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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Difference_Between_K_And_Q |
Sometimes it is necessary to determine in which direction a reaction will progress based on initial activities or concentrations. In these situations, the relationship between the reaction quotient, \(Q_c\), and the equilibrium constant, \(K_c\), is essential in solving for the net change. With this relationship, the direction in which a reaction will shift to achieve chemical equilibrium, whether to the left or the right, can be easily calculated. \(K_c\) can be used calculate the final concentrations at for a reaction using an and the natural progression of the reaction, from left to right or from right to left. However, what if you do not know which way the reaction will progress? A simple relationship between \(K_c\) and the reaction quotient, known as \(Q_c\), can help. The reaction quotient, \(Q\), expresses the relative ratio of products to reactants at a given instant. Using either the initial concentrations or initial activities of all the components of the reaction, the progression of an reaction can easily be determined. Given the general chemical reaction \[ aA +bB \rightleftharpoons gG +hH \] \(Q\) may be expressed as the following equations: \[ Q= \dfrac{a_{init}^ga_{init}^h}{a_{init}^a a_{init}^b} \] or \[ Q_{c}= \dfrac{[G]_{init}^g[H]_{init}^h}{[A]_{init}^a[B]_{init}^b} \] Remember that the concentrations of liquids and solids do not change, so they are excluded from the expression. As shown above, the value of \(Q\) can be found by raising the products to the power of their coefficients, or stoichiometric factors, divided by the reactants raised to their coefficients. If the concentration of products in the numerator is much larger than that of reactants in the denominator, \(Q\) will be a large value. On the other hand, a small amount of products (small numerator) divided by a large value for the concentration of reactants (large denominator) would result in a small value for Q. The expressions for Q are very similar to those for \(K\): \[ K= \dfrac{a_G^g a_{H}^h}{a_{A}^a a_{B}^b} \] or \[ K_c = \dfrac{[G]^g [H]^h}{[A]^a [B]^b} \] To determine which direction a reaction will go towards, simply compare \(Q_c\), the initial concentration ratio, to \(K_c\), the equilibrium constant, and evaluate the results. When you set \(Q\) against \(K\), there are five possible relationships: To properly predict which way a reaction will progress, you must know these relationships. When Q=K, the system is at equilibrium and there is no shift to either the left or the right. Take, for example, the reversible reaction shown below: \[ CO_{(g)}+2H_{2 \; (g)} \rightleftharpoons CH_{3}OH_{(g)} \] The value of K at 483 K is 14.5. If Q=14.5, the reaction is in equilibrium and will be no evolution of the reaction either forward or backwards. When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right. Consider again: \[ CO_{(g)}+2H_{2 \; (g)} \rightleftharpoons CH_{3}OH_{(g)} \] For Q<K: \[ CO_{(g)}+2H_{2 \; (g)} \longrightarrow CH_{3}OH_{(g)} \] so that equilibrium may be established. If Q=0, then Q is less than K. Therefore, when Q=0, the reaction shifts to the right (forward). An easy way to remember this relationship is by thinking, “once you have nothing, the only thing left to do is to move forward.” If Q equals to zero, the reaction will shift forward (to the right): \[ CO_{(g)}+2H_{2 \; (g)} \longrightarrow CH_{3}OH_{(g)} \] When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K: \[ CO_{(g)}+2H_{2 \; (g)} \longleftarrow CH_{3}OH_{(g)} \] When Q=∞, the reaction shifts to the left (backwards). This is a variation of when Q>>>K. \[ CO_{(g)}+2H_{2 \; (g)} \longleftarrow CH_{3}OH_{(g)} \] An easy way to remember these relationships is by thinking of the > or < as the mouth of an alligator. The alligator will "eat" in the direction that the reaction shifts as long as \(Q\) is written before \(K\). Remembering these simple relationships will aid you to solving for the progression of a reaction. A chart outlining them can be found below. Depending on what a problem asks of you, sometimes it is unnecessary to make any calculations at all. Take, for example, the now familiar reversible reaction listed below \[ CO_{(g)}+2H_{2 \; (g)} \rightleftharpoons CH_{3}OH_{(g)} \] What do you think will happen if more of the product, methanol (CH3OH), is added? Equilibrium will be disrupted, and the increase in products mean that Q>K. In order to re-establish equilibrium, the reaction will progress to the left, towards the reactants. This means some of the added methanol will break down into carbon monoxide and hydrogen gas. \[ CO_{(g)}+2H_{2 \; (g)} \longleftarrow CH_{3}OH_{(g)} \] Now, what if more of the reactants, carbon monoxide and hydrogen gas, are a? You should realize that this would upset the equilibrium. Q<K, because the value for the amount of reactants, or the denominator of the Q expression, has increased. To establish equilibrium again, the reaction will favor the product, so the reaction will progress to the right. \[ CO_{(g)}+2H_{2 \; (g)} \longrightarrow CH_{3}OH_{(g)} \] The ideas illustrated above show whereby when an equilibrated system is subjected to a change in temperature, pressure, or concentration of a species in the reaction, the system responds by achieving a new equilibrium that partially offsets the impact of the change. Predicting which way a reaction will go can be the easiest thing that you will ever do in chemistry! To properly use the relationship between Q and K, you must know how to set it up. Take, for example, the reaction below: If you start with 4.00M CH , 2.00M C H , and 3.00M H , which direction will the reaction progress to reach equilibrium? 1) Consider this reaction: \(2NOBr_{(g)} \leftrightharpoons 2NO_{(g)}+Br_{2}\) If K = 0.0142 and the initial concentrations are 1.0 M NOBr, 0.2M NO, and 0.8M Br , which way will the reaction progress to reach equilibrium? 2) What is Q and its purpose? 3) Consider the reaction following in equilibrium: \(N_{2}O_{4 \; (g)} \leftrightharpoons 2NO_{2 \; (g)}\) If more N O is added, which way will the reaction proceed? 4) Consider the following reaction: \(CO_{(g)}+Cl_{2 \; (g)} \leftrightharpoons COCl_{2 \; (g)}\) With a K of 1.2 x 10 at 668 K, is the reaction in equilibrium when there are 5.00 mol CO(g), 2.00 mol Cl (g), and 6.00 mol of COCl (g) in a 3.00L flask? If not, which direction will the reaction progress to reach equilibrium? 5) Consider the following reaction: \(H_{2 \; (g)}+I_{2 \; (g)} \leftrightharpoons 2HI_{(g)}\) If K =50.2 at 718 K and the initial concentrations are 0.5 M H , 0.15M I and 0.05M HI, which way will the reaction progress? 6) Consider the following reaction: \(2COF_{2 \; (g)} \leftrightharpoons CO_{2 \; (g)}+CF_{4 \; (g)}\) If K = 2.00 at 473 K and the initial concentrations are 2.0 M CO , 4.0 M CF , and 0.5 M COF , which way will the reaction progress? 7) Consider the following reaction: \(2SO_{2 \; (g)}+O_{2 \: (g)} \leftrightharpoons 2SO_{3 \; (g)}\) K =100. With the initial masses of 20 g SO , 13 g O , and 25 g SO in a 5.0 L container, which way will the reaction progress. 1) The reactions shifts to the left, towards the reactants. NOBr= 1M, NO= 0.2M, Br = 0.8M \(Q_c= \dfrac {[0.2]^2 [0.8]}{[1]^2}\) \(Q_c= 0.032\) Therefore, Q > K and the reactions shifts towards the reactants. 2) Q is a reaction quotient, which helps determine if a reaction will shift forward or backwards. As a system approaches towards equilibrium, Q approaches towards K. 3) The reaction will proceed to the right. 4) No, it is not at equilibrium. Since Q<K, the reaction will shift to the right to reach equilibrium. 5) Q = 0.033, so Q<K. The reaction will shift to the right. 6) Q = 32.0, so Q>K. The reaction will shift to the left. 7) Q = 12, so Q<K. The reaction will shift to the right. | 8,013 | 1,670 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Mass_Spec/Mass_Spectrometry_-_Fragmentation_Patterns |
Following are examples of compounds listed by functional group, which demonstrate patterns which can be seen in mass spectra of compounds ionized by electron impact ionization. These examples do not provide information about the fragmentation mechanisms that cause these patterns. Additional information can be found in mass spectrometry reference books. An alcohol's molecular ion is small or non-existent. Cleavage of the C-C bond next to the oxygen usually occurs. A loss of H O may occur as in the spectra below. 3-Pentanol (C H O) with MW = 88.15 Cleavage of bonds next to the carboxyl group results in the loss of hydrogen (molecular ion less 1) or the loss of CHO (molecular ion less 29). 3-Phenyl-2-propenal (C H O) with MW = 132.16 Molecular ion peaks are present, possibly with low intensity. The fragmentation pattern contains clusters of peaks 14 mass units apart (which represent loss of (CH2)nCH3). Hexane (C6H14) with MW = 86.18 Primary amides show a base peak due to the McLafferty rearrangement. 3-Methylbutyramide (C H NO) with MW = 101.15 Molecular ion peak is an odd number. Alpha-cleavage dominates aliphatic amines. n-Butylamine (C4H11N) with MW = 73.13 Another example is a secondary amine shown below. Again, the molecular ion peak is an odd number. The base peak is from the C-C cleavage adjacent to the C-N bond. n-Methylbenzylamine (C H N) with MW = 121.18 Molecular ion peaks are strong due to the stable structure. Naphthalene (C H ) with MW = 128.17 In short chain acids, peaks due to the loss of OH (molecular ion less 17) and COOH (molecular ion less 45) are prominent due to cleavage of bonds next to C=O. 2-Butenoic acid (C H O ) with MW = 86.09 Fragments appear due to bond cleavage next to C=O (alkoxy group loss, -OR) and hydrogen rearrangements. Ethyl acetate (C H O ) with MW = 88.11 Fragmentation tends to occur alpha to the oxygen atom (C-C bond next to the oxygen). Ethyl methyl ether (C H O) with MW = 60.10 The presence of chlorine or bromine atoms is usually recognizable from isotopic peaks. 1-Bromopropane (C H Br) with MW = 123.00 Major fragmentation peaks result from cleavage of the C-C bonds adjacent to the carbonyl. 4-Heptanone (C H O) with MW = 114.19 Dr. Linda Breci, Associate Director | 2,267 | 1,673 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Arrhenius_Concept_of_Acids_and_Bases |
The Arrhenius definition can be summarized as "Arrhenius acids form hydrogen ions in aqueous solution with Arrhenius bases forming hydroxide ions." In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. As defined by Arrhenius, acid-base reactions are characterized by acids, which dissociate in aqueous solution to form hydrogen ions (H ) and bases, which form hydroxide (OH ) ions. Acids are defined as a compound or element that releases hydrogen (H ) ions into the solution (mainly water). \[ NHO_3 (aq) + H_2O(l) \rightarrow H_3O^+ + NO_3^- (aq)\] In this reaction nitric acid (HNO ) disassociates into hydrogen (H ) and nitrate (NO ) ions when dissolved in water. Bases are defined as a compound or element that releases hydroxide (OH ) ions into the solution. In this reaction lithium hydroxide (LiOH) dissociates into lithium (Li ) and hydroxide (OH ) ions when dissolved in water.
| 1,061 | 1,676 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/07._Angular_Momentum/7.5%3A_Rigid_Rotor |
Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the of the molecules with the electromagnetic field of the exciting microwave photon. To probe the pure rotational transitions for molecules, scientists use microwave rotational spectroscopy. This spectroscopy utilizes photons in the microwave range to cause transitions between the quantum rotational energy levels of a gas molecule. The reason why the sample must be in the gas phase is due to intermolecular interactions hindering rotations in the liquid and solid phases of the molecule. For microwave spectroscopy, molecules can be broken down into 5 categories based on their shape and the inertia around their 3 orthogonal rotational axes. These 5 categories include diatomic molecules, linear molecules, spherical tops, symmetric tops and asymmetric tops. The to the rigid rotor is \[H = T\] since, \[H = T + V\] Where \(T\) is kinetic energy and \(V\) is potential energy. Potential energy, \(V\), is 0 because there is no resistance to the rotation (similar to a model). Since \(H = T\), we can also say that: \[{T = }\dfrac{1}{2}\sum{m_{i}v_{i}^2}\] However, we have to determine \(v_i\) in terms of rotation since we are dealing with rotation. Since, \[\omega = \dfrac{v}{r}\] where \(\omega\) = angular velocity, we can say that: \[v_{i} = \omega{X}r_{i}\] Thus we can rewrite the T equation as: \[T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)}\] Since \(\omega\) is a scalar constant, we can rewrite the T equation as: \[T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2}\] where \(l_i\) is the of the particle, and is the angular momentum of the . Also, we know from physics that, \[L = I\omega\] where is the moment of inertia of the rigid body relative to the axis of rotation. W \[T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2\] \[E_n = \dfrac{J(J+1)h^2}{8\pi^2I} \] and \(\hbar\) However, if we let: \[B = \dfrac {h}{8 \pi^2I} \] where \(B\) is a rotational constant, then we can substitute it into the \(E_n\) equation and get: \[E_{n} = J(J+1)Bh\] Considering the transition energy between two energy levels, the difference is a multiple of 2. That is, from \(J = 0\) to \(J = 1\), the \(\Delta{E_{0 \rightarrow 1}}\) is 2Bh and from J = 1 to J = 2, the \(\Delta{E}_{1 \rightarrow 2}\) is 4Bh. When a gas molecule is irradiated with microwave radiation, a photon can be absorbed through the interaction of the photon’s electronic field with the electrons in the molecules. For the microwave region this energy absorption is in the range needed to cause transitions between rotational states of the molecule. However, only molecules with a permanent dipole that changes upon rotation can be investigated using microwave spectroscopy. This is due to the fact that their must be a charge difference across the molecule for the oscillating electric field of the photon to impart a torque upon the molecule around an axis that is perpendicular to this dipole and that passes through the molecules center of mass. This interaction can be expressed by the transition dipole moment for the transition between two rotational states \[ \text{Probability of Transition}=\int \psi_{rot}(F)\hat\mu \psi_{rot}(I)d\tau\] Where Ψ is the complex conjugate of the wave function for the final rotational state, Ψ is the wave function of the initial rotational state , and μ is the dipole moment operator with Cartesian coordinates of μ , μ , μ . For this integral to be nonzero the integrand must be an even function. This is due to the fact that any odd function integrated from negative infinity to positive infinity, or any other symmetric limits, is always zero. In addition to the constraints imposed by the transition moment integral, transitions between rotational states are also limited by the nature of the photon itself. A photon contains one unit of angular momentum, so when it interacts with a molecule it can only impart one unit of angular momentum to the molecule. This leads to the selection rule that a transition can only occur between rotational energy levels that are only one quantum rotation level (J) away from another . \[ \Delta\textrm{J}=\pm 1 \] The transition moment integral and the selection rule for rotational transitions tell if a transition from one rotational state to another is allowed. However, what these do not take into account is whether or not the state being transitioned from is actually populated, meaning that the molecule is in that energy state. This leads to the concept of the Boltzmann distribution of states. The is a statistical distribution of energy states for an ensemble of molecules based on the temperature of the sample . \[ \dfrac{n_J}{n_0} = \dfrac{e^{(-E_{rot}(J)/RT)}}{\sum_{J=1}^{J=n} e^{(-E_{rot}(J)/RT)}}\] where E is the molar energy of the J rotational energy state of the molecule, This distribution of energy states is the main contributing factor for the observed absorption intensity distributions seen in the microwave spectrum. This distribution makes it so that the absorption peaks that correspond to the transition from the energy state with the largest population based on the Boltzmann equation will have the largest absorption peak, with the peaks on either side steadily decreasing. A molecule can have three types of degrees of freedom and a total of 3N degrees of freedom, where N equals the number of atoms in the molecule. These degrees of freedom can be broken down into 3 categories . Each of these degrees of freedom is able to store energy. However, In the case of rotational and vibrational degrees of freedom, energy can only be stored in discrete amounts. This is due to the quantized break down of energy levels in a molecule described by quantum mechanics. In the case of rotations the energy stored is dependent on the rotational inertia of the gas along with the corresponding quantum number describing the energy level. To analyze molecules for rotational spectroscopy, we can break molecules down into 5 categories based on their shapes and their moments of inertia around their 3 orthogonal rotational axes: The rotations of a diatomic molecule can be modeled as a rigid rotor. This rigid rotor model has two masses attached to each other with a fixed distance between the two masses. It has an inertia (I) that is equal to the square of the fixed distance between the two masses multiplied by the reduced mass of the rigid rotor. \[ \large I_e= \mu r_e^2\] \[ \large \mu=\dfrac{m_1 m_2} {m_1+m_2} \] Using quantum mechanical calculations it can be shown that the energy levels of the rigid rotator depend on the inertia of the rigid rotator and the quantum rotational number J . \[ E(J) = B_e J(J+1) \] \[ B_e = \dfrac{h}{8 \pi^2 cI_e}\] However, this rigid rotor model fails to take into account that bonds do not act like a rod with a fixed distance, but like a spring. This means that as the angular velocity of the molecule increases so does the distance between the atoms. This leads us to the in which a centrifugal distortion term (\(D_e\)) is added to the energy equation to account for this stretching during rotation. \[ E(J)(cm^{-1}) = B_e J(J+1) – D_e J^2(J+1)^2\] This means that for a diatomic molecule the transitional energy between two rotational states equals \[ E=B_e[J'(J'+1)-J''(J''+1)]-D_e[J'^2(J'+1)^2-J''^2(J'+1)^2]\label{8} \] Where J’ is the quantum number of the final rotational energy state and J’’ is the quantum number of the initial rotational energy state. Using the selection rule of \(\Delta{J}= \pm 1\) the spacing between peaks in the microwave absorption spectrum of a diatomic molecule will equal \[ E_R =(2B_e-4D_e)+(2B_e-12D_e){J}''-4D_e J''^3\] Linear molecules behave in the same way as diatomic molecules when it comes to rotations. For this reason they can be modeled as a non-rigid rotor just like diatomic molecules. This means that linear molecule have the same equation for their rotational energy levels. The only difference is there are now more masses along the rotor. This means that the inertia is now the sum of the distance between each mass and the center of mass of the rotor multiplied by the square of the distance between them . \[ \large I_e=\sum_{j=1}^{n} m_j r_{ej}^2 \] Where m is the mass of the jth mass on the rotor and r is the equilibrium distance between the j mass and the center of mass of the rotor. Spherical tops are molecules in which all three orthogonal rotations have equal inertia and they are highly symmetrical. This means that the molecule has no dipole and for this reason spherical tops do not give a microwave rotational spectrum. Examples: Symmetrical tops are molecules with two rotational axes that have the same inertia and one unique rotational axis with a different inertia. Symmetrical tops can be divided into two categories based on the relationship between the inertia of the unique axis and the inertia of the two axes with equivalent inertia. If the unique rotational axis has a greater inertia than the degenerate axes the molecule is called an oblate symmetrical top. If the unique rotational axis has a lower inertia than the degenerate axes the molecule is called a prolate symmetrical top. For simplification think of these two categories as either frisbees for oblate tops or footballs for prolate tops. In the case of linear molecules there is one degenerate rotational axis which in turn has a single rotational constant. With symmetrical tops now there is one unique axis and two degenerate axes. This means an additional rotational constant is needed to describe the energy levels of a symmetrical top. In addition to the rotational constant an additional quantum number must be introduced to describe the rotational energy levels of the symmetric top. These two additions give us the following rotational energy levels of a prolate and oblate symmetric top \[E_{(J,K)}(cm^{-1})=B_e*J(J+1)+(A_e-B_e)K^2 \] Where B is the rotational constant of the unique axis, A is the rotational constant of the degenerate axes, \(J\) is the total rotational angular momentum quantum number and K is the quantum number that represents the portion of the total angular momentum that lies along the unique rotational axis. This leads to the property that \(K\) is always equal to or less than \(J\). Thus we get the two selection rules for symmetric tops \[\Delta J = 0, \pm1 \] \[\Delta K=0\] when \( K\neq 0 \) \[\Delta J = \pm1 \] \[\Delta K=0 \] when \( K=0 \) However, like the rigid rotor approximation for linear molecules, we must also take into account the elasticity of the bonds in symmetric tops. Therefore, in a similar manner to the rigid rotor we add a centrifugal coupling term, but this time we have one for each quantum number and one for the coupling between the two. \[E_{(J,K)}(cm^{-1})=B_e J(J+1)-D_{eJ} J^2(J+1)^2+(A_e-B_e)*K^2 \label{13}\] \[-D_{ek} K^4-D_{ejk} J(J +1)K^2 \label{14}\] Asymmetrical tops have three orthogonal rotational axes that all have different moments of inertia and most molecules fall into this category. Unlike linear molecules and symmetric tops these types of molecules do not have a simplified energy equation to determine the energy levels of the rotations. These types of molecules do not follow a specific pattern and usually have very complex microwave spectra. In addition to microwave spectroscopy, IR spectroscopy can also be used to probe rotational transitions in a molecule. However, in the case of IR spectroscopy the rotational transitions are coupled to the vibrational transitions of the molecule. One other spectroscopy that can probe the rotational transitions in a molecule is Raman spectroscopy, which uses UV-visible light scattering to determine energy levels in a molecule. However, a very high sensitivity detector must be used to analyze rotational energy levels of a molecule. | 12,069 | 1,678 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Stability_of_Metal_Complexes_and_Chelation |
Ligands like chloride, water, and ammonia are said to be monodentate (one-toothed, from the Greek mono, meaning “one,” and the Latin dent-, meaning “tooth”): they are attached to the metal via only a single atom. Ligands can, however, be bidentate (two-toothed, from the Greek di, meaning “two”), tridentate (three-toothed, from the Greek tri, meaning “three”), or, in general, polydentate (many-toothed, from the Greek poly, meaning “many”), indicating that they are attached to the metal at two, three, or several sites, respectively. Ethylenediamine (H NCH CH NH , often abbreviated as en) and diethylenetriamine (H NCH CH NHCH CH NH , often abbreviated as dien) are examples of a bidentate and a tridentate ligand, respectively, because each nitrogen atom has a lone pair that can be shared with a metal ion. When a bidentate ligand such as ethylenediamine binds to a metal such as Ni , a five-membered ring is formed. A metal-containing ring like that shown is called a chelate ring (from the Greek chele, meaning “claw”). Correspondingly, a polydentate ligand is a chelating agent, and complexes that contain polydentate ligands are called chelate complexes. Experimentally, it is observed that metal complexes of polydentate ligands are significantly more stable than the corresponding complexes of chemically similar monodentate ligands; this increase in stability is called the chelate effect. For example, the complex of Ni with three ethylenediamine ligands, [Ni(en) ] , should be chemically similar to the Ni complex with six ammonia ligands, [Ni(NH ) ] . In fact, the equilibrium constant for the formation of [Ni(en) ] is almost 10 orders of magnitude larger than the equilibrium constant for the formation of [Ni(NH ) ] ( ): Chelate complexes are more stable than the analogous complexes with monodentate ligands. The stability of a chelate complex depends on the size of the chelate rings. For ligands with a flexible organic backbone like ethylenediamine, complexes that contain five-membered chelate rings, which have almost no strain, are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings. For example, the complex of nickel (II) with three ethylenediamine ligands is about 363,000 times more stable than the corresponding nickel (II) complex with trimethylenediamine (H NCH CH CH NH , abbreviated as tn): \(\begin{align} & \mathrm{[Ni(H_2O)_6]^{2+}+3en} & & \rightleftharpoons \mathrm{[Ni(en)_3]^{2+}+6H_2O(l)} & K_\mathrm f=6.76 \times 10^{17} \\ & \mathrm{[Ni(H_2O)_6]^{2+}+3tn} & & \rightleftharpoons \mathrm{[Ni(tn)_3]^{2+}+6H_2O(l)} & K_\mathrm f=1.86 \times 10^{12}\end{align} \label{23.11}\) *The above measurements were done in a solution of ionic strength 0.15 at 25º C. Arrange [Cr(en) ] , [CrCl ] , [CrF ] , and [Cr(NH ) ] in order of increasing stability. : four Cr(III) complexes relative stabilities A Determine the relative basicity of the ligands to identify the most stable complexes. B Decide whether any complexes are further stabilized by a chelate effect and arrange the complexes in order of increasing stability. A The metal ion is the same in each case: Cr . Consequently, we must focus on the properties of the ligands to determine the stabilities of the complexes. Because the stability of a metal complex increases as the basicity of the ligands increases, we need to determine the relative basicity of the four ligands. Our earlier discussion of acid–base properties suggests that ammonia and ethylenediamine, with nitrogen donor atoms, are the most basic ligands. The fluoride ion is a stronger base (it has a higher charge-to-radius ratio) than chloride, so the order of stability expected due to ligand basicity is [CrCl ] < [CrF ] < [Cr(NH ) ] ≈ [Cr(en) ] . B Because of the chelate effect, we expect ethylenediamine to form a stronger complex with Cr than ammonia. Consequently, the likely order of increasing stability is [CrCl ] < [CrF ] < [Cr(NH ) ] < [Cr(en) ] . Arrange [Co(NH ) ] , [CoF ] , and [Co(en) ] in order of decreasing stability. : [Co(en) ] > [Co(NH ) ] > [CoF ] The chelate effect can be seen by comparing the reaction of a chelating ligand and a metal ion with the corresponding reaction involving comparable monodentate ligands. For example, comparison of the binding of 2,2'-bipyridine with pyridine or 1,2-diaminoethane (ethylenediamine=en) with ammonia. It has been known for many years that a comparison of this type always shows that the complex resulting from coordination with the chelating ligand is . This can be seen by looking at the values for adding two monodentates compared with adding one bidentate, or adding four monodentates compared to two bidentates, or adding six monodentates compared to three bidentates. The Chelate Effect is that complexes resulting from coordination with the chelating ligand is much more thermodynamically stable than complexes with non-chelating ligands. A number of points should be highlighted from the formation constants in . In Table \(\Page {1}\), it can be seen that the ΔH° values for the formation steps are almost identical, that is, heat is evolved to about the same extent whether forming a complex involving monodentate ligands or bidentate ligands. What is seen to vary significantly is the ΔS° term which changes from negative (unfavorable) to positive (favorable). Note as well that there is a dramatic increase in the size of the ΔS° term for adding two compared to adding four monodentate ligands. (-5 to -35 JK mol ). What does this imply, if we consider ΔS° to give a measure of disorder? In the case of complex formation of Ni with ammonia or 1,2-diaminoethane, by rewriting the equilibria, the following equations are produced. Using the equilibrium constant for the reaction (3 above) where the bidentate ligands replace the monodentateligands, we find that at a temperature of 25° C: \[\Delta G^° = -2.303 \, RT \, \log_{10} K\] \[= -2.303 \,RT \,(18.28 - 8.61)\] \[= -54 \text{ kJ mol}^{-1}\] Based on measurements made over a range of temperatures, it is possible to break down the \(\Delta G^°\) term into the enthalpy and entropy components. \[\Delta G^° = \Delta H^° - T \Delta S^°\] The result is that: \(\Delta H^° = -29 kJ mol^{-1}\) - TΔS° = -25 kJ mol
and at 25C (298K)
ΔS° = +88 J K mol For many years, these numbers have been in textbooks. For example, the third edition of "Basic Inorganic Chemistry" by F.A. Cotton, G. Wilkinson and P.L. Gaus, John Wiley & Sons, Inc, 1995, on page 186 gives the values as: ΔG° = -67 kJ mol
ΔH° = -12 kJ mol
