARS), which includes skin burns, nausea, vomiting, and eventually death within days if the exposure and dosage of radiation is high. There are currently a number of nuclear productions that result in radioactive waste. There are four different types of radioactive waste that result from nuclear power: | 2,086 | 762 |
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The Wave-Particle Duality theory states that waves can exhibit particle-like properties while particles can exhibit wave-like properties. This definition opposes classical mechanics or Newtonian Physics. In the 17 century, Newton demonstrated that, similar to wave, beams of light can also diffract and interfere with one another by shining white light into a prism to collect seven different colors and recombining them with a second prism to produce white light. This wave theory of light (classical physics) was confirmed by Young's in 1801 (figure 1). This classical theory was also proven by Davisson and Germer in 1925, when they aimed a beam of electrons at nickel, and the diffraction of the electrons produced fringes (Figure 2A). Fringes are properties of waves, and the diffraction is explained using the interference properties of waves. The dark fringes are produced when the waves are in phase, and light fringes are produced when the waves are out of phase ( ). Based on the classical theory, light's energy will follow Rayleigh-Jeans Law: \[ \rho = \dfrac{8 \pi T}{\lambda^4} \] where According to this equation, radiance energy is continous and will increase to infinity if the wavelength gets very small. However, in 1899, Otto R. Lummer and Ernst Pringsheim discovered which showed that radiant energy is discreet and has a max value. The energy doesn't go to infinity as the classical physics had predicted but declines after reaching a max value. The first experiments towards Wave-Particle duality were done by German Physicist Max Planck (1858-1947). Using blackbody radiator (equal emitter and absorber of radiation at all wavelengths), Planck derived the equation for the smallest amount of energy that can be changed into light \[E=h \nu\] where h is Planck’s constant 6.626x10 J.S and is the frequency. He also formulated the quantum theory by saying that light that was emitted had discrete levels of energy, and that energy that was radiated was quantized; (where n is an integer, and can be zero or a positive number). Quantization of energy states that there are discrete values or states, and energies in between the values of n are forbidden. Hence, he stated that if x number of particles were present with a certain frequency value, than energy would be Frequency is related to the wavelength where or Replace into the above equation, we have In 1905, Einstein assumed that Planck’s discrete energies are packets of energy called photons. The total energy of a system is equal to the kinetic energy plus the potential energy, and as always the Law of Conservation of energy applies. Einstein explained that in the photoelectric effect energy each photon's energy is absorbed by one electrons in a given metal, and as a result the electron was able to eject if the photon's energy is equal or greater than the threshold energy (Figure 2). The threshold energy is the amount of energy needed to eject an electron, and is called . Since we can rewrite the equation to show that the total energy is equal to plus the kinetic energy The photoelectric effect shows that light behaves like a photon or a particle packed with energy, in other words light waves behave like particles. According to Particle theory of light, light energy will increase to a discreet and finite value unless goes to zero, which will never happen according to the theory. This helps explain the blackbody radiation observation. The number of electrons ejected from the metal increases as the intensity of the light increases. An electron that is not held strongly will have more kinetic energy. The threshold energy must be absorbed in order for an electron to be ejected. Due to conservation of energy, the kinetic energy (T) of the electron is dependent on the frequency of the wavelength of incident light. Remember, high frequencies have short wavelengths therefore photons with short wavelengths will be higher in energy. There is a linear relationship between kinetic energy of the ejected electron and frequency. After the work function energy is absorbed by the electron, the rest of the energy that was provided by the photon changes into kinetic energy , and hence the equation . An electron that only absorbs the threshold energy has no kinetic once it has traveled outside the metal. Since both particle and wave theories of light seem to explain a portion of light properties correctly, which is the correct one? In 1924, de Broglie (1892-1987) proposed an answer to this question. He assumed that all moving objects have wave-like properties. He combined Planck's constant and linear momentum so \[\lambda = \dfrac{hc}{E} \tag{1}\] and ( is the momentum of the object, is the object's mass, and is the velocity of the object) so Plug this into (1), we have This equation postulates that all moving object with a mass will have a wavelength which is called , but these wavelengths are only seen with objects that have a very small mass. Since is very small (6.626 x 10 Js), any object that has a large mass will have its wavelength close to zero. That is why we cannot see a walking human's wavelength. This relationship was confirmed by the Davisson and Germer diffraction experiments, where the wavelengths of the electrons, that gave diffraction patterns were same as the predicted wavelength using de Broglie relationship. or we can write n = λE/hc = 334J(6 x 10 m)/[(6.626 x 10 Js)(3 x 10 )] = 1.01 x 10 photons t = 1.01 x 10 photons/(3 x 10 photons/s) = 3.37 x 10 s | 5,539 | 763 |
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A reaction intermediate is transient species within a multi-step that is produced in the preceding step and consumed in a subsequent step to ultimately generate the final reaction product. Intermediate reactions are common in the biological world; a prime example can be seen in the of metabolites and nutrients. The lifetime of an intermediate is usually short because it is usually consumed to make the next product of the reaction sequence. In a biochemical pathway, the overall reaction may seem to form only but product but may require multiple smaller steps to achieve that goal. It must be kept “in mind that an intermediate is always formed in an early elementary step and consumed in a later ” (1). The general setup of a multi-step equation is: \[ CH_4 + Cl_2 \rightarrow X \;\;\; \text{Step 1} \nonumber \] \[ X \rightarrow CH_3Cl + HCl \;\;\; \text{Step 2} \nonumber \] The first step yields a substance X, which is the intermediate of the chlorination reaction "formed on the pathway between reactants and products" (2). An example of a biological process is . The overall balanced equation of glycolysis is D- Glucose + NAD + 2 ADP + 2 Pi --> 2 Pyruvate + 2NADH + 2H +2ATP + 2H O
GLC --> PYR Glucose (GLC) is the beginning reactant and is converted through a series of steps to form the final product, pyruvate (PYR). However, the intermediate “appears in the mechanism of the reaction but is not in the overall balanced equation.” One example of an intermediate in a glycolysis step reaction is the conversion of: Glucose --> Glucose-6-phosphate = intermediate The first step of glycolysis yields glucose-6-phosphate from a glucose molecule. However, in the overall balanced reaction of glycolysis, there is no presence of glucose-6-phosphate written because it exists for a short time before it is consumed in the next reaction. Identify and circle the intermediates of the glycolysis pathway and explain why they are considered intermediates. NOTE: *- DHAP is isomerized into G-3-P therefore the downstream products should be doubled. | 2,077 | 765 |
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G.N. Lewis used dots to represent the valence electrons in his teaching of chemical bonding. He eventually published his theory of chemical bonding in 1916. He put dots around the symbols so that we the valence electrons for the main group elements. Formation of chemical bonds to complete the requirement of eight electrons for the atom becomes a natural tendency. Lewis dot symbols of the first two periods are given here to illustrate this point. In fact, the entire group (column) of elements have the same Lewis dot symbols, because they have the same number of valence electrons. Lewis dot structures are useful in explaining the chemical bonding in molecules or ions. When several dot structures are reasonable for a molecule or ion, they all contribute to the molecular or ionic structure making it more stable. The representation of a molecular or ionic structure by several structures is called . The more stable the dot structure is, the more it contributes to the electronic structure of the molecule or ion. You need to know what dot structures represent, how to draw them, and what the formal charges for the atoms in the structure are. When several dot structures are possible, consider the resonance structures to interpret the real structure. Apply some simple rules to explain which of the resonance structures are major contributors to the electronic structure. Follow these simple steps to draw Lewis dot structures: Draw Lewis dot structures for \(\ce{CH4}\), \(\ce{NH3}\), \(\ce{HF}\), \(\ce{OF2}\), \(\ce{F2}\), \(\ce{O2}\), \(\ce{N2}\), \(\ce{Cl-}\) and some compounds you know. The on any atom in a Lewis structure is a number assigned to it according to the number of valence electrons of the atom and the number of electrons around it. The of an atom is equal to the number of valence electrons, minus the number of unshared electrons, and half of the bonding electrons, ½ . \(Formal\: charge = N_{\large{v.e.}} - N_{\large{us.e.}} - \dfrac{1}{2} N_{\large{b.e.}}\) Some practice of assigning formal charge is necessary before you master this technique. Some examples of drawing Lewis structure and assigning formal charge are given below. The is a hypothetical charge from the dot structure. The formal charges in a structure tell us the quality of the dot structure. Often, many Lewis dot structures are possible. These are possible resonance structures, but often we should write a reasonable one, which is stable. The formal charge guides us about the stability of the dot structure. The guidance is called : Draw Lewis dot structure for \(\ce{SO2}\). Put down number of valence electrons: \(\mathrm{
:\overset{\Large{..}}O :
:\overset{\Large{..}}S :
:\overset{\Large{..}}O :}\) Put all atoms together to make a molecule and check to see if it satisfies the octet rule. \(\begin{alignat}{1}
:&\overset{\Large{..}}{\ce O} :
:&&\overset{\Large{..}}{\ce S} :
:&&\overset{\Large{..}}{\ce O} : &&\textrm{ <= octet rule not satisfied}\\
&\,0 &&\,0 &&\,0 &&\textrm{ formal charge}
\end{alignat}\) Adjust bonding electrons so that octet rules apply to all the atoms. \(\begin{alignat}{1}
&:\underset{\Large{..}}{\overset{\Large{..}}{\ce O}}
&&:\overset{\Large{..}}{\ce S} :
:&&\overset{\Large{..}}{\ce O} : &&\textrm{ <- octet rule satisfied}\\
&\,{-1} &&\,{+1} &&\,0 &&\textrm{ formal charge}
\end{alignat}\) Since the left \(\ce{O}\) has 6 unshared plus 2 shared electrons, it effectively has 7 electrons for a 6-valence-electron \(\ce{O}\), and thus its formal charge is -1. Formal charge for \(\ce{O}\) = 6 - 6 - (2/2) = -1.
Formal charge for \(\ce{S}\) = 6 - 2 - (6/2) = +1. There is yet another structure that does not satisfy the octet rule, but it's a reasonable structure: \(\begin{alignat}{1}
&:\underset{\Large{..}}{\overset{\Large{..}}{\ce O}}
&&:\overset{\Large{..}}{\ce S} :
&&\underset{\Large{..}}{\overset{\Large{..}}{\ce O}} : &&\textrm{ <- octet rule not satisfied}\\
&\,{-1} &&\,{+2} &&{-1} &&\textrm{ formal charge}
\end{alignat}\) When several structures with different electron distributions among the bonds are possible, all structures contribute to the electronic structure of the molecule. These structures are called resonance structures. A combination of all these resonance structures represents the real or observed structure. The Lewis structures of some molecules do not agree with the observed structures. For such a molecule, several dot structures may be drawn. All the dot structures contribute to the . The more stable structures contribute more than less stable ones. For resonance structures, the skeleton of the molecule (or ion) stays in the same relative position, and only distributions of electrons in the resonance structures are different. Let us return to the \(\ce{SO2}\) molecule. The molecule has a bent structure due to the lone pair of electrons on \(\ce{S}\). In the last structure that has a formal charge, there is a single \(\ce{S-O}\) bond and a double \(\ce{S=O}\) bond. These two bonds can switch over giving two resonance structures as shown below. In structure , the formal charges are +2 for \(\ce{S}\), and -1 for both \(\ce{O}\) atoms. In structures and , the formal charges are +1 for \(\ce{S}\), and -1 for the oxygen atom with a single bond to \(\ce{S}\). The low formal charges of \(\ce{S}\) make structures and more stable or more important contributors. The formal charges for all atoms are zero for structure 4, given earlier. This is also a possible resonance structure, although the octet rule is not satisfied. Combining resonance structures and results in the following structure: Draw the Lewis dot structures and resonance structures for the following. Some hints are given. \(\ce{CO2}\) - \(\textrm{:O::C::O:}\) (plus two more dots for each of \(\ce{O}\))
\(\ce{NO2}\) - \(\ce{.NO2}\) (bent molecule due to the odd electron)
\(\ce{NO2-}\) - \(\ce{:NO2-}\) (same number of electron as \(\ce{SO2}\))
\(\ce{HCO2-}\) - \(\ce{H-CO2}\)
\(\ce{O3}\) - (ozone, \(\ce{OO2}\); same number of electrons as \(\ce{SO2}\))
\(\ce{SO3}\) - (consider \(\ce{O-SO2}\), and the resonance structures)
\(\ce{NO3-}\) (see Example 2 below)
\(\ce{CO3^2-}\) (ditto) Notice that some of the resonance structures may not satisfy the octet rule. The \(\ce{NO2}\) molecule has an odd number of electrons, and the octet rule cannot be satisfied for the nitrogen atom. Draw the resonance structures of \(\ce{NO3-}\) The resonance structure is shown on the right here. Note that only the locations of double and single bonds change here. What are the formal charges for the \(\ce{N}\) atoms? What are the formal charges for the oxygen atoms that are single bonded and double bonded to \(\ce{N}\) respectively? Please work these numbers out. The more stable the structure, the more it contributes to the resonance structure of the molecule or ion. All three structures above are the same, only the double bond rotates. Draw the Lewis dot structures and resonance structures for \(\ce{HNO3}\)
\(\ce{H2SO4}\)
\(\ce{H2CO3}\)
\(\ce{HClO4}\)
\(\ce{C5H5N}\)
\(\ce{NO3-}\)
\(\ce{SO4^2-}\)
\(\ce{CO3^2-}\)
\(\ce{ClO4-}\)
\(\ce{C6H6}\) (Benzene)
\(\ce{Cl2CO}\) You have to do these on paper, because putting dots around the symbols is very difficult using a word processor. The octet rule should be applied to \(\ce{HNO3}\), \(\ce{NO3-}\), \(\ce{H2CO3}\), \(\ce{CO3^2-}\), \(\ce{C5H5N}\), \(\ce{C6H6}\), and \(\ce{Cl2CO}\). We can write Lewis dot structures that satisfy the octet rule for many molecules consisting of main-group elements, but the octet rule may not be satisfied for a number of compounds. For example, the dot structures for \(\ce{NO}\), \(\ce{NO2}\), \(\ce{BF3}\) (\(\ce{AlCl3}\)), and \(\ce{BeCl2}\) do not satisfy the octet rule. N:::O: :C:::O: The above are structures for the gas molecules. The solids of \(\ce{AlCl3}\) and \(\ce{BeCl2}\) are polymeric with bridged chlorides. Polymeric
solid
structures :Cl: :Cl: :Cl:
/ \ / \ / \
Be Be
\ / \ / \ /
:Cl: :Cl: :Cl: Alumunum chloride, \(\ce{AlCl3}\), is a white, crystalline solid, and an ionic compound. However, it has a low melting point of 465 K (192°C), and the liquid consists of dimers, \(\ce{Al2Cl6}\), whose structure is shown above. It vaporizes as dimers, but further heating gives a monomer that has the same structure as the \(\ce{BF3}\). In compounds \(\ce{PF5}\), \(\ce{PCl5}\), \(\textrm{:SF}_4\), \(\textrm{::ClF}_3\), \(\textrm{:::XeF}_2\) and \(\textrm{:::I}_3^-\), the center atoms have more than 10 electrons instead of 8. In compounds \(\ce{SF6}\), \(\ce{IOF5}\), \(\textrm{:IF}_5\), \(\ce{BrF5}\), \(\textrm{::XeF}_4\), \(\ce{PF6-}\) etc, the center atoms have 12 electrons. The formulas given above follow a systematic pattern according to the positions of the elements on the periodic table. As the number of atoms bonded to it decreases, the number of unshared electrons increase. Hint: Number of valence electrons = 4 + 6 + 6 What about \(\ce{NO2}\)? The oxygen double bonded to \(\ce{S}\) has a formal charge of 0. What is the formal charge of \(\ce{C}\) in \(\ce{O-C}\textrm{:::O}\)? The oxygen double bonded to \(\ce{S}\) has formal charge of -1. What is the formal charge of \(\ce{C}\) in \(\ce{O-C}\textrm{:::O}\)? Hint: The one with formal charge = 0 for all atoms. and which one you think is the best? You can write two resonance structures for each of the two to give 4 resonance structures for \(\ce{NO2}\). What about \(\ce{B(-F)3}\), all single bonds? What about \(\ce{Cl}\) in the same structure? Hint: ozone | 9,637 | 766 |
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The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by . In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place. A simple example of a substitution reaction is the formation of chloromethane and chlorine: \[ \ce{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}\] The equation for the reaction is simple, the ingredients are cheap, and the product is useful. However, if we want to decide in advance whether such a reaction is actually feasible, we have to know more. Particularly, we have to know whether the reaction proceeds in the direction it is written and, if so, whether conditions can be found under which it proceeds at a convenient rate. Obviously, if one were to mix methane and chlorine and find that, at most, only \(1 \%\) conversion to the desired product occurred and that the \(1 \%\) conversion could be achieved only after a day or so of strong heating, this reaction would be both too unfavorable and too slow for an industrial process. One way of visualizing the problems involved is with energy diagrams, which show the energy in terms of some arbitrary that is a measure of progress between the initial and final states (Figure 4-4). Diagrams such as Figure 4-4 may not be familiar to you, and a mechanical analogy may be helpful to provide better understanding of the very important ideas involved. Consider a two-level box containing a number of tennis balls. An analog to an energetically favorable reaction would be to have all of the balls on the upper level where any disturbance would cause them to roll down to the lower level under the influence of gravity, thereby losing energy. If the upper level is modified and a low fence added to hold the balls in place, it will be just as energetically favorable as when the fence is not there for the balls to be at the lower level. The difference is that the process will not occur without some major disturbance. We can say there is an to occurrence of the favorable process. Now, if we shake the box hard enough, the balls on the upper level can acquire enough energy to bounce over the barrier and drop to the lower level. The balls then can be said to acquire enough to surmount the barrier. At the molecular level, the activation energy must be acquired either by collisions between molecules as the result of their thermal motions, or from some external agency, to permit the reactants to get over the barrier and be transformed into products. We shortly will discuss this more, but first we wish to illustrate another important concept with our mechanical analogy, that of and . With gentle shaking of our two-level box, all of the balls on the upper level are expected to wind up on the lower level. There will not be enough activation to have them go from the lower to the upper level. In this circumstance, we can say that the balls are not equilibrated between the lower and upper levels. However, if we shake the box and , no matter whether we start with all of the balls on the lower or upper level, an will be set up with, on the average, most of the balls in the energetically more favorable lower level, but some in the upper level as well. To maintain a constant average fraction of the balls at each level with vigorous and continued shaking, the at which balls go from the upper to the lower level must be equal to the that they go in the opposite direction. The balls now will be between the two levels. At equilibrium, the fraction of the balls on each of the two levels is wholly independent of the height of the barrier, just as long as the activation (shaking) is sufficient to permit the balls to go ways. The diagrams of Figure 4-4 are to be interpreted in the same general way. If thermal agitation of the molecules is sufficient, then equilibrium can be expected to be established between the reactants and the products, whether the overall reaction is energetically favorable (left side of Figure 4-4) or energetically unfavorable (right side of Figure 4-4). But as with our analogy, when equilibrium is established we expect the major portion of the molecules to be in the more favorable energy state. What happens when methane is mixed with chlorine? No measurable reaction occurs when the gases are mixed and kept in the dark at room temperature. Clearly, either the reaction is energetically unfavorable or the energy barrier is high. The answer as to which becomes clear when the mixture is heated to temperatures in excess of \(300^\text{o}\) or when exposed to strong violet or ultraviolet light, whereby a rapid or even explosive reaction takes place. Therefore the reaction is energetically favorable, but the activation energy is greater than can be attained by thermal agitation alone at room temperature. Heat or light therefore must initiate a pathway for the reactants to be converted to products that has a low barrier or activation energy. Could we have predicted the results of this experiment ahead of time? First, we must recognize that there really are several questions here. Could we have decided whether the reaction was energetically favorable? That the dark reaction would be slow at room temperature? That light would cause the reaction to be fast? We consider these and some related questions in detail because they are questions and the answers to them are relevant in one way or another to the study of reactions in organic chemistry. Presumably, methane could react with chlorine to give chloromethane and hydrogen chloride, or chloromethane could react with hydrogen chloride to give methane and chlorine. If conditions were found for which both reactions proceeded at a finite rate, equilibrium finally would be established when the rates of the reactions in each direction became equal: \[ \ce{CH_4 + Cl_2 \rightleftharpoons CH_3Cl + HCl}\] At equilibrium, the relationship among the amounts of reactants and products is given by the equilibrium constant expression \[K_{eq} = \dfrac{[CH_3Cl,HCl]}{[CH_4,Cl_2]} \label{4-1}\] in which \(K_\text{eq}\) is the equilibrium constant. The quantities within the brackets of Equation \(\ref{4-1}\) denote either concentrations for liquid reactants or partial pressures for gaseous substances. If the equilibrium constant \(K_\text{eq}\) is \(1\), then on mixing equal volumes of each of the participant substances (all are gases above \(-24^\text{o}\)), reaction to the will be initially faster than reaction to the left, until equilibrium is established; at this point there will be more chloromethane and hydrogen chloride present than methane and chlorine. However, if the equilibrium constant were \(1\), the reaction initially would proceed faster to the and, at equilibrium, there would be more methane and chlorine present than chloromethane and hydrogen chloride.\(^4\) For methane chlorination, we know from experiment that the reaction goes to the right and that \(K_\text{eq}\) is much greater than unity. Naturally, it would be helpful in planning other organic preparations to be able to estimate \(K_\text{eq}\) in advance. It is a common experience to associate chemical reactions with equilibrium constants greater than one with the evolution of heat, in other words, with negative \(\Delta H^\text{0}\) values. There are, in fact, many striking examples. Formation of chloromethane and hydrogen chloride from methane and chlorine has a \(K_\text{eq}\) of \(10^{18}\) and \(\Delta H^\text{0}\) of \(-24 \: \text{kcal}\) per mole of \(CH_3Cl\) formed at \(25^\text{o}\). Combustion of hydrogen with oxygen to give water has a \(K_\text{eq}\) of \(10^{40}\) and \(\Delta H^\text{0} = -57 \: \text{kcal}\) per mole of water formed at \(25^\text{o}\). However, this correlation between \(K_\text{eq}\) and \(\Delta H^\text{0}\) is neither universal nor rigorous. Reactions are known that absorb heat (are endothermic) and yet have \(K_\text{eq} > 1\). Other reactions have large \(\Delta H^\text{0}\) values and equilibrium constants much less than \(1\). The problem is that the energy change that correlates with \(K_\text{eq}\) is not \(\Delta H^\text{0}\) but \(\Delta G^\text{0}\) (the so-called change of " ")\(^5\), and if we know \(\Delta G^\text{0}\), we can calculate \(K_\text{eq}\) by the equation \[ \Delta G^o =-2.303 RT \log_{10} K_{eq} \label{4-2}\] in which \(R\) is the gas constant and \(T\) is the absolute temperature in degrees Kelvin. For our calculations, we shall use \(R\) as \(1.987 \: \text{cal} \: \text{deg}^{-1} \: \text{mol}^{-1}\) and you should not forget to convert \(\Delta G^\text{0}\) to \(\text{cal}\). Tables of \(\Delta G^\text{0}\) values for formation of particular compounds (at various temperatures and states) from the elements are available in handbooks and the literature. With these, we can calculate equilibrium constants quite accurately. For example, handbooks give the following data, which are useful for methane chlorination: Combining these with proper regard for sign gives and \(\text{log} \: K_\text{eq} = -\left( -24.7 \times 1000 \right)/ \left(2.303 \times 1.987 \times 298.2 \right)\), so \(K_\text{eq} = 1.3 \times 10^{18}\). Unfortunately, insufficient \(\Delta G^\text{0}\) values for formation reactions are available to make this a widely applicable method for calculating \(K_\text{eq}\) values. The situation is not wholly hopeless, because there is a relationship between \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) that also involves \(T\) and another quantity, \(\Delta S^\text{0}\), the standard of the process: \[ \Delta G^o = \Delta H^o -T \Delta S^o \label{4-3}\] This equation shows that \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) are equal when \(\Delta S^\text{0}\) is zero. Therefore the sign and magnitude of \(T \Delta S^\text{0}\) determine how well \(K_\text{eq}\) correlates with \(\Delta H^\text{0}\). Now, we have to give attention to whether we can estimate \(T \Delta S^\text{0}\) values well enough to decide whether the \(\Delta H^\text{0}\) of a given reaction (calculated from bond energies or other information) will give a good or poor measure of \(\Delta G^\text{0}\). To decide whether we need to worry about \(\Delta S^\text{0}\) with regard to any particular reaction, we have to have some idea what physical meaning entropy has. To be very detailed about this subject is beyond the scope of this book, but you should try to understand the physical basis of entropy, because if you do, then you will be able to predict at least qualitatively whether \(\Delta H^\text{0}\) will be about the same or very different from \(\Delta G^\text{0}\). Essentially, the entropy of a chemical system is a measure of its or . Other things being the same, the more random the system is, the more favorable the system is. Different kinds of molecules have different degrees of translational, vibrational, and rotational freedom and, hence, different average degrees of molecular disorder or randomness. Now, if for a chemical reaction the degree of molecular disorder is different for the products than for the reactants, there will be a change in entropy and \(\Delta S^\text{0} \neq 0\). A spectacular example of the effect of molecular disorder in contributing to the difference between \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\) is afforded by the formation of liquid nonane, \(C_9H_{20}\), from solid carbon and hydrogen gas at \(25^\text{o}\): \[\ce{9C(s) + 10H_2(g) \rightarrow C_910_{20}(l)}\] with \(\Delta H^o = -54.7 \, kcal\) and \(\Delta S^o = 5.0 \, kcal\). Equations \(\ref{4-2}\) and \(\ref{4-3}\) can be rearranged to calculate \(\Delta S^\text{0}\) and \(K_\text{eq}\) from \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\): and \[K_{eq} = 10^{-\Delta G^o/2.303 \,RT} = 10^{-5.900/(2.303 \times 1.987 \times 298.2)} = 4.7 \times 10^{-5}\] These \(\Delta H^\text{0}\), \(\Delta S^\text{0}\), and \(K_\text{eq}\) values can be compared to those for \(H_2 + \frac{1}{2} O_2 \longrightarrow H_2O\), for which \(\Delta H^\text{0}\) is \(-57 \: \text{kcal}\), \(\Delta S^\text{0}\) is \(8.6 \: \text{e.u.}\), and \(K_\text{eq}\) is \(10^{40}\). Obviously, there is something about the entropy change from carbon and hydrogen to nonane. The important thing is that there is a great in the constraints on the atoms on each side of the equation. In particular, hydrogen molecules in the gaseous state have great translational freedom and a high degree of disorder, the greater part of which is lost when the hydrogen atoms become attached to a chain of carbons. This makes for a large \(\Delta S^\text{0}\), which corresponds to a in \(K_\text{eq}\). The differences in constraints of the carbons are less important. Solid carbon has an ordered, rigid structure with little freedom of motion of the individual carbon atoms. These carbons are less constrained in nonane, and this would tend to make \(\Delta S^\text{0}\) more positive and \(\Delta G^\text{0}\) more negative, corresponding to an increase in \(K_\text{eq}\) (Equations \(\ref{4-2}\) and \(\ref{4-3}\)). However, this is a effect on \(\Delta S^\text{0}\) compared to the enormous difference in the degree of disorder of hydrogen between hydrogen gas and hydrogen bound to carbon in nonane. Negative entropy effects usually are observed in ring-closure reactions such as the formation of cyclohexane from 1-hexene, which occur with substantial loss of rotational freedom (disorder) about the \(C-C\) bonds: There is an even greater loss in entropy on forming cyclohexane from ethene because substantially more freedom is lost in orienting three ethene molecules to form a ring: For simple reactions, with the same number of molecules on each side of the equation, with no ring formation or other unusual changes in the constraints between the products and reactants, \(\Delta S^\text{0}\) usually is relatively small. In general, for such processes, we know from experience that \(K_\text{eq}\) \(\Delta H^\text{0}\) \(-15 \: \text{kcal}\) \(\Delta H^\text{0}\) \(+15 \: \text{kcal}\). We can use this as a "rule of thumb" to predict whether \(K_\text{eq}\) should be greater or less than unity for vapor-phase reactions involving simple molecules. Some idea of the degree of success to be expected from this rule may be inferred from the examples in Table 4-5, which also contains a further comparison of some experimental \(\Delta H^\text{0}\) values with those calculated from bond energies. Suppose \(\Delta G^\text{0}\) is positive, what hope do we have of obtaining a useful conversion to a desired product? There is no simple straightforward and general answer to this question. When the reaction is reversible the classic procedure of removing one or more of the products to prevent equilibrium from being established has many applications in organic chemistry, as will be seen later. When this approach is inapplicable, a change in reagents is necessary. Thus, iodine does not give a useful conversion with 2,2-dimethylpropane, \(1\), to give 1-iodo-2,2-dimethylpropane, \(2\), because the position of equilibrium is too far to the left (\(K_\text{eq} \cong 10^{-5}\)): Alternative routes with favorable \(\Delta G^\text{0}\) values are required. Development of ways to make indirectly, by efficient processes, what cannot be made directly is one of the most interesting and challenging activities of organic chemists. To reach an understanding of why methane and chlorine do not react in the dark, we must consider the details of the reaction occurs - that is, the . The simplest mechanism would be for a chlorine molecule to collide with a methane molecule in such a way as to have chloromethane and hydrogen chloride formed directly as a result of a breaking of the \(Cl-Cl\) and \(C-H\) bonds and making of the \(C-Cl\) and \(H-Cl\) bonds (see Figure 4-5). The failure to react indicates that there must be an energy barrier too high for this mechanism to operate. Why should this be so? First, this mechanism involves a very precisely oriented "four-center" collision between chlorine and methane that would have a low probability of occurrence (i.e., a large decrease in entropy because a precise orientation means high molecular ordering). Second, it requires pushing a chlorine molecule sufficiently deeply into a methane molecule so one of the chlorine atoms comes close enough to the carbon to form a bond and yield chloromethane. Generally, to bring nonbonded atoms to near-bonding distances (\(1.2 \: \text{A}\) to \(1.8 \: \text{A}\)) requires a large expenditure of energy, as can be seen in Figure 4-6. Interatomic repulsive forces increase rapidly at short distances, and pushing a chlorine molecule into a methane molecule to attain distances similar to the \(1.77\)-\(\text{A}\) carbon-chlorine bond distance in chloromethane would require a considerable amount of compression (see Figure 4-7). Valuable information about interatomic repulsions can be obtained with space-filling models of the CPK type ( ), which have radii scaled to correspond to actual atomic interference radii, that is, the interatomic distance at the point where curves of the type of Figure 4-6 start to rise steeply. With such models, the degree of atomic compression required to bring the nonbonded atoms to within near-bonding distance is more evident than with the ball-and-stick models. It may be noted that four-center reactions of the type postulated in Figure 4-5 are encountered only rarely. If the concerted four-center mechanism for formation of chloromethane and hydrogen chloride from chlorine and methane is discarded, all the remaining possibilities are . A slow stepwise reaction is dynamically analogous to the flow of sand through a succession of funnels with different stem diameters. The funnel with the smallest stem will be the most important bottleneck and, if its stem diameter is much smaller than the others, it alone will determine the flow rate. Generally, a multistep chemical reaction will have a slow (analogous to the funnel with the small stem) and other relatively , which may occur either before or after the slow step. A possible set of steps for the chlorination of methane follows: Reactions (1) and (2) involve dissociation of chlorine into chlorine atoms and the breaking of a \(C-H\) bond of methane to give a methyl radical and a hydrogen atom. The methyl radical, like chlorine and hydrogen atoms, has one election not involved in bonding. Atoms and radicals usually are highly reactive, so formation of chloromethane and hydrogen chloride should proceed readily by Reactions (3) and (4). The crux then will be whether Steps (1) and (2) are reasonable under the reaction conditions. In the absence of some , only collisions due to the usual thermal motions of the molecules can provide the energy needed to break the bonds. At temperatures below \(100^\text{o}\), it is very rare indeed that the thermal agitation alone can supply sufficient energy to break any significant number of bonds stronger than \(30\) to \(35 \: \text{kcal mol}^{-1}\). The \(Cl-Cl\) bond energy from Table 4-3 is \(58.1 \: \text{kcal}\), which is much too great to allow bond breaking from thermal agitation at \(25^\text{o}\) in accord with Reaction (1). For Reaction (2) it is not advisable to use the \(98.7 \: \text{kcal} \: C-H\) bond energy from Table 4-3 because this is one fourth of the energy required to break all four \(C-H\) bonds ( ). More specific are given in Table 4-5, and it will be seen that to break one \(C-H\) bond of methane requires \(104 \: \text{kcal}\) at \(25^\text{o}\), which again is too much to be gained by thermal agitation. Therefore we can conclude that Reactions (1)-(4) can not be an important mechanism for chlorination of methane at room temperature. One might ask whether dissociation into ions would provide viable mechanisms for methane chlorination. Part of the answer certainly is: Not in the vapor phase, as the following thermochemical data show: Ionic dissociation simply does not occur at ordinarily accessible temperatures by collisions between molecules in the vapor state. What is needed for formation of ions is either a highly energetic external stimulus, such as bombardment with fast-moving electrons, or an ionizing solvent that will assist ionization. Both of these processes will be discussed later. The point here is that ionic dissociation is not a viable step for the vapor-phase chlorination of methane. First, we should make clear that the light does more than provide energy merely to lift the molecules of methane and chlorine over the barrier of Figure 4-4. This is evident from the fact that very little light is needed, far less than one light photon per molecule of chloromethane produced. The light could activate either methane or chlorine, or both. However, methane is colorless and chlorine is yellow-green. This indicates that chlorine, not methane, interacts with visible light. A photon of near-ultraviolet light, such as is absorbed by chlorine gas, provides more than enough energy to split the molecule into two chlorine atoms: Once produced, a chlorine atom can remove a hydrogen atom from a methane molecule and form a methyl radical and a hydrogen chloride molecule. The bond-dissociation energies of \(CH_4\) (\(104 \: \text{kcal}\)) and \(HCl\) (\(103.1 \: \text{kcal}\)) suggest that this reaction is endothermic by about \(1 \: \text{kcal}\): Use of bond-dissociation energies gives a calculated \(\Delta H^\text{0}\) of \(-26 \: \text{kcal}\) for this reaction, which is certainly large enough, by our rule of thumb, to predict that \(K_\text{eq}\) will be greater than 1. Attack of a methyl radical on molecular chlorine is expected to require somewhat more oriented collision than for a chlorine atom reacting with methane (the chlorine molecule probably should be endwise, not sidewise, to the radical) but the interatomic repulsion probably should not be much different. The net result of \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) and \(CH_3 \cdot + Cl_2 \longrightarrow CH_3Cl + Cl \cdot\) is formation of chloromethane and hydrogen chloride from methane and chlorine. Notice that the chlorine atom consumed in the first step is replaced by another one in the second step. This kind of sequence of reactions is called a because, in principle, one atom can induce the reaction of an infinite number of molecules through operation of a "chain" or cycle of reactions. In our example, chlorine atoms formed by the action of light on \(Cl_2\) can induce the chlorination of methane by the : In practice, chain reactions are limited by so-called processes. In our example, chlorine atoms or methyl radicals are destroyed by reacting with one another, as shown in the following equations: Chain reactions may be considered to involve three phases: First, must occur, which for methane chlorination is activation and conversion of chlorine molecules to chlorine atoms by light. Second, steps convert reactants to products with no net consumption of atoms or radicals. The propagation reactions occur in competition with steps, which result in destruction of atoms or radicals. Putting everything together, we can write: The chain-termination reactions are expected to be exceedingly fast because atoms and radicals have electrons in unfilled shells that normally are bonding. As a result, bond formation can begin as soon as the atoms or radicals approach one another closely, without need for other bonds to begin to break. The evidence is strong that bond-forming reactions between atoms and radicals usually are , that there is almost no barrier or activation energy required, and the rates of combination are simply the rates at which encounters between radicals or atoms occur. If the rates of combination of radicals or atoms are so fast, you might well wonder how chain propagation ever could compete. Of course, competition will be possible if the propagation reactions themselves are fast, but another important consideration is the fact that the . Suppose that the concentration of \(Cl \cdot\) is \(10^{-11} \: \text{M}\) and the \(CH_4\) concentration \(1 \: \text{M}\). The probability of encounters between two \(Cl \cdot\) atoms will be proportional to \(10^{-11} \times 10^{-11}\), and between \(CH_4\) and \(Cl \cdot\) atoms it will be \(10^{-11} \times 1\). Thus, other things being the same, \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) (propagation) would be favored over \(2Cl \cdot \longrightarrow Cl_2\) (termination) by a factor of \(10^{11}\). Under favorable conditions, the methane-chlorination chain may go through 100 to 10,000 cycles before termination occurs by radical or atom combination. Consequently the efficiency (or ) of the reaction is very high in terms of the amount of chlorination that occurs relative to the amount of the light absorbed. The overall rates of chain reactions usually are slowed very much by substances that can combine with atoms or radicals and convert them into species incapable of participating in the chain-propagation steps. Such substances are called , or . Oxygen acts as an inhibitor in the chlorination of methane by rapidly combining with a methyl radical to form the comparatively stable (less reactive) peroxymethyl radical, \(CH_3OO \cdot\). This effectively terminates the chain: To a considerable degree, we can predict reactivities, provided we use common sense to limit our efforts to reasonable situations. In the preceding section, we argued that reactions in which atoms or radicals combine can well be expected to be extremely fast because each entity has a potentially bonding electron in an outer unfilled shell, and bringing these together to form a bond does not require that other bonds be broken: The difference between the average energy of the reactants and the energy of the transition state is called the (Figure 4-4). We expect this energy to be smaller (lower barrier) if a weak bond is being broken and a strong bond is being made. The perceptive reader will notice that we are suggesting a parallel between reaction rate and \(\Delta H^\text{0}\) because \(\Delta H^\text{0}\) depends on the difference in strengths of the bonds being broken and formed. Yet previously ( ), we pointed out that the energy barrier for a reaction need bear no relationship to how energetically feasible the reaction is, and this is indeed true for complex reactions involving many steps. But our intuitive parallel between rate and \(\Delta H^\text{0}\) usually works quite well for the rates of steps. This is borne out by experimental data on rates of removal of a hydrogen atom from methane by atoms or radicals (\(X \cdot\)), such as \(F \cdot\), \(Cl \cdot\), \(Br \cdot\), \(HO \cdot\), \(H_2N \cdot\), which generally parallel the strength of the new bond formed: Similarly, if we look at the \(H-C\) bond-dissociation energies of the hydrocarbons shown in Table 4-6, we would infer that \(Cl \cdot\) would remove a hydrogen most rapidly from the carbon forming the weakest \(C-H\) bond and, again, this is very much in accord with experience. For example, the chlorination of methylbenzene (toluene) in sunlight leads to the substitution of a methyl hydrogen rather than a ring hydrogen for the reason that the methyl \(C-H\) bonds are weaker and are attacked more rapidly than the ring \(C-H\) bonds. This can be seen explicitly in the \(\Delta H^\text{0}\) values for the chain-propagation steps calculated from the bond-dissociation energies of Table 4-6. The \(\Delta H^\text{0}\) of ring-hydrogen abstraction is unfavorable by \(+7 \: \text{kcal}\) because of the high \(C-H\) bond energy (\(110 \: \text{kcal}\)). Thus this step is not observed. It is too slow in comparison with the more favorable reaction at the methyl group even though the second propagation step is energetically favorable by \(-37 \: \text{kcal}\) and presumably would occur very rapidly. Use of bond-dissociation energies to predict relative reaction rates becomes much less valid when we try to compare different kinds of reactions. To illustrate, ethane might react with \(F \cdot\) to give fluoromethane or hydrogen fluoride: It is not a good idea to try to predict the relative rates of these two reactions on the basis of their overall \(\Delta H^\text{0}\) values because the nature of the bonds made and broken is too different. Faced with proposing a mechanism for a reaction that involves overall making or breaking of more than two bonds, the beginner almost invariably tries to concoct a process wherein, with a step, all of the right bonds break and all of the right bonds form. Such mechanisms, called , have three disadvantages. First, they are almost impossible to prove correct. Second, prediction of the relative rates of reactions involving concerted mechanisms is especially difficult. Third, concerted mechanisms have a certain sterility in that one has no control over what happens while they are taking place, except an overall control of rate by regulating concentrations, temperature, pressure, choice of solvents, and so on. To illustrate, suppose that methane chlorination appeared to proceed by way of a one-step concerted mechanism: At the instant of reaction, the reactant molecules in effect would disappear into a dark closet and later emerge as product molecules. There is no way to prove experimentally that all of the bonds were made and formed simultaneously. All one could do would be to use the most searching possible tests to probe for the existence of discrete steps. If these tests fail, the reaction still would not be concerted because other, still more searching tests might be developed later that would give a different answer. The fact is, once you accept that a particular reaction is concerted, you, in effect, accept the proposition that further work on its is futile, no matter how important you might feel that other studies would be regarding the factors affecting the reaction rate. The experienced practitioner in reaction mechanisms accepts a concerted mechanism for a reaction involving the breaking and making of more than two bonds as a last resort. He first will try to analyze the overall transformation in terms of discrete steps that are individually simple enough surely to be concerted and that also involves energetically reasonable intermediates. Such an analysis of a reaction in terms of discrete mechanistic steps offers many possibilities for experimental studies, especially in development of procedures for detecting the existence, even if highly transitory, of the proposed intermediates. We shall give many examples of the fruitfulness of this kind of approach in subsequent discussions. \(^4\)If calculations based on chemical equilibrium constants are unfamiliar to you, we suggest you study one of the general chemistry texts listed for supplemental reading at the end of Chapter 1. \(^5\)Many books and references use \(\Delta F^\text{0}\) instead of \(\Delta G^\text{0}\). The difference between standard Gibbs energy \(\Delta G^\text{0}\) and the Gibbs energy \(\Delta G\) is that \(\Delta G^\text{0}\) is defined as the value of the free energy when all of the participants are in standard states. The free energy for \(\Delta G\) for a reaction \(\text{A} + \text{B} + \cdots \longrightarrow \text{X} + \text{Y} + \cdots\) is equal to \(\Delta G^\text{0} - 2.303 RT \: \text{log} \: \frac{\left[ \text{X} \right] \left[ \text{Y} \right] \cdots}{\left[ \text{A} \right] \left[ \text{B} \right] \cdots}\) where the products, \(\left[ \text{X} \right], \left[ \text{Y} \right] \cdots\), and the reactants, \(\left[ \text{A} \right], \left[ \text{B} \right] \cdots\), do not have to be in standard states. We shall use only \(\Delta G^\text{0}\) in this book. \(^6\)The entropy unit \(\text{e.u.}\) has the dimensions calorie per degree or \(\text{cal deg}^{-1}\). and (1977) | 32,267 | 767 |
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Capillary action can be defined as the ascension of liquids through slim tube, cylinder or permeable substance due to adhesive and cohesive forces interacting between the liquid and the surface. When intermolecular bonding of a liquid itself is substantially inferior to a substances’ surface it is interacting, capillarity occurs Also, the diameter of the container as well as the gravitational forces will determine amount of liquid raised. While, water possesses this unique property, a liquid like mercury will not display the same attributes due to the fact that it has higher cohesive force than adhesive force. Three main variables that determine whether a liquid possesses capillary action are: Capillary action only occurs when the adhesive forces are stronger than the cohesive forces, which invariably becomes , in the liquid. A good way to remember the difference between adhesive and cohesive forces is that with hesive forces you another set of molecules, the molecules of the surface, for the liquid to bond with. With hesive forces, the molecules of the liquid will only operate with their own kind. Decreased surface tension also increases capillary action. This is because decreased surface tension means that the intermolecular forces are decreased, thus decreasing cohesive forces. As a result, capillary action will be even greater. Practical use of capillary action is evident in all forms of our daily lives. It makes performing our tasks efficiently and effectively. Some applications of this unique property include: When measuring the level of liquid of a test tube or buret, it is imperative to measure at the meniscus line for an accurate reading. It is possible to measure the height (represented by h) of a test tube, buret, or other liquid column using the formula: \[ h = \dfrac{2\gamma \cos\theta}{\rho\;g\;r} \] In this formula, At optimum level, in which a glass tube filled with water is present in air, this formula can determine the height of a specific column of water in meters (m): \[ h\approx\dfrac{1.4 \times 10^{-5}}{r}\] When certain objects that are porous encounter a liquid medium, it will begin to absorb the liquid at a rate which actually decreases over a period of time. This formula is written as: \[ V = S*A\sqrt{t} \] In this specific formula, | 2,322 | 768 |
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The name for zinc is of German origins, zink. It has been known since pre-historic times and compounds as well as the metal had been used for many years before anyone ever thought about elements at all! is a blue-white metal of moderate strength, hardness and ductility. Zinc is one of the least common elements and is mostly produced through of aqueous zinc sulfate. About one third of all metallic zinc is used to manufacture galvanized nails. Because of its low melting point and its ability to form bonds with iron or steel, it serves to coat the metal and protect it from corrosion. Metallic zinc is also used to make dry cell batteries. Pure zinc is a bluish-silver and ductile metal with a low melting and boiling point. Most zinc today is obtained from ZnS, extracted from zinc blende ore and roasted to remove the sulfur. Zinc can also be obtained by electrolysis of aqueous zinc sulfate, a common laboratory exercise. Zinc is reasonably resistant to corrosion and is used as a covering for baser metals like iron ("galvanizing"). Zinc can be readily cast or molded. Zinc has many unique characteristics. For example, its vapor burns in air with a green flame, forming zinc oxide. Zinc oxide is a common zinc compound that is used in paints, cosmetics, plastics and more. Metallic zinc reacts with weak acids very slowly. Sulfur has a strong affinity for zinc. When heated, the two powders react explosively to form zinc sulfide. Zinc sulfide is used to make television screens and fluorescent light bulbs. Zinc also reacts with halogens. However, as the electronegativity decreases among the halogen group, the reactivity with zinc decreases. Thus, the most electronegative of the halogens (Fluorine) reacts with zinc violently, while the less electronegative halogen (Iodine) only generates a small amount of heat. Interestingly, properties of zinc are strongly affected by impurities such as lead, cadmium and iron. Also, zinc is most often used as a reducing agent in chemical reactions and it forms complex ions with ammonia and cyanide ions. | 2,080 | 769 |
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This is an introduction to some of the experimental methods used in school laboratories to find orders of reaction. There are two fundamentally different approaches to this: investigating what happens to the initial rate of the reaction as concentrations change, and following a particular reaction to completion and processing the results from that single reaction. This could be the time required for 5 cm of gas to be produced, for a small, measurable amount of precipitate to form, or for a dramatic color change to occur. Examples of these three indicators are discussed below. The concentration of one of the components of the reaction could be changed, holding everything else constant: the concentrations of other reactants, the total volume of the solution and the temperature. The time required for the event to occur is then measured. This process is repeated for a range of concentrations of the substance of interest. A reasonably wide range of concentrations must be measured.This process could be repeated by altering a different property. The slope of this linear region is V/t. \[\text{initial rate} \propto \frac{1}{t} \nonumber \] \[ \text{rate} \propto [A]^n \nonumber \] \[ \log (\text{rate}) \propto n\log [A] \nonumber \] he temperature, the total volume of the solution, and the mass of manganese(IV) oxide. The manganese(IV) oxide must also always come from the same bottle so that its state of division is always the same. Mixing dilute hydrochloric acid with sodium thiosulphate solution causes the slow formation of a pale yellow precipitate of sulfur. \[ Na_2S_2O_{2(aq)} + 2HCl_{(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + S_{(s)} + SO_{2(g)} \nonumber \] thiosulphate \[ 2S_2O^{2-}_{3(aq)} + I_{2(aq)} \rightarrow S_2O_{6(aq)}^{2-} + 2I^-_{(aq)} \nonumber \] \[ CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^- \nonumber \] \[ CH_3COCH_3 + I_2 \overset{H^+}{\longrightarrow} CH_3COCH_2I + H^+ + I^- \nonumber \] | 1,967 | 770 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Equilibrium_Constant |
For the hypothetical chemical reaction: \[aA + bB \rightleftharpoons cC + dD\] the equilibrium constant is defined as: \[K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\] The notation [A] signifies the molar concentration of species A. An alternative expression for the equilibrium constant involves partial pressures: \[K_P = \dfrac{P_C^c P_D^d}{P_A^aP_B^b}\] Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction \[CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g)\] is the following: \[K_C = \dfrac{[H_2]^2}{[H_2O]^2}\] H The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action. Determine the equilibrium constant for a chemical reaction. In this part of the experiment, the following reaction is studied: \[NaHCO_3 (s) \rightleftharpoons NaOH (s) + CO_2 (g) \] The steps in the experiment are those that would be performed in practice in the laboratory: At room temperature, the reaction does not proceed at a measurable rate. At 800 K, however, the rate of reaction is relatively fast and equilibrium is quickly achieved. A manometer is attached to the bulb to measure the partial pressure of carbon dioxide in the bulb. Assume the sodium hydrogen carbonate occupies a negligible fraction of the volume of the 500 mL bulb. Also assume that the connections to the manometer have a negligible volume. (The stopcock to the bulb is always open, because the bulb is always open to the manometer so that the pressure may be continually monitored. Both air and carbon dioxide are colorless gases.) Use the experimental data to calculate and for this reaction. In this part of the experiment, the value and are provided for verification. | 2,007 | 771 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Atomic_and_Ionic_Radius |
Unlike a ball, an atom does not have a fixed radius. The radius of an atom can only be found by measuring the distance between the nuclei of two touching atoms, and then halving that distance. As you can see from the diagrams, the same atom could be found to have a different radius depending on what was around it. The left hand diagram shows bonded atoms. The atoms are pulled closely together and so the measured radius is less than if they are just touching. This is what you would get if you had metal atoms in a metallic structure, or atoms covalently bonded to each other. The type of atomic radius being measured here is called the metallic radius or the covalent radius depending on the bonding. The right hand diagram shows what happens if the atoms are just touching. The attractive forces are much less, and the atoms are essentially "unsquashed". This measure of atomic radius is called the van der Waals radius after the weak attractions present in this situation. The exact pattern you get depends on which measure of atomic radius you use - but the trends are still valid. The following diagram uses metallic radii for metallic elements, covalent radii for elements that form covalent bonds, and van der Waals radii for those (like the noble gases) which don't form bonds. It is fairly obvious that the atoms get bigger as you go down groups. The reason is equally obvious - you are adding extra layers of electrons. You have to ignore the noble gas at the end of each period. Because neon and argon don't form bonds, you can only measure their van der Waals radius - a case where the atom is pretty well "unsquashed". All the other atoms are being measured where their atomic radius is being lessened by strong attractions. You aren't comparing like with like if you include the noble gases. Leaving the noble gases out, atoms get smaller as you go across a period. If you think about it, the metallic or covalent radius is going to be a measure of the distance from the nucleus to the electrons which make up the bond. (Look back to the left-hand side of the first diagram on this page if you aren't sure, and picture the bonding electrons as being half way between the two nuclei.) From lithium to fluorine, those electrons are all in the 2-level, being screened by the 1s electrons. The increasing number of protons in the nucleus as you go across the period pulls the electrons in more tightly. The amount of screening is constant for all of these elements. In the period from sodium to chlorine, the same thing happens. The size of the atom is controlled by the 3-level bonding electrons being pulled closer to the nucleus by increasing numbers of protons - in each case, screened by the 1- and 2-level electrons. Although there is a slight contraction at the beginning of the series, the atoms are all much the same size. The size is determined by the 4s electrons. The pull of the increasing number of protons in the nucleus is more or less offset by the extra screening due to the increasing number of 3d electrons. Ionic radii are difficult to measure with any degree of certainty, and vary according to the environment of the ion. For example, it matters what the co-ordination of the ion is (how many oppositely charged ions are touching it), and what those ions are. There are several different measures of ionic radii in use, and these all differ from each other by varying amounts. It means that if you are going to make reliable comparisons using ionic radii, they have to come from the same source. What you have to remember is that there are quite big uncertainties in the use of ionic radii, and that trying to explain things in fine detail is made difficult by those uncertainties. What follows will be adequate for UK A level (and its various equivalents), but detailed explanations are too complicated for this level. : This is the easy bit! As you add extra layers of electrons as you go down a group, the ions are bound to get bigger. The two tables below show this effect in Groups 1 and 7. Let's look at the radii of the simple ions formed by elements as you go across Period 3 of the Periodic Table - the elements from Na to Cl. The table misses out silicon which does not form a simple ion. The phosphide ion radius is in brackets because it comes from a different data source, and I am not sure whether it is safe to compare it. The values for the oxide and chloride ions agree in the different source, so it is probably OK. The values are again for 6-co-ordination, although I can't guarantee that for the phosphide figure. First of all, notice the big jump in ionic radius as soon as you get into the negative ions. Is this surprising? Not at all - you have just added a whole extra layer of electrons. Notice that, within the series of positive ions, and the series of negative ions, that the ionic radii fall as you go across the period. We need to look at the positive and negative ions separately. The difference between the size of similar pairs of ions actually gets even smaller as you go down Groups 6 and 7. For example, the Te ion is only 0.001 nm bigger than the I ion. As far as I am aware there is no simple explanation for this - certainly not one which can be used at this level. This is a good illustration of what I said earlier - explaining things involving ionic radii in detail is sometimes very difficult. This is only really a variation on what we have just been talking about, but fits negative and positive isoelectronic ions into the same series of results. Remember that isoelectronic ions all have exactly the same electron arrangement. The nitride ion value is in brackets because it came from a different source, and I don't know for certain whether it relates to the same 6-co-ordination as the rest of the ions. This matters. My main source only gave a 4-coordinated value for the nitride ion, and that was 0.146 nm. You might also be curious as to how the neutral neon atom fits into this sequence. Its van der Waals radius is 0.154 or 0.160 nm (depending on which source you look the value up in) - bigger than the fluoride ion. You can't really sensibly compare a van der Waals radius with the radius of a bonded atom or ion. | 6,221 | 772 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/08%3A_Nucleophilic_Substitution_and_Elimination_Reactions/8.04%3A_General_Considerations_of_Substitution_Reactions |
We now wish to discuss displacements by reagents (\(Y:\)) on alkyl derivatives (\(RX\)). These are or reactions involving attack by a nucleophile at . A typical example is the reaction of hydroxide ion with bromomethane to displace bromide ion: The electron pair of the \(C-O\) bond can be regarded as having been donated by the hydroxide ion, while the electron pair of the \(C-Br\) bond departs with the leaving bromide ion. The name for this type of reaction is abbreviated \(\text{S}_\text{N}\), \(\text{S}\) for substitution and \(\text{N}\) for nucleophilic. Reactions of this type are very useful. They can lead to compounds in which the new bond to carbon in the alkyl group, \(R\) is to chlorine, bromine, iodine, oxygen, sulfur, carbon, nitrogen, or phosphorus, depending on the nature of the nucleophile used. Nucleophilic substitutions are especially important for alkyl halides, but they should not be considered to be confined to alkyl halides. Many other alkyl derivatives such as alcohols, ethers, esters, and "onium ions"\(^3\) also can undergo \(\text{S}_\text{N}\) reactions if conditions are appropriate. The scope of \(\text{S}_\text{N}\) reactions is so broad that it is impossible to include all the various alkyl compounds and nucleophiles that react in this manner. Rather we shall approach the subject here through consideration of the mechanisms of \(\text{S}_\text{N}\) reactions, and then develop the scope of the reactions in later chapters. The mechanism of an \(\text{S}_\text{N}\) reaction and the reactivity of a given alkyl compound \(RX\) toward a nucleophile \(Y\) depend upon the nature of \(R\), \(X\), and \(Y\), and upon the nature of the solvent. For an \(\text{S}_\text{N}\) reaction to occur at a reasonable rate, it is very important to select a solvent that will dissolve both the alkyl compound and the nucleophilic reagent; considerable assistance may be required from both the solvent and the nucleophile to break what usually is a slightly polar \(C-X\) bond. However, the solvents that best dissolve slightly polar organic compounds seldom will dissolve the common, rather highly polar, nucleophilic agents such as \(NaBr\), \(NaCN\), and \(H_2O\). In practice, relatively polar solvents, or solvent mixtures, such as 2-propanone (acetone), aqueous 2-propanone, ethanol, aqueous 1,4-dioxacyclohexane (dioxane), and so on, provide the best compromise for reactions between alkyl compounds and saltlike nucleophilic reagents. The importance of the solvent in stabilizing ions can be evaluated from the estimated thermochemistry of ionic reactions discussed in Section 8-2. \(^3\)Examples of -onium cations are and (1977) | 2,695 | 775 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.04%3A_Colligative_Properties_of_a_Dilute_Solution |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The of a solution are usually considered to be: Figure 12.3 illustrates the freezing-point depression and boiling-point elevation of an aqueous solution. At a fixed pressure, pure liquid water is in equilibrium with ice at the freezing point and with steam at the boiling point. These are the temperatures at which H\(_2\)O has the same chemical potential in both phases at this pressure. At these temperatures, the chemical potential curves for the phases intersect, as indicated by open circles in the figure. The presence of dissolved solute in the solution causes a lowering of the H\(_2\)O chemical potential compared to pure water at the same temperature. Consequently, the curve for the chemical potential of H\(_2\)O in the solution intersects the curve for ice at a lower temperature, and the curve for steam at a higher temperature, as indicated by open triangles. The freezing point is depressed by \(\Del T\subs{f}\), and the boiling point (if the solute is nonvolatile) is elevated by \(\Del T\bd\). Although these expressions provide no information about the activity coefficient of a solute, they are useful for estimating the solute molar mass. For example, from a measurement of any of the colligative properties of a dilute solution and the appropriate theoretical relation, we can obtain an approximate value of the solute molality \(m\B\). (It is only approximate because, for a measurement of reasonable precision, the solution cannot be extremely dilute.) If we prepare the solution with a known amount \(n\A\) of solvent and a known mass of solute, we can calculate the amount of solute from \(n\B=n\A M\A m\B\); then the solute molar mass is the solute mass divided by \(n\B\). As in Sec. 12.2.1, we assume the solid that forms when a dilute solution is cooled to its freezing point is pure component A. Equation 12.3.6 gives the general dependence of temperature on the composition of a binary liquid mixture of A and B that is in equilibrium with pure solid A. We treat the mixture as a solution. The solvent is component A, the solute is B, and the temperature is the freezing point \(T\f\): \begin{equation} \Pd{T\f}{x\A}{\!p} = \frac{T\f^2}{\Delsub{sol,A}H} \bPd{(\mu\A/T)}{x\A}{T,p} \tag{12.4.1} \end{equation} Consider the expression on the right side of this equation in the limit of infinite dilution. In this limit, \(T\f\) becomes \(T\f^*\), the freezing point of the pure solvent, and \(\Delsub{sol,A}H\) becomes \(\Delsub{fus,A}H\), the molar enthalpy of fusion of the pure solvent. To deal with the partial derivative on the right side of Eq. 12.4.1 in the limit of infinite dilution, we use the fact that the solvent activity coefficient \(\g\A\) approaches \(1\) in this limit. Then the solvent chemical potential is given by the Raoult’s law relation \begin{gather} \s{ \mu\A = \mu\A^* + RT\ln x\A } \tag{12.4.2} \cond{(solution at infinite dilution)} \end{gather} where \(\mu\A^*\) is the chemical potential of A in a pure-liquid reference state at the same \(T\) and \(p\) as the mixture. (At the freezing point of the mixture, the reference state is an unstable supercooled liquid.) If the solute is an electrolyte, Eq. 12.4.2 can be derived by the same procedure as described in Sec. 9.4.6 for an ideal-dilute binary solution of a nonelectrolyte. We must calculate \(x\A\) from the amounts of all species present at infinite dilution. In the limit of infinite dilution, any electrolyte solute is completely dissociated to its constituent ions: ion pairs and weak electrolytes are completely dissociated in this limit. Thus, for a binary solution of electrolyte B with \(\nu\) ions per formula unit, we should calculate \(x\A\) from \begin{equation} x\A = \frac{n\A}{n\A + \nu n\B} \tag{12.4.3} \end{equation} where \(n\B\) is the amount of solute formula unit. (If the solute is a nonelectrolyte, we simply set \(\nu\) equal to \(1\) in this equation.) From Eq. 12.4.2, we can write \begin{equation} \bPd{(\mu\A/T)}{x\A}{T,p} \ra R \quad \tx{as} \quad x\A \ra 1 \tag{12.4.4} \end{equation} In the limit of infinite dilution, then, Eq. 12.4.1 becomes \begin{equation} \lim_{x\A\rightarrow 1}\Pd{T\f}{x\A}{\!p} = \frac{R(T\f^*)^2}{\Delsub{fus,A}H} \tag{12.4.5} \end{equation} It is customary to relate freezing-point depression to the solute concentration \(c\B\) or molality \(m\B\). From Eq. 12.4.3, we obtain \begin{equation} 1-x\A=\frac{\nu n\B}{n\A+\nu n\B} \tag{12.4.6} \end{equation} In the limit of infinite dilution, when \(\nu n\B\) is much smaller than \(n\A\), \(1-x\A\) approaches the value \(\nu n\B/n\A\). Then, using expressions in Eq. 9.1.14, we obtain the relations \begin{gather} \s{ \begin{split} \dx\A & = -\dif(1- x\A) = -\nu\dif(n\B/n\A) \cr & = -\nu V\A^*\dif c\B \cr & = -\nu M\A\dif m\B \end{split} } \tag{12.4.7} \cond{(binary solution at} \nextcond{infinite dilution)} \end{gather} which transform Eq. 12.4.5 into the following (ignoring a small dependence of \(V\A^*\) on \(T\)): \begin{gather} \lim_{c\B\rightarrow 0} \tag{12.4.8} \Pd{T\f}{c\B}{\!p} = -\frac{\nu V\A^* R(T\f^*)^2}{\Delsub{fus,A}H} \cr \lim_{m\B\rightarrow 0}\Pd{T\f}{m\B}{\!p} = -\frac{\nu M\A R(T\f^*)^2}{\Delsub{fus,A}H} \end{gather} We can apply these equations to a nonelectrolyte solute by setting \(\nu\) equal to \(1\). As \(c\B\) or \(m\B\) approaches zero, \(T\f\) approaches \(T\f^*\). The freezing-point depression (a negative quantity) is \(\Del T\subs{f}=T\f-T\f^*\). In the range of molalities of a dilute solution in which \(\pd{T\f}{m\B}{p}\) is given by the expression on the right side of Eq. 12.4.8, we can write \begin{equation} \Del T\subs{f} = -\frac{\nu M\A R(T\f^*)^2}{\Delsub{fus,A}H}m\B \tag{12.4.9} \end{equation} The or cryoscopic constant, \(K\subs{f}\), is defined for a binary solution by \begin{equation} K\subs{f} \defn -\lim_{m\B\rightarrow 0} \frac{\Del T\subs{f}}{\nu m\B} \tag{12.4.10} \end{equation} and, from Eq. 12.4.9, has a value given by \begin{equation} K\subs{f} = \frac{M\A R(T\f^*)^2}{\Delsub{fus,A}H} \tag{12.4.11} \end{equation} The value of \(K\subs{f}\) calculated from this formula depends only on the kind of solvent and the pressure. For H\(_2\)O at \(1\br\), the calculated value is \(K\bd=1.860\units{K kg mol\(^{-1}\)}\) (Prob. 12.4). In the dilute binary solution, we have the relation \begin{gather} \s{ \Del T\subs{f} = - \nu K\subs{f} m\B } \tag{12.4.12} \cond{(dilute binary solution)} \end{gather} This relation is useful for estimating the molality of a dilute nonelectrolyte solution (\(\nu{=}1\)) from a measurement of the freezing point. The relation is of little utility for an electrolyte solute, because at any electrolyte molality that is high enough to give a measurable depression of the freezing point, the mean ionic activity coefficient deviates greatly from unity and the relation is not accurate. We can apply Eq. 12.3.6 to the boiling point \(T\bd\) of a dilute binary solution. The pure phase of A in equilibrium with the solution is now a gas instead of a solid. (We must assume the solute is nonvolatile or has negligible partial pressure in the gas phase.) Following the procedure of Sec. 12.4.1, we obtain \begin{equation} \lim_{m\B\rightarrow 0}\Pd{T\bd}{m\B}{\!p} = \frac{\nu M\A R(T\bd^*)^2}{\Delsub{vap,A}H} \tag{12.4.13} \end{equation} where \(\Delsub{vap,A}H\) is the molar enthalpy of vaporization of pure solvent at its boiling point \(T\bd^*\). The or ebullioscopic constant, \(K\bd\), is defined for a binary solution by \begin{equation} K\bd \defn \lim_{m\B\rightarrow 0}\frac{\Del T\bd}{\nu m\B} \tag{12.4.14} \end{equation} where \(\Del T\bd=T\bd-T\bd^*\) is the boiling-point elevation. Accordingly, \(K\bd\) has a value given by \begin{equation} K\bd = \frac{M\A R(T\bd^*)^2}{\Delsub{vap,A}H} \tag{12.4.15} \end{equation} For the boiling point of a dilute solution, the analogy of Eq. 12.4.12 is \begin{gather} \s{ \Del T\bd = \nu K\bd m\B } \tag{12.4.16} \cond{(dilute binary solution)} \end{gather} Since \(K\subs{f}\) has a larger value than \(K\bd\) (because \(\Delsub{fus,A}H\) is smaller than \(\Delsub{vap,A}H\)), the measurement of freezing-point depression is more useful than that of boiling-point elevation for estimating the molality of a dilute solution. In a binary two-phase system in which a solution of volatile solvent A and nonvolatile solute B is in equilibrium with gaseous A, the vapor pressure of the solution is equal to the system pressure \(p\). Equation 12.3.7 gives the general dependence of \(p\) on \(x\A\) for a binary liquid mixture in equilibrium with pure gaseous A. In this equation, \(\Delsub{sol,A}V\) is the molar differential volume change for the dissolution of the gas in the solution. In the limit of infinite dilution, \(-\Delsub{sol,A}V\) becomes \(\Delsub{vap,A}V\), the molar volume change for the vaporization of pure solvent. We also apply the limiting expressions of Eqs. 12.4.4 and 12.4.7. The result is \begin{equation} \lim_{c\B\rightarrow 0}\Pd{p}{c\B}{T} = -\frac{\nu V\A^* RT}{\Delsub{vap,A}V} \qquad \lim_{m\B\rightarrow 0}\Pd{p}{m\B}{T} = -\frac{\nu M\A RT}{\Delsub{vap,A}V} \tag{12.4.17} \end{equation} If we neglect the molar volume of the liquid solvent compared to that of the gas, and assume the gas is ideal, then we can replace \(\Delsub{vap,A}V\) in the expressions above by \(V\A^*\gas =RT/p\A^*\) and obtain \begin{equation} \lim_{c\B\rightarrow 0}\Pd{p}{c\B}{T} \approx -\nu V\A^* p\A^* \qquad \lim_{m\B\rightarrow 0}\Pd{p}{m\B}{T} \approx -\nu M\A p\A^* \tag{12.4.18} \end{equation} where \(p\A^*\) is the vapor pressure of the pure solvent at the temperature of the solution. Thus, approximate expressions for vapor-pressure lowering in the limit of infinite dilution are \begin{equation} \Del p \approx -\nu V\A^* p\A^* c\B \qquad \tx{and} \qquad \Del p \approx -\nu M\A p\A^* m\B \tag{12.4.19} \end{equation} We see that the lowering in this limit depends on the kind of solvent and the solution composition, but not on the kind of solute. The osmotic pressure \(\varPi\) is an intensive property of a solution and was defined in Sec. 12.2.2. In a dilute solution of low \(\varPi\), the approximation used to derive Eq. 12.2.11 (that the partial molar volume \(V\A\) of the solvent is constant in the pressure range from \(p\) to \(p+\varPi\)) becomes valid, and we can write \begin{equation} \varPi = \frac{\mu\A^*-\mu\A}{V\A} \tag{12.4.20} \end{equation} In the limit of infinite dilution, \(\mu\A^*-\mu\A\) approaches \(-RT\ln x\A\) (Eq. 12.4.2) and \(V\A\) becomes the molar volume \(V\A^*\) of the pure solvent. In this limit, Eq. 12.4.20 becomes \begin{equation} \varPi = -\frac{RT\ln x\A}{V\A^*} \tag{12.4.21} \end{equation} from which we obtain the equation \begin{equation} \lim_{x\A\rightarrow 1}\Pd{\varPi}{x\A}{T,p} = -\frac{RT}{V\A^*} \tag{12.4.22} \end{equation} The relations in Eq. 12.4.7 transform Eq. 12.4.22 into \begin{equation} \lim_{c\B\rightarrow 0}\Pd{\varPi}{c\B}{T,p} = \nu RT \tag{12.4.23} \end{equation} \begin{equation} \lim_{m\B\rightarrow 0}\Pd{\varPi}{m\B}{T,p} = \frac{\nu RTM\A}{V\A^*} = \nu \rho\A^* RT \tag{12.4.24} \end{equation} Equations 12.4.23 and 12.4.24 show that the osmotic pressure becomes independent of the kind of solute as the solution approaches infinite dilution. The integrated forms of these equations are \begin{gather} \s{ \varPi=\nu c\B RT } \tag{12.4.25} \cond{(dilute binary solution)} \end{gather} \begin{gather} \s{ \varPi=\frac{RTM\A}{V\A^*}\nu m\B = \rho\A^* RT\nu m\B } \tag{12.4.26} \cond{(dilute binary solution)} \end{gather} Equation 12.4.25 is for osmotic pressure. If there is more than one solute species, \(\nu c\B\) can be replaced by \(\sum_{i\ne\tx{A}}c_i\) and \(\nu m\B\) by \(\sum_{i\ne\tx{A}}m_i\) in these expressions. In Sec. 9.6.3, it was stated that \(\varPi/m\B\) is equal to the product of \(\phi_m\) and the limiting value of \(\varPi/m\B\) at infinite dilution, where \(\phi_m = (\mu\A^*-\mu\A)/RTM\A\sum_{i \ne \tx{A}}m_i\) is the osmotic coefficient. This relation follows directly from Eqs. 12.2.11 and 12.4.26. | 19,460 | 777 |
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As can be seen from the Arrhenius equation, the magnitude of the activation energy, \(E_a\), determines the value of the rate constant, \(k\), at a given temperature and thus the overall reaction rate. Catalysts provide a means of reducing \(E_a\) and increasing the reaction rate. Catalysts are defined as substances that participate in a chemical reaction but are not changed or consumed. Instead they provide a new mechanism for a reaction to occur which has a lower activation energy than that of the reaction without the catalyst. refers to reactions in which the catalyst is in solution with at least one of the reactants whereas refers to reactions in which the catalyst is present in a different phase, usually as a solid, than the reactants. Figure 29.8.1
shows a comparison of energy profiles of a reaction in the absence and presence of a catalyst. Consider a non-catalyzed elementary reaction \[\text{A} \overset{k}{\longrightarrow} \text{P} \nonumber \] which proceeds at rate \(k\) at a certain temperature. The reaction rate can be expressed as \[\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \nonumber \] In the presence of a catalyst \(\text{C}\), we can write the reaction as \[\text{A} + \text{C} \overset{k_\text{cat}}{\longrightarrow} \text{P} + \text{C} \nonumber \] and the reaction rate as \[\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] - k_\text{cat} \left[ \text{A} \right] \left[ \text{C} \right] \nonumber \] where the first term represents the uncatalyzed reaction and the second term represents the catalyzed reaction. Because the reaction rate of the catalyzed reaction is often magnitudes larger than that of the uncatalyzed reaction (i.e. \(k_\text{cat} \gg k\)), the first term can often be ignored. A common example of homogeneous catalysts are acids and bases. For example, given an overall reaction is \(\text{S} \rightarrow \text{P}\). If \(k\) is the rate, then \[\dfrac{d \left[ \text{P} \right]}{dt} = k \left[ \text{S} \right] \nonumber \] The purpose of an enzyme is to enhance the rate of production of the product \(\text{P}\). The equations of the acid-catalyzed reaction are \[\begin{align} \text{S} + \text{A}H &\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{S}H^+ + \text{A}^- \\ \text{S}H^+ + H_2 O &\overset{k_2}{\rightarrow} \text{P} + H_3 O^+ \\ H_3 O^+ + \text{A}^- &\overset{k_3}{\underset{k_{-3}}{\rightleftharpoons}} \text{A}H + H_2 O \end{align} \nonumber \] The full set of kinetic equations is \[\begin{align} \dfrac{d \left[ \text{S} \right]}{dt} &= -k_1 \left[ \text{S} \right] \left[ \text{A} H \right] + k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] \\ \dfrac{ d \left[ \text{A} H \right]}{dt} &= -k_1 \left[ \text{S} \right] \left[ \text{A} H \right] + k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] - k_{-3} \left[ \text{A} H \right] + k_3 \left[ H_3 O^+ \right] \left[ \text{A}^- \right] \\ \dfrac{d \left[ \text{S} H^+ \right]}{dt} &= k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] - k_2 \left[ \text{S} H^+ \right] \\ \dfrac{d \left[ \text{A}^- \right]}{dt} &= k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] -k_2 \left[ \text{A}^- \right] \left[ H_3 O^+ \right] + k_{-3} \left[ \text{A} H \right] \\ \dfrac{d \left[ \text{P} \right]}{dt} &= k_2 \left[ \text{S} H^+ \right] \\ \dfrac{d \left[ H_3 O^+ \right]}{dt} &= -k_2 \left[ \text{S} H^+ \right] - k_3 \left[ H_3 O^+ \right] \left[ \text{A}^- \right] + k_{-3} \left[ \text{A} H \right] \end{align} \nonumber \] We cannot easily solve these, as they are nonlinear. However, let us consider two cases \(k_2 \gg k_{-1} \left[ \text{A}^- \right]\) and \(k_2 \ll k_{-1} \left[ \text{A}^- \right]\). In both cases, \(\text{S} H^+\) is consumed quickly, and we can apply a steady-state approximation: \[\dfrac{d \left[ \text{S} H^+ \right]}{dt} = k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{A}^- \right] \left[ \text{S} H^+ \right] - k_2 \left[ \text{S} H^+ \right] = 0 \nonumber \] Rearranging in terms of \(\text{S} H^+\) yields \[\left[ \text{S} H^+ \right] = \dfrac{k_1 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right] + k_2} \nonumber \] and the rate of production of \(\text{P}\) can be written as \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{S} H^+ \right] = \dfrac{k_1 k_2 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right] + k_2} \nonumber \] In the case where \(k_2 \gg k_{-1} \left[ \text{A}^- \right]\), Equation 29.8.17 can be written as \[\dfrac{d \left[ \text{P} \right]}{dt} = k_1 \left[ \text{S} \right] \left[ \text{A} H \right] \nonumber \] which is known as a general acid-catalyzed reaction. On the other hand, if \(k_2 \ll k_{-1} \left[ \text{A}^- \right]\), we can use an equilibrium approximation to write the rate of production of \(\text{P}\) as \[\dfrac{d \left[ \text{P} \right]}{dt} = \dfrac{k_1 k_2 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right]} = \dfrac{k_1 k_2}{k_{-1} K} \left[ \text{S} \right] \left[ H^+ \right] \nonumber \] where \(K\) is the acid dissociation constant: \[K = \dfrac{ \left[ \text{A}^- \right] \left[ H^+ \right]}{\left[ \text{A} H \right]} \nonumber \] In this case, the reaction is hydrogen ion-catalyzed. Many gas-phase reactions are catalyzed on a solid surface. For a first-order, unimolecular reaction, the reaction mechanism can be written as \[\text{A} \left( g \right) + \text{S} \left( s \right) \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{AS} \left( s \right) \nonumber \] \[\text{AS} \left( s \right) \overset{k_2}{\rightarrow} \text{P} \left( g \right) + \text{S} \left( g \right) \nonumber \] where the first step is reversible adsorption of the gas molecule, \(\text{A}\), onto active sites on the catalyst surface, \(\text{S}\), to form a transition state, \(\text{AS}\), and the second step is the conversion of adsorbed \(\text{A}\) molecules to species \(\text{P}\). Applying the steady-state approximation to species \(\text{AS}\), we can write \[\dfrac{d \left[ \text{AS} \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{S} \right] - k_{-1} \left[ \text{AS} \right]_{ss} - k_2 \left[ \text{AS} \right]_{ss} = 0 \nonumber \] Because the concentration of total active sites on the catalyst surface is fixed at \(\left[ \text{S} \right]_0\), the concentration of adsorbed species on the catalyst surface, \(\left[ \text{AS} \right]\) can be written as \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{AS} \right]_{ss} = k_2 \theta \left[ \text{S} \right]_0 = \dfrac{k_1 k_2}{k_1 \left[ \text{A} \right] + k_{-1} + k_2} \left[ \text{A} \right] \left[ \text{S} \right]_0 \nonumber \] Langmuir-Hinshelwood Eley-Rideal Eley-Rideal Eley-Rideal \[0 = k_1 \left[ \text{A} \right] \left( 1 - \theta \right) \left[ \text{S} \right]_0 - k_{-1} \left[ \text{S} \right]_0 - k_2 \theta \left[ \text{S} \right]_0 \left[ \text{B} \right] \nonumber \] \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{AS} \right]_{ss} \left[ \text{B} \right] = k_2 \theta \left[ \text{S} \right]_0 \left[ \text{B} \right] = \dfrac{k_1 k_2 \left[ \text{A} \right]}{k_1 \left[ \text{A} \right] + k_{-1}} \left[ \text{S} \right]_0 \left[ \text{B} \right] \nonumber \] \[\dfrac{d \left[ \text{P} \right]}{dt} = K k_2 \left[ \text{B} \right] \dfrac{K \left[ \text{A} \right]}{K \left[ \text{A} \right] + 1} \nonumber \] ( ) | 7,623 | 778 |
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The biological functions of proteins are extremely diverse. Some act as hormones that regulate various metabolic processes. An example is insulin, which regulates blood-sugar levels. Enzymes act as catalysts for biological reactions, and other proteins serve as biological structural materials - for example, collagen and elastin in connective tissue and keratin in hair. Iron-containing proteins (hemoglobin and myoglobin in mammals) and copper-containing proteins (hemocyanins in shellfish) transport molecular oxygen. Some blood proteins form antibodies, which provide resistance to disease, while the so-called nucleoproteins are important constituents of the genes that supply and transmit genetic information in cell division. Motion by means of muscle contraction and the generation and transmission of nerve impulses also involve proteins. How can a group of compounds, made from a common basis set of amino acids, be so remarkably heterogeneous and exhibit such varied yet specific functions? Clearly, the primary structure and presence or absence of special functional groups, metals, and so on, are of paramount importance. Of complementary importance are the of proteins, which are dictated not just by the primary structure but by the way the primary structure is put together biochemically. The polypeptide chains are seldom, if every fully extended, but are coiled and folded into more or less stable conformations. As a result, amino-acid side chains in distant positions in the linear sequence are brought into close proximity, and this juxtaposition often is crucial for the protein to fulfill its specific biological function. The elucidation of the detailed of protein molecules - in fact, the spatial locations of the individual atoms in a protein - is accomplished primarily by x-ray crystallography. The three-dimensional structures of more than twenty proteins have now been established by this technique. The importance of x-ray crystallography to structural and biological chemistry has been recognized in the award of six Nobel Prizes in this area.\(^6\) A number of important proteins and their properties are listed in Table 25-3. There are several other points to notice about the \(\alpha\) helix shown in Figure 25-11. The amide groups are planar and normally retain the stable configuration in the helical structure; bond lengths and bond angles are normal, and the \(\ce{-NH} \cdots \ce{O=C}-\) hydrogen bonds are nearly linear. However, the hydrogen bonds are not quite parallel to the long axis of the coil, so there are 3.6 rather than 4 amino-acid units per helical turn, and the spacing between turns is about \(5.4 \: \text{Å}\). The \(\alpha\) helix in proteins has turns like a right-hand screw thread. The amino acids of the side chains lie outside the coil of the \(\alpha\) helix and are in close proximity to the side chains three and four amino-acid units apart. Because of this proximity, steric hindrance between large side chains can be sufficient to reduce the stability of the normal \(\alpha\) helix. When such hindrance occurs, there is a discontinuity in the helical structure, and the peptide chain may assume more random arrangements about the \(\ce{C-C}_\alpha\) and \(\ce{N-C}_\alpha\) bonds (see Figure 25-12), thereby allowing the molecule to fold back on itself and form new hydrogen bonds. The helical structure apparently is always interrupted at proline or hydroxyproline residues because the \(\ce{N-C}_\alpha\) bonds of these amino acids are not free to rotate (they are incorporated in five-membered rings) and also because the proline and hydroxyproline amide nitrogens have no hydrogens to participate in hydrogen bonding to carbonyl groups. Pauling and Corey recognized a second stable conformation of polypeptide chains - the extended chain or \(\beta\)-pleated sheet (Figure 25-13). In this conformation the chains are fully extended with trans amide configurations. In this arrangement the distance is maximized between adjacent amino-acid side chains. Hydrogen bonding of the type \(\ce{-N-H} \cdots \ce{O=C}-\) is now rather than between amino acids in a single chain (as in the \(\alpha\) helix). This type of structure is not as common as the \(\alpha\) helix and is found extensively only in silk fibroin. However, a number of proteins with a single polypeptide chain can form short sections of "antiparallel" \(\beta\)-pleated sheets by folding back on themselves, as illustrated in Figure 25-14. Another very important factor in protein architecture is the disulfide \(\ce{-S-S}-\) link. Remote parts of the polypeptide chain can be held close together through the oxidative coupling of two cysteine thiol groups to form a disulfide bridge: The disulfide bridges in some proteins are between different peptide chains. Insulin, for instance, has two interchain as well as one intrachain \(\ce{S-S}\) bridges (Figure 25-8). Some idea of the complexity of protein conformations can be gained from the structure of myoglobin. This protein is responsible for the storage and transport of molecular oxygen in the muscle tissue of mammals. It is a compact molecule of 153 amino-acid units in a chain that is extensively coiled as an \(\alpha\) helix. There are eight regions of discontinuity in the helical structure, and in these regions the chain folds on itself as shown in Figure 15-16. Four of the eight nonhelical regions occur at proline residues; the reason for the discontinuity at the other regions is not entirely clear. With the exception of two histidine units, the interior regions of myoglobin accommodate only the nonpolar side chains; the interior, therefore, is mostly hydrocarbonlike and repellant to water and other polar molecules. In contrast, the polar side chains are on the exterior of the protein. A number of proteins, including myoglobin, possess one or more nonpeptide components associated with specific sites on the polypeptide chain. These components are called and are essential to the biological activity. When the prosthetic group is removed, the residual protein is referred to as an . In myoglobin the prosthetic group is a molecule of . The heme group belongs to a class of interesting compounds called , which are metal complexes of a highly conjugated ring system composed of four azacyclopentadiene (pyrrole) rings linked by \(\ce{-CH=}\) bridges between the 2 and 5 positions. The parent compound is known as . Porphyrins have highly stabilized electronic excited states and absorb visible light. As a result they usually are brightly colored compounds (e.g., chlorophyll, Figure 20-6). The porphyrin of heme is known as protoporphyrin IX, and the associated metal is iron [as \(\ce{Fe}\)(II) or \(\ce{Fe}\)(III)]. You will notice that the porphyrin ring carries methyl, ethenyl, and propanoic acid side chains: A major effort on the part of several eminent chemists in the early part of the century led to the elucidation of the structure of heme. The German chemist Hans Fischer successfully synthesized heme in 1929, a feat for which, in 1930, he received the Nobel Prize in chemistry. [Some years earlier (1915), Richard Willstatter received a Nobel Prize for structural studies of chlorophyll and plant pigments.] A very important question is, how does the particular combination of protein and iron-porphyrin allow myoglobin to reversibly bind molecular oxygen? The answer to this question is not known in all its details, but it is well established that \(\ce{Fe}\)(II)-porphyrins will complex readily and reversibly with oxygen. There are two additional coordination sites around the iron in heme besides the four ring nitrogens. These are indicated below as the general ligands \(\ce{L}\): The disclike heme molecule fits into a cleft in the protein structure and is bound to it through one of the \(\ce{L}\) coordination sites to a histidine nitrogen. The remaining coordination site on the other side of the ring is occupied by molecular oxygen. In the absence of the coordination by histidine, the porphyrin iron would be oxidized rapidly to the ferric state, which does not bind oxygen. A number of model compounds have been synthesized which have \(\ce{Fe}\)(II)-porphyrin rings carrying a side chain with histidine arranged to be able to coordinate with the metal on one side. Several of these substances show promise as oxygen carriers with properties similar to myoglobin. Hemoglobin is related to myoglobin in both its structure and function. It reversibly binds molecular oxygen which it transports in the red corpuscles of blood rather than in muscle tissue. However, hemoglobin is made up of polypeptide chains, in contrast to myoglobin which has only chain. Two of the hemoglobin chains are of one kind with 141 amino acid residues, called the \(\alpha\) chains, and two are of another kind with 146 amino acids, called the \(\beta\) chains. Each chain, or , contains one heme group identical with the heme in myoglobin. The subunits are held in the hemoglobin by noncovalent interactions and provide four hemes and hence four binding sites for molecular oxygen. The \(\alpha\) and \(\beta\) hemes have different affinities for oxygen but function in a cooperative way to increase oxygen availability to the cells. In spite of the fact that the \(\alpha\) and \(\beta\) chains of hemoglobin are nonidentical with the myoglobin chain, the three-dimensional structures of all three chains are strikingly similar; myoglobins and hemoglobins differ slightly in amino acid composition, depending on the species but the protein shape remains essentially the same. Many factors contribute to the three-dimensional structures of proteins. We already have mentioned hydrogen bonding between amide groups, location and character of prosthetic groups, and disulfide bonds. Other important influences include electrostatic interactions between ionic groups (\(\ce{-NH_3^+}\), \(\ce{-CO_2^-}\)), hydrogen-bonding involving side-chain substituents \(\left( \ce{-CH_2OH} \right)\), and nonbonded interactions. Except for the disulfide linkages, most of these interactions are weak compared to covalent bond strengths, and the conformations of many proteins can be altered rather easily. In fact, several have conformations that clearly are in dynamic equilibrium under physiological conditions. Such structural flexibility may be necessary for the protein to be functional, but if the conformation is altered - that is, if it is denatured - its biological activity usually is destroyed. In many cases there are important interactions between protein molecules that may lead to highly organized structures such as the pleated sheet of silk fibroin (Figure 25-13) or the coiling of \(\alpha\) helices, as found in \(\alpha\)-keratins, the fibrous proteins of hair, horn, and muscles (Figure 25-17). This sort of organization of protein molecules is called structure and is an important feature of many proteins that associate into dimers, tetramers, and so on. The tetrameric structure of hemoglobin is an important example. \(^6\)The following Nobel laureates received their awards for contributions to the use of x-ray crystallography for structure determination: 1914, Max von Laue (physics), diffraction of x-rays in crystals; 1915, William Bragg and Lawrence Bragg (physics), study of crystal structure by means of x-rays; 1954, Linus Pauling (chemistry), study of structure of proteins; 1962, Max Perutz and John Kendrew (chemistry), structures of myoglobin and hemoglobin; 1962, Francis Crick, James Watson, and Maurice Wilkins (physiology and medicine), double helix of DNA; 1964, Dorothy Hodgkin (chemistry), determination of structure of vitamin B\(_{12}\) and penicillin by x-ray methods. She later determined the three-dimensional structure of insulin. and (1977) | 11,842 | 779 |
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Pasteur showed that sugar fermentation in the presence of oxygen ( ) leads to a maximum rate of yeast growth, but minimum alcohol production. Excluding air (so the process continues ) slows yeast growth, but increases alcohol production. Anaerobic processes usually produce less energy than aerobic ones. We notice the same thing in our muscles when all the blood oxygen is used up, and lactic acid is produced (see below). Let's see how thermochemical equations help explain the "Pasteur Effect" and energy associated with lactic acid buildup. In case you're wondering, the heat energy values below can be obtained from the National Institute of Standars and Technology website NIST, from the home page use a search for "ethanol". Perhaps the most useful feature of thermochemical equations is that they can be combined to determine Δ values for other chemical reactions. Consider, for example, the following two-step sequence. Step 1 is anaerobic fermentation of glucose, C H O , to make 2 mol of ethanol, C H OH and 2 mol CO ( ): \[\ce{C H O (l) → 2 C H OH(l) + 2 CO (g) Δ = –74.4 kJ = Δ } \nonumber \] Note that a small amount of energy is produced anaerobically. \ If oxygen becomes available, the C H OH reacts with 6 mol O yielding 4 mol CO : \[\ce{2 C H OH + 6 O ( ) → 4 CO ( ) + 6H O(l) Δ = –2734 kJ = Δ } \nonumber \] The net result of this two-step process is production of 6 mol CO from the original 1 mol C H O and 6 mol O . All the ethanol produced in step 1 is used up in step 2. The overall effect is the same as the aerobic fermentation of glucose: \[\ce{C H O (s) + 6 O (g) → 6 CO (g) + 6 H O (25 , 1 Atm) Δ = –2808 kJ} \nonumber \] Now we see an explanation for the "Pasteur Effect". If yeast grows in air, it can produce 2808 kJ/mol sugar, just like we do. That's energy that can be used to synthesize compounds and grow (we could use it to move around, but yeast can't do that!). If yeast ferments the sugar anaerobically, it can only produce 74 kJ/mol sugar, so growth is retarded, but it produces a lot of alcohol! The difference is the energy that comes from the aerobic metabolism of alcohol, producing 2734 kJ (for 2 mol ethanol). On paper this net result can be obtained by the two chemical equations as though they were algebraic equations. The ethanol produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction \[\ce{C H O (l) → + 2 CO (g) Δ = –74.4 kJ = Δ } \nonumber \] + 6 O ( ) → 4 CO ( ) + 6H O Δ = –2734 kJ = Δ \(^{\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}\) \[\ce{C H O (s) + 6 O (g) → 6 CO (g) + 6 H O (25 , 1 Atm) Δ = –2808 kJ} \nonumber \] Experimentally it is found that the enthalpy change for the net reaction is the of the enthalpy changes for steps 1 and 2: Δ = –74.4 kJ + (–2734 kJ) + –2808 kJ = Δ + Δ } \nonumber \] That is, the thermochemical equation \[\ce{C H O (s) + 6 O (g) → 6 CO (g) + 6 H O Δ = –2808 kJ} \nonumber \] is the correct one for the overall reaction. Note that this is the same equation, and same heat of reaction, that we used [Weight of Food and Energy Production |before]. In the general case it is always true that . This principle is known as . If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain Δ values for reactions which cannot be carried out experimentally, as the next example shows. During exercise, glucose is first metabolized according to equation (1) above because there is plenty of oxygenated blood around muscle tissue. But as the oxygen is depleted, glucose is metabolized anaerobically to lactate ion to produce energy (we'll simplify by showing it as solid lactic acid): When blood once more supplies oxygen to the muscle, the lactic acid is metabolized according to the equation below: \[\ce{C H O (s) + 3 O ( ) → 3 CO ( ) + 3 H O (l)Δ = –1344 kJ (1)} \nonumber \] Use Δ for this reaction,and for the aerobic metabolism of glucose: \[\ce{C H O (s) + 6 O (g) → 6 CO (g) + 6 H O Δ = –2808 kJ (2) } \nonumber \] to calculate Δ for the reaction, \[\ce{2 C H O (s) → C H O (s) Δ = ? kJ } \label{3} \] We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1): Since Eq. (3) has 2 mol C H O (s) on the left, we multiply Eq. (1) by 2. We also double the heat energy produced. Since Eq. (3) no H O or CO , we need to cancel these molecules. Since Eq. (1) has mol CO and 6 mol H O on the , whereas there also CO and H O on the of Eq. (2), we write Eq. (2) in reverse so they'll cancel. We also change the sign on the heat energy, indicating that it is absorbed rather than released. Reversing (2) also puts C H O (s) on the right in (2'), where it appears in (3), so we can combine the equations, cancelling molecules that appear on both sides. We then have a. \[\ce{2 C H O (s) + 6 O ( ) → 6 CO ( ) + 6 H O (la) Δ = 2 x –1344 kJ = -2688 kJ (1')} \nonumber \] b. \[\ce{6 CO (g) + 6 H O → C H O (s) + 6 O (g) Δ = +2808 kJ (2')} \nonumber \] c. \[\ce{2 C H O (s)→ C H O (s) Δ = +2808 kJ - 2688kJ = +120 kJ (3)} \nonumber \] The result (3) is interesting; in reverse, it's C H O (s) → 2 C H O (s) Δ = -120 kJ (3) which is the amount of energy from anaerobic conversion of glucose to lactic acid. We can see that it isn't much, but the reaction proceeds rapidly again and again, producing significant amounts of energy. The alternative would be no energy at all in the absence of oxygen. Since lactic acid is normally metabolized as soon as oxygen is again available, it isn't the cause of "day after" muscle ache, as people erroneously say. It's actually the cause of burning muscles that you feel during exercise. But even then, it's an indirect effect; the lactic acid itself doesn't cause the burning, but causes formation of a flood of ATP, which hydrolyzes to give the acid which causes the |pain | 6,114 | 780 |
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The skeleton structure of a covalent molecule can often be determined by considering the valences of the constituent atoms. Usually the atom which forms the largest number of bonds is found in the center of the skeleton, where it can connect to the maximum number of other atoms. There are several possible ways to link the atoms together The usual valence of H is , and so structures 3 and 4, which have bonds to H, may be eliminated. The usual valence of Cl is also 1, and so structure 2 may also be ruled out. Structure 1 shows H forming one bond, Cl forming one, and O forming two, in agreement with the usual valences, and so it is chosen. The total number of valence electrons available is 1 from H plus 6 from O plus 7 from Cl, or 14. Filling these into the skeleton we have Note that O, which had the largest valence, is in the center of the skeleton. Draw a structural formula for hydroxylamine, NH O. In this case N has the largest valence (3), followed by O (2) and H (1). Both N and O can form “bridges” between other atoms, but H cannot. Therefore we place N and O in the center of the skeleton to give by addition of the three H atoms. There are a total of 5 + 3 + 6 = 14 valence electrons from N, 3H’s and O. These can be placed as follows: Once the Lewis diagram has been determined, the molecular formula is often rewritten to remind us of what the structural formula is. For example, the molecular formula for hydroxylamine is usually written NH OH instead of NH O to remind us that two H’s are bonded to N and one to O. It is assumed that the person reading the formula will realize that N and O each have one valence electron left to share with each other, connecting —NH with —OH. In some cases more than one skeleton structure will satisfy the valence of each atom and the octet rule as well. For example, you can verify that the molecular formula C H O corresponds to both of the following: In such a case we can only decide which molecular structure we have by experiment. The properties of ethyl alcohol when diluted with water and consumed are well known. Dimethyl ether is a gas. Like the diethyl ether used in operating rooms, it is highly explosive and can put you to sleep. Two molecules, such as dimethyl ether and ethyl alcohol, which have the same molecular formula but different structural formulas are said to be isomers. The C H O molecules and their skeletal structures were created using MolView. | 2,462 | 781 |
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