-TΔS° = -55 kJ mol The conclusion they drew from these incorrect numbers was that the chelate effect was essentially an entropy effect, since the TΔS° contribution was nearly 5 times bigger than ΔH°. In fact, the breakdown of the ΔG° into ΔH° and TΔS° shows that the two terms are nearly equal (-29 and -25 kJ mol , respectively) with the ΔH° term a little bigger! The entropy term found is still much larger than for reactions involving a non-chelating ligand substitution at a metal ion. How can we explain this enhanced contribution from entropy? One explanation is to count the number of species on the left and right hand side of the equation above. It will be seen that on the left-hand-side there are 4 species, whereas on the right-hand-side there are 7 species, that is a net gain of 3 species occurs as the reaction proceeds. This can account for the increase in entropy since it represents an increase in the disorder of the system. An alternative view comes from trying to understand how the reactions might proceed. To form a complex with 6 monodentates requires 6 separate favorable collisions between the metal ion and the ligand molecules. To form the tris-bidentate metal complex requires an initial collision for the first ligand to attach by one arm but remember that the other arm is always going to be nearby and only requires a rotation of the other end to enable the ligand to form the chelate ring. If you consider dissociation steps, then when a monodentate group is displaced, it is lost into the bulk of the solution. On the other hand, if one end of a bidentate group is displaced the other arm is still attached and it is only a matter of the arm rotating around and it can be reattached again. conditions favor the formation of the complex with bidentate groups rather than monodentate groups. | 8,477 | 1,679 |
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The previous Essential Skills sections introduced some fundamental operations that you need to successfully manipulate mathematical equations in chemistry. This section describes how to convert between temperature scales and further develops the topic of unit conversions started in Essential Skills 2 . The concept of temperature may seem familiar to you, but many people confuse temperature with heat. is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas is the flow of thermal energy between objects with different temperatures. Science only uses SI and derived units with few if any exceptions. Therefore this text avoids conversions between customary units (foot, pound, Farenheit) that are simply inappropriate and only used in a few countries. Two different scales are commonly used to measure temperature: Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes. The Celsius scale was developed in 1742 by the Swedish astronomer Anders Celsius. It is based on the melting and boiling points of water under normal atmospheric conditions. The current scale is an inverted form of the original scale, which was divided into 100 increments. Because of these 100 divisions, the Celsius scale is also called the . Lord Kelvin, working in Scotland, developed the Kelvin scale in 1848. His scale uses molecular energy to define the extremes of hot and cold. Absolute zero, or 0 K, corresponds to the point at which molecular energy is at a minimum. The Kelvin scale is preferred in scientific work, although the Celsius scale is also commonly used. Temperatures measured on the Kelvin scale are reported simply as K, not °K. The kelvin is the same size as the Celsius degree, so measurements are easily converted from one to the other. The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K. The Kelvin and Celsius scales are related as follows: \[ T \left ( in ^{o}C \right ) + 273.15 = T \left( in\; K \right ) \] \[ T \left( in\; K \right ) - 273.15 = T \left ( in ^{o}C \right ) \] Liquid Gold: 3080 K = (3080 -273.15) C =2807 C Liquid N 77.36 K = (77.36 -274.15) C = -195.79 C In Essential Skills 2, you learned a convenient way of converting between units of measure, such as from grams to kilograms or seconds to hours. The use of units in a calculation to ensure that we obtain the final proper units is called . For example, if we observe experimentally that an object’s potential energy is related to its mass, its height from the ground, and to a gravitational force, then when multiplied, the units of mass, height, and the force of gravity must give us units corresponding to those of energy. Energy is typically measured in joules, calories, or electron volts (eV), defined by the following expressions: \[ 1\;J = 1\;\left ( kg\cdot m^{2} \right )/s^{2} = 1\;coulomb\cdot volt \] \[ 1\;cal = 4.184\;J \] \[ 1 eV = 1.602 \times 10^{-19} J \] To illustrate the use of dimensional analysis to solve energy problems, let us calculate the kinetic energy in joules of a 320 g object traveling at 123 cm/s. To obtain an answer in joules, we must convert grams to kilograms and centimeters to meters. Using Equation 5.1.5, the calculation may be set up as follows: \[ KE=\dfrac{1}{2}mv^{2} =\dfrac{1}{2}\left ( g \right )\left [ \left ( \dfrac{cm}{s} \right )\left ( \dfrac{cm}{s} \right ) \right ]^{2} \] \[ =\left ( \cancel{g} \right )\left ( \dfrac{kg}{\cancel{g}} \right )\left ( \dfrac{\cancel{cm^{2}}}{s^{2}} \right ) \left ( \dfrac{m^{2}}{\cancel{cm^{2}}} \right )=\dfrac{kg\cdot m^{2}}{s^{2}} \] \[ =\dfrac{1}{2}320 \; \cancel{g} \left ( \dfrac{1 \; kg}{1000 \;\cancel{g}} \right )\left [ \left ( \dfrac{123 \cancel{cm}}{1 \; s} \right ) \left ( \dfrac{1 \; m}{100 \;\cancel{cm}} \right ) \right ]^{2} =\dfrac{0.320 \; kg}{2}\left [ \dfrac{123 \; m}{s\left ( 100 \right )} \right ]^{2} \] \[ = 0.242 \; kg\cdot m^{2}/s^{2} = 0.242 \; J \] Alternatively, the conversions may be carried out in a stepwise manner: \[ 320 \; \cancel{g}\left ( \dfrac{1 \; kg}{1000 \; \cancel{g}} \right )= 0.320 \; kg\] \[ 123 \; \cancel{cm}\left ( \dfrac{1 \; m}{100 \; \cancel{cm}} \right )= 1.23 \; m\] \[ KE=\dfrac{1}{2}mv^{2} =\dfrac{1}{2}\left ( g \right )\left [ \left ( \dfrac{cm}{s} \right )\left ( \dfrac{cm}{s} \right ) \right ]^{2} \] \[ KE=\dfrac{1}{2} \left ( 0.320 kg \right )\left ( \dfrac{1.23 \; m}{s} \right )^{2} = 0.242 \left ( kg\cdot m^{2}/s^{2} \right )=0.242 \; J \] However, this second method involves an additional step A small discrepancy between answers using the different methods might be expected due to rounding to the correct number of significant figures for each step when carrying out the calculation in a stepwise manner. Recall that all digits in the calculator should be carried forward when carrying out a calculation using multiple steps. In this problem, we first converted kilocalories to kilojoules and then converted ounces to grams. Skill Builder ES2 allows you to practice making multiple conversions between units in a single step. | 5,382 | 1,681 |
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Using , calculate the standard reaction Gibbs functions (\(\ G^o\)) for the following reactions at 298 K. Estimate \(\ G\) at 1000 K from its value at 298 K for the reaction \[C(s) + 2 H_2(g) \rightarrow CH_4(g)\] with \(\Delta G = -50.75\, kJ \,at\, 298\, K\)/ The standard Gibbs function for formation (\(\Delta G_f^o\)) of \(PbO_2(s)\) is -217.4 kJ/mol at 298 K. Assuming \(O_2\) is an ideal gas, find the standard Helmholtz function for formation (\(\Delta A_f^o\) for \(PbO_2\) at 298K. Calculate the entropy change for 1.00 mol of an ideal monatomic gas (C = 3/2 R) undergoing an expansion and simultaneous temperature increase from 10.0 L at 298 K to 205.0 L at 455 K. Consider a gas that obeys the equation of state \[ p =\dfrac{nRT}{V-nb}\] Show that \[\left( \dfrac{\partial C_p}{\partial p} \right)_T=0\] for an ideal gas. Derive the thermodynamic equation of state \[\left( \dfrac{\partial H}{\partial p} \right)_T = V( 1- T \alpha)\] Derive the thermodynamic equation of state \[\left( \dfrac{\partial U}{\partial V} \right)_T = T \dfrac{ \alpha}{\kappa_T} -p\] The “Joule Coefficient” is defined by \[ \mu_J = \left( \dfrac{\partial T}{\partial V} \right)_U \] Show that \[ \mu_J = \dfrac{1}{C_V} \left( p - \dfrac{T \alpha}{\kappa_T }\right)\] and evaluate the expression for an ideal gas. Derive expressions for the pressure derivatives \[ \left( \dfrac{\partial X}{\partial p} \right)_T\] where\(X\) is \(U\), \(H\), \(A\), \(G\), and \(S\) at constant temperature in terms of measurable properties. (The derivation of \( \left( \dfrac{\partial H}{\partial p} \right)_T\) was done in problem Q6.7). Evaluate the expressions for for a van der Waals gas. Derive expressions for the volume derivatives \[ \left( \dfrac{\partial X}{\partial V} \right)_T\] where \(X\) is \(U\), \(H\), \(A\), \(G\), and \(S\) at constant temperature in terms of measurable properties. (The derivation of \( \left( \dfrac{\partial U}{\partial V} \right)_T\) was done in problem Q8.8.) Evaluate the expressions for for a van der Waals gas. Evaluate the difference between \(C_p\) and \(C_V\) for a gas that obeys the equation of state \[ p =\dfrac{nRT}{V-nb}\] The adiabatic compressibility (\(k_S\)) is defined by \[ \kappa_S = \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_S\] Show that for an ideal gas, \[ \kappa_S = \dfrac{1}{p \gamma}\] | 2,365 | 1,682 |
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Since the equilibrium constant \(K_p\) is a function of \(\Delta G^o_{rxn}\) which is defined for a specific composition (all reactants in their standard states and at unit pressure (or fugacity), changes in pressure have no effect on equilibrium constants for a fixed temperature. However, changes in pressure can have profound effects on the compositions of equilibrium mixtures. To demonstrate the relationship, one must recall Dalton’s law of partial pressures. According to this relationship, the partial pressure of a component of a gas-phase mixture can be expressed \[ p_i = \chi_t p_{tot} \nonumber \] It is the combination of mole fractions that describes the composition of the equilibrium mixture. Substituting the above expression into the expression for \(K_p\) yields \[ K_p = \prod_i (\chi_ip_{tot})^{\nu_i} \nonumber \] This expression can be factored into two pieces – one containing the mole fractions and thus describing the composition, and one containing the total pressure. \[ K_p =\left( \prod_i \chi_i^{\nu_i} \right) \left( \prod_i p_{tot}^{\nu_i} \right) \nonumber \] The second factor is a constant for a given total pressure. If the first term is given the symbol \(K_x\), the expression becomes \[ K_p=K_x (p_{tot})^{\sum_i \nu_i} \nonumber \] In this expression, \(K_x\) has the same form as an equilibrium constant \[ K_x = \prod \chi_i^{\sum_i \nu_i} \nonumber \] but is not itself a constant. The value of\(K_x\) will vary with varying composition, and will to vary with varying total pressure (in most cases) in order to maintain a constant value of \(K_p\). Consider the following reaction at equilibrium. \[A(g) + 2 B(g) \rightleftharpoons C(g) + D(g) \nonumber \] In which direction will the equilibrium shift if the volume of the reaction vessel is decreased? A decrease in the volume will lead to an increase in total pressure. Since the equilibrium constant can be expressed as \[ K_p = \dfrac{p_c p_D}{p_A p_B^2} = \dfrac{\chi_p \chi_D}{\chi_A \chi_B^2} (p_{tot})^{-1} \nonumber \] An increase in pressure will lead to an increase in \(K_x\) to maintain a constant value of \(K_p\). So the reaction will shift to form more of the products \(C\) and \(D\). : This should make some sense, since a shift to the side of the reaction with fewer moles of gas will lower the total pressure of the reaction mixture, and thus relieving the stress introduced by increasing the pressure. This is exactly what is expected according to Le Chatelier's principle. It should be noted that there are several ways one can affect the total pressure of a gas-phase equilibrium. These include the introduction or removal of reactants or products (perhaps through condensation or some other physical process), a change in volume of the reaction vessel, or the introduction of an inert gas that does not participate in the reaction itself. (Changes in the temperature will be discussed in a later section.) The principle of Le Chatelier's can be used as a guide to predict how the equilibrium composition will respond to a change in pressure. Le Chatelier's principle: When a stress is introduced to a system at equilibrium, the system will adjust so as to reduce the stress. Le Chatlier’s principle is fairly clear on how to think about the addition or removal of reactants or products. For example, the addition of a reactant will cause the system to shift to reduce the partial pressure of the reactant. It can do this by forming more products. An important exception to the rule that increasing the total pressure will cause a shift in the reaction favoring the side with fewer moles of gas occurs when the total pressure is increased by introducing an inert gas to the mixture. The reason is that the introduction of an inert gas will affect the total pressures the partial pressures of each individual species. A 1.0 L vessel is charged with 1.00 atm of A, and the following reaction is allowed to come to equilibrium at 298 K. \[A(g) \rightleftharpoons 2 B(g) \nonumber \] with \(K_p = 3.10\). First, we can use an ICE table to solve part a). So (for convenience, consider \(K_p\) to have units of atm) \[ 3.10 \,atm = \dfrac{(2x)^2}{1.00 \,atm - x} \nonumber \] Solving for \(x\) yields values of \[x_1= -1.349 \,atm \nonumber \] \[x_1= 0.574 \,atm \nonumber \] Clearly, \(x_1\), while a solution to the mathematical problem, is not physically meaningful since the equilibrium pressure of B cannot be negative. So the equilibrium partial pressures are given by \[ p_A = 1.00 \,atm - 0.574\, atm = 0.426 \,atm \nonumber \] \[ p_B = 2(0.574 \,atm) = 1.148 \,atm \nonumber \] So the mole fractions are given by \[\ chi_A = \dfrac{0.426 \,atm}{0.426\,atm + 1.148\,atm} = 0.271 \nonumber \] \[ \chi_B=1-\chi_A = 1-0.271 = 0.729 \nonumber \] The volume is doubled. Again, an ICE table is useful. The initial pressures will be half of the equilibrium pressures found in part a). So the new equilibrium pressures can be found from \[ 3.10 \,atm = \dfrac{(0.574\,atm + 2x)^2}{0.213\,atm - x} \nonumber \] And the values of \(x\) that solve the problem are \[x_1= -1.4077 \,atm \nonumber \] \[x_1= 0.05875 \,atm \nonumber \] We reject the negative root (since it would cause both of the partial pressures to become negative. So the new equilibrium partial pressures are \[ p_A = 0.154\, atm \nonumber \] \[ p_B = 0.0692\, atm \nonumber \] And the mole fractions are \[\chi_A = 0.182 \nonumber \] \[\chi_B = 0.818 \nonumber \] We can see that the mole fraction of \(A\) decreased and the mole fraction \(B\) increased. This is the result expected by Le Chatlier’s principle since the lower total pressure favors the side of the reaction with more moles of gas. We introduce 1.000 atm of an inert gas. The new partial pressures are \[p_A = 0.154 \,atm \nonumber \] \[p_B = 0.692 \,atm \nonumber \] \[p_{Ar} = 1.000\, atm \nonumber \] And because the partial pressures of A and B are unaffected, the equilibrium does not shift! What affected is the composition, and so the mole fractions will change. \[ \chi_A = \dfrac{0.154 \,atm}{0.154 atm + 0.692 \,atm + 1.000\, atm} = 0.08341 \nonumber \] \[ \chi_B = \dfrac{0.692 \,atm}{0.154 atm + 0.692 \,atm + 1.000\, atm} = 0.08341 \nonumber \] \[ \chi_{Ar} = \dfrac{1.000\, atm}{0.154 atm + 0.692 \,atm + 1.000\, atm} = 0.08341 \nonumber \] And since \[ K_p = K_x(p_{tot}) \nonumber \] \[\dfrac{(0.3749)^2}{0.08342} (1.846\,atm) = 3.1 \nonumber \] Within round-off error, the value obtained is the equilibrium constant. So the conclusion is that the introduction of an inert gas, even though it increases the total pressure, does not induce a change in the partial pressures of the reactants and products, so it does not cause the equilibrium to shift. ICE is an acronym for “Initial, Change, Equilibrium”. An ICE table is a tool that is used to solve equilibrium problems in terms of an unknown number of moles (or something proportional to moles, such as pressure or concentration) will shift for a system to establish equilibrium. See (Tro, 2014) or a similar General Chemistry text for more background and information. | 7,101 | 1,683 |
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Draw resonance structures for NO3- and ClO3-, please show formal charges in drawings. Calculate the dipole moment (\(\mu\)) of a two ions of +2e and -2e that are separated by 100 pm. Express dipole moment in both units of \(C\,m\) and \(D\). \[ \mu = Q \times r = 2 ( 1.602 \times 10^{-19})( 8 \times 10^{-10}) = 2.56 \times 10^{-28} C\,m\] \[(2.56 \times 10^{-28} \cancel{C\,m}) \left( \dfrac{1\,D}{3.336 \times 10^{-30} \cancel{C\,m}} \right) = 76.8 \;D\] What are the resonance of \(NO_3^0\)? Resonance structure in basic means that there are multiple way to draw a Lewis structure. For example: *Source from Michael Blaber Draw resonance structures for the carbonate ion (\(CO_3^{-2}\)). An ionic bond wave function takes into account the probability that the electrons of a multiatomic molecule might exist on the same atom; the covalent bond wave function assumes they exist on separate atoms. What is the hybridization of HCl? \[\mu = 2.5 D \times \dfrac{3.3356\times 10^{30}C m}{D} = 8.39 \times 10^{-30}Cm\] \[r = 5.5 \; angstroms= 5.5 \times 10^{-10}m\] \[V = \dfrac{q\mu}{4\pi \varepsilon _{o}r^{2}} = \dfrac{(+1.602\times10^{-19}C)(8.39\times 10^{-30}Cm)}{4\pi (8.851\times 10^{-12}C^{-2}N^{-1}m^{-2})(5.5\times 10^{-10})} = 2.196\times 10^{-11} J\] \(F_2^+\) has bond order 1.5, \(F_2\) has bond order 1.0. The bond lengths are inverse to the bond order, so the order is F2+ < F2 .There is 1 unpaired electron in F2+, 0 unpaired electrons in F2 BOND ORDER OF F = (8-6)/2 =1 BOND ORDER OF F (8-5)/2 =1.5 Based on bond order F is stable than F Diatomic carbon (C ) exists at very high temperatures. What type or types of bonds do you expect a molecule of diatomic carbon to have? 1. Draw a molecular orbital diagram. Answer: The molecular orbital diagram predicts that there will be a double bond consisting of two π bonds. Based on general diagram for molecular orbital of second row of O molecule, Though there are some exceptions, the double bond rule states that atoms of period 3 and greater do not form double or triple bonds. Explain why this is generally true. Pi bonds are usually weaker than sigma bonds because there is less orbital overlap. The larger atomic radius of period 3 and greater elements means that there is even less overlap between the orbitals that would form pi bonds. There is not enough overlap to form the pi bonds that would make for double or triple bonds. Myohemerythrin is an iron-containing protein that binds oxygen in marine invertebrates. The protein is monomeric (has one subunit) and contains 0.8107% iron by mass. Estimate the molar mass of myohemerythrin. Is your answer the same as the known molar mass of myohemerythrin, 13,780 grams per mole? If it is not, account for the difference between your estimate and the known value. 1. Find the mass of the protein if there is one atom of iron per protein. \[ 55.85\ g\ Fe/mol \times \dfrac{100.0\ g\ protein}{0.8107\ g\ Fe}\] \[ 6889 g/mol \] 2. This is much less than the known molar mass. In fact, it is half. \[ \dfrac{13780\ g/mol}{6889\ g/mol} = 2.000 \] We know the protein is monomeric, so there must be two atoms of iron in each protein. Based on the molecular orbital diagram, there are more anti-bonding than bonding state. The bonding order of Ne is: BO = 1/2 (8-8) = 0 Therefore, \(Ne_2\) does not exist in nature. Will the dipole moment of a cis-dichloroethylene molecule increase or decrease upon heating? Why? The dipole moment will decrease because cis-trans isomerization will occur when heated. This is because the trans isomer of the molecule does not have a dipole moment present. Draw the \(\sigma\) and \(\sigma^*\) molecular orbital of \(CO\). Draw the MO energy level diagram and write the electron coefficient. Calculate the Bond order. Electron Configuration: ( 1s) ( 1s) ( 2s) ( 2s) ( x) ( y) ( 2sp) | 3,855 | 1,684 |
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In the examples in this text, the exponents in the rate law are almost always the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is rate = . We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
\( rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=-k\left [ reactant \right ]^{0}=k(1)=k \tag{14.3.1} \) Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of − . The value of is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of , a positive value. The change in concentration of reactant and product with time produces a straight line. The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
\( \left [ A \right ]=\left [ A_{0} \right ]-kt \tag{14.3.2} \) where [A] is the initial concentration of reactant A. ( .2 has the form of the algebraic equation for a straight line, = + , with = [A], = − , and = [A] .) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second. Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N O on a platinum (Pt) surface to produce N and O , which occurs at temperatures ranging from 200°C to 400°C:
\( 2N_{2}O \left ( g \right )\overset{Pt}{\rightarrow} 2N_{2} \left ( g \right ) + O_{2} \left ( g \right ) \tag{14.3.3} \) Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N O decomposes is the amount of Pt available (not the of Pt). As long as there is enough N O to react with the entire Pt surface, doubling or quadrupling the N O concentration will have no effect on the reaction rate. The reaction rate is as follows:
\( rate =-\dfrac{1}{2}\left (\dfrac{\Delta \left [N_{2}O \right ]}{\Delta t} \right )=\dfrac{1}{2}\left ( \dfrac{\Delta \left [N_{2} \right ]}{\Delta t} \right )=\dfrac{\Delta \left [O_{2} \right ]}{\Delta t} k\left [N_{2}O \right ]^{0}=k \tag{14.3.4} \) Thus the rate at which N O is consumed and the rates at which N and O are produced are independent of concentration. As shown in , the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N O concentration in the experimentally derived rate law is the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, . This graph shows the concentrations of reactants and products versus time for the zeroth-order catalyzed decomposition of N O to N and O on a Pt surface. The change in the concentrations of all species with time is linear. If a plot of reactant concentration versus time is linear, then the reaction is zeroth order in that reactant. A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is where NAD (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in ). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in ). These examples illustrate two important points: In a first-order reaction , the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
\( rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=k\left [ A \right ]^{1}=k\left [ A \right ] \tag{14.3.5} \) If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s ). The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
\( rate= \left [ A \right ]= \left [ A_{0} \right ] e^{-kt} \tag{14.3.6} \) where [A] is the initial concentration of reactant A at = 0; is the rate constant; and is the base of the natural logarithms, which has the value 2.718 to three decimal places. (Essential Skills 6 in , discusses natural logarithms.) Recall that an integrated rate law gives the relationship between reactant concentration and time. predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and :
\( rate= ln \left [ A \right ]= ln \left [ A_{0} \right ] -kt \tag{14.3.7} \) Because has the form of the algebraic equation for a straight line, = + , with = ln[A] and = ln[A] , a plot of ln[A] versus for a first-order reaction should give a straight line with a slope of − and an intercept of ln[A] . Either the differential rate law ( ) or the integrated rate law ( .7) can be used to determine whether a particular reaction is first order. First-order reactions are very common. In this chapter, we have already encountered two examples of first-order reactions: the hydrolysis of aspirin ( ) and the reaction of -butyl bromide with water to give -butanol (Equation ). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows: Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in is important because cisplatin, the form in which the drug is administered, is the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in have been studied extensively to find ways of maximizing the concentration of the active species. If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order. The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in . The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in shows that the reaction rate doubles [(1.8 × 10 M/min) ÷ (9.0 × 10 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10 M/min) ÷ (9.0 × 10 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = [cisplatin] . Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of . For example, substituting the values for Experiment 3 into , Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug. At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction: \( CH_{3}CH_{2}Cl\left ( g \right )\overset{\Delta }{\rightarrow} HCl\left ( g \right )+C_{2}H_{4}\left ( g \right ) \) Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction. balanced chemical equation, initial concentrations of reactant, and initial rates of reaction reaction order and rate constant Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction. Use measured concentrations and rate data from any of the experiments to find the rate constant. The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate. Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH CH Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH CH Cl]. This behavior is characteristic of a first-order reaction, for which the rate law is rate = [CH CH Cl]. We can calculate the rate constant ( ) using any row in the table. Selecting Experiment 1 gives the following: Exercise Sulfuryl chloride (SO Cl ) decomposes to SO and Cl by the following reaction: Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction. first order; = 2.2 × 10 s We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in for = 100 min ([cisplatin] = 0.0086 M) and = 1000 min ([cisplatin] = 0.0022 M), \( slope = \dfrac{ln\left [ cisplatin \right ]_{1000}- ln \left [ cisplatin \right ]_{100}}{1000\;min-100\;min} \) \( -k = \dfrac{ln 0.0022-ln 0.0086}{1000\;min-100\;min}=\dfrac{-6.12-\left ( -4.76 \right )}{900\; min}=-1.51\times 10^{-3} min^{-1} \) \( k = 1.51\times 10^{-3} \; min^{-1} \) The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min because the times plotted on the horizontal axes in parts (a) and (b) in are in minutes rather than seconds. The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions. Refer back to Example 4. If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M? (Recall that we calculated the rate constant for this reaction in Example 4.) initial concentration, rate constant, and time interval concentration at specified time and time required to obtain particular concentration Substitute values for the initial concentration ([A] ) and the calculated rate constant for the reaction ( ) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time . Given a concentration [A], solve the integrated rate law for time . The exponential form of the integrated rate law for a first-order reaction ( 3.6) is [A] = [A] . Having been given the initial concentration of ethyl chloride ([A] ) and having calculated the rate constant in Example 4 ( = 1.6 × 10 s ), we can use the rate law to calculate the concentration of the reactant at a given time . Substituting the known values into the integrated rate law, \( \left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=\left [ CH_{3} CH_{2}Cl \right ]_{0}e^{-kt} \) \( =0.0200\;M \; exp \left [ \left (-1.6\times 10^{-6}\;s^{-1} \right ) \left ( 10\;h \right ) \left ( 60\;min/h \right )\left ( 60 \;s/min \right ) \right ) ] \) \( = 0.0189\;M We could also have used the logarithmic form of the integrated rate law ( ): \( ln \left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -kt \) \( = ln 0.0200\;M - \left (-1.6\times 10^{-6}\;s^{-1} \right ) \left ( 10\;h \right ) \left ( 60\;min/h \right )\left ( 60 \;s/min \right ) \) \( = -3.912--0.0576=-3.970 \) \( \left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=e^{-3.970}\;M \) =0.0189\;M To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for . gives the following: \( ln \left [ CH_{3} CH_{2}Cl \right ]_{t}=ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -kt \) \( kt= ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -ln \left [ CH_{3} CH_{2}Cl \right ]_{t}= ln \; \dfrac{\left [ CH_{3} CH_{2}Cl \right ]_{0}}{ \left [ CH_{3} CH_{2}Cl \right ]_{t}} \) \( t= \dfrac{1}{k}\; ln\dfrac{\left [ CH_{3} CH_{2}Cl \right ]_{0}}{\left [ CH_{3} CH_{2}Cl \right ]_{t}} = \dfrac{1}{1.6\times 10^{-6}\;s^{-1}}\; ln\dfrac{0.0200\;M}{0.0050\;M} \) \( t= \dfrac{ln\;4.0}{k}\; =8.7\times 10^{5}\;s=240\;h=2.4\times 10^{2}\;h \) Exercise In the exercise in Example 4, you found that the decomposition of sulfuryl chloride (SO Cl ) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO Cl that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO Cl to decompose? 0.0252 M; 29 h The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a , in which two smaller molecules, each called a , combine to form a larger molecule (a ). The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
\( rate=-\dfrac{\Delta \left [ A \right ]}{2 \Delta t}=k\left [ reactant \right ]^{2} \tag{14.3.8} \) Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M ·s ). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s). For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
\( \dfrac{1 }{\left [ A \right ]}= \dfrac{1 }{\left [ A_{0} \right ]} +kt \tag{14.3.9} \) Because has the form of an algebraic equation for a straight line, = + , with = 1/[A] and = 1/[A] , a plot of 1/[A] versus for a simple second-order reaction is a straight line with a slope of and an intercept of 1/[A] . Second-order reactions generally have the form 2A → products or A + B → products. Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO to NO and O and the decomposition of HI to I and H . Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture. Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows: For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law ( ) or the integrated rate law ( ). To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7: Because (1.7) = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration. \( rate \;\;\alpha\;\;\left [ monomer \right ]^{2} \) This means that the reaction is second order in the monomer. Using and the data from any row in , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following: \( rate = k\left [ A^{2} \right ] \) \( 8.0\times 10^{-5}\;M/min = k\left (4.4\times 10^{-3}\;M \right ) \) \( 4.1\;min^{-1} = k \) We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in . The measurements show that the concentration of the monomer (initially 5.4 × 10 M) decreases with increasing time. This graph also shows that the reaction decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus should be a straight line, as shown in part (b) in . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, = 4.1 M ·min , which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. For two or more reactions of the , the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. The differential and integrated rate laws for zeroth-, first-, and second-order reactions and their corresponding graphs are shown in in . At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen. \( 2NO_{2}\left ( g \right )\overset{\Delta }{\rightarrow} 2NO\left ( g \right )+O_{2}\left ( g \right )\) Experimental data for the reaction at 300°C and four initial concentrations of NO are listed in the following table: Determine the reaction order and the rate constant. balanced chemical equation, initial concentrations, and initial rates reaction order and rate constant From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant ( ). We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10 ) ÷ (1.35 × 10 ) = 4.0], which means that the reaction rate is proportional to [NO ] . Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO ] . This behavior is characteristic of a second-order reaction. We have rate = [NO ] . We can calculate the rate constant ( ) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following: \( rate = k\left [ NO_{2} \right ]^{2} \) \( 5.40\times 10^{-5}\; M/s = k\left (0.010\;M \right )^{2} \) \( 0.540\; M^{-1}s^{-1} = k \) Exercise When the highly reactive species HO forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows: The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table: Determine the reaction order and the rate constant. second order in HO ; = 1.4 × 10 M ·s If a plot of reactant concentration versus time is not linear but a plot of 1/reaction concentration versus time is linear, then the reaction is second order. If a flask that initially contains 0.056 M NO is heated at 300°C, what will be the concentration of NO after 1.0 h? How long will it take for the concentration of NO to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in Example 6. balanced chemical equation, rate constant, time interval, and initial concentration final concentration and time required to reach specified concentration Given , , and [A] , use the integrated rate law for a second-order reaction to calculate [A]. Setting [A] equal to 1/10 of [A] , use the same equation to solve for . We know and [NO ] , and we are asked to determine [NO ] at = 1 h (3600 s). Substituting the appropriate values into , \( \dfrac{1}{\left [NO_{2} \right ]_{3600}} = \dfrac{1}{\left [NO_{2} \right ]_{0}} +kt=\dfrac{1}{0.056\;M} +\left ( 0.54\;M^{-1}s^{-1} \right )\left (3600\; s \right ) \) \( =2.0\times 10^{-3} \; M^{-1} \) Thus [NO ] = 5.1 × 10 M. In this case, we know and [NO ] , and we are asked to calculate at what time [NO ] = 0.1[NO ] = 0.1(0.056 M) = 0.0056 M. To do this, we solve for , using the concentrations given. \( \dfrac{\dfrac{1}{\left [NO_{2} \right ]_{3600}} - \dfrac{1}{\left [NO_{2} \right ]_{0}}}{k} =\dfrac{\left ( 1/0.0056\;M \right )-\left ( 1/0.056\;M \right )}{0.54\;M^{-1}s^{-1}}=3.0\times10^{2}\;s=5\;min \) NO decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min. Exercise In the exercise in Example 6, you calculated the rate constant for the decomposition of HO as = 1.4 × 10 M ·s . This high rate constant means that HO decomposes rapidly under the reaction conditions given in the problem. In fact, the HO molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO , calculate the concentration of HO that remains after 1.0 h at 25°C. How long will it take for 90% of the HO to decompose? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in the exercise in Example 6. 2.0 × 10 M; 6.4 × 10 s In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form A + B → products, in which the reaction is first order in A and first order in B. The differential rate law for this reaction is as follows:
\( rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=-\dfrac{\Delta \left [ B \right ]}{\Delta t}=k\left [ A \right ]\left [ B \right ] \tag{14.3.10}\) Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. We presented one example at the end of , the reaction of CH Br with OH to produce CH OH. The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanisms can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction A + B → products, for example, we need to determine and the exponents and in the following equation:
\( rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.3.11} \) To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Rate data for a hypothetical reaction of the type A + B → products are given in . The general rate law for the reaction is given in . We can obtain or directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in . \( \dfrac{rate^{1}}{rate^{3}}= \dfrac{k\left [ A^{1} \right ]^{m}\left [ B^{1} \right ]^{n}}{k\left [ A^{3} \right ]^{m}\left [ B^{3} \right ]^{n}} \) Inserting the appropriate values from , \( \dfrac{8.5\times10^{-3}\;\cancel{M/min}}{34.\times10^{-3}\;\cancel{M/min}}= \dfrac{k\left [ 0.50\;\cancel{M} \right ]^{m}\cancel{\left [ 05.0\;M \right ]^{n}}}{k\left [ 1.00\;\cancel{M} \right ]^{m}\cancel{\left [ 0.50\;M \right ]^{n}}} \) Because 1.00 to any power is 1, [1.00 M] = 1.00 M. We can cancel like terms to give 0.25 = [0.50] , which can also be written as 1/4 = [1/2] . Thus we can conclude that = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for . Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for . \( \dfrac{rate^{1}}{rate^{3}}= \dfrac{k\left [ A^{1} \right ]^{m}\left [ B^{1} \right ]^{n}}{k\left [ A^{5} \right ]^{m}\left [ B^{5} \right ]^{n}} \) Substituting the appropriate values from , \( \dfrac{8.5\times10^{-3}\;\cancel{M/min}}{8.5\times10^{-3}\;\cancel{M/min}}= \dfrac{\cancel{k\left [ 0.50\;\cancel{M} \right ]^{m}}\left [ 05.0\;\cancel{M} \right ]^{n}}{\cancel{k\left [ 0.50\;M \right ]^{m}}\left [ 1.00\;\cancel{M} \right ]^{n}} \) Canceling leaves 1.0 = [0.50] , which gives = 0; that is, the reaction is zeroth order in B. The experimentally determined rate law is therefore \( rate=k\left [ A \right ]^{2}\left [ B \right ]^{0} \) We can now calculate the rate constant by inserting the data from any row of into the experimentally determined rate law and solving for . Using Experiment 2, we obtain \( 19.\times10^-3\; M/min=k\left ( 0.75\;M \right )^{2} \) \( 3.4\times10^-2\; M^{-1}min^{-1}=k \) You should verify that using data from any other row of gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same. Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with O to give NO , which then reacts rapidly with water to give nitrous acid and nitric acid: These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O at 25°C: Determine the rate law for the reaction and calculate the rate constant. balanced chemical equation, initial concentrations, and initial rates rate law and rate constant Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction. Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for . Comparing Experiments 1 and 2 shows that as [O ] is doubled at a constant value of [NO ], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O ] , so the reaction is first order in O . Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O ] is held constant. That is, the reaction rate is proportional to [NO] , which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction: \( rate =k\left [ NO \right ]^{2}\left [ O_{2} \right ] \) The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives \( k=\dfrac{k}{\left [ NO \right ]^{2}\left [ O_{2} \right ]} = \dfrac{7.98\times10{-3}\;M/s}{\left ( 0.0235\;M \right )^{2}\left ( 0.0125\;M \right )}=1.16\times10{3}\;M^{-2}s^{-1} \) The overall reaction order ( + ) is 3, so this is a third-order reaction, a reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases. Exercise The peroxydisulfate ion (S O ) is a potent oxidizing agent that reacts rapidly with iodide ion in water: \( S_{2}O_{8}^{2-}\left ( aq \right )+3I^{-}\left ( aq \right ) \rightarrow 2SO_{4}^{2-}\left ( aq \right )+I_{3}^{-}\left ( aq \right ) \) The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant. rate = [S O ,I ]; = 20 M ·s The reaction rate of a is independent of the concentration of the reactants. The reaction rate of a is directly proportional to the concentration of one reactant. The reaction rate of a simple is proportional to the square of the concentration of one reactant. Knowing the of a reaction gives clues to the . \( rate = -\dfrac{\Delta \left [ A \right ] }{\Delta t} = k \) : \( \left [ A \right ]=\left [ A_{0} \right ]-kt \) : \( rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=k\left [ A \right] \) : \( \left [ A \right ]= \left [ A_{0} \right ] e^{-kt} \) : \( ln \left [ A \right ]= ln \left [ A_{0} \right ] -kt \) : \( ln \left [ A \right ]= ln \left [ A_{0} \right ] +kt \) : \( \dfrac{1 }{\left [ A \right ]}= \dfrac{1 }{\left [ A_{0} \right ]}+kt \) What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order? Predict whether the following reactions are zeroth order and explain your reasoning. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form? If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order? The reaction of NO with O is found to be second order with respect to NO and first order with respect to O . What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate? Iodide reduces Fe(III) according to the following reaction: Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order? Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows: What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction? 1-Bromopropane is a colorless liquid that reacts with S O according to the following reaction: The reaction is first order in 1-bromopropane and first order in S O , with a rate constant of 8.05 × 10 M ·s . If you began a reaction with 40 mmol/100 mL of C H Br and an equivalent concentration of S O , what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be? The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δ = [A] [B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally. First order in Fe ; second order in I ; third order overall; rate = [Fe ,I ] . 1.29 × 10 M/s; 3.22 × 10 M/s | 36,456 | 1,685 |
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If we extend the structure of corannulene by adding similar cycles of five benzene rings, the curvature of the resulting molecule should increase, and eventually close into a sphere of carbon atoms. The archetypical compound of this kind (C ) has been named because of its resemblance to the geodesic structures created by Buckminster Fuller. It is a member of a family of similar carbon structures that are called . These materials represent a third class of carbon allotropes. Alternating views of the C fullerene structure are shown on the right, together with a soccer ball-like representation of the 12 five and 20 six-membered rings composing its surface. Precise measurement by Atomic Force Microscopy (AFM) has shown that the C-C bond lengths of the six-membered rings are not all equal, and depend on whether the ring is fused to a five or six-membered beighbor. . Although C is composed of fused benzene rings its chemical reactivity resembles that of the cycloalkenes more than benzene. Indeed, exposure to light and oxygen slowly degrade fullerenes to cage opened products. Most of the reactions thus far reported for C involve addition to, rather than substitution of, the core structure. These reactions include hydrogenation, bromination and hydroxylation. Strain introduced by the curvature of the surface may be responsible for the enhanced reactivity of C . . Larger fullerenes, such as C , C , C & C have ellipsoidal or distorted spherical structures, and fullerene-like assemblies up to C have been detected. A fascinating aspect of these structures is that the space within the carbon cage may hold atoms, ions or small molecules. Such species are called . The cavity of C is relatively small, but encapsulated helium, lithium and atomic nitrogen compounds have been observed. Larger fullerenes are found to encapsulate lanthanide metal atoms. Interest in the fullerenes has led to the discovery of a related group of carbon structures referred to as nanotubes. As shown in the following illustration, nanotubes may be viewed as rolled up segments of graphite. The chief structural components are six-membered rings, but changes in tube diameter, branching into side tubes and the capping of tube ends is accomplished by fusion with five and seven-membered rings. Many interesting applications of these unusual structures have been proposed. | 2,393 | 1,686 |
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Many unsaturated cyclic compounds have exceptional properties that we now consider characteristic of "aromatic" systems. The following cases are illustrative: The first three compounds (cyclic polyenes) have properties associated with alkenes in general. Each reacts readily with bromine to give addition products, as do most alkenes. The thermodynamic change on introducing double bonds into the carbon atom ring is also typical of alkenes (a destabilization of ca. 26 kcal/mol for each double bond). Conjugation offsets this increase in energy by a small amount (4-6 kcal/mol). The remaining four compounds exhibit very different properties, and are considered aromatic. Benzene and pyridine are relatively unreactive with bromine, requiring heat and/or catalysts to force reaction, the result of which is substitution rather than addition. Furan and pyrrole react more rapidly with bromine, but they also give substitution products. This tendency to favor substitution rather than addition suggests that the parent unsaturated ring system has exceptional stability. Thermodynamic measurements support this conclusion. The enhanced stability, often referred to as , ranges (in the above cases) from a low of 16 kcal/mol for furan to 36 kcal/mol for benzene. 1,3-Cyclopentadiene and 1,3,5-cycloheptatriene both fail to meet the first requirement, since one carbon atom of each ring is sp hybridized and has no p-orbital. Cyclooctatetraene fails both requirements, although it has a ring of sp hybridized atoms. This molecule is not planar ( a geometry that would have 135º bond angles ). Angle strain is relieved by adopting a tub-shaped conformation; consequently, the p-orbitals can only overlap as isolated pairs, not over the entire ring. Furthermore, cyclooctatetraene has 8 π-electrons, a number not consistent with the Hückel Rule. Benzene is the archetypical aromatic compound. It is planar, bond angles=120º, all carbon atoms in the ring are sp hybridized, and the pi-orbitals are occupied by 6 electrons. The aromatic heterocycle pyridine is similar to benzene, and is often used as a weak base for scavenging protons. Furan and pyrrole have heterocyclic five-membered rings, in which the heteroatom has at least one pair of non-bonding valence shell electrons. By hybridizing this heteroatom to a sp state, a p-orbital occupied by a pair of electrons and oriented parallel to the carbon p-orbitals is created. The resulting planar ring meets the first requirement for aromaticity, and the π-system is occupied by 6 electrons, 4 from the two double bonds and 2 from the heteroatom, thus satisfying the Hückel Rule. Four illustrative examples of aromatic compounds are shown above. The sp hybridized ring atoms are connected by brown bonds, the π-electron pairs and bonds that constitute the aromatic ring are colored blue. Electron pairs that are not part of the aromatic π-electron system are black. The first example is azulene, a blue-colored 10 π-electron aromatic hydrocarbon isomeric with naphthalene. The second and third compounds are heterocycles having aromatic properties. Pyridine has a benzene-like six-membered ring incorporating one nitrogen atom. The non-bonding electron pair on the nitrogen is not part of the aromatic π-electron sextet, and may bond to a proton or other electrophile without disrupting the aromatic system. In the case of thiophene, a sulfur analog of furan, one of the sulfur electron pairs (colored blue) participates in the aromatic ring π-electron conjugation. The last compound is imidazole, a heterocycle having two nitrogen atoms. Note that only one of the nitrogen non-bonding electron pairs is used for the aromatic π-electron sextet. The other electron pair (colored black) behaves similarly to the electron pair in pyridine. Monocyclic compounds made up of alternating conjugated double bonds are called . Benzene and 1,3,5,7-cyclooctatetraene are examples of annulenes; they are named [6]annulene and [8]annulene respectively, according to a general nomenclature system in which the number of pi-electrons in an annulene is designated by a number in brackets. Some annulenes are aromatic (e.g. benzene), but many are not due to non-planarity or a failure to satisfy the Hückel Rule. Compounds classified as [10]annulenes (a Hückel Rule system) serve to illustrate these factors. As shown in the following diagram, 1,3,5,7,9-cyclodecapentaene fails to adopt a planar conformation, either in the all cis-configuration or in its 1,5-trans-isomeric form. The transannular hydrogen crowding that destabilizes the latter may be eliminated by replacing the interior hydrogens with a bond or a short bridge (colored magenta in the diagram). As expected, the resulting 10 π-electron annulene derivatives exhibit aromatic stability and reactivity as well as characteristic in the nmr. Naphthalene and azulene are [10]annulene analogs stabilized by a transannular bond. Although the CH bridged structure to the right of naphthalene in the diagram is not exactly planar, the conjugated 10 π-electron ring is sufficiently close to planarity to achieve aromatic stabilization. The bridged [14]annulene compound on the far right, also has . Formulation of the Hückel rule prompted organic chemists to consider the possible aromaticity of many unusual unsaturated hydrocarbons. One such compound was the 6 π-electron bicyclic structure, now known as barrelene. Although the π-bonds in barrelene are not coplanar, it was believed that transannular overlap might still lead to aromatic stabilization. A synthesis of barrelene (bicyclo[2.2.2]-2,5,7-octatriene) was accomplished nearly fifty years ago by H. Zimmerman (Wisconsin), using a double Hofmann elimination. As shown in the following diagram, the chemical behavior of this triene confirmed it was not aromatic in the accepted sense of this term. Bromine addition took place rapidly with transannular bond formation, in the same fashion as with norbornadiene (bicyclo[2.2.1]-2,5-heptadiene). Pyrolysis of barrelene gave the expected products benzene and acetylene. The heat of hydrogenation of barrelene reflects its thermodynamic stability. The value for cyclohexene is -28 kcal/mol, significantly less than one third of the barrelene number. Furthermore, the first double bond of barrelene is reduced with the release of 36.7 kcal/mol heat, indicating destabilization rather than stabilization. The electronic spectrum of barrelene shows a π-electron interaction similar to that in related homoconjugated dienes. (λ ≅220-230 nm). An explanation for the lack of aromatic behavior in the case of barrelene may be found by comparing the orbital symmetry of the six component p-orbitals with those of benzene. Benzene is an annulene in which all six p-orbitals may be oriented with congruent overlapping phases. The cylindrical array of p-orbitals in barrelene cannot be so arranged, as shown in the diagram on the right. There will always be one region (a nodal plane) in which the transannular overlap is incongruent. , a Jmol model of barrelene will be displayed in a separate window. This model may be moved about for viewing. The p-orbitals of the double bonds may also be displayed. | 7,210 | 1,688 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX9._Peptides_and_Proteins%3A_Laboratory_Synthesis |
Peptides and proteins are very important in biology. As a result, synthesis of these molecules has become very important, allowing for the laboratory study of model compounds that can give us insight into how proteins work, as well as pharmaceutically important compounds. Structurally, amide or peptide bonds are very stable and resistant to carboxyl substitution. That stability makes optimal structures from which to construct proteins. Despite being composed of very long chains of linked amino acids, proteins actually have some limits on their conformational flexibility (their "floppiness"). That allows proteins to more reliably hold a particular shape. Both the stability and the structural rigidity of peptides arises from the nature of the peptide bond. The pi donation that hinders nucleophiles from substituting at the carbonyl is pronounced enough that it can be considered to form an additional bond. Thus, peptides behave as though they contain C=N bonds rather than C-N bonds. X-ray structure determinations show that the peptide nitrogens in proteins are trigonal planar, not pyramidal. In addition, many peptides exhibit cis-trans isomerism. For every peptide bond, two different isomers can occur, depending on whether a substituent attached to nitrogen is on the same side of the C=N bond as the carbonyl oxygen or the opposite side. The great stability of these structures does not mean they are easy to make. Part of the difficulty stems from the fact that amino acids are difunctional. In order to form long chain structures, amino acids must be able to react twice: once with an amine, to grow in one direction, and once with a carboxylic acid to grow in the other direction. In other words, an amino acid contains both a nucleophile and an electrophile. Suppose we were to try to make the dipeptide, ala-phe. This peptide contains an alanine connected to a phenylalanine through a peptide bond. The peptide bond is formed between the carboxylic acid of alanine and the amine of phenylalanine. Assuming the amino acids do react together to form the peptide, combining these two reactants would likely produce a mixture of four dipeptides: ala-phe ala-ala phe-phe phe-ala In other words, peptide formation from amino acids is non-selective. Draw structures for the four peptides formed by combining phe and ala. What tripeptides would be produced by mixing ala, gly and val? Simply combining these peptides might not result in any peptide formation at all. Why not? In laboratory syntheses, a number of techniques have been used to make peptide synthesis selective. Most frequently, protecting groups are used. A protecting group "masks" one of the two functional groups on an amino acid, but leaves the other one open. If one amino acid has its amine protected, it can only react via its carboxylic acid. If the other amino acid has its carboxylic acid protected, it can only react via its amino group. Only one combination will result. The key to protecting groups is that the reaction used to mask one of the functional groups must be reversible. You must be able to take the protecting group back off when it is no longer needed. Carboxylic acids are normally protected as esters. Esters can be removed via acid- or base-catalyzed hydrolysis (as can amides, but esters are more reactive, being farther up the ski hill). Amines are normally protected as amides. However, we need to be able to remove specific amides her: the ones that mask the amines, not the ones that we have formed to link two amino acids together. As a result, in peptide synthesis, amines are usually protected as carbamates. Carbamates can be cleaved more easily than amides. Propose a reason for the relative reactivity of carbamates and amides. An additional complication in peptide synthesis is that amines and carboxylic acids do not really exist together. Instead, a proton is transferred from the carboxylic acid to the amine, forming a salt. The carboxylate is no longer very electrophilic, due to its negative charge. Because of its positive charge, the ammonium ion is no longer very nucleophilic. To get around this problem, a number of coupling agents have been developed. A coupling agent can temporarily convert the carboxylate anion into a more reactive electrophile. To do so, it exploits the nucleophilicity of the carboxylate anion. After donating to the coupling agent, the carbonyl compound becomes more electrophilic. , | 4,452 | 1,689 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Mass_Spectrometers_(Instrumentation)/Injection_Stage |
The principle components of a mass spectrometer are an inlet, ion source, mass analyzer, detector, and data analysis. The function of an inlet system is to introduce a small amount of sample into the ion source with minimal loss of vacuum. However, many mass spectrometers contain more than one inlet system to accommodate a variety of samples. These include batch inlets, direct probe inlets, and chromatographic and capillary electrophoretic inlet systems. The batch inlet system is considered the most common and simplest inlet system. Normally, the inside of the system is lined with glass to elude losses of polar analytes by adsorption. This system externally volatizes the sample which leaks into an empty ionization region. Boiling points up to 500 degrees C of gaseous and liquid samples can be used on typical systems. The system's vacuum contains a sample pressure of 10 to 10 Torr. Liquids are introduced using a microliter syringe into a reservoir; gases are enclosed in a metering area that is confined between two valves before being expanded into a reservoir container. Liquids that have boiling points lower than 500 degrees C can not be used in the system because the reservoir and tubing need to be kept at high temperatures by ovens and heating tapes. This is to ensure that the liquid samples are transformed to the gaseous phase and then leaked through a metal or glass diaphragm containing pinholes to the ionization area. A direct probe inlet is for small quantities of sample, solids, and nonvolatile liquids. Solids and nonvolatile liquids are injected through a probe, or sample holder. The probe is inserted through a vacuum lock which is designed to limit the volume of air needed to pump from the system after the probe has been inserted into the ionization section. Unlike the batch inlet, the sample will need to be cooled and/or heated on the probe. The probe is placed extremely close (a few millimeters) to the ionization source, where the slit leads to the spectrometer, and the sample is held in place on the surface of a glass or aluminum capillary tube or a small cup. This position makes it possible for thermally unstable compounds to be analyzed before decomposition because of the low pressure in the ionization area which is in close proximity to the sample. Do to the probe, nonvolatile samples such as carbohydrates, steroids, and metal-organic species can be studied because the low pressures lead to increased concentrations of the nonvolatile samples. The principle sample requirement is attainment of an analyte partial pressure of at least 10 torr before the onset of decomposition. Chromatographic systems and Capillary Electrophoretic units are often coupled with mass spectrometers in order to allow separation and identification of the components in the sample. If these systems and units are linked with a mass spectrometer, then other specialized inlets, electrokinetic and pressure injection, are required. Electrokinetic and pressure injection controls the amount of volume injected by the duration of the injection, which typically range between 5 to 50 nL. The electrokinetic injection method involves one end of the capillary and electrode in a small cup removed from the buffer. Ionic migration and electroosmotic flow of the sample in the capillary are a result of a voltage applied for a recorded time. After this voltage is applied, the capillary end and electrode are placed back into the regular buffer solution for the remainder of the separation. The electrokinetic Injection technique injects larger quantities of the mobile ions versus the slower moving ions. The pressure injection method is similar to the electrokinetic injection because the capillary end and electrode are removed from the buffer and placed into a small cup. However, instead of a voltage being applied, a pressure difference drives the sample into the capillary. A vacuum is applied at the detector end which produces the potential difference by pressurizing the sample or elevating the sample end. Pressure injection does not discriminate because of ion mobility, but it cannot be used in gel-filled capillaries. | 4,187 | 1,690 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.04%3A__Heats_of_Formation |
and presented a wide variety of chemical reactions, and you learned how to write balanced chemical equations that include all the reactants and the products except heat. One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data starting from the elemental forms of each atom at 25 C and 1 atm pressure.
\[ elements \rightarrow compound \;\;\;\;\ \Delta H_{rxn} = \Delta H_{f} \] For example, The sign convention for Δ is the same as for any enthalpy change: Δ < 0 if heat is released when elements combine to form a compound and Δ > 0 if heat is absorbed. The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. The magnitude of Δ for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane. The standard conditions for which most thermochemical data are tabulated are a of 1 atmosphere (atm) for all gases and a of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state . This is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ( ) (which is pronounced “delta H eff naught”). . For example, although oxygen can exist as ozone (O ), atomic oxygen (O), and molecular oxygen (O ), O is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H (g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure \(\Page {1}\) ). Therefore, O (g), H (g), and graphite have values of zero. The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:
\[ 6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \] It is not possible to measure the value of for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, O , and H and measuring the heat evolved as glucose is formed; the reaction shown in Equation \(\Page {3}\) does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of for an extensive list of compounds are given in the Reference Tables. Note that values are always reported in kilojoules per mole of the substance of interest. Also notice in the Reference Tables that the standard enthalpy of formation of O (g) is zero because it is the most stable form of oxygen in its standard state. For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. compound balanced chemical equation for its formation from elements in standard states Use to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made. To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in the by a Fractional coefficients are required in this case because values are reported for of the product, HCl. Multiplying both H (g) and Cl (g) by 1/2 balances the equation: This equation can be balanced by inspection to give There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for reaction involving substances whose values are known. The standard enthalpy of reaction ( ) is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction
\[ aA + bB \rightarrow cC + dD \] where A, B, C, and D are chemical substances and , , , and are their stoichiometric coefficients. The magnitude of is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
\[ \begin{matrix} \Delta H_{rxn}^{o} = \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] - & \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ] \\ reactants & products \end{matrix} \] More generally, we can write \[ \Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \] where the symbol Σ means “sum of” and and are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation \(\Page {6}\) arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters. Products minus reactants” summations are typical of state functions. To demonstrate the use of tabulated values, we will use them to calculate for the combustion of glucose, the reaction that provides energy for your brain:
\[ C_{6}H_{12}O_{6} \left ( s \right ) + O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right ) + 6H_{2}O\left ( l \right ) \] Using Equation \(\Page {6}\), we write
\[ \Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ 6\Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \] From the , the relevant values are , , and . Because O (g) is a pure element in its standard state, Inserting these values into and changing the subscript to indicate that this is a combustion reaction, we obtain \[ \begin{matrix} \Delta H_{comb}^{o} = \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] \\ - \left [-1273.3 + 6\left ( 0 \; kJ/mol \right ) \right ] = -2802.5 \; kJ/mol \end{matrix} \] As illustrated in Figure \(\Page {3}\), we can use Equation \(\Page {9}\) to calculate for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled The alternative hypothetical pathway consists of four separate reactions that convert the reactants in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the values of the reactants. Consequently, the enthalpy changes are \[ \begin{matrix} \Delta H_{1}^{o} = \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] = -1 \; \cancel{mol \; glucose}\left ( \frac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) = +1273.3 \; kJ \\ \Delta H_{2}^{o} = 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; O_{2}}\left ( \frac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) = 0 \; kJ \end{matrix} \] (Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change.) The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O ) to the elements is therefore +1273.3 kJ. The reactions that convert the elements to final products (downward purple arrows in Figure \(\Page {2}\) ) are identical to those used to define the values of the products. Consequently, the enthalpy changes (from the \[ \begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \\ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix} \] The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ):
\[ \Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \] This is the same result we obtained using the “products minus reactants” rule and values. The two results must be the same because Equation \(\Page {11}\) is just a more compact way of describing the thermochemical cycle shown in Figure \(\Page {2}\). Long-chain fatty acids such as palmitic acid [CH (CH ) CO H] are one of the two major sources of energy in our diet ( ). Use the data in the to calculate for the combustion of palmitic acid. Based on the energy released in combustion , which is the better fuel — glucose or palmitic acid? compound and values per mole and per gram After writing the balanced chemical equation for the reaction, use Equation \(\Page {6}\) and the values from the to calculate the energy released by the combustion of 1 mol of palmitic acid. Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation \(\Page {9}\) for the combustion of glucose to determine which is the better fuel. To determine the energy released by the combustion of palmitic acid, we need to calculate its As always, the first requirement is a balanced chemical equation: \[ C_{16}H_{32}O_{2}\left (s \right )+23O_{2}\left ( g \right ) \rightarrow 16CO_{2} \left ( g \right )+16H_{2})\left ( l \right ) \] Using Equation 9.4.5 (“products minus reactants”) with values from the (and omitting the physical states of the reactants and products to save space) gives \[ \Delta H_{comb}^{o} = \sum m \Delta {H^o}_f\left( {products} \right) - \sum n \Delta {H^o}_f\left( {reactants} \right) \notag \] \[ = \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \notag \] \[ - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \notag \] \[ = -9977.3 \; kJ/mol \notag \] This is the energy released by the combustion of 1 mol of palmitic acid. The energy released by the combustion of 1 g of palmitic acid is \[ \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \notag \] As calculated in Equation \(\Page {9}\), of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore \[ \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \notag \] The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight. Use the data in to calculate for the , which is used industrially on an enormous scale to obtain H (g): \[ \begin{pmatrix}
CO\left ( g \right )+H_{2}O\left ( g \right )\rightarrow CO_{2} \left ( g \right )+H_{2}\left ( g \right ) \\
water-gas \; shift \; reaction
\end{pmatrix} \notag \] −41.2 kJ/mol We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s which we cannot obtain otherwise. This procedure is illustrated in Example 3. Beginning in 1923, tetraethyllead [(C H ) Pb] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are CO (g), H O(l), and red PbO(s). What is the standard enthalpy of formation of tetraethyllead, given that is −19.29 kJ/g for the combustion of tetraethyllead and of red PbO(s) is −219.0 kJ/mol? reactant, products, and values of the reactants Write the balanced chemical equation for the combustion of tetraethyl lead. Then insert the appropriate quantities into to get the equation for of tetraethyl lead. Convert per gram given in the problem to per mole by multiplying per gram by the molar mass of tetraethyllead. Use the to obtain values of for the other reactants and products. Insert these values into the equation for of tetraethyl lead and solve the equation. The balanced chemical equation for the combustion reaction is as follows: \[ \left ( C_{2}H_{5} \right )_{4} Pb\left ( l \right ) + 27 O_{2}\left ( g \right ) \rightarrow 2 PbO \left ( s \right ) +16 CO_{2}\left ( g \right ) +20 H_{2}O \left ( l \right ) \] Using Equation \(\Page {6}\) gives \[ \Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \notag \] Solving for \[ \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \frac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \frac{\Delta H_{comb}^{o}}{2} \notag \] The values of all terms other than are given in the The magnitude of is given in the problem in kilojoules per of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get for 1 mol of tetraethyl lead: \[ \Delta H_{comb}^{o} = \left ( \dfrac{-1929 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) = -6329 \; kJ/mol \notag \] Because the balanced chemical equation contains 2 mol of tetraethyllead, is \[ \Delta H_{rxn}^{o} = 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \frac{-6929 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) = -12,480 \; kJ \notag \] Inserting the appropriate values into the equation for \[ \begin{matrix}
\Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = & \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] \\
& & +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \\
& & \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\\
& = & -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol
\end{matrix} \notag \] Ammonium sulfate [(NH ) SO ] is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid: The value of kilojoules −1181 kJ/mol Equation \(\Page {6}\) : The ) is the enthalpy change that accompanies the formation of a compound from its elements. ( ) are determined under : a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The ( ) can be calculated from the sum of the of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. \(\Page {9}\) Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state? What is the difference between ΔH and ΔH ? How can ΔH of a compound be determined if the compound cannot be prepared by the reactions used to define its standard enthalpy of formation? For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. a. HBr b. CH OH c. NaHCO Describe the distinction between Δ and Δ . The following table lists ΔH values for some ionic compounds. If 1 mol of each solute is dissolved in 500 mL of water, rank the resulting solutions from warmest to coldest. \(\Page {9}\) Using , calculate for each chemical reaction. a. 2Mg(s) + O (g) → 2MgO(s) b. CaCO (s, calcite) → CaO(s) + CO (g) c. AgNO (s) + NaCl(s) → AgCl(s) + NaNO (s) Using , determine for each chemical reaction. a. 2Na(s) + Pb(NO ) (s) → 2NaNO (s) + Pb(s) b. Na CO (s) + H SO (l) → Na SO (s) + CO (g) + H O(l) c. 2KClO (s) → 2KCl(s) + 3O (g) Calculate for each chemical equation. If necessary, balance the chemical equations. a. Fe(s) + CuCl (s) → FeCl (s) + Cu(s) b. (NH ) SO (s) + Ca(OH) (s) → CaSO (s) + NH (g) + H O(l) c. Pb(s) + PbO (s) + H SO (l) → PbSO (s) + H O(l) Calculate for each reaction. If necessary, balance the chemical equations. a. 4HBr(g) + O (g) → 2H O(l) + 2Br (l) b. 2KBr(s) + H SO (l) → K SO (s) + 2HBr(g) c. 4Zn(s) + 9HNO (l) → 4Zn(NO ) (s) + NH (g) + 3H O(l) Use the data in to calculate for the reaction Sn(s, white) + 4HNO (l) → SnO (s) + 4NO (g) + 2H O(l). Use the data in to calculate for the reaction P O (s) + 6H O(l) → 4H PO (l). How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas? How much energy is released or consumed if 10.0 g of N O is completely decomposed to produce gaseous nitrogen dioxide and oxygen? In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction: Calculate ΔH for this reaction. Is the reaction exothermic or endothermic? Would you expect heat to be evolved during each reaction? How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate? a. −1203 kJ/mol O b. 179.2 kJ c. −59.3 kJ −174.1 kJ/mol −20.3 kJ −34.3 kJ/mol Cl ; exothermic Δ = −2.86 kJ CaCl : −4.6 kJ; K CO , −0.65 kJ; mixing, −0.28 kJ | 20,949 | 1,692 |
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The space shuttle used 2 Solid Propellant Boosters (SRBs, white) and a tank of LOX/LH2 (large orange tank) Jet aircraft designed to fly in Earth's atmosphere carry just fuel, and rely on atmospheric O , which is supplied in excess, to burn the fuel. Rockets designed for interplanetary flight need to supply both a fuel and an oxidant for their propulsion, and the ratio of the two has to be exactly right for maximum propulsion and minimum mass. If the wrong size tanks for fuel and oxidizer are designed, some portion of such a reactant 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 . Newton's Third Law of Motion is "every action is accompanied by an equal and opposite reaction." A rocket operates on this principle. The force exerted by the rocket on the ejected reaction products equals the force of the ejected particles on the rocket. So the (the product of the force and time that it acts) are equal in magnitude and opposite in direction. Since the impulse (F × t) equals the momentum (m × v), the momentum of the spent fuel in one direction equals the momentum of the rocket in the other. The fuel and oxidant must be designed to rapidly heat and eject combustion products in one direction, causing motion of the rocket in the opposite direction. The gauge of efficiency for rocket propellants is specific impulse, stated in seconds. The higher the number, the "hotter" the propellant. Specific Impulse (I is the period in seconds for which a 1-pound (0.45-kilogram) mass of propellant (total of fuel and oxidizer) will produce a thrust of 1 pound (0.45- kilogram) of force. The specific impulse for a fuel may vary somewhat due to conditions NASA and commercial launch vehicles use four types of propellants: (1) petroleum; (2) cryogenics; (3) hypergolics; and (4) solids . Examples involving cryogenics (liquid hydrogen and liquid oxygen) and solids (aluminum and ammonium perchlorate, or Ammonium Perchlorate Composite Propellants, APCPs) are give below. Hypergolics are compounds that react upon mixing (without an ignition source), like nitrogen tetroxide and hydrazine which we'll discuss later. Petroleum derivatives include RP-1 (Rocket Propellant or Refined Petroleum-1), which is similar to kerosene and used with an oxidant like liquid oxygen. The Space Shuttle's large rust-orange booster fuel tank shown in the picture above holds liquid oxygen (LOX, 629,340 kg) and liquid hydrogen (LH2, 106 261 kg) . a. Which is the limiting reagent? b. What mass of product will be formed? c. Did NASA make a mistake? The balanced equation 2 H + O → H O tells us that according to the atomic theory, 2 mol H is required for each mole of O . That is, the stoichiometric ratio S(H /O ) = 2 mol H / 1 mol O . Let us see how many moles of each we actually have \(\text{n}_{\text{H}_{2}}=\text{1.06261}\times10^{8}\text{g}~\times~\) \(\frac{\text{1 mol H}_{2}}{\text{2.016 g}}\) \(~=~5.271\times~10^{7}\text{ mol H}_{2}\)
\(\text{n}_{\text{O}_{2}}\) \(~=~\text{6.293}~\times~10^{8}\text{g}\times \frac{\text{1 mol O}_{2}}{\text{31.999 g}}\) \(=\text{1.967}~\times~10^{7}\text{ mol O}_{\text{2}} \) If all the H were to react, the stoichiometric ratio allows us to calculate the amount of O that would be required: \(\text{n}_{\text{O}_{2}}~=~ \text{n}_{\text{H}_{2}}~~\times~~\) \(~~\frac{\text{1 mol O}_{2}}{\text{2 mol H}_{2}}~~\) \(~=~2.63\times10^{7}\text{mol O}_{2}\) This is more than the amount of oxygen present,so oxygen is the and H is present in excess. If all the O reacts, the stoichiometric ratio allows us to calculate the amount of H that would be required: \(\text{n}_{\text{H}_{2}}~=~ \text{n}_{\text{O}_{2}}~~\times~~\) \(~~\frac{\text{2 mol H}_{2}}{\text{1 mol O}_{2}}~~\)
\(~=~1.967\times10^{7}\text{mol O}_{2}~\times~\) \(\frac{\text{2 mol H}_{2}}{\text{1 mol O}_{2}}~=~\) \(3.934\times10^{7}\text{mol H}_{2}\) We require less than the amount of H present, so it is the . When the reaction ends, 3.934 x 10 mol of H will have reacted with 1.967 x 10 mol O and there will be (5.271 x 10 mol - 3.934 x 10 mol) = 1.337 x 10 mol H left over. Oxygen is therefore the limiting reagent. b. Since the hydrogen doesn't all react, we need to calculate the amount of water produced from the amount of oxygen consumed, by using the stoichiometric ratio: \(\text{n}_{\text{H}_{2}\text{O}}~=~ \text{n}_{\text{O}_{2}}~~\times~~\) \(~~\frac{\text{2 mol H}_{2}\text{O}}{\text{1 mol O}_{2}}~~\) \(~=~3.934\times10^{7}\text{mol H}_{2}\text{O}\) The mass of water is then calculated by using the molar mass: \(3.934\times10^{7}\text{mol H}_{2}\text{O}~\times~\) \(\frac{\text{18.01 g}}{\text{1 mol H}_{2}\text{O}}\) \(~=~7.08\times10^{8}\text{g H}_{2}\text{O}\) c. The excess hydrogen is not a mistake. The reaction is so exothermic that it ejects some of the unreacted hydrogen. This doesn't matter, since any ejected mass contributes to the backward momentum of fuel, and the forward momentum of the rocket. Since the mass of hydrogen is small, its velocity is large, and it can contribute to a large forward velocity of the rocket. These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed. In the end, 1.063 x 10 - 7.930 x 10 = 2.70 x 10 g of H will remain, along with 7.08 x 10 g of water, for a total of 7.35 x 10 g. The mass of the reactants was also 1.063 x 10 + 6.29 x 10 = 7.35 x 10 g. 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 \(\frac{n_{\text{H}_{2}}\text{(initial)}}{n_{\text{O}_{2}}\text{(initial)}}\) \(~=~\frac{5.271\times10^{7}\text{mol H}_{2}}{1.967 \times 10^{7}\text{ mol O}_{2}}\) \(~=~\frac{\text{2.68 mol H}_{2}}{\text{1 mol O}_{\text{2}}}\) was less than the stoichiometric ratio \(\text{S}\left( \frac{\text{H}_{2}}{\text{O}_{2}} \right)=\frac{\text{2 mol H}_{2}}{\text{1 mol O}_{2}}\) This indicated that there was not enough O to react with all the hydrogen, and oxygen 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. 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. Solid Rocket Boosters (SRBs) are the type implicated in the infamous Challenger Disaster , where a poor seal allowed escape of hot gases. Each SRB motor contains a propellant mixture (weighting about 590,000kg) consisting of ammonium perchlorate (oxidizer, 69.6% by weight), and aluminum (fuel,16%). A catalyst (iron oxide, 0.4%), a binder (which also acts as secondary fuel, 12.04%), and a curing agent (1.96%). This propellant is commonly referred to as Ammonium Perchlorate Composite Propellant (APCP). a. An amateur rocketeer wants to make 100g of propellant. He mixes the propellant with 69.6 g of ammonium perchlorate (the required 69.6%), but he has no binder, so uses 16.0 g of Al plus another 12.0 g Al in place of the binder (which is also a fuel), or 28% aluminum in the mixture. Is this ratio correct? b. If not, which is the limiting reactant? c. Calculate the mass of Al O produced by the first reaction below. 6 NH ClO + 10 Al → 5 Al O + 6 HCl + 3 N + 9 H O or 6 NH ClO + 10 Al → 4 Al O + 2 AlCl + 3 N + 12 H O The stoichiometric ratio connecting Al and NH ClO is \(\text{S}\left( \frac{\text{Al}}{\text{NH}_{4}\text{ClO}_{3}} \right)\) \(~=~\frac{\text{10 mol Al}}{\text{6 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}\) \(~=~\frac{\text{1.67 mol Al}}{\text{1 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}\) The initial amounts of Al and NH ClO are calculated using appropriate molar masses \(~=~28.0\text{g}~\times~ \frac{\text{1 mol Al}}{\text{26.982 g}}\) \(~=~\text{1.04 mol Al}\)
\(\text{n}_{\text{NH}_{\text{4}}\text{ClO}_{\text{4}}\text{(initial)}}\) \(~=~\text{69.6 g}~\times~ \frac{\text{1 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}{\text{117.489 g}}\) \(~=~\text{0.592 mol NH}_{\text{4}}\text{ClO}_{\text{4}}\) Their ratio is \(\frac{n_{\text{Al}\text{(initial)}}}{n_{\text{NH}_{\text{4}}\text{ClO}_{\text{4}}\text{(initial)}}}\) \(~=~\frac{\text{1.04 mol Al}}{\text{0.592 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}\) \(~=~\frac{\text{1.76 mol Al}}{\text{1 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}\) Since this ratio is larger than the stoichiometric ratio, you have more than enough Al to react with all the NH ClO . NH ClO is the limiting reagent, so the ratio isn't stoichiometric. 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 Al will be left over, but all the initial amount of NH ClO will be consumed. Therefore we use to calculate how much Al O can be obtained \(n_{\text{NH}_{4}\text{ClO}_{4}}~~\) \(\xrightarrow{S\text{(Al}_{2}\text{O}_{3}\text{/NH}_{4}\text{ClO}_{4}\text{)}}\) \(~~n_{\text{Al}_{2}\text{O}_{3}}\) \(\xrightarrow{M_{\text{Al}_{2}\text{O}_{3}}}\) \(~~\text{m}_{\text{Al}_{2}\text{O}_{3}}\) \(m_{\text{Al}_{2}\text{O}_{3}}\) \(~=~\text{0.592 mol NH}_{4}\text{ClO}_{4}~~\) \(\times~~\frac{\text{5 mol Al}_{2}\text{O}_{3}}{\text{6 mol NH}_{4}\text{ClO}_{4}}~~\) \(\times ~~\frac{\text{101.961 g}}{\text{mol Al}_{2}\text{O}_{3}}\) \(~=~\text{50.3 g Al}_{2}\text{O}_{3}\) You may want to verify the rest of the values in the table: | 10,624 | 1,693 |
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In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region \(Δx\); e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position \(Δx\) and the uncertainty in momentum \(Δp\). \[\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}\] You can see from Equation \(\ref{5-22}\) that as \(Δp\) approaches 0, \(Δx\) must approach ∞, which is the case of the free particle discussed . This uncertainty principle, which also is discussed in , is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section. Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons. | 1,643 | 1,694 |
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The of a system in is intimately related to the ; two systems having to have the same temperature if they are to be in thermal equilibrium (i.e. there is no net flow between them). However, it is most useful to have a temperature scale. By making use of the one can define an absolute temperature \[T = \frac{pV}{Nk_B}\] however, perhaps a better definition of temperature is \[\frac{1}{T(E,V,N)} = \left. \frac{\partial S}{\partial E}\right\vert_{V,N}\] where is the entropy. Temperature has the SI units of (K) (named in honour of ) The kelvin is the fraction 1/273.16 of the thermodynamic temperature of the of . \[T = \frac{2}{3} \frac{1}{k_B} \overline {\left(\frac{1}{2}m_i v_i^2\right)}\] where is the . The kinematic temperature so defined is related to the theorem. | 818 | 1,695 |
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On the graph of ionization energies, it can be seen that ionization energies increase much more slowly across the transition and inner transition elements than for the representative elements. For example, the ionization energy of the representative-element boron is 800 kJ mol . Five elements later we find neon, whose ionization energy is 2080 kJ mol , an increase of 160 percent. In the fourth period, the transition-element scandium has an ionization energy of 631 kJ mol . Five elements later we find iron at 759 kJ mol , an increase of only 20 percent. All the lanthanoids have ionization energies from 500 to 600 kJ mol , and the actinoids are all between 580 and 680 kJ mol . These similarities among the transition and especially the inner transition elements illustrate statements made about . The distinguishing electron for a transition element enters a subshell in the shell, while for an inner transition element it usually enters an subshell in the shell. Thus the distinction between an element and the one preceding it in the periodic table is much smaller than among the representative elements. Furthermore, experimental measurements show that for transition and inner transition elements the electrons lost when ionization occurs are the ones which were added to build up the atomic electron configuration. Instead, . Determine the electron configuration of the Fe ion. Since the charge on the ion is +3, three electrons must have been removed from a neutral iron atom (Fe). The electron configuration of Fe is Fe: 1 2 2 3 3 3 4 or [Ar]3 4 We now remove electrons successively from subshells having the largest principal quantum number: Fe : [Ar]3 4 one 4s electron removed Fe [Ar]3 a second 4 electron removed Fe [Ar]3 since no electrons are left in the = 4 shell, one 3 electron is removed The behavior described in the previous paragraph and the example may be better understood by comparing the 3d and 4s shells, as in the following figure. Electrons in the subshell having the largest principal quantum number (4 in the example and the figure above) are, on the average, farther from the nucleus, and they are first to be removed. The first ionization energy of iron is not much larger than that of scandium because in each case a 4 electron is being removed. The iron atom has five more protons in the nucleus, but it also has five more 3 electrons which spend most of their time the nucleus and the 4 electrons. The screening effect of such 3 electrons causes the nuclear charge to increase very slowly from one transition element to the next. The attraction for 4 electrons, and hence ionization energy, also increases very slowly. Macroscopic properties such as high thermal and electric conductivity, malleability, and ductility were mentioned in a as characteristics of . In addition, most metals have low ionization energies, usually below 800 kJ mol . In other words, a metal consists of atoms, each of which has at least one loosely held electron. When such atoms pack close together in a solid metal, the loosely held electrons are relatively free to move from one atom to another. If excess electrons are forced into one end of a metal wire, it is relatively easy for electrons to flow out of the other end. Thus an electric current may be carried through the wire, and the high conductivity of all metals may be understood. More detailed microscopic interpretations of metallic properties are given for later, but for the time being we are primarily interested in the location of metallic elements in the periodic table. Ionization energies are smallest near the bottom and on the left of the periodic table, and so this is where metals are found. Moreover, ionization energies increase slowly from one transition element to the next and hardly at all across the inner transition elements. Therefore transition and inner transition elements are metals. In periodic groups IIIA, IVA, and VA elements near the top of the table have large ionization energies and little metallic character. Ionization energies decrease as one moves downward, however. For example, Al is quite metallic, although the element above it, B, is not. A heavy “stairstep” line is usually drawn on the periodic table to separates the nonmetals (above and to the right) from the metals. Elements such as B, Si, Ge, As, Sb, and Te, which are adjacent to the stairstep, have intermediate properties and are called . This same class is also referred to as . | 4,519 | 1,696 |
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Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, zeroth o know how to determine the reaction order from experimental data. A is one whose rate is independent of concentration; its differential rate law is \[\text{rate} = k. \nonumber \] We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1} \] Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of \(−k\). The value of \(k\) is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of \(k\), a positive value. The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form \[[A] = [A]_0 − kt \label{14.4.2} \] where \([A]_0\) is the initial concentration of reactant \(A\). has the form of the algebraic equation for a straight line, \[y = mx + b, \nonumber \] with \(y = [A]\), \(mx = −kt\), and \(b = [A]_0\).) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second. Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N O on a platinum (Pt) surface to produce N and O , which occurs at temperatures ranging from 200°C to 400°C: \[\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3} \] Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N O to react with the entire Pt surface, doubling or quadrupling the N O concentration will have no effect on the reaction rate. At very low concentrations of N O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N O concentration. The reaction rate is as follows: \[\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4} \] Thus the rate at which N O is consumed and the rates at which N and O are produced are independent of concentration. As shown in , the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally. A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is where NAD (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero ( \(\Page {3a}\)). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in \(\Page {3}\)). These examples illustrate two important points: A Discussing Zero-Order Reactions. Link: In a , the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5} \] If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s ). The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows: \[[A] = [A]_0e^{−kt} \label{14.4.6} \] where \([A]_0\) is the initial concentration of reactant \(A\) at \(t = 0\); \(k\) is the rate constant; and is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of \(A\) and \(t\): \[\ln[A] = \ln[A]_0 − kt \label{14.4.7} \] Because has the form of the algebraic equation for a straight line, \[y = mx + b, \nonumber \] with \(y = \ln[A]\) and \(b = \ln[A]_0\), a plot of \(\ln[A]\) versus \(t\) for a first-order reaction should give a straight line with a slope of \(−k\) and an intercept of \(\ln[A]_0\). Either the differential rate law ( ) or the integrated rate law ( ) can be used to determine whether a particular reaction is first order. Discussing the The First-Order Integrated Rate Law Equation: First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows: Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in \(\Page {4}\) have been studied extensively to find ways of maximizing the concentration of the active species. If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order. The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in ble \(\Page {1}\). The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in \(\Page {1}\) shows that the reaction rate doubles [(1.8 × 10 M/min) ÷ (9.0 × 10 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10 M/min) ÷ (9.0 × 10 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = [cisplatin] . Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table \(\Page {1}\). For example, substituting the values for Experiment 3 into , 3.6 × 10 M/min = (0.024 M) 1.5 × 10 min = Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug. At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction: \[\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber \] Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction. balanced chemical equation, initial concentrations of reactant, and initial rates of reaction reaction order and rate constant Use measured concentrations and rate data from any of the experiments to find the rate constant. The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate. Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH CH Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH CH Cl]. This behavior is characteristic of a first-order reaction, for which the rate law is rate = [CH CH Cl]. We can calculate the rate constant ( ) using any row in the table. Selecting Experiment 1 gives the following: 1.60 × 10 M/s = (0.010 M) 1.6 × 10 s = Sulfuryl chloride (SO Cl ) decomposes to SO and Cl by the following reaction: \[SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber \] Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction. We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in for = 100 min ([cisplatin] = 0.0086 M) and = 1000 min ([cisplatin] = 0.0022 M), \[\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}}
\\[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}}
\\[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber \] The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min because the times plotted on the horizontal axes in parts (a) and (b) in are in minutes rather than seconds. The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions. Example Using the First-Order Integrated Rate Law Equation: If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M ( = 1.6 × 10 s ) ? initial concentration, rate constant, and time interval concentration at specified time and time required to obtain particular concentration The exponential form of the integrated rate law for a first-order reaction ( ) is [A] = [A] . Having been given the initial concentration of ethyl chloride ([A] ) and having the rate constant of = 1.6 × 10 s , we can use the rate law to calculate the concentration of the reactant at a given time . Substituting the known values into the integrated rate law, \[\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt}
\\[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}})
\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \] We could also have used the logarithmic form of the integrated rate law ( ): \[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt
\\[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})]
\\[4pt] &=-3.912-0.0576=-3.970 \nonumber
\\[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber
\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \] To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for \(t\). Eq gives the following: \[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt
\\[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t}
\\[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right)
\\[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber \] In the exercise in Example \(\Page {1}\), you found that the decomposition of sulfuryl chloride (\(\ce{SO2Cl2}\)) is first order, and you calculated the rate constant at 320°C. The simplest kind of is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form \[\ce{2A → products.}\nonumber \] A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer). The differential rate law for the simplest second-order reaction in which 2A → products is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8} \] Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M ·s ). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s). For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time: \[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9} \] Because has the form of an algebraic equation for a straight line, = + , with = 1/[A] and = 1/[A] , a plot of 1/[A] versus for a simple second-order reaction is a straight line with a slope of and an intercept of 1/[A] . Second-order reactions generally have the form 2A → products or A + B → products. Discussing the Second-Order Integrated Rate Law Equation: Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO to NO and O and the decomposition of to I and H . Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture. Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows: Figure \(\Page {7}\) For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law ( ) or the integrated rate law ( ). To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7: \[\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber \] Because (1.7) = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration. rate ∝ [monomer] This means that the reaction is second order in the monomer. Using and the data from any row in , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following: \[\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber \] We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in . The measurements show that the concentration of the monomer (initially 5.4 × 10 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus should be a straight line, as shown in . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, = 4.1 M ·min , which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally. For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen. \[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber \] Experimental data for the reaction at 300°C and four initial concentrations of NO are listed in the following table: Determine the reaction order and the rate constant. balanced chemical equation, initial concentrations, and initial rates reaction order and rate constant We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10 ) ÷ (1.35 × 10 ) = 4.0], which means that the reaction rate is proportional to [NO ] . Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO ] . This behavior is characteristic of a second-order reaction. We have rate = [NO ] . We can calculate the rate constant ( ) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following: \[\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2
\\5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2
\\0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber \] When the highly reactive species HO forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows: \[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber \] The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table: Determine the reaction order and the rate constant. If a plot of reactant concentration versus time is linear, but a plot of concentration versus time is linear, then the reaction is second order. If a flask that initially contains 0.056 M NO is heated at 300°C, what will be the concentration of NO after 1.0 h? How long will it take for the concentration of NO to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated above. balanced chemical equation, rate constant, time interval, and initial concentration final concentration and time required to reach specified concentration We know and [NO ] , and we are asked to determine [NO ] at = 1 h (3600 s). Substituting the appropriate values into , \[\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \\[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \\[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber \] Thus [NO ] = 5.1 × 10 M. In this case, we know and [NO ] , and we are asked to calculate at what time [NO ] = 0.1[NO ] = 0.1(0.056 M) = 0.0056 M. To do this, we solve for , using the concentrations given. \[ \begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \\[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \\[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber \] NO decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min. In the previous exercise, you calculated the rate constant for the decomposition of HO as = 1.4 × 10 M ·s . This high rate constant means that HO decomposes rapidly under the reaction conditions given in the problem. In fact, the HO molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO , calculate the concentration of HO that remains after 1.0 h at 25°C. How long will it take for 90% of the HO to decompose? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in the exercise in Example \(\Page {3}\). In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form \(A + B \rightarrow products\), in which the reaction is first order in \(A\) and first order in \(B\). The differential rate law for this reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A,\textrm B] \label{14.4.10} \] Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism. | 26,220 | 1,697 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Fundamentals_of_Spectroscopy/Selection_Rules |
A describes how the probability of transitioning from one level to another cannot be . It has two sub-pieces: a and a . A gross selection rule illustrates characteristic requirements for atoms or molecules to display a spectrum of a given kind, such as an IR spectroscopy or a microwave spectroscopy. Once the atom or molecules follow the gross selection rule, the specific selection rule must be applied to the atom or molecules to determine whether a certain transition in quantum number may happen or not. Selection rules specify the possible transitions among quantum levels due to absorption or emission of electromagnetic radiation. Incident electromagnetic radiation presents an oscillating electric field \(E_0\cos(\omega t)\) that interacts with a transition dipole. The dipole operator is \(\mu = e \cdot r\) where \(r\) is a vector pointing in a direction of space. A dipole moment of a given state is \[\mu_z=\int\psi_1 \,^{*}\mu_z\psi_1\,d\tau\] A transition dipole moment is a transient dipolar polarization created by an interaction of electromagnetic radiation with a molecule \[(\mu_z)_{12}=\int\psi_1 \,^{*}\mu_z\psi_2\,d\tau\] In an experiment we present an electric field along the z axis (in the laboratory frame) and we may consider specifically the interaction between the transition dipole along the x, y, or z axis of the molecule with this radiation. If \(\mu_z\) is zero then a transition is forbidden. The selection rule is a statement of when \(\mu_z\) is non-zero. We can consider selection rules for electronic, rotational, and vibrational transitions. We consider a hydrogen atom. In order to observe emission of radiation from two states \(mu_z\) must be non-zero. That is \[(\mu_z)_{12}=\int\psi_1^{\,*}\,e\cdot z\;\psi_2\,d\tau\neq0\] For example, is the transition from \(\psi_{1s}\) to \(\psi_{2s}\) allowed? \[(\mu_z)_{12}=\int\psi_{1s}\,^{\,*}\,e\cdot z\;\psi_{2s}\,d\tau\] Using the fact that z = r cosq in spherical polar coordinates we have \[(\mu_z)_{12}=e\iiint\,e^{-r/a_0}r\cos \theta \biggr(2-\frac{r}{a_0}\biggr)e^{-r/a_0}r^2\sin\theta drd\theta\,d\phi\] We can consider each of the three integrals separately. \[\int_{0}^{\infty}e^{-r/a_0}r\biggr(2-\frac{r}{a_0}\biggr)e^{-r/a_0}r^2dr\int_{0}^{\pi}\cos\theta\sin\theta\,d\theta\int_{0}^{2\pi }d\phi\] If any one of these is non-zero the transition is not allowed. We can see specifically that we should consider the q integral. We make the substitution \(x = \cos q, dx = -\sin\; q\; dq\) and the integral becomes \[-\int_{1}^{-1}x dx=-\frac{x^2}{2}\Biggr\rvert_{1}^{-1}=0\] which is zero. The result is an even function evaluated over odd limits. In a similar fashion we can show that transitions along the x or y axes are not allowed either. This presents a selection rule that transitions are forbidden for \(\Delta{l} = 0\). For electronic transitions the selection rules turn out to be \(\Delta{l} = \pm 1\) and \(\Delta{m} = 0\). These result from the integrals over spherical harmonics which are the same for rigid rotator wavefunctions. We will prove the selection rules for rotational transitions keeping in mind that they are also valid for electronic transitions. We can use the definition of the transition moment and the to derive selection rules for a rigid rotator. Once again we assume that radiation is along the z axis. \[(\mu_z)_{J,M,{J}',{M}'}=\int_{0}^{2\pi } \int_{0}^{\pi }Y_{J'}^{M'}(\theta,\phi )\mu_zY_{J}^{M}(\theta,\phi)\sin\theta\,d\phi,d\theta\] Notice that m must be non-zero in order for the transition moment to be non-zero. This proves that a molecule must have a permanent dipole moment in order to have a rotational spectrum. The spherical harmonics can be written as \[Y_{J}^{M}(\theta,\phi)=N_{\,JM}P_{J}^{|M|}(\cos\theta)e^{iM\phi}\] where \(N_{JM}\) is a normalization constant. Using the standard substitution of \(x = \cos q\) we can express the rotational transition moment as \[(\mu_z)_{J,M,{J}',{M}'}=\mu\,N_{\,JM}N_{\,J'M'}\int_{0}^{2 \pi }e^{I(M-M')\phi}\,d\phi\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx\] The integral over f is zero unless M = M' so \(\Delta M = \) 0 is part of the rigid rotator selection rule. Integration over \(\phi\) for \(M = M'\) gives \(2\pi \) so we have \[(\mu_z)_{J,M,{J}',{M}'}=2\pi \mu\,N_{\,JM}N_{\,J'M'}\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx\] We can evaluate this integral using the identity \[(2J+1)x\,P_{J}^{|M]}(x)=(J-|M|+1)P_{J+1}^{|M|}(x)+(J-|M|)P_{J-1}^{|M|}(x)\] Substituting into the integral one obtains an integral which will vanish unless \(J' = J + 1\) or \(J' = J - 1\). \[\int_{-1}^{1}P_{J'}^{|M'|}(x)\Biggr(\frac{(J-|M|+1)}{(2J+1)}P_{J+1}^{|M|}(x)+\frac{(J-|M|)}{(2J+1)}P_{J-1}^{|M|}(x)\Biggr)dx\] This leads to the selection rule \(\Delta J = \pm 1\) for absorptive rotational transitions. Keep in mind the physical interpretation of the quantum numbers \(J\) and \(M\) as the total angular momentum and z-component of angular momentum, respectively. As stated above in the section on electronic transitions, these selection rules also apply to the orbital angular momentum (\(\Delta{l} = \pm 1\), \(\Delta{m} = 0\)). The harmonic oscillator wavefunctions are \[\psi_{\,v}(q)=N_{\,v}H_{\,v}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}\] where \(H_v(a1/2q)\) is a and a = (km/á2)1/2. The transition dipole moment for electromagnetic radiation polarized along the z axis is \[(\mu_z)_{v,v'}=\int_{-\infty}^{\infty}N_{\,v}N_{\,v'}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}H\mu_z(\alpha^{1/2}q)e^{-\alpha\,q^2/2}dq\] Note that we continue to use the general coordinate q although this can be z if the dipole moment of the molecule is aligned along the z axis. The transition moment can be expanded about the equilibrium nuclear separation. \[\mu_z(q)=\mu_0+\biggr({\frac{\partial\mu }{\partial q}}\biggr)q+.....\] where m0 is the dipole moment at the equilibrium bond length and q is the displacement from that equilibrium state. From the first two terms in the expansion we have for the first term \[(\mu_z)_{v,v'}=\mu_0\int_{-\infty}^{\infty}N_{\,v}N_{\,v'}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}H_v(\alpha^{1/2}q)e^{-\alpha\,q^2/2}dq\] This term is zero unless v = v’ and in that case there is no transition since the quantum number has not changed. \[(\mu_z)_{v,v'}=\biggr({\frac{\partial\mu }{\partial q}}\biggr)\int_{-\infty}^{\infty}N_{\,v}N_{\,v'}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}H_v(\alpha^{1/2}q)e^{-\alpha\,q^2/2}dq\] This integral can be evaluated using the Hermite polynomial identity known as a recursion relation \[xH_v(x)=vH_{v-1}(x)+\frac{1}{2}H_{v+1}(x)\] where x = Öaq. If we now substitute the recursion relation into the integral we find \[(\mu_z)_{v,v'}=\frac{N_{\,v}N_{\,v'}}{\sqrt\alpha}\biggr({\frac{\partial\mu }{\partial q}}\biggr)\] \[\int_{-\infty}^{\infty}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}\biggr(vH_{v-1}(\alpha^{1/2}q)+\frac{1}{2}H_{v+1}(\alpha^{1/2}q)\biggr)dq\] which will be non-zero if v’ = v – 1 or v’ = v + 1. Thus, we see the origin of the vibrational transition selection rule that v = ± 1. We also see that vibrational transitions will only occur if the dipole moment changes as a function nuclear motion. | 7,180 | 1,698 |
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So far, we have only discussed systems that are comprised of one component. Because a lot of chemistry occurs in mixtures or produces a mixture, chemists need to consider the thermodynamics of mixtures. A mixture can consist of many different components, however, for the sake of simplicity, we will restrict ourselves for now to two-component mixtures. Two-component mixtures can consist of two gases, two liquids, two solids, or even a liquid and a gas. Let's consider a two-component system where the volume and number of moles are changing. For example, we could have a system of a certain size of 1 and reduce its size stepwise in successive steps by taking half of it and throwing the other half away. The number of moles of each component, \(n_1\) and \(n_2\), will change as the volume of the system, \(V\), changes: \[dn_1=n_1dV \nonumber \] \[dn_2=n_2dV \nonumber \] The extensive Gibbs free energy will be affected the same way: \[dG = GdV \nonumber \] At constant \(T\) and \(P\) we can write: \[dG = \cancel{-SdT} + \cancel{VdP} + \mu_1dn_1+\mu _2dn_2 \nonumber \] So: \[dG = \mu _1dn_1+\mu _2dn_2 \nonumber \] \[GdV = \mu _1n_1dV+\mu _2n_2dV \nonumber \] If we integrate this from the original size, 1, down to 0 (or 0 to 1, it does not matter). We get: \[ \int_0^1 GdV= \int_0^1 \mu _1n_1dV+ \int_0^1 \mu _2n_2dV \nonumber \] \[G \int_0^1 dV= \mu _1n_1 \int_0^1 dV+\mu _2n_2 \int_0^1 dV \nonumber \] \[G=\mu _1n_1+\mu _2n_2 \nonumber \] By the same argument we have: \[V=\bar{V}_{1}n_1+\bar{V}_{2}n_2 \nonumber \] where \(\bar{V}_i\) is the partial molar volume for component \(i\). These partial molar volumes are generally a function of composition (and \(P\), \(T\)) and have been tabulated for a number of liquid systems. They allow us to calculate the real volume of a binary mixture. Volumes are generally speaking . This fact is typically ignored in volumetric analysis and the use of molarities. Fortunately the deviations are often negligible in dilute solutions. For phase diagrams, molarity (moles per liter) is not a very suitable quantity to use for concentration due to its volume dependence. Usually we work with mole fractions or molalities (moles per kilogram), where there are no volume dependencies. Gases can always mix in any ratio and mixtures typically act close to ideal unless heavily compressed or brought to low temperatures. The only exception is if the gases react (e.g, HCl and NH ). Gas molecules experience little interaction with each other and, therefore, it does not matter much whether the molecules are different or the same. The total pressure can be computed by adding the partial pressures of the two components (Dalton's Law of Partial Pressures): \[P_{total} = P_1 + P_2 \label{PreDalton} \] There are binary liquid systems that are fully miscible and are said to act as . Liquid molecules typically experience interactions with their neighbors. For the solution to be ideal, the interactions must remain even when the neighboring substance is different. This means they must be chemically similar. For this reason, liquid binaries are often ideal. The next nearest thing are . Even these systems can display phase segregation and limited mutual solubilities at low temperatures. Many liquid-liquid binaries diverge from ideality even more than the regular solutions and many of them are hardly miscible at all. Solid binaries tend to be even less miscible than liquid binaries to the point that immiscibility is the rule and miscibility is the exception. Even totally miscible systems like electrum (the alloys of silver and gold) are far from ideal. Another point of practical (kinetic rather than thermodynamic) importance is that even if two compounds are able to form a homogeneous solid solution, it usually takes heating for prolonged periods to get them to mix because solid diffusion is typically very slow. Nevertheless, solid solubility is an important issue for many systems, particularly for metal alloys. Two molecular solid substances that differ vastly in shape, size, polarity and or hydrogen bonding (e.g. organic compounds) typically have mutual solid solubility. The latter fact is frequently exploited in organic chemistry to purify compounds through recrystallization. Solid solutions are relatively infrequent and never ideal. Let's mix two liquids together. Liquids typically have different boiling points, with one being more volatile than the other. The vapor pressure of a component scales simply with the equilibrium vapor pressure of the pure component. In the gas phase, is applicable: \[y_i= \dfrac{P_i}{P_{total}} \label{Dalton} \] This is a consequence of the fact that ideal gases do not interact. The latter implies that the total pressure is simply the sum of the partial ones: \[P_{total} = \sum_i^N P_i \nonumber \] If the liquid solution is ideal, then the vapor pressure of both components follow , which states that the equilibrium vapor pressure above the mixture is the equilibrium pressure of the pure component times the mole fraction: \[P_i = x_iP^*_i \label{Raoult} \] Note that values for pure components are typically indicated by adding an asterisk * superscript. The idea behind Raoult's law is that if the interactions are similar, it is a matter of random chance which component sits at the interface at any given moment. The equilibrium vapor pressure has to do with the probability that a molecule escapes from the interface into the gas phase and is dependent on both the substances volatility and the number that cover the surface. This leads to Raoult's Law, where we must multiply the vapor pressure of the pure liquid (volatility) by the mole fraction (number on the surface). Raoult's law seldom holds completely, which is more applicable if the two components are almost chemically identical like two isomers, e.g., 1-propanol and 2-propanol. If we assume that temperature is constant, we can plot the total pressure for both Dalton and Raoult's laws versus composition (of gas: \(y_1\) and liquid: \(x_1\) on the same axis). \[P_{total} = P_1 + P_2 = x_1P^*_1 +x_2P^*_2 = x_1P^*_1 +(1-x_1)P^*_2= P^*_2- x_1(P^*_2-P^*_1) \label{liquidus} \] Clearly this is a straight line going from \(P^*_2\) at \(x=0\) to \(P^*_1\) at \(x=1\). However the composition of the vapor in equilibrium with a liquid at a given mole fraction \(x\) is than that of the liquid. So \(y\) is not \(x\). If we take Dalton's law (Equation \(\ref{Dalton}\)) and substitute Raoult's Law (Equation \(\ref{Raoult}\)) in the numerator and the straight line in the denominator we get: \[y_1 = \dfrac{x_1P^*_1}{P^*_2- x_1(P^*_2-P^*_1)} \label{vaporus1} \] Suppose \(P^*_1 = 50\) Torr and \(P^*_2 = 25\) Torr. If (X_1= 0.6\) what is the composition of the vapor? We can rearrange Equations \(\ref{liquidus}\) and \(\ref{vaporus1}\) to plot the total pressure as function of \(y_1\): \[P_{total}= \dfrac{P^*_1P^*_2 }{P^*_1 + (P^*_2-P^*_1)y_1} \label{vaporus2} \] This is a straight line. As you can see when we plot both lines we get a diagram with three regions. At high pressures we just have a liquid. At low pressures we just have a gas. In between we have a or . Points inside this region represent states that the system cannot achieve homogeneously. The horizontal shows which phases coexists. I used the same 25 and 50 Torr values for the pure equilibrium pressures as in the question above. If you try to make a system with composition x and impose a pressure that falls in the forbidden zone you get two phases: a gaseous one that is richer in the more volatile component and a liquid one that is poorer in the volatile component than the overall composition would indicate. Note that the question: is really a function of , so that if we want to represent our knowledge in a diagram we should make it a three dimensional picture. This is not so easy to draw and not easy to comprehend visually either. This is why we usually look at a 2D cross section of the 3D space. The above diagram is : we vary \(P\), keeping \(T\) constant. It is, however, more usual (and easier) to do it the other way around. We keep pressure constant (say 1 bar, that's easy) and start heating things up isobarically. The boiling points of our mixtures can also be plotted against \(x\) (the liquid composition) and \(y\) (the gaseous one) on the same horizontal axis. Again because in general \(y\) is not equal \(x\) we get two different curves. Neither of them are straight lines in this case and we end up with a lens-shaped two phase region: What happens to a mixture with a given overall composition x(=x ) when it is brought to a temperature where it boils can be seen at the intersection of a vertical line (an isopleth) at \(x_{overall}\) and a horizontal one (an isotherm) at \(T_{boil}\). If the intersection points in inside the two phase region a vapor phase and a liquid phase result that have a different composition from the overall one. The vapor phase is always richer in the more volatile component (the one with the lowest boiling point, on the left in the diagram). The liquid phase is enriched in the less volatile one. How much of each phase is present is represented by the arrows in the diagram. The amount of liquid is proportional to the left arrow, the amount of gas to the right one (i.e. it works crosswise). The composition of the liquid in equilibrium with the vapor is: \[x_2 = \dfrac{n^{liq}_2}{n^{liq}_{1+2}} \nonumber \] \[x_2^*n^{liq}_{1+2} = n^{liq}_2 \nonumber \] The composition of the vapor is: \[y_2 = \dfrac{n^{gas}_2}{n^{gas}_{1+2}} \nonumber \] \[y_2^*n^{gas}_{1+2} = n^{gas}_2 \nonumber \] The overall composition is: \[x_{all} = \dfrac{n^{liq+gas}_2 }{n^{liq+gas}_{1+2}} \nonumber \] \[x_{all}^*n^{liq+gas}_{1+2} = n^{gas}_2+n^{liq}_2 \nonumber \] \[x_{all}^*n^{liq+gas}_{1+2} = y_2^*n^{gas}_{1+2}+x_2^*n^{liq}_{1+2} \nonumber \] \[x_{all}^*n^{gas}_{1+2}+x_{all}^*n^{liq}_{1+2} = y_2^*n^{gas}_{1+2}+x_2^*n^{liq}_{1+2} \nonumber \] Thus: \[ \dfrac{n^{liq}_{1+2} }{n^{gas}_{1+2}} = \dfrac{y_2-x_{all}}{ x_{all}-x_2} \nonumber \] The difference in composition between the gas and the liquid can be exploited to separate the two components, at least partially. We could trap the vapor and cool it down to form a liquid with a different composition. We could then boil it again and repeat the process. Each time the vapor will be more enriched in the volatile phase whereas the residual liquid is more enriched in the less volatile one. This process is known as . In practice the process is done one a fractionation column which makes it possible to have a series of vapor-liquid equilibria at once. A good degree of purity can be reached this way, although 100% purity would take an infinite number of distillation steps. | 10,788 | 1,699 |
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Suppose you are asked to develop an analytical method to determine the concentration of lead in drinking water. How would you approach this problem? To provide a structure for answering this question, it is helpful to consider four levels of analytical methodology: techniques, methods, procedures, and protocols [Taylor, J. K. , , 600A–608A]. A is any chemical or physical principle that we can use to study an analyte. There are many techniques for that we can use to determine the concentration of lead in drinking water [Fitch, A.; Wang, Y.; Mellican, S.; Macha, S. , , 727A–731A]. In graphite furnace atomic absorption spectroscopy (GFAAS), for example, we first convert aqueous lead ions into free atoms—a process we call atomization. We then measure the amount of light absorbed by the free atoms. Thus, GFAAS uses both a chemical principle (atomization) and a physical principle (absorption of light). See for a discussion of graphite furnace atomic absorption spectroscopy. A is the application of a technique for a specific analyte in a specific matrix. As shown in Figure 3.2.1
, the GFAAS method for determining the concentration of lead in water is different from that for lead in soil or blood. A is a set of written directions that tell us how to apply a method to a particular sample, including information on how to collect the sample, how to handle interferents, and how to validate results. A method may have several procedures as each analyst or agency adapts it to a specific need. As shown in Figure 3.2.1
, the American Public Health Agency and the American Society for Testing Materials publish separate procedures for determining the concentration of lead in water. Finally, a is a set of stringent guidelines that specify a procedure that an analyst must follow if an agency is to accept the results. Protocols are common when the result of an analysis supports or defines public policy. When determining the concentration of lead in water under the Safe Drinking Water Act, for example, the analyst must use a protocol specified by the Environmental Protection Agency. There is an obvious order to these four levels of analytical methodology. Ideally, a protocol uses a previously validated procedure. Before developing and validating a procedure, a method of analysis must be selected. This requires, in turn, an initial screening of available techniques to determine those that have the potential for monitoring the analyte. | 2,479 | 1,701 |
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The formation of ionic compounds is very exothermic Removing an electron from an atom, such as Na, is endothermic because energy needs to be used to overcome the attractive forces within the atom. Adding an electron is the opposite process and releases lots of energy The principal reason that ionic compounds are stable is the attraction between ions of unlike charge. This attraction draws the ions together, releasing energy and causing the ions to form a solid array (lattice) : energy required to separate completely a mole of a solid ionic compounds into its gaseous ions Large values of lattice energy mean that the ions are strongly attracted to one another Energy released by the attraction between the ions of unlike charges more than makes up for the endothermic nature of ionization energies, making the formation of ionic compounds an exothermic process \[E = k \dfrac{Q_1Q_2}{ d} \nonumber \] For a given arrangement of ions, the lattice energy increases as the charges of ions increase and as their radii decrease. The magnitude of lattice energies depends primarily on the ionic charges because ionic radii do not vary over a wide range. Many ions tend to have noble gas electron configurations. This is why Na can have a +1 charge, but not a +2 one. Once an ion has reached noble gas configuration, it wants to stay there. Similarly, addition of electrons to nonmetals is either exothermic or slightly endothermic as long as electrons are being added to the valence shell. Further addition of electrons requires tremendous amounts of energy; more than is available form the lattice energy The lattice energies of ionic compounds are generally large enough to compensate for the loss of up to only 3 electrons from atoms. Thus we find cations only having charges of +1, +2, or +3. Because most transition metals have more than 3 electrons beyond a noble gas core, attainment of a noble gas configuration for these ions is usually impossible. When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest value of . Thus, a transition metal always loses the outer electrons before it loses electrons from the underlying subshell. Sizes of ions are important in determining both the way in which the ions pack in a solid and the lattice energy of the solid. It is also a major factor governing the properties of ions in solution The size of an atom depends on its nuclear charge, the number of electrons it possesses, and the orbitals in which the outer-shell electrons reside Positive ions are formed by removing 1 or more electrons from the outermost region of the atom. Thus, the formation of a cation not only vacates the most spatially extended orbitals, it also decreases the total electron-electron repulsions. Hence, cations are smaller than the original atoms from which they came. The opposite happens when speaking of negative ions. An added electron increases electron-electron repulsions and causes the electrons to spread out more in space. For ions of the same charge, size increases as we go down a group Ionic substances are usually brittle with high melting points. They are usually crystalline, meaning that they have flat surfaces that make characteristic angles with one another. For nonmetals, the number of valence electrons is the same as the group number Knowing this, we can predict that an element in Group 7A would need one covalent bond in order to get an octet, an element in Group 6A would need two, and so on. Distance between bonded atoms decreases as the number of shared electron pairs increases Used to estimate whether a bond will be nonpolar, polar covalent, or ionic : ability of an atom to attract electrons to itself An atom with a very negative electron affinity and high ionization energy will both attract electrons from other atoms and resist having its electrons attracted away; it will be highly electronegative Highest electronegativity = 4.0 (Fluorine), lowest = 0.7 (Cesium) Electronegativity increases form left to right, and usually decreases with increasing atomic number in any one group Differences in electronegativities: Nonpolar = 0 – 0.4 Polar covalent = 0.4 – 1.6 Ionic = > 1.6 (> 50% = ionic) : "delta sign"; represent partial positive and negative charge. The atom with the is the more electronegative one are individual Lewis structures in cases where two or more Lewis structures are equally good descriptions of a single molecule. If a molecule (or ion) has two or more resonance structures, the molecule is a blend of these structures. The molecule oscillate rapidly between two or more different forms. In a few molecules, such as ClO , , and NO , the number of electrons is odd. In NO for example, there are 5+6 = 11 valence electron. Hence, complete pairing of these electrons is impossible and an octet around each atom cannot be achieved. Second type of exception occurs when there are fewer than eight electrons around an atom in a molecule or ion. Relatively rare situation; most often encountered in compounds of Boron and Beryllium. For example, let’s consider Boron Trifluoride, BF There are 6 electrons around the Boron atom. We can form a double bond between Boron and any of the 3 Fluorine atoms (3 possible resonance structures) However, by doing so, we forced a Fluorine atom to share additional electrons with Boron. This would make the F atom to have a +1 charge, and the Boron atom to have a –1 charge, which is extremely unfavorable. We then conclude that the structures containing the double bonds are less important than the one illustrated on the right. The octet rule works as well as it does because the representative elements usually employ only an and three valence shell orbitals in bonding, and these hold eight electrons. Because elements of the second period have only 2s and 2p orbitals, they cannot have more than an octet of electrons in their valence shells. However, from the third period on, the elements have unfilled orbitals that can be used in bonding. Size also plays an important role in determining whether an atom can accommodate more than eight electrons. The larger the central atom, the larger the number of atoms that can surround it. The size of the surrounding atoms is also important. Expanded valence shells occur most often when the central atom is bonded to the smallest and most electronegative atoms. : aka substance. For polyatomic molecules, we must often utilize average bond energies. Bond energy is positive, the greater the bond energy, the stronger the bond A molecule with strong bonds generally has less tendency to undergo chemical change than does one with weak bonds Cl – Cl + H – CH → H – Cl + Cl – CH Bonds broken: 1 mol Cl – Cl, 1 mol C – H Bonds made: 1 mol H – Cl, 1 mol C – Cl \[∆H = [D (Cl – Cl) + D(C – H)] [D (H – Cl) + D (Cl – Cl)] \nonumber \] = (242 kJ + 413 kJ) – (431kJ + 328 kJ) = 104 kJ As the number of bonds between a given element increase, the bond energy increases and the bond length decreases. Hence, the atoms are held more tightly and closely together. In general, : aka ; a positive or negative whole number assigned to an element in a molecule or ion on the basis of a set of normal rules; to some degree it reflects the positive or negative character of an atom Oxidation numbers do correspond to real charges on the atoms, in the special case of simple ionic substances Group 1A elements are +1, Group 2A elements are +2, and Aluminum is +3. The most electronegative element, F, is always found in the –1 oxidation state. Oxygen is usually in the –2 state, however, it can be –1 in peroxides. Hydrogen has an oxidation number of +1 when it is bonded to a more electronegative element (most nonmetals), and of –1 when bonded to less electronegative elements (most metals) Name of the less electronegative element is given first, followed by the name of the more electronegative element modified to have an ending Compounds of metals in higher oxidation states tend to be molecular rather than ionic | 8,074 | 1,702 |
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Previously, we described the two idealized extremes of chemical bonding: Most compounds, however, have polar covalent bonds, which means that electrons are shared between the bonded atoms. Figure \(\Page {1}\) compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta (\(\delta\)) is used to indicate that a bonded atom possesses a partial positive charge, indicated by \(\delta^+\), or a partial negative charge, indicated by \(\delta^-\), and a bond between two atoms that possess partial charges is a . The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. (\(\chi\)) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is toward the more electronegative atom. A bond in which the electronegativity of B (\(\chi_B\)) is greater than the electronegativity of A (\(\chi_A\)), for example, is indicated with the partial negative charge on the more electronegative atom: \[ \begin{matrix}
_{less\; electronegative}& & _{more\; electronegative}\\
A\; &-& B\; \; \\
^{\delta ^{+}} & & ^{\delta ^{-}}
\end{matrix} \label{9.3.1} \] One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms: \[Δ\chi = \chi_B − \chi_A. \nonumber \] To predict the polarity of the bonds in Cl , HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms ( ): \(\chi_{Cl} = 3.16\), \(\chi_H = 2.20\), and \(\chi_{Na} = 0.93\). \(\ce{Cl2}\) must be nonpolar because the electronegativity difference (\(Δ\chi\)) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, \(Δ\chi\) is 2.23. This high value is typical of an ionic compound (\(Δ\chi ≥ ≈1.5\)) and means that the valence electron of sodium has been completely transferred to chlorine to form Na and Cl ions. In HCl, however, \(Δ\chi\) is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is \[ \begin{matrix}
_{\delta ^{+}}& & _{\delta ^{-}}\\
H\; \; &-& Cl
\end{matrix} \nonumber \] Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated. Bond polarity and ionic character increase with an increasing difference in electronegativity. As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in \(\ce{NaCl}\), \(\ce{Cl2}\), \(\ce{ClF5}\), and \(\ce{HClO4}\) would be exactly the same. The asymmetrical charge distribution in a polar substance such as \(\ce{HCl}\) produces a dipole moment \( Qr \) is abbreviated by the Greek letter mu (\(\mu\)). The dipole moment is defined as the product of the partial charge \(Q\) on the bonded atoms and the distance \(r\) between the partial charges: \[ \mu=Qr \label{9.3.2} \] where \(Q\) is measured in coulombs (\(C\)) and \(r\) in meters. The unit for dipole moments is the debye (D): \[ 1\; D = 3.3356\times 10^{-30}\; C\cdot m \label{9.7.3} \] When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure \(\Page {2}\)). We can measure the partial charges on the atoms in a molecule such as \(\ce{HCl}\) using Equation \ref{9.3.2}. If the bonding in \(\ce{HCl}\) were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of \(\ce{HCl}\) is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is \[ \begin{align*} Q &=\dfrac{\mu }{r} \\[4pt] &=1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right ) \\[4pt] &=2.901\times 10^{-20}\;C \label{9.7.4} \end{align*} \] By dividing this calculated value by the charge on a single electron (1.6022 × 10 C), we find that the electron distribution in\(\ce{HCl}\) is asymmetric and that effectively it appears that there is a net negative charge on the \(\ce{Cl}\) of about −0.18, effectively corresponding to about 0.18 e . This certainly does not mean that there is a fraction of an electron on the \(\ce{Cl}\) atom, but that the distribution of electron probability favors the Cl atom side of the molecule by about this amount. \[ \dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{9.7.5} \] To form a neutral compound, the charge on the \(\ce{H}\) atom must be equal but opposite. Thus the measured dipole moment of \(\ce{HCl}\) indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing\(\ce{HCl}\) as \( \begin{matrix}
_{\delta ^{+}}& & _{\delta ^{-}}\\
H\; \; &-& Cl
\end{matrix} \) we can therefore indicate the charge separation quantitatively as \( \begin{matrix}
_{0.18\delta ^{+}}& & _{0.18\delta ^{-}}\\
H\; \; &-& Cl
\end{matrix} \) Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine (\(\chi_H = 2.20\) and \(\chi_{Cl} = 3.16\)) so \[\chi_{Cl} − \chi_H = 0.96 \nonumber \] This is a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule. In HCl, for example, the dipole moment is indicated as follows: The arrow shows the direction of electron flow by pointing toward the more electronegative atom. As the figure above shows, we represent dipole moments by an arrow with a length proportional to \(μ\) and pointing from the positive charge to the negative charge. However, the convention is still widely used especially among physicists. The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure \(\Page {2}\) shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as \(NaCl_{(g)}\) and \(CsF_{(g)}\) is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure \(\Page {2}\) show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example \(\Page {1}\). In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl. chemical species, dipole moment, and internuclear distance percent ionic character The charge on each atom is given by \[ \begin{align*} Q &=\dfrac{\mu }{r} \\[4pt] &=9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right ) \\[4pt] &=1.272\times 10^{-19}\;C \end{align*} \nonumber \] Thus NaCl behaves as if it had charges of 1.272 × 10 C on each atom separated by 236.1 pm. The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron): \[ \begin{align*} \text{ ionic character} &=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right ) \\[4pt] &=79.39\%\simeq 79\% \end{align*} \nonumber \] In the gas phase, silver chloride (\(\ce{AgCl}\)) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride? 55.5% Bond polarity and ionic character increase with an increasing difference in electronegativity. \[ \mu = Qr \nonumber \] Compounds with have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a , which is the product of the partial charges on the bonded atoms and the distance between them. ( ) | 9,345 | 1,703 |
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Let us turn our attention from the macroscopic to the microscopic level. According to the first law of thermodynamics, the heat energy absorbed as we raise the temperature of a substance cannot be destroyed. But where does it go? In the case of a monatomic gas, like neon, this question is easy to answer. All the energy absorbed is converted into the kinetic energy of the neon molecules (atoms). , we found that the kinetic energy of the molecules in a sample of gas is given by the expression \[E_{k}=\tfrac{\text{3}}{\text{2}}nRT \nonumber \] Thus if the temperature of a sample of neon gas is raised from to , the kinetic energy of the molecules increases from / to / , a total change of Inserting the value of in appropriate units, we obtain This is the same quantity that is obtained by substituting the experimental value of for neon (calculated in ) into . In other words the quantity of heat found experimentally exactly matches the increase in kinetic energy of the molecules required by the kinetic theory of gases. Table \(\Page {1}\) lists the values not only for neon but for some other gases as well. We immediately notice that only the noble gases and other mon-atomic gases such as Hg and Na have molar heat capacities equal to / , or 12.47 J K mol . All other gases have higher molar heat capacities than this. Moreover, as the table shows, the more complex the molecule, the higher the molar heat capacity of the gas. There is a simple reason for this behavior. A molecule which has two or more atoms is not only capable of moving from one place to another ( ), it can also about itself, and it can change its shape by . When we heat a mole of Cl molecules, for example, we not only need to supply them with enough energy to make them move around faster (increase their translational kinetic energy), we must also supply an additional quantity of energy to make them rotate and vibrate more strongly than before. For a mole of more complex molecules like -butane even more energy is required since the molecule is capable of changing its shape in all sorts of ways. In the butane molecule there are three C—C bonds around which segments of the molecule can rotate freely. All the bonds can bend or stretch, and the whole molecule can rotate as well. Such a molecule is constantly flexing and writhing at room temperature. As we raise the temperature, this kind of movement occurs more rapidly and extra energy must be absorbed in order to make this possible. When we heat and , the situation is somewhat different than for . The rapid increase of vapor pressure with temperature makes it virtually impossible to heat a solid or liquid in a closed container, and so heat capacities are always measured at constant pressure rather than at constant volume. Some values for selected simple liquids and solids at the melting point are shown in Table \(\Page {2}\). In general the of solids and liquids are higher than those of gases. This is because of the intermolecular forces operating in solids and liquids. When we heat solids and liquids, we need to supply them with potential energy as well as kinetic energy. Among the solids, the heat capacities of the metals are easiest to explain since the solid consists of individual atoms. Each atom can only vibrate in three dimensions. According to a theory first suggested by Einstein, this vibrational energy has the value 3 , while the heat capacity is given by 3 = 24.9 J K mol . As can be seen from the table, most monatomic solids have values slightly larger than this. This is because solids expand slightly on heating. The atoms get farther apart and thus increase in potential as well as vibrational energy. Solids which contain molecules rather than atoms have much higher heat capacities than 3 . In addition to the vibration of the whole molecule about its site in the crystal lattice, the individual atoms can also vibrate with respect to each other. Occasionally molecules can rotate in the crystal, but usually rotation is only possible when the solid melts. As can be seen from the values for molecular liquids in Table \(\Page {2}\), this sudden ability to rotate causes a sharp increase in the heat capacity. For monatomic substances, where there is no motion corresponding to the rotation of atoms around each other, the heat capacity of the liquid is only very slightly higher than that of the solid. | 4,437 | 1,704 |
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Alkylmagnesium reagents, alkylcuprates and complex hydrides can all react with carboxyloids. When they do, a carbon or hydrogen nucleophile bonds to the carbonyl carbon, usually replacing the leaving group at that position. Another common carbon nucleophile is an enolate ion. Enolate ions can also react with carboxyloids, although not typically with amides. Probably the most common enolate reaction involving carboxyloids is the reaction of esters. If a strong base is added to solution of ester, some of the esters will become deprotonated, forming enolate anions. These ions will be nucleophiles. Some esters will remain protonated. These esters will be electrophiles. Donation of the enolate to the ester, with subsequent loss of the leaving group, leads to a beta-ketoester. In principle, this reaction could conceivably go backwards. The enolate ion could potentially be displaced by an alkoxide to get back to an ester and an enolate ion. That's because the enolate ion is a relatively stable ion, and a moderately good leaving group. However, that generally doesn't happen. Under basic conditions, the beta-ketoester is usually deprotonated, forming a particularly stable ion. This ion formation acts as a "thermodynamic sink" for the reaction, pulling it forward until all of the ester has been consumed. Show why the ion that results from deprotonation of the beta-ketoester is particularly stable. The formation of a beta-ketoester from two esters is called a "Claisen condensation". It is often followed by another important reaction: decarboxylation. If a beta-ketoester is treated with aqueous acid and heated, a couple of reactions take place. First, the ester portion of the molecule is converted into a carboxylic acid. Second, the carboxylic acid is decarboxylated. Carbon dioxide is formed, and the organic molecule becomes a ketone. Decarboxylation is related to the retro-aldol reaction; formally, it can be thought of as leading to an enolate leaving group. Decarboxylation most commonly occurs in beta-ketoacids, rather than in other carboxylic acids. Otherwise, that leaving group could not occur. The ease of decarboxylation in beta-ketoacids is related to the stability of the enolate anion. Under acidic conditions, of course, an enolate anion does not occur; instead, an enol is formed. Draw a mechanism for: , | 2,353 | 1,705 |
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We will be restricting ourselves to the canonical ensemble (constant temperature and constant pressure). Consider a collection of \(N\) molecules. The probability of finding a molecule with energy \(E_i\) is equal to the fraction of the molecules with energy \(E_i\). That is, in a collection of \(N\) molecules, the probability of the molecules having energy \(E_i\): \[P_i = \dfrac{n_i}{N} \nonumber \] This is the directly obtained from the Boltzmann distribution, where the fraction of molecules \(n_i /N\) having energy \(E_i\) is: \[P_i = \dfrac{n_i}{N} = \dfrac{e^{-E_i/kT}}{Q} \label{BD1} \] The average energy is obtained by multiplying \(E_i\) with its probability and summing over all \(i\): \[ \langle E \rangle = \sum_i E_i P_i \label{Mean1} \] Equation \(\ref{Mean1}\) is the standard average over a distribution commonly found in quantum mechanics as . The quantum mechanical version of this Equation is \[ \langle \psi | \hat{H} | \psi \rangle \nonumber \] where \(\Psi^2\) is the distribution function that the Hamiltonian operator (e.g., energy) is averaged over; this equation is also the starting point in the Variational method approximation. Equation \(\ref{Mean1}\) can be solved by plugging in the Boltzmann distribution (Equation \(\ref{BD1}\)): \[ \langle E \rangle = \sum_i{ \dfrac{E_ie^{-E_i/ kT}}{Q}} \label{Eq1} \] Where \(Q\) is the partition function: \[ Q = \sum_i{e^{-\dfrac{E_i}{kT}}} \nonumber \] We can take the derivative of \(\ln{Q}\) with respect to temperature, \(T\): \[ \left(\dfrac{\partial \ln{Q}}{\partial T}\right) = \dfrac{1}{kT^2}\sum_i{\dfrac{E_i e^{-E_i/kT}}{Q}} \label{Eq2} \] Comparing Equation \(\ref{Eq1}\) with \(\ref{Eq2}\), we obtain: \[ \langle E \rangle = kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right) \nonumber \] It is common to write these equations in terms of \(\beta\), where: \[ \beta = \dfrac{1}{kT} \nonumber \] The partition function becomes: \[ Q = \sum_i{e^{-\beta E_i}} \nonumber \] We can take the derivative of \(\ln{Q}\) with respect to \(\beta\): \[ \left(\dfrac{\partial \ln{Q}}{\partial\beta}\right) = -\sum_i{\dfrac{E_i e^{-\beta E_i}}{Q}} \nonumber \] And obtain: \[ \langle E \rangle = -\left(\dfrac{\partial \ln{Q}}{\partial\beta}\right) \nonumber \] Replacing \(1/kT\) with \(\beta\) often simplifies the math and is easier to use. It is not uncommon to find the notation changes: \(Z\) instead of \(Q\) and \(\bar{E}\) instead of \( \langle E \rangle \). | 2,468 | 1,706 |
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The most readily available alkenyl halide is chloroethene (vinyl chloride), which can be prepared by a number of routes: The most economical commercial preparation is high-temperature chlorination of ethene. A useful modification of this process uses hydrogen chloride in place of chlorine. An oxidizing agent is required to raise the oxidation state of chlorine in \(\ce{HCl}\) to that of \(\ce{Cl_2}\); molecular oxygen is used for this purpose along with cupric salts as catalysts. General methods of preparation for alkenyl and alkynyl halides are listed in Table 14-5. By the alkynyl halides we mean 1-halo-alkynes. One interesting method by which they may be prepared employs 1-alkynes with hypohalites: This kind of reaction does not proceed with either alkanes or alkenes. Chloroethene is produced in vast quantities for the production of polymers (polyvinyl chloride) and copolymers: These polymers commonly are described as PVC plastics or less specifically as “vinyl.” They are materials that may be either flexible or rigid according to what they are mixed with, and they are used in the manufacture of many familiar articles such as plastic curtains, rainwear, floor tile, synthetic leather goods, upholstery, table mats, phonograph records, insulation, plastic pipes, tubing, and packaging materials. Recently, it has been found that persons working in plants that manufacture and use chloroethene have an unusually high incidence of an unusual type of liver cancer. As a result, strict safety regulations and pollution standards have been set for plants where chloroethene is made or used. The once widespread use of chloroethene as a propellant for aerosol cans has been curtailed. Polyvinyl chloride itself seems to be quite safe, but there are possible problems with its incorporation into interior building materials, clothing, and upholstery because heat, such as fire, causes polyvinyl chloride to decompose, thereby producing hydrogen chloride as one decomposition product. In closed areas the toxicity of hydrogen chloride gas may be as serious a hazard as the fire itself. Other polymers may give off similarly toxic products on strong heating. The outstanding chemical characteristic of alkenyl halides is their general inertness in \(S_\text{N}1\) and \(S_\text{N}2\) reactions. Thus chloroethene fails to react with silver nitrate in ethanol (i.e., low \(S_\text{N}1\) reactivity), fails to react with potassium iodide in acetone (i.e., low \(S_\text{N}2\) reactivity), and only reacts slowly with sodium hydroxide to give ethyne (low \(E2\) reactivity). The haloalkynes, such as \(\ce{RC \equiv C-Cl}\), are similarly unreactive. It is not surprising that \(\ce{=C-X}\) and \(\ce{\equiv C-X}\) bonds are hard to break heterolytically. In general, \(\ce{C-X}\) bonds are strong in alkenyl halides (cf. Table 4-6) and this property tends to make them less reactive than alkyl halides. Furthermore, double- and triple-bonded carbons are more strongly electron-attracting than saturated \(sp^3\) carbons, which is the reason why 1-alkynes and alkenes are stronger acids ( ) than alkanes. Consequently it is easier to break a \(\ce{\equiv C-H}\) bond in the sense \(\ce{C}^\ominus \ce{H}^\oplus\) than as \(\ce{\equiv C}^\oplus \ce{H}^\ominus\). It also will be more difficult to ionize a carbon-halogen bond to \(\ce{C}^\oplus \ce{X}^\ominus\) if the carbon is unsaturated. Therefore ethenyl and ethynyl cations, such as \(\ce{CH_2=CH}^\oplus\) and \(\ce{HC \equiv C}^\oplus\), are difficult to generate from the corresponding halides. Superior leaving groups are required, such as trifluoromethanesulfonate, \(\ce{-OSO_3CF_3}\) (Section 8-7C): The reason for the lack of \(S_\text{N}2\) reactivity in ethenyl or ethynyl halides may be that the attacking nucleophile is unable to react by the concerted inversion mechanism that invariably is observed with alkyl halides: Nevertheless, substitution of the halogen does occur under some circumstances. In such cases, the nucleophile first to the multiple bond, and in a subsequent step the halogen leaves as halide ion. This is an “addition-elimination” mechanism, of which we will have more examples later: and (1977) | 4,211 | 1,707 |
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In the attempt to measure the heat equivalent of mechanical work (as Joule did in his famous experiment) a student uses an apparatus similar to that shown below: The 1.50 kg weight is lifted 30.0 cm against the force due to gravity (9.8 N). If the specific heat of water is 4.184 J/(g °C), what is the expected temperature increase of the 1.5 kg of water in the canister? 1.00 mol of an ideal gas, initially occupying 12.2 L at 298 K, expands isothermally against a constant external pressure of 1.00 atm until the pressure of the gas is equal to the external pressure. Calculate \(\Delta p\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the expansion. Consider 1.00 mol of an ideal gas expanding isothermally at 298 K from an initial volume of 12.2 L to a final volume of 22.4 L. Calculate \(\Delta p\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the expansion. Consider 1.00 mol of an ideal gas (C = 3/2 R) Occupying 22.4 L that undergoes an isochoric (constant volume) temperature increase from 298 K to 342 K. Calculate \(\Delta p\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the change. Consider 1.00 mol of an ideal gas (C = 5/2 R) initially at 1.00 atm that undergoes an isobaric expansion from 12.2 L to 22.4 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the change. Consider 1.00 mol of an ideal gas (C = 3/2 R) initially at 12.2 L that undergoes an adiabatic expansion to 22.4 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the change. Derive an expression for the work of an isothermal, reversible expansion of a gas that follows the equation of state (in which \(a\) is a parameter of the gas) \[ pV = nRT -\dfrac{an^2}{V}\] from \(V_1\) to \(V_2\). Use the following data [Huff, Squitieri, and Snyder, J. Am. Chem. Soc., , 3380 (1948)] to calculate the standard enthalpy of formation of tungsten carbide, \(WC(s)\). The standard molar enthalpy of combustion (\(\Delta H_c\)) of propane gas is given by \[C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l)\] with \(\Delta H_c = -2220 \,kJ/mol\) The standard molar enthalpy of vaporization (\(\Delta H_{vap}\)) for liquid propane \[C_3H_8(l) \rightarrow C_3H_8(g)\] with \(\Delta H_{vap} = 15\, kJ/mol\) The enthalpy of combustion (\(\Delta H_c\)) of aluminum borohydride, \(Al(BH_4)_3(l)\), was measured to be -4138.4 kJ/mol [Rulon and Mason, , , 5491 (1951)]. The combustion reaction for this compound is given by \[ Al(BH_4)_3(l) + 6 O_2(g) \rightarrow ½ Al_2O_3(s) + 3/2 B_2O_3(s) + 6 H_2O(l)\] Given the following additional data, calculate the enthalpy of formation of \(Al(BH_4)_3(g)\). The standard enthalpy of formation (\(\Delta H_f^o\)) for water vapor is -241.82 kJ/mol at 25 °C. Use the data in the following table to calculate the value at 100 °C. \(\Delta C_p = (1.00 + 2.00 \times 10^{-3} T)\, J/K\) and \(\Delta H_{298} = -5.00\, kJ\) for a dimerization reaction \[2 A \rightarrow A_2\] Find the temperature at which \(\ H = 0\). From the following data, determine the lattice energy of \(BaBr_2\). \[Ca(s) \rightarrow Ca(g)\] with \(\Delta H_{sub} = 129\, kJ/mol\) \[Br_2(l) \rightarrow Br_2(g)\] with \(\Delta H_{vap} = 31\, kJ/mol\) \[Br_2(g) \rightarrow 2 Br(g)\] with \(D(Br-Br) = 193 \, kJ/mol\) \[Ca(g) \rightarrow Ca^+(g) + e^-\] with \(1^{st} \, IP(K) = 589.8 \, kJ/mol\) \[Ca^+(g) \rightarrow Ca^{2+}(g) + e^-\] with \(2^{nd} IP(K) = 1145.4 \,kJ/mol\) \[Br(g) + e^- \rightarrow Br-(g) \] with \(1^{st} EA(Br) = 194 \, kJ/mol\) \[Ca(s) + Br_2^-(l) \rightarrow CaBr_2(s)\] with \(\Delta H_f = -675 \, kJ/mol\) Using average bond energies ( ) estimate the reaction enthalpy for the reaction \[C_2H_4 + HBr \rightarrow C_2H_5Br\] | 3,722 | 1,708 |
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Until now, we have discussed bonding only in terms of electron pair associated with two nuclei. These we may call electrons. In fact, bonding electrons can be associated with more than two nuclei, and there is a measure of stability to be gained by this because the degree of bonding increases when the electrons can distribute themselves over a greater volume. This effect often is called or . It is important only if the component atomic orbitals overlap significantly, and this will depend in large part on the molecular geometry. The classic example of resonance is provided by the \(\pi\) bonding of benzene. This compound was shown in Chapter 1 to have the molecular formula \(C_6H_6\), to be planar, and hexagonal with bond angles of \(120^\text{o}\), and to possess six equivalent \(C-C\) bonds and six equivalent \(C-H\) bonds. Benzene usually is written with a structural formula proposed by Kekulé: That benzene is more stable than a single Kekulé, or 1,3,5-cyclohexatriene, structure can be gauged by comparing the experimental heat of combustion of benzene with the calculated value based on the average bond energies of Table 4-3: \[\ce{C6H6(g) + 15/2 O2 -> 6CO2(g) + 3H2O(g)}\] with \[\begin{aligned}
&\Delta H_{\exp }^{0}=-789 \mathrm{kcal} \\
&\Delta H_{\text {calc }}^{0}=-827 \mathrm{kcal}
\end{aligned}\] About \(38 \: \text{kcal}\) energy is released on combustion than calculated. Benzene, therefore, is \(38 \: \text{kcal mol}^{-1}\) than the cyclohexatriene structure predicts. Atomic-orbital models, like that shown for benzene, are useful descriptions of bonding from which to evaluate the potential for electron delocalization. But they are cumbersome to draw routinely. We need a simpler representation of electron delocalization. The method that commonly is used is to draw a set of structures, each of which represents a reasonable way in which the electrons (usually in \(p\) orbitals) could be paired. If more than one such structure can be written, the actual molecule, ion, or radical will have properties corresponding to some hybrid of these structures. A double-headed arrow \(\leftrightarrow\) is written between the structures that we consider to contribute to the hybrid. For example, the two Kekulé forms are two possible electron-pairing schemes or that could contribute to the resonance hybrid of benzene: It is very important to know what attributes a reasonable set of valence-bond structures has to have to contribute to a hybrid structure. It is equally important to understand what is and what is not implied in writing a set of structures. Therefore we shall emphasize the main points to remember in the rest of this section. 1. The members of a set of structures, as the two Kekulé structures for benzene, have no individual reality. They are hypothetical structures representing different electron-pairing schemes. We are not to think of benzene as a 50:50 mixture of equilibrating Kekulé forms. 2. To be reasonable, all structures in a set representing a resonance hybrid must have exactly the same locations of the atoms in space. For example, formula \(7\) does represent a valid member of the set of valence-bond structures of benzene, because the atoms of \(7\) have different positions from those of benzene (e.g., \(7\) is not planar): Structure \(7\) actually represents a known \(C_6H_6\) isomer that has a very different chemistry from that of benzene. 3. All members of the set must have the same number of paired or unpaired electrons. For the normal state of benzene, the six \(\pi\) electrons have three of one spin and three of the other. Structures such as \(8\), with four electrons of one spin and two of the other, are not valid contributors to the ground state of benzene: 4. The importance of resonance in any given case will depend on the energies of the contributing structures. The lower and more nearly equivalent the members of the set are in energy, the more important resonance becomes. That is to say, (as for the two Kekulé structures of benzene). As a corollary, the structure of a molecule is least likely to be satisfactorily represented by a conventional structural formula when two (or more) energetically equivalent, low-energy structures may be written. 5. If there is only low-energy structure in the set then, to a first approximation, the resonance hybrid may be assigned properties like those expected for that structure. As an example, we show three possible pairing schemes for ethene, \(9\), \(10\), and \(11\): Although \(10\) and \(11\) are equivalent, they are much higher in energy than \(9\) (see discussion in ). Therefore they do not contribute substantially to the structure of ethene that is best represented by \(9\). Resonance is by no means restricted to organic molecules. The following sets of valence-bond structures represent the hybrid structures of nitrate ion, \(NO_3^\ominus\), carbonate ion \(CO_3^{2 \ominus}\), and nitrous oxide, \(N_2O\). These are only representative examples. We suggest that you check these structures carefully to verify that each member of a set conforms to the general rules for resonance summarized above. A shorthand notation of hybrid structures frequently is used in which the delocalized \(\pi\)-bonding is shown as a broken line. For benzene, an inscribed circle also is used to indicate continuous \(\pi\) bonding: Electron delocalization is an important factor in the reactivity (or lack of it) of organic molecules. As an example, recall from Chapter 4 that the bond energies of various types of \(C-H\) bonds differ considerably (see Table 4-6). In particular, the methyl \(C-H\) bond in propene is about \(9 \: \text{kcal}\) than the methyl \(C-H\) bond of ethane or propane, and this difference can be explained by the use of the resonance concept. The following bond dissociations are involved: delocalization is possible for the propyl radical, propane, or propene. Accordingly, the methyl \(C-H\) bond strength in propene is less than in propane because of stabilization of the 2-propenyl radical. The foregoing discussion adds further to our understanding of the selectivity observed in the halogenation reactions discussed in Chapter 4. When propene is chlorinated in sunlight, the product is 3-chloropropene, and we may explain this on the basis that the radical-chain reaction involves propagation steps in which a chlorine atom attacks the hydrogen corresponding to the \(C-H\) bond: The resonance theory is very useful in accounting for, and in many cases predicting, the behavior of substances with \(\pi\) bonds. However, it is not omnipotent. One example where it fails is cyclobutadiene, for which we can write two equivalent valence-bond structures corresponding to the Kekulé structures for benzene: Despite this, cyclobutadiene is an extremely unstable substance, reacting with itself almost instantly at temperatures above \(-250^\text{o}\). For better understanding of this and some related problems, we provide a more detailed discussion of electron delocalization in Chapter 21. and (1977) | 7,108 | 1,709 |
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One of the pioneers in the field of modern thermodynamics was James P. Joule (1818 - 1889). Among the experiments Joule carried out, was an attempt to measure the effect on the temperature of a sample of water that was caused by doing work on the water. Using a clever apparatus to perform work on water by using a falling weight to turn paddles within an insulated canister filled with water, Joule was able to measure a temperature increase in the water. Thus, Joule was able to show that work and heat can have the same effect on matter – a change in temperature! It would then be reasonable to conclude that heating, as well as doing work on a system will increase its energy content, and thus it’s ability to perform work in the surroundings. This leads to an important construct of the : The capacity of a system to do work is increased by heating the system or doing work on it. The (U) of a system is a measure of its capacity to supply energy that can do work within the surroundings, making U the ideal variable to keep track of the flow of heat and work energy into and out of a system. Changes in the internal energy of a system (\(\Delta U\)) can be calculated by \[\Delta U = U_f - U_i \label{FirstLaw} \] where the subscripts \(i\) and \(f\) indicate initial and final states of the system. \(U\) as it turns out, is a state variable. In other words, the amount of energy available in a system to be supplied to the surroundings is independent on how that energy came to be available. That’s important because the manner in which energy is transferred is path dependent. There are two main methods energy can be transferred to or from a system. These are suggested in the previous statement of the first law of thermodynamics. Mathematically, we can restate the first law as \[\Delta U = q + w \nonumber \] or \[dU = dq + dw \nonumber \] where q is defined as the amount of energy that flows into a system in the form of and w is the amount of energy lost due to the system doing on the surroundings. Heat is the kind of energy that in the absence of other changes would have the effect of changing the temperature of the system. A process in which heat flows into a system is from the standpoint of the system (\(q_{system} > 0\), \(q_{surroundings} < 0\)). Likewise, a process in which heat flows out of the system (into the surroundings) is called (\(q_{system} < 0\), \(q_{surroundings} > 0\)). In the absence of any energy flow in the form or work, the flow of heat into or out of a system can be measured by a change in temperature. In cases where it is difficult to measure temperature changes of the system directly, the amount of heat energy transferred in a process can be measured using a change in temperature of the soundings. (This concept will be used later in the discussion of calorimetry). An infinitesimal amount of heat flow into or out of a system can be related to a change in temperature by \[dq = C\, dT \nonumber \] where C is the and has the definition \[ C = \dfrac{dq}{\partial T} \nonumber \] Heat capacities generally have units of (J mol K ) and magnitudes equal to the number of J needed to raise the temperature of 1 mol of substance by 1 K. Similar to a heat capacity is a which is defined per unit mass rather than per mol. The specific heat of water, for example, has a value of 4.184 J g K (at constant pressure – a pathway distinction that will be discussed later.) How much energy is needed to raise the temperature of 5.0 g of water from 21.0 °C to 25.0 °C? \[\begin{align*} q &=mC \Delta T \\[4pt] &= (5.0 \,\cancel{g}) (4.184 \dfrac{J}{\cancel{g} \, \cancel{°C}}) (25.0 \cancel{°C} - 21.0 \cancel{°C}) \\[4pt] &= 84\, J \end{align*} \] A partial derivative, like a total derivative, is a slope. It gives a magnitude as to how quickly a function changes value when one of the dependent variables changes. Mathematically, a partial derivative is defined for a function \(f(x_1,x_2, \dots x_n)\) by \[\left( \dfrac{ \partial f}{\partial x_i} \right)_{x_j \neq i} = \lim_{\Delta _i \rightarrow 0} \left( \dfrac{f(x_1+ \Delta x_1 , x_2 + \Delta x_2, \dots, x_i +\Delta x_i, \dots x_n+\Delta x_n) - f(x_1,x_2, \dots x_i, \dots x_n) }{\Delta x_i} \right) \nonumber \] Because it measures how much a function changes for a change in a given dependent variable, infinitesimal changes in the in the function can be described by \[ df = \sum_i \left( \dfrac{\partial f}{\partial x_i} \right)_{x_j \neq i} \nonumber \] So that each contribution to the total change in the function \(f\) can be considered separately. For simplicity, consider an ideal gas. The pressure can be calculated for the gas using the ideal gas law. In this expression, pressure is a function of temperature and molar volume. \[ p(V,T) = \dfrac{RT}{V} \nonumber \] The partial derivatives of p can be expressed in terms of \(T\) and \(V\) as well. \[ \left( \dfrac{\partial p}{ \partial V} \right)_{T} = - \dfrac{RT}{V^2} \label{max1} \] and \[ \left( \dfrac{\partial p}{ \partial T} \right)_{V} = \dfrac{R}{V} \label{max2} \] So that the change in pressure can be expressed \[ dp = \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \label{eq3} \] or by substituting Equations \ref{max1} and \ref{max2} \[ dp = \left( - \dfrac{RT}{V^2} \right ) dV + \left( \dfrac{R}{V} \right) dT \nonumber \] Macroscopic changes can be expressed by integrating the individual pieces of Equation \ref{eq3} over appropriate intervals. \[ \Delta p = \int_{V_1}^{V_2} \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \int_{T_1}^{T_2} \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \nonumber \] This can be thought of as two consecutive changes. The first is an (constant temperature) expansion from \(V_1\) to \(V_2\) at \(T_1\) and the second is an (constant volume) temperature change from \(T_1\) to \(T_2\) at \(V_2\). For example, suppose one needs to calculate the change in pressure for an ideal gas expanding from 1.0 L/mol at 200 K to 3.0 L/mol at 400 K. The set up might look as follows. \[ \Delta p = \underbrace{ \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV}_{\text{isothermal expansion}} + \underbrace{ \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT}_{\text{isochoric heating}} \nonumber \] or \[ \begin{align*} \Delta p &= \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV + \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT \\[4pt] &= \left[ \dfrac{R(200\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} + \left[ \dfrac{RT}{3.0 \, L/mol} \right]_{ 200\,K}^{400\,K} \\[4pt] &= R \left[ \left( \dfrac{200\,K}{3.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{200\,K}{3.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \] Alternatively, one could calculate the change as an isochoric temperature change from \(T_1\) to \(T_2\) at \(V_1\) followed by an isothermal expansion from \(V_1\) to \(V_2\) at \(T_2\): \[ \Delta p = \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT + \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV \nonumber \] \[ \begin{align*} \Delta p &= \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT + \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV \\[4pt] &= \left[ \dfrac{RT}{1.0 \, L/mol} \right]_{ 200\,K}^{400\,K} + \left[ \dfrac{R(400\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} \\[4pt] &= R \left[ \left( \dfrac{400\,K}{1.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{400\,K}{1.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \] Work can take several forms, such as expansion against a resisting pressure, extending length against a resisting tension (like stretching a rubber band), stretching a surface against a surface tension (like stretching a balloon as it inflates) or pushing electrons through a circuit against a resistance. The key to defining the work that flows in a process is to start with an infinitesimal amount of work defined by what is changing in the system. The pattern followed is always an infinitesimal displacement multiplied by a resisting force. The total work can then be determined by integrating along the pathway the change follows. What is the work done by 1.00 mol an ideal gas expanding from a volume of 22.4 L to a volume of 44.8 L against a constant external pressure of 0.500 atm? \[dw = -p_{ext} dV \nonumber \] since the pressure is constant, we can integrate easily to get total work \[ \begin{align*} w &= -p_{exp} \int_{V_1}^{V_2} dV \\[4pt] &= -p_{exp} ( V_2-V_1) \\[4pt] &= -(0.500 \,am)(44.8 \,L - 22.4 \,L) \left(\dfrac{8.314 \,J}{0.08206 \,atm\,L}\right) \\[4pt] &= -1130 \,J = -1.14 \;kJ \end{align*} \] : The ratio of gas law constants can be used to convert between atm∙L and J quite conveniently! | 8,915 | 1,710 |
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Atoms can join together by forming a chemical bond, which is a very strong attraction between two atoms. Chemical bonds are formed when electrons in different atoms interact with each other to make an arrangement that is more stable than when the atoms are apart. What causes atoms to make a chemical bond with other atoms, rather than remaining as individual atoms? A clue comes by considering the noble gas elements, the rightmost column of the periodic table. These elements—helium, neon, argon, krypton, xenon, and radon—do not form compounds very easily, which suggests that they are especially stable as lone atoms. What else do the noble gas elements have in common? Except for helium, they all have eight valence electrons. Chemists have concluded that atoms are especially stable if they have eight electrons in their outermost shell. This useful rule of thumb is called the , and it is a key to understanding why compounds form. Of the noble gases, only krypton, xenon, and radon have been found to make compounds. There are two ways for an atom that does not have an octet of valence electrons to obtain an octet in its outer shell. One way is the transfer of electrons between two atoms until all atoms have octets. Because some atoms will lose electrons and some atoms will gain electrons, there is no overall change in the number of electrons, but individual atoms acquire a nonzero electric charge. Those that lose electrons become positively charged, and those that gain electrons become negatively charged. Charged atoms are called ions. Because opposite charges attract (while like charges repel), these oppositely charged ions attract each other, forming . The resulting compounds are called and are the primary subject of this chapter. The second way for an atom to obtain an octet of electrons is by sharing electrons with another atom. These shared electrons simultaneously occupy the outermost shell of more than one atom. The bond made by electron sharing is called a . Despite our focus on the octet rule, we must remember that for small atoms, such as hydrogen, helium, and lithium, the first shell is, or becomes, the outermost shell and hold only two electrons. Therefore, these atoms satisfy a “ ” rather than the octet rule. A sodium atom has one valence electron. Do you think it is more likely for a sodium atom to lose one electron or gain seven electrons to obtain an octet? Although either event is possible, a sodium atom is more likely to lose its single valence electron. When that happens, it becomes an ion with a net positive charge. This can be illustrated as follows: A fluorine atom has seven valence electrons. Do you think it is more likely for a fluorine atom to lose seven electrons or gain one electron to obtain an octet? Write the formula of the resulting ion. The process that involves less number of electrons is more favorable. Fluorine would gain one electron. The formula of the resulting ion is F 1. What is the octet rule? 2. How are ionic bonds formed? 3. Why is an ionic compound unlikely to consist of two positively charged ions? 4. Why is an ionic compound unlikely to consist of two negatively charged ions? 5. A calcium atom has two valence electrons. Do you think it will lose two electrons or gain six electrons to obtain an octet in its outermost electron shell? Write the formula of the resulting ion. 6. An aluminum atom has three valence electrons. Do you think it will lose three electrons or gain five electrons to obtain an octet in its outermost electron shell? Write the formula of the resulting ion. 7. A selenium atom has six valence electrons. Do you think it will lose six electrons or gain two electrons to obtain an octet in its outermost electron shell? Write the formula of the resulting ion. 8. An iodine atom has seven valence electrons. Do you think it will lose seven electrons or gain one electron to obtain an octet in its outermost electron shell? Write the formula of the resulting ion. 1. The octet rule is the concept that atoms tend to have eight electrons in their valence electron shell. 2. Ionic bonds are formed by the attraction between oppositely charged ions. 3. Positive charges repel each other, so an ionic compound is not likely between two positively charged ions. 4. Negative charges repel each other also. 5. Ca atom is more likely to lose two electrons. It will become Ca ion. 6. An Al atom is more likely to lose three electrons. It will become Al ion. 7. Selenium is more likely to gain two electrons. It will become Se ion. 8. Iodine is more likely to gain one electron. It will become I ion. | 4,624 | 1,711 |
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This page explains the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, K and pK . We are going to use the Bronsted-Lowry definition of an acid. When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to produce a hydroxonium ion and a negative ion depending on what acid you are starting from. In the general case . . . \[HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{1}\] These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen ions that we can think of the reaction as being one-way. The acid is virtually 100% ionized. For example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the reverse reaction happens that we can write: \[ H_2O_{(l)} + HCl_{(g)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)} \tag{2}\] At any one time, virtually 100% of the hydrogen chloride will have reacted to produce hydroxonium ions and chloride ions. Hydrogen chloride is described as a strong acid. A strong acid is one which is virtually 100% ionized in solution. Other common strong acids include sulphuric acid and nitric acid. You may find the equation for the ionization written in a simplified form: \[ HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \tag{3}\] This shows the hydrogen chloride dissolved in the water splitting to give hydrogen ions in solution and chloride ions in solution. This version is often used in this work just to make things look easier. If you use it, remember that the water is actually involved, and that when you write H what you really mean is a hydroxonium ion, H O . pH is a measure of the concentration of hydrogen ions in a solution. Strong acids like hydrochloric acid at the sort of concentrations you normally use in the lab have a pH around 0 to 1. The lower the pH, the higher the concentration of hydrogen ions in the solution. Suppose you had to work out the pH of 0.1 mol dm hydrochloric acid. All you have to do is work out the concentration of the hydrogen ions in the solution, and then use your calculator to convert it to a pH. With strong acids this is easy. Hydrochloric acid is a strong acid - virtually 100% ionized. Each mole of HCl reacts with the water to give 1 mole of hydrogen ions and 1 mole of chloride ions That means that if the concentration of the acid is 0.1 mol dm , then the concentration of hydrogen ions is also 0.1 mol dm . Use your calculator to convert this into pH. My calculator wants me to enter 0.1, and then press the "log" button. Yours might want you to do it in a different order. You need to find out! log [0.1] = -1 But pH = - log [0.1] - (-1) = 1 The pH of this acid is 1. A weak acid is one which doesn't ionize fully when it is dissolved in water. Ethanoic acid is a typical weak acid. It reacts with water to produce hydroxonium ions and ethanoate ions, but the back reaction is more successful than the forward one. The ions react very easily to reform the acid and the water. \[ CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+ \tag{4}\] At any one time, only about 1% of the ethanoic acid molecules have converted into ions. The rest remain as simple ethanoic acid molecules. Most organic acids are weak. Hydrogen fluoride (dissolving in water to produce hydrofluoric acid) is a weak inorganic acid that you may come across elsewhere. The position of equilibrium of the reaction between the acid and water varies from one weak acid to another. The further to the left it lies, the weaker the acid is. \[HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{5}\] You can get a measure of the position of an equilibrium by writing an equilibrium constant for the reaction. The lower the value for the constant, the more the equilibrium lies to the left. The dissociation (ionization) of an acid is an example of a homogeneous reaction. Everything is present in the same phase - in this case, in solution in water. You can therefore write a simple expression for the equilibrium constant, K . Here is the equilibrium again: \[HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{5}\] You might expect the equilibrium constant to be written as: \[]K_c = \dfrac{[H_3O^+,A^-]}{[HA,H_2O]} \] However, if you think about this carefully, there is something odd about it. At the bottom of the expression, you have a term for the concentration of the water in the solution. That's not a problem - except that the number is going to be very large compared with all the other numbers. In 1 dm of solution, there are going to be about 55 moles of water. If you had a weak acid with a concentration of about 1 mol dm , and only about 1% of it reacted with the water, the number of moles of water is only going to fall by about 0.01. In other words, if the acid is weak the concentration of the water is virtually constant. In that case, there isn't a lot of point in including it in the expression as if it were a variable. Instead, a new equilibrium constant is defined which leaves it out. This new equilibrium constant is called K . You may find the K expression written differently if you work from the simplified version of the equilibrium reaction: This may be written with or without state symbols. It is actually exactly the same as the previous expression for K ! Remember that although we often write H for hydrogen ions in solution, what we are actually talking about are hydroxonium ions. This second version of the K expression is not as precise as the first one. To take a specific common example, the equilibrium for the dissociation of ethanoic acid is properly written as: \[CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+ \tag{7}\] The K expression is: If you are using the simpler version of the equilibrium . . . \[CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \tag{8}\] . . . the K expression is: The table shows some values of K for some simple acids: These are all weak acids because the values for K are very small. They are listed in order of decreasing acid strength - the K values get smaller as you go down the table. However, if you aren't very happy with numbers, that isn't immediately obvious. Because the numbers are in two parts, there is too much to think about quickly! To avoid this, the numbers are often converted into a new, easier form, called pK . pK bears exactly the same relationship to K as pH does to the hydrogen ion concentration: \[pK_a = -\log_{10} K_a \tag{9}\] If you use your calculator on all the K values in the table above and convert them into pK values, you get: Notice that the weaker the acid, the larger the value of pK . It is now easy to see the trend towards weaker acids as you go down the table. | 6,720 | 1,712 |
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It has been demonstrated that water, in the presence of an acid or a base, adds rapidly to the carbonyl function of aldehydes and ketones establishing a reversible equilibrium with a (geminal-diol or -diol). The word germinal or gem comes from the Latin word for twin, . Isolation of -diols is difficult because the reaction is reversibly. Removal of the water during a reaction can cause the conversion of a gem-diol back to the corresponding carbonyl. In most cases the resulting -diol is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde where the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. The addition of electron donating alkyl groups stabilized the partial positive charge on the carbonyl carbon and decreases the amount of -diol product at equilibrium. Because of this ketones tend to form less than 1% of the hydrate at equilibrium. Likewise, the addition of strong electron-withdrawing groups destabilizes the carbonyl and tends to form stable -diols. Two examples of this are chloral, and 1,2,3-indantrione. It should be noted that chloral hydrate is a sedative and has been added to alcoholic beverages to make a “Knock-out” drink also called a Mickey Finn. Also, ninhydrin is commonly used by forensic investigators to resolve finger prints. The mechanism is catalyzed by the addition of an acid or base. Note! This may speed up the reaction but is has not effect on the equilibriums discussed above. Basic conditions speed up the reaction because hydroxide is a better nucleophilic than water. Acidic conditions speed up the reaction because the protonated carbonyl is more electrophilic. 1) Nucleophilic attack by hydroxide 2) Protonation of the alkoxide 1) Protonation of the carbonyl 2) Nucleophilic attack by water 3) Deprotonation 1) Draw the expected products of the following reactions. 2) Of the following pairs of molecules which would you expect to form a larger percentage of -diol at equilibrium? Please explain your answer. 3) Would you expect the following molecule to form appreciable amount of -diol in water? Please explain your answer.
1) 2) The compound on the left would. Fluorine is more electronegative than bromine and would remove more electron density from the carbonyl carbon. This would destabilize the carbonyl allowing for more -diol to form. 3) Although ketones tend to not form -diols this compound exists almost entirely in the -diol form when placed in water. Ketones tend to not form -diols because of the stabilizing effect of the electron donating alkyl group. However, in this case the electron donating effects of alkyl group is dominated by the presence of six highly electronegative fluorines. | 2,970 | 1,713 |
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As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R C=CR ) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas: Some representative alkenes—their names, structures, and physical properties—are given in Table \(\Page {1}\). We used only condensed structural formulas in Table \(\Page {1}\). Thus, CH =CH stands for The double bond is shared by the two carbons and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C H , whereas that for ethane is C H . The first two alkenes in Table \(\Page {1}\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(\Page {1}\)). Ethylene is a major commercial chemical. The chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products. Although there is only one alkene with the formula C H (ethene) and only one with the formula C H (propene), there are several alkenes with the formula C H . Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC): Name each compound. Name each compound. Just as there are cycloalkanes, there are . These compounds are named like alkenes, but with the prefix - attached to the beginning of the parent alkene name. Draw the structure for each compound. Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds. Draw the structure for each compound. | 2,086 | 1,714 |
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This page explains the terms strong and weak as applied to bases. As a part of this it defines and explains K and pK . The usual way of comparing the strengths of bases is to see how readily they produce hydroxide ions in solution. This may be because they already contain hydroxide ions, or because they take hydrogen ions from water molecules to produce hydroxide ions. A strong base is something like sodium hydroxide or potassium hydroxide which is fully ionic. You can think of the compound as being 100% split up into metal ions and hydroxide ions in solution. Each mole of sodium hydroxide dissolves to give a mole of hydroxide ions in solution. Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation. Working out the pH of a strong base Remember that: \[pH = -\log_{10} [H^+]\] Since pH is a measure of hydrogen ion concentration, how can a solution which contains hydroxide ions have a pH? To understand this, you need to know about the ionic product for water. Wherever there is water, an equilibrium is set up. Using the simplified version of this equilibrium: \[ H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}\] In the presence of extra hydroxide ions from, say, sodium hydroxide, the equilibrium is still there, but the position of equilibrium has been shifted well to the left according to Le Chatelier's Principle. There will be far fewer hydrogen ions than there are in pure water, but there will still be hydrogen ions present. The pH is a measure of the concentration of these. What is the pH of 0.500 mol dm sodium hydroxide solution:? Because the sodium hydroxide is fully ionic, each mole of it gives that same number of moles of hydroxide ions in solution. [OH ] = 0.500 mol dm Now you use the value of K at the temperature of your solution. You normally take this as 1.00 x 10 mol dm . [H ] [OH ] = 1.00 x 10 This is true whether the water is pure or not. In this case we have a value for the hydroxide ion concentration. Substituting that gives: [H ] x 0.500 = 1.00 x 10 If you solve that for [H ], and then convert it into pH, you get a pH of 13.7. Ammonia is a typical weak base. Ammonia itself obviously doesn't contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. \[ NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)}\] However, the reaction is reversible, and at any one time about 99% of the ammonia is still present as ammonia molecules. Only about 1% has actually produced hydroxide ions. A weak base is one which doesn't convert fully into hydroxide ions in solution. When a weak base reacts with water, the position of equilibrium varies from base to base. The further to the left it is, the weaker the base. You can get a measure of the position of an equilibrium by writing an equilibrium constant for the reaction. The lower the value for the constant, the more the equilibrium lies to the left. In this case the equilibrium constant is called K . This is defined as: \[ K_b = \dfrac{[B:H^+,OH^-]}{[B:]}\] The relationship between K and pK is exactly the same as all the other "p" terms in this topic: \[pK_b = -\log_{10} K_b\] The table shows some values for \(K_b\) and \(pK_b\) for some weak bases. As you go down the table, the value of K is increasing. That means that the bases are getting stronger. As K gets bigger, pK gets smaller. The lower the value of pK , the stronger the base. This is exactly in line with the corresponding term for acids, pK - the smaller the value, the stronger the acid. | 3,730 | 1,715 |
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In 1923, chemists Johannes Nicolaus Brønsted and Thomas Martin Lowry independently developed definitions of acids and bases based on the compounds' abilities to either donate or accept protons (\(H^+\) ions). In this theory, acids are defined as proton donors; whereas bases are defined as proton acceptors. A compound that acts as both a Brønsted-Lowry acid and base together is called amphoteric. Brønsted-Lowry theory of acid and bases took the Arrhenius definition one step further, as a substance no longer needed to be composed of hydrogen (H ) or hydroxide (OH ) ions in order to be classified as an acid or base. For example , consider the following chemical equation: \[ HCl \; (aq) + NH_3 \; (aq) \rightarrow NH_4^+ \; (aq) + Cl^- \; (aq) \] Here, hydrochloric acid (HCl) "donates" a proton (H ) to ammonia (NH ) which "accepts" it , forming a positively charged ammonium ion (NH ) and a negatively charged chloride ion (Cl ). Therefore, HCl is a Brønsted-Lowry acid (donates a proton) while the ammonia is a Brønsted-Lowry base (accepts a proton). Also, Cl is called the of the acid HCl and NH is called the of the base NH . In this theory, an (like in the Arrhenius theory) and a . A basic salt, such as Na F , generates OH ions in water by taking protons from water itself (to make HF): \[F^-_{(aq)} + H_2O_{(l)} \rightleftharpoons HF_{(aq)} + OH^-\] When a Brønsted acid dissociates, it increases the concentration of hydrogen ions in the solution, \([H^+]\); conversely, Brønsted bases dissociate by taking a proton from the solvent (water) to generate \([OH^-]\). \[HA_{(aq)} \rightleftharpoons A^-_{(aq)} + H^+_{(aq)}\] \[K_a=\dfrac{[A^-,H^+]}{[HA]}\] \[B_{(aq)} + H_2O_{(l)} \rightleftharpoons HB^+_{(aq)} + OH^-_{(aq)}\] \[K_b = \dfrac{[HB^+,OH^-]}{[B]}\] The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case. To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the number of hydrogens has decreased that substance is the acid (donates hydrogen ions). If the number of hydrogens has increased that substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is viewed in reverse a new acid and base can be identified. The substances on the right side of the equation are called conjugate acid and conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over. Acids are Proton Donors and Bases are Proton Acceptors For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a made up of and . \[ HA + Z \rightleftharpoons A^- + HZ^+ \] A stands for an Acidic compound and Z stands for a Basic compound How can H O be a base? I thought it was neutral? | 3,119 | 1,716 |
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The concept of the atom (Greek: atomos, "indivisible"), an indivisible particle of matter, goes back to ancient Greece and a man named Democritus, a rival of Aristotle. Democritus held that all matter could be subdivided only until some finite particle was reached. Aristotle had a quite different idea, that matter was a continuous substance, not composed of any fundamental units. And Aristotle was much more famous and admired in his time and so his error persisted into the late 1700's. But eventually work by men such as Lavoisier began to suggest that Aristotle had been seriously wrong. In 1808 the first statement of a modern atomic theory was published by John Dalton, a Quaker schoolmaster from Manchester. It may not seem like much, but such a theory was used to explain two of the major laws in chemistry: and the . And the third postulate led Dalton to formulate the Law of Multiple Proportions. John Dalton apparently thought that atoms were pretty much little solid spheres without internal structure. His experiments with gases eventually led him to the idea that matter came in small indivisible units but his use of the word "atom" was variable. He described what we now call molecules as "atoms" and was convinced that the simplest binary compound of any two elements would have those atoms in a 1:1 ratio. According to Dalton, water would be HO. This led to a relative mass scale for atoms that was mostly in error. Dalton also had his own notation for some of the elements he recognized. Dalton's "modern" atomic theory, in todays terms, would be something like this: Some contemporaries of Dalton remained unimpressed. An eminent organic chemist of the time, Adolf Kolbe, said in 1877, "Dalton's atoms are no more than stupid hallucinations...mere table-tapping and supernatural explanations." The first evidence for sub-atomic particles came from experiments with the conduction of electricity through gases in sealed glass tubes at low pressures. Figure : J.J. Thomson (seen here in his lab with one of his many hand-made cathode ray tubes) Associated with the flow of electricity in such a tube are rays which originate from the negative electrode--the cathode. Thus these tubes have been called cathode ray tubes. From careful experiments with cathode ray tubes J.J. Thomson demonstrated in 1897 that the rays consist of a stream of negatively charged particles which he called electrons. He was able to measure the charge/mass ratio of these particles and found this to be the same regardless of what gas was in the tube or what metal the electrodes were made from. In 1909 Robert Millikan used the classic oil drop experiment to determine the charge on these particles. Using the smallest charge obtained and Thomson's charge/mass ratio the electron mass is roughly 1/2000 the mass of the lightest atom. Thus there are obviously particles smaller than atoms. Millikan used his apparatus to make many measurements of the effect of the electric field vs. gravity on the charged oil droplets. His eventual conclusion was that the charges on the drops were either equal to or multiples of one number which he decided was the charge of a single electron. With this value, and the charge/mass ratio that Thomson had measured earlier it was possible to calculate the mass of the electron. Thomson's dilemma: how could matter containing electrons be neutral and where was all the mass? Other experiments with discharge tubes suggested the existence of a positive particle with much greater mass (the ). Based on this evidence Thomson proposed the first atomic model with sub-atomic particles. Thomson's work (and the work of others) with cathode ray tubes eventually identified a much more massive positive particle apparently common to all matter. These were called protons. With the ingredients needed to keep matter neutral, Thomson speculated and calculated just how it might all hold together. His conclusion has become known as the "plum pudding" model of the atom and was the first to include sub-atomic particles with charges. Thomson's calculations indicated that to remain stable atoms might need to have hundreds, if not thousands, of electrons. However, the developing techniques for measuring relative atomic masses seemed to contradict this, suggesting instead that less than 100 electrons would be more likely. Finally in the years between 1908 and 1911 Ernest Rutherford and his students Hans Geiger and Ernest Marsden performed the famous gold-foil experiment in which the nuclear arrangement of the atom was discovered. Rutherford's vision of the atom as a kind of "solar system" in miniature is what most people probably carry around in their heads. As appealingly simple as it is, Rutherford's model contained a serious flaw. Rutherford's model of the atom was no doubt appealing in part because it seemed to mimic the larger world people could see. But just as the planets slowly lose orbital energy as they circle the sun, the energy supply of the tiny electron should be degraded too. Physicists had already established in Rutherford's era that a charged particle moving in a circle, but keeping a constant orbital radius must radiate energy. Hence the nuclear model of the atom should render all matter unstable as electrons gradually spiral into nuclei when their orbits "decay" due to loss of energy. Obviously, this does not happen. The next step in the development of the atomic model changed the way in which scientists looked at and thought about matter, especially very small pieces like electrons and protons. | 5,582 | 1,718 |
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“The first few of what are now known to be organic chemicals to be discovered were produced by living organisms and were therefore called ‘organic.’ One such compound is urea, which occurs in urine. In 1828 Friedrich Wöhler disproved the idea that all organic compounds must come from living organisms when he accidentally discovered that urea could be made by the reaction of cyanic acid (HOCN) and ammonia, both simple organic compounds. This discovery established the science of organic chemistry based upon the unique bonding capabilities of the carbon atom leading to the synthesis and discovery of tens of millions of unique organic compounds. In 2009 the American Chemical Society Chemical Abstract Service registered the 60 millionth compound(most of which are organic compounds) only 9 months after the registration of the 40 millionth, apace of discovery of more than one new compound per minute.”
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A vacuum distillation apparatus is shown in Figure 5.50, using a setup. A can also be used. It is assumed that readers have previously performed a simple distillation under atmospheric pressure, so in this section are described differences between atmospheric and reduced pressure distillations. Always use a stir bar, not boiling stones. Grease all joints. Use a Claisen adapter, as solutions tend to bump under vacuum. Connect thick-walled hosing at the vacuum adapter to a trap, then to the vacuum source (water aspirator or vacuum pump). Turn on the vacuum first, before heating, to remove very volatile components. When confident the apparatus is maintaining a reduced pressure, then heat the sample. Use glass wool or foil insulation if the sample is stubbornly refluxing instead of distilling. Record temperature and pressure if using a manometer during active distillation. Pure compounds may not distill over a constant temperature due to changes in pressure. If multiple fractions will be collected, the system needs to be vented and cooled in between (or a cow receiving flask used). To stop the distillation, remove the heat and cool the flask in a tap water bath. open the apparatus to the atmosphere by opening the pinch clamp on the trap, or removing the tubing on the vacuum adapter. Lastly turn off the vacuum. Correct order: cool, open to atmosphere, then turn off vacuum. | 1,407 | 1,721 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Ligand_Exchange_Reactions_(Thermodynamics) |
This page explains what is meant by a stability constant for a complex ion, and goes on to look at how its size is governed in part by the entropy change during a ligand exchange reaction. If you add ammonia solution to a solution containing hexaaquacopper(II) ions, [Cu(H O) ] , four of the water molecules are eventually replaced by ammonia molecules to give [Cu(NH ) (H O) ] . This can be written as an equilibrium reaction to show the overall effect: \[\ce{[Cu(H2O)6]^{2+} + 4NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O} \label{all}\] In fact, the water molecules get replaced one at a time, and so this is made up of a series of part-reactions: \[\ce{[Cu(H2O)6]^{2+} + NH3 <=> [Cu(NH3)(H2O)5]^{2+} + H2O}\label{step1}\] \[\ce{ [Cu(NH3)(H2O)5]^{2+} + NH3 <=> [Cu(NH3)2(H2O)4]^{2+} + H2O} \label{step2}\] \[\ce{[Cu(NH3)2(H2O)4]^{2+} + NH3 <=> [Cu(NH3)3(H2O)3]^{2+} + H2O} \label{step3}\] \[\ce{[Cu(NH3)3(H2O)3]^{2+} + NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + H2O} \label{step4}\] Although this can look a bit daunting at first sight, all that is happening is that first you have one, then two, then three, then four water molecules in total replaced by four ammine ligands. Let's take a closer look at the first of these equilibria (Equation \ref{step1}). Like any other equilibrium, this one has an equilibrium constant, \(K_c\) - except that in this case, we call it a stability constant. Because this is the first water molecule to be replaced, we call it \(K_1\) and is given by this expression: \[K_1 = \dfrac{[\ce{[Cu(NH3)(H2O)5]^{2+}}]}{\ce{[[Cu(H2O)6]^{2+}},\ce{NH3]}}\] There are two points of possible confusion here - one minor, one more important! With that out of the way, let's go back to where we were - but introduce a value for \(K_1\): \[\ce{ [Cu(H2O)6]^{2+} + NH3 <=> [Cu(NH3)(H2O)5]^{2+} + H2O} \nonumber\] The value of the equilibrium constant is fairly large, suggesting that there is a strong tendency to form the ion containing an ammonia molecule. A high value of a stability constant shows that the ion is easily formed. Each of the other equilibria above also has its own stability constant, K , K and K . For example, K is given by: \[\ce{ [Cu(NH3)(H2O)5]^{2+} + NH3 <=> [Cu(NH3)2(H2O)4]^{2+} + H2O} \nonumber\] The ion with two ammonias is even more stable than the ion with one ammonia. You could keep plugging away at this and come up with the following table of stability constants: You will often find these values quoted as log K . All this does is tidy the numbers up so that you can see the patterns more easily. The ions keep on getting more stable as you replace up to 4 water molecules, but notice that the equilibrium constants are gradually getting less big as you replace more and more waters. This is common with individual stability constants. The overall stability constant is simply the equilibrium constant for the total reaction: \[\ce{ [Cu(H2O)6]^{2+} + 4NH3 <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O} \nonumber\] It is given by this expression: You can see that overall this is a very large equilibrium constant, implying a high tendency for the ammonias to replace the waters. The "log" value is 13.1. This overall value is found by multiplying together all the individual values of K , K and so on. To find out why that works, you will need a big bit of paper and some patience! Write down expressions for all the individual values (the first two are done for you above), and then multiply those expressions together. You will find that all the terms for the intermediate ions cancel out to leave you with the expression for the overall stability constant. Whether you are looking at the replacement of individual water molecules or an overall reaction producing the final complex ion, a stability constant is simply the equilibrium constant for the reaction you are looking at. The the value of the stability constant, the further the reaction lies to the . That implies that complex ions with large stability constants are more stable than ones with smaller ones. Stability constants tend to be very large numbers. In order to simplify the numbers a "log" scale is often used. Because of the way this works, a difference of 1 in the log value comes from a 10 times difference in the stability constant. A difference of 2 comes from a 100 (in other words, 10 ) times difference in stability constant - and so on. The chelate effect is an effect which happens when you replace water (or other simple ligands) around the central metal ion by multidentate ligands like 1,2-diaminoethane (often abbreviated to "en") or EDTA. Compare what happens if you replace two water molecules around a [Cu(H O) ] ion with either 2 ammonia molecules or one molecule of 1,2-diaminoethane. This second structure is known as a chelate from a Greek word meaning a crab's claw. You can picture the copper ion as being nipped by the claw of the 1,2-diaminoethane molecule. Chelates are much more stable than complex ions formed from simple monodentate ligands. The overall stability constants for the two ions are: The reaction with the 1,2-diaminoethane could eventually go on to produce a complex ion [Cu(en) ] . Simplifying the structure of this: The overall stability constant for this (as log K) is 18.7. Another copper-based chelate comes from the reaction with EDTA. This also has a high stability constant - log K is 18.8. However many examples you take, you always find that a chelate (a complex ion involving multidentate ligands) is more stable than ions with only monodentate ligands. This is known as the chelate effect. If you compare the two equilibria below, the one with the 1,2-diaminoethane ("en") has the higher equilibrium (stability) constant (for values, see above). The enthalpy changes of the two reactions are fairly similar. You might expect this because in each case you are breaking two bonds between copper and oxygen atoms and replacing them by two bonds between copper and nitrogen atoms. If the enthalpy changes are similar, what causes the difference in the extent to which the two reactions happen? You need to think about the entropy change during each reaction. Entropy is most easily thought of as a measure of disorder. Any change which increases the amount of disorder increases the tendency of a reaction to happen. If you look again at the two equiilbria, you might notice that the 1,2-diaminoethane equilibrium does lead to an increase in the entropy. There are only two species on the left-hand side of the equation, but three on the right. You can obviously get more entropy out of three species than out of only two. Compare that with the other equilibrium. In this case, there is no change in the total number of species before and after reaction, and so no useful contribution to an increase in entropy. In the case of the complex with EDTA, the increase in entropy is very pronounced. Here, we are increasing the number of species present from two on the left-hand side to seven on the right. You can get a major amount of increase in disorder by making this change. Reversing this last change is going to be far more difficult in entropy terms. You would have to move from a highly disordered state to a much more ordered one. That isn't so likely to happen, and so the copper-EDTA complex is very stable. Complexes involving multidentate ligands are more stable than those with only monodentate ligands. The underlying reason for this is that each multidentate ligand displaces more than one water molecule. This leads to an increase in the number of species present in the system, and therefore an increase in entropy. An increase in entropy makes the formation of the chelated complex more favorable. Jim Clark ( ) | 7,706 | 1,722 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/Reactions_of_Alkenes_with_Hydrogen_Halides |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between the hydrogen halides and alkenes like ethene and cyclohexene. Hydrogen halides include hydrogen chloride and hydrogen bromide. Alkenes react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other. In the case of ethene, bromoethane is formed. \[ \ce{CH_2=CH_2 + HBr \rightarrow CH_3CH_2Br}\] With cyclohexene you get bromocyclohexane. The structures of the cyclohexene and the bromocyclohexane are often simplified: Be sure that you understand the relationship between these simplified diagrams and the full structures. The reactions are examples of electrophilic addition. With ethene and HBr: and with cyclohexene: Hydrogen chloride and the other hydrogen halides add on in exactly the same way. For example, hydrogen chloride adds to ethene to make chloroethane: \[ \ce{CH_2=CH_2 + HCl \rightarrow CH_3CH_2Cl}\] The only difference is in how fast the reactions happen with the different hydrogen halides. The rate of reaction as you go from HF to HCl to HBr to HI. HF > HCl > HBr > HI The reason for this is that as the halogen atoms get bigger, the strength of the hydrogen-halogen bond falls. Bond strengths (measured in kiloJoules per mole) are: H-F (569 kJ) > HCl (432 kJ) > HBr (366 kJ) > HI (298 kJ) As you have seen in the HBr case, in the first step of the mechanism the hydrogen-halogen bond gets broken. If the bond is weaker, it will break more readily and so the reaction is more likely to happen. The reactions are still examples of electrophilic addition. With ethene and HCl, for example: This is exactly the same as the mechanism for the reaction between ethene and HBr, except that we've replaced Br by Cl. All the other mechanisms for symmetrical alkenes and the hydrogen halides would be done in the same way. Jim Clark ( ) | 1,967 | 1,724 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/13%3A_Intermolecular_Forces/13.06%3A_Hydrophobic_Interaction |
Hydrophobic interactions describe the relations between water and (low water-soluble molecules). Hydrophobes are nonpolar molecules and usually have a long chain of carbons that do not interact with water molecules. The mixing of fat and water is a good example of this particular interaction. The common misconception is that water and fat doesn’t mix because the that are acting upon both water and fat molecules are too weak. However, this is not the case. The behavior of a fat droplet in water has more to do with the enthalpy and entropy of the reaction than its intermolecular forces. American chemist discovered that nonpolar substances like fat molecules tend to clump up together rather than distributing itself in a water medium, because this allow the fat molecules to have minimal contact with water. The image above indicates that when the hydrophobes come together, they will have less contact with water. They interact with a total of 16 water molecules before they come together and only 10 atoms after they interact. When a hydrophobe is dropped in an aqueous medium, hydrogen bonds between water molecules will be broken to make room for the hydrophobe; however, water molecules do not react with hydrophobe. This is considered an endothermic reaction, because when bonds are broken heat is put into the system. Water molecules that are distorted by the presence of the hydrophobe will make new hydrogen bonds and form an ice-like cage structure called a cage around the hydrophobe. This orientation makes the system (hydrophobe) more structured with an decrease of the total entropy of the system; therefore . The change in enthalpy (\( \Delta H \)) of the system can be negative, zero, or positive because the new hydrogen bonds can partially, completely, or over compensate for the hydrogen bonds broken by the entrance of the hydrophobe. The change in enthalpy, however, is insignificant in determining the spontaneity of the reaction (mixing of hydrophobic molecules and water) because the change in entropy ( ) is large. According to the formula with a small unknown value of and a large negative value of , the value of will turn out to be positive. A positive indicates that the mixing of the hydrophobe and water molecules is not spontaneous. The mixing hydrophobes and water molecules is not spontaneous; however, hydrophobic interactions between hydrophobes are spontaneous. When hydropobes come together and interact with each other, enthalpy increases ( \( \Delta H \) is positive) because some of hydrogen bonds that form the clathrate cage will be broken. Tearing down a portion of the clathrate cage will cause the entropy to increase ( \( \Delta S \) is positive), since forming it decreases the entropy. According to the Equation \(\ref{eq1}\) Result: negative and hence hydrophobic interactions are spontaneous. Hydrophobic interactions are relatively stronger than other weak intermolecular forces (i.e., interactions or Hydrogen bonds). The strength of Hydrophobic Interactions depend on several factors including (in order of strength of influence): Hydrophobic Interactions are important for the folding of proteins. This is important in keeping a protein stable and biologically active, because it allow to the protein to decrease in surface are and reduce the undesirable interactions with water. Besides from proteins, there are many other biological substances that rely on hydrophobic interactions for its survival and functions, like the phospholipid bilayer membranes in every cell of your body! | 3,580 | 1,725 |
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Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes. In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane derivatives are different from cyclopentane and cyclohexane, because their \(\ce{C-C-C}\) angles cannot have the tetrahedral value of \(109.5^\text{o}\). At the same time, Baeyer hypothesized that the difficulties encountered in synthesizing cycloalkane rings from \(\ce{C_7}\) upward was the result of the angle strain that would be expected if the large rings were regular planar polygons (see Table 12-3). Baeyer also believed that cyclohexane had a planar structure like that shown in Figure 12-2, which would mean that the bond angles would have to deviate \(10.5^\text{o}\) from the tetrahedral value. However, in 1895, the then unknown chemist H. Sachse suggested that cyclohexane exists in the strain-free chair and boat forms discussed in Section 12-3. This suggestion was not accepted at the time because it led to the prediction of several possible isomers for compounds such as chlorocyclohexane. The idea that such isomers might act as a single substance, as the result of rapid equilibration, seemed like a needless complication, and it was not until 1918 that E. Mohr proposed a definitive way to distinguish between the Baeyer and Sachse cyclohexanes. As will be discussed in Section 12-9, the result, now known as the Sachse-Mohr theory, was complete confirmation of the idea of nonplanar large rings. Because cyclopentane and cyclobutane (Sections 12-3E and 12-3F) also have nonplanar carbon rings, it is clear that the Baeyer postulate of planar rings is not correct. Nonetheless, the idea of angle strain in small rings is important. There is much evidence to show that such strain produces thermodynamic instability and usually, but not always, enhanced chemical reactivity. The strain in ring compounds can be evaluated quantitatively by comparing the heats of combustion per \(\ce{CH_2}\) group, as in Table 12-3. The data indicate that cyclohexane is virtually strain-free, because the heat of combustion per \(\ce{CH_2}\) is the same as for alkanes \(\left( 157.4 \: \text{kcal mol}^{-1} \right)\). The increase for the smaller rings clearly reflects increasing angle strain and, to some extent, unfavorable interactions between nonbonded atoms. For rings from \(\ce{C_7}\) to \(\ce{C_{12}}\) there appears to be a residual strain for each additional \(\ce{CH_2}\) of \(1\) to \(1.5 \: \text{kcal mol}^{-1}\). These rings can be puckered into flexible conformations with normal \(\ce{C-C-C}\) angles, but as will be shown in Section 12-6, from \(\ce{C_7}\) to \(\ce{C_{13}}\) such arrangements all have pairs of partially eclipsed or interfering hydrogens. The larger cycloalkanes such as cyclopentadecane appear to be essentially strain-free. We expect that the total strain in cycloalkanes of the type \(\ce{(CH_2)}_n\) should decrease rapidly in the order \(n = 2 > n = 3 > n = 4\). However, the data of Table 12-3 show that the order actually is \(3 \cong 4 > 2\). This difference in order often is disguised by dividing the heats of combustion by the numbers of \(\ce{CH_2}\) groups and showing that the heats of combustion per \(\ce{CH_2}\) are at least in the order expected from bond-angle strain. This stratagem does not really solve the problem. It is important to recognize that when we evaluate strain from the heats of combustion per \(\ce{CH_2}\) group, we are assuming that the \(\ce{C-H}\) bonds have the strength, independent of \(n\). However, the bond-dissociation energies of each of the \(\ce{C-H}\) bonds of ethene and cyclopropane are greater than of the \(\ce{C_2-H}\) bonds of propane (Table 4-6). Any amount that these bonds are than normal will make the strain energies judged from heats of combustion appear to be . If we take the \(\ce{C-H}\) bonds to be on average \(2 \: \text{kcal mol}^{-1}\) stronger in cyclobutane, \(6 \: \text{kcal mol}^{-1}\) stronger in cyclopropane, and \(13 \: \text{kcal mol}^{-1}\) in ethene, we can correct the carbon-carbon strain energies accordingly. For cyclobutane the corrected strain then is \(8 \times 2\) (for the eight \(\ce{C-H}\) bonds) \(+ \: 26.3\) (total strain from Table 12-3) \(= 42.3 \: \text{kcal mol}^{-1}\). The corresponding figures for cyclopropane are \(6 \times 6 + 27.6 = 63.6 \: \text{kcal mol}^{-1}\), and for ethene, \(4 \times 13 + 22.4 = 74.4 \: \text{kcal mol}^{-1}\). The results support the intuitive expectations by giving larger differences in the right direction for the strain energies of cyclobutane, cyclopropane, and ethene. Whether this analysis is quantitatively correct or not, it does give some indication of why is not a very precise concept - unless we can reliably estimate the effects of a strain. and (1977) | 4,870 | 1,726 |
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\[I_{2(g)} + H_{2 (g)} \rightarrow 2HI_{(aq)}\] rate of reaction = k[Br ,H ] 0.005 Ms =k ( 0.15M) (0.2M) \[A + B + C \rightarrow P \] rate of reaction = k[A,B,C] 0.05Ms =k (0.05M)(0,01M)(0.25M) k= 0.05Ms /( 0.05) (0.01)( 0.25) M k= 400M s What are the units of the rate constant for a second-order reaction? Calculate the fraction of the starting quantity of A that will be used up after 60 s. Given the reaction below which is found to be the first order in A and \(t_{1/2} = 40\;s\) \[ A \rightarrow B + C\] with \(t_{1/2} = 40 \; s\] The remain of fraction A after 60 s: The fraction will be used up after 60s: Given the first order reaction is completed 90% in 30 mins at 298 K. Calculate the rate constant. 90%= 0.9 Assume the half life of the first order decay of radioactive isotope takes about 1 year (365 days). How long will it take the radioactivity of that isotope to decay by 60%? The decomposition of dinitrogen peroxide (\(N_2O_5\)) is a first-order reaction with a rate constant of 0.045 min at 300 K. \[2 N_{2}O_{5(g)} \rightarrow 4NO_{2}(g)+O_{2}\] The integrated rate equation of a first-order reaction is: \[[A] = [A]_{0}e^{-kt}\] Substituting concentration for moles of reactant and plugging in the known values: \[n_{A} = 0.040_{0}e^{-0.045*5}\] \[n_{A} = 0.032\] First, you need to figure the half life of your compound. We do this by solving for : \[t_{1/2}=\dfrac{0.693}{k}\] \[k=\dfrac{0.693}{t\dfrac{1}{2}}\] \[k=\dfrac{0.693}{2 hrs} =0.3465\] with this half life, we can find the time it will take by solving for \[ln\dfrac{[A]}{[A]^{_{o}}}=-kt\] We do not have the initial and final concentration, but that is okay. Assuming the initial concentration is 100 g, we can assume that the final concentration is 10 g because that would be reduced 90% as stated in the problem. \[ln\dfrac{[10]}{[100]}=-kt\] \[\dfrac{ln[0.1]}{-0.3465}=t=6.65\; hours\] So 5 hours will be definitely not enough time to reduce the compound \[t_{1/2}=\dfrac{0.693}{k}\] \[t_{1/2}=\dfrac{0.693}{1.5x10^{^{-4}}}\] \[=4.62 \times 10^3\; mins\] The half-life of a second order reaction \[2A\rightarrow P\] is given by: \[t_{1/2}=\dfrac{1}{k[A]_{o}}\] Calculate the order of the reaction and the rate constant of Cytobutane decompose to ethylene based on equation \[C H_{8 g)} \rightarrow 2C_ H_{ (g)}\] Given the following data of concentration [A] over a period of time, decide if the data represents first order or second order. Solve for K. Show graphs. The data best fits a Second order graph. The equation for a second order reaction is: 1/[A]= kt+ 1/[A ] When solving for K, the equation is rearranged to: =k t Plug in the numbers given: = k (0-54) .025=k One, can also obtain the slop of the graph and k=.025 If a compound’s ½ life is 15.6 days. What is the value of k? How long will it take to decompose to 10%. Use first order reaction. Equation for half-life of a First-Order Reaction is: \[ t_½ = \dfrac{\ln (2)}{k}\] If we plug in the information given: \[15.6 days= \dfrac{\ln(2)}{k}\] \[k=4.4 \times 10^{-2}\, day^{-1}\] The second equation needed is the first order reaction, which is: ln([A]/[A ])= -kt so, ln ([A]/[A ])=.1 t=-1/k x ln([A]/[A ]) t= -1/(4.4 x 10 day ) x (.1) t=52 days 12) In a second order reaction 2A--> products, the final concentration is .28M. What is the initial concentration if k=.32M s and the times is 5 seconds. Second order reaction equation is: 1/[A]= kt+ 1/[A ] Filling the information given: 1/[.28M]=.32M s (5)+ 1/[A ] 1/[A ]= 1.97 [A ]= .50M [A]=[A] e^-kt [A] = 90% [A] = 10% t = 30min * 60s/1min = 1800s [10%]/[90%]=e^ -k*1800 ln[10%/90%]/1800 = k = 3.95E-4 t1/2= ln(2)/k t1/2 = .693/3.95E-4= 1754s 1754s*1min/60s = 29 min Given rate constant for second order reaction \[2 NO_{2(g)} \rightarrow 2NO_{(g)} + O_{2(g)}\] is 1.08 M s at 600 C. Find the time that would take for the concentration of NO from 1.24 M to 0.56 M? 1/ [A] – 1/[A] = kt t= 1/k ( 1/[A] – 1/[A] ) = (1/ 1.08 M s ) (1/0.56 – 1/1.24) = 7.2 s Is which order of reaction half-life is independent of initial concentration? First order because the half-life equation for first order is t ln(2)/k, it does not have [A ] The decomposition of N2) is the first order. At 365 C, t1/2 is 1.79 x 10 min. given intial pressure of 1.05 atm.. Calculate total pressure. \[\begin{align} P &= P_{N_2O} + P_{N_2} + P_{O_2} \\ &= 0.525 + 0.525 +0.2625 = 1.31 \;atm \end{align}\] The integrated rate law for the zero-order reaction A → B is [A] = [A] - a) skektch the following plots: (i) rate vs. [A] (ii) [A] vs. t rate = k rate is independent of [A] (ii) [A] vs. t [A] = [A] - b) Derive an expression for the half-life of the reaction. At t = t , [A] = [A] /2 so, [A] /2 = [A] - t = 1/2 [A} c) Calculate the time in half-lives when the integrated rate law is no longer valid (that is, when [A] = 0) [A] = 0 = [A] - t = 1/2 [A} ⇒ k = 1/2t [A} t = [A} /k = [A] /2 so, [A] / (1/(2t )) [A] = 2t integrated rate law is no longer valid after 2 half-lives Jack, Jill, and you are in a physical chemistry class. The professor writes the following equations on the black board. \[A\rightarrow B\] \[[A]=[A_{0}]e^{-kt}\] a) \[\int ^{[A]}_{[A]_{0}} \dfrac{d[A]}{[A]}=\int^{t}_{0} -kdt\] \[\ln[A] \mid ^{[A]}_{[A]_{0}} = -kt\] \[\ln[A] -\ln[A]_{0}=-kt\] \[\ln\dfrac{[A]}{[A]_{0}} =-kt\] \[\dfrac{[A]}{[A]_{0}}=e^{-kt}\] \[[A]=[A_{0}]e^{-kt}\] b) \[t_{1/2}=\dfrac{\ln 2}{k}\] \[10 hours = \dfrac{\ln2}{k}\] \[k=\dfrac{\ln2}{10 hours}\times\dfrac{1hour}{60min}\times\dfrac{1min}{60secs}\] \[k=\] In the nuclear industry, workers use a rule of thumb that the readioactivity from any sample will be relatively harmless after 10 half-lives. Calculate the fraction of a radioactive sample that remains after this time period (hint: Radioactive decay obeys first-order kinetics) [A] = [A] e t = ln2 / k [A] = (100)e =.00454 % remaining after 10 half-lives a) Radioactive decay occurs in first order. b) \[\dfrac{[A]}{[A]_{0}}=e^{-kt}\] \[0.004=e^{-k\,30years}\] \[\ln[0.004]=-k\,30years\] \[k=\dfrac{-30}{\ln[0.004]}\] Many reactions involving heterogeneous catalysis are zero order; that is, rate = k. An example is the decomposition of phosphine (\(\ce{PH3}\)) over tungsten (W): \[\ce{4PH3(g) →P4(g) + 6H2(g)}\] The rate for this reaction is independent of [PH ] as long as phosphine's pressure is sufficiently high (>= 1 atm). Explain. With sufficient PH , all of the catalytic sites on the tungsten surface are occupied. Further increases in the amount of phosphine cannot affect the reaction, and the rate is independent of [PH ]. A mixture of compounds M and N whose half-lives are 40 minutes and 17 minutes, respectively. They decompose by first-order kinetics. If their concentrations are equal initially, how long does concentration of N to be half that of M? [I] is the initial concentration of M and N -> t = 30.1 min Compounds A and B both decay by first-order kinetics. The half-life of A is 20 minutes and the half-life of B is 48 minutes. If a container initially contains equal concentrations of compounds A and B, after how long will the concentration of B be twice that of A? 1. Write, in mathematical terms, the information given in the problem and what the problem is asking for. \[ t_{1/2, A} = 20.0 min \] \[ t_{1/2, B} = 48.0 min \] \[ [A]_0 = [B]_0 \] A and B decay by first-order, so \[ -\dfrac{d[A]}{dt} = k_A [A] \] \[ -\dfrac{d[B]}{dt} = k_B [B] \] (Note: the rate constants for A and B are not equal, so indicate which is which with subscripts.) Want find t at which the following is true: \[ [B] = 2[A] \] 2. Substitute the integrated rate equations for [A] and [B] \[ [B]_0 e^{-k_B t} = 2 [A]_0 e^{-k_A t} \] 3. Write expressions for the rate constants in terms of half-lives, and substitute into the equation. \[ t_{1/2, A} = \dfrac{ln2}{k_A} \Rightarrow k_A = \dfrac{ln2}{t_{1/2, A}} \] \[ t_{1/2, B} = \dfrac{ln2}{k_B} \Rightarrow k_B = \dfrac{ln2}{t_{1/2, B}} \] \[ [B]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 [A]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} \] 4. Solve for t Since initial concentrations of A and B are equal: \[ e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 e^{-\dfrac{ln2}{t_{1/2, B}} t} \] Take natural log of both sides: \[ -\dfrac{t}{t_{1/2, B}} ln2 = \left ( 1 - \dfrac{t}{t_{1/2, A}} \right ) ln2 \] \[ t = \dfrac{1}{\dfrac{1}{t_{1/2, A}} - \dfrac{1}{t_{1/2, B}}} \] 5. Plug in values for half-lives \[ t = \dfrac{1}{\dfrac{1}{20.0 min} - \dfrac{1}{48.0 min}} = 34.2\ min \] Answer: 34.2 minutes In Q3 thermodynamics and in Q9 chemical kinetics, the term “reversible “ is used. How do you understand this term? (it has a same meaning in these two chapters) Actually, this word is used to describe a “reversible” reaction in which both forward and backward reactions can happen in kinetics. In thermodynamics, “reversible” is used to describe a process that is in equilibrium along the pathway from the initial to final states. If a reaction has come to thermodynamic equilibrium, can we say anything in particular about the system's kinetics? Equilibrium occurs when all reactants and products are being consumed at the same rate that they are created. Take the simple example: \[ A\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}B \] When the system is at equilibrium, \[ \dfrac{d[A]}{dt} = \dfrac{d[B]}{dt} = 0 \] \[ k_{-1} [A] = k_1 [B] \] \[ \dfrac{k_{-1}}{k_1} = \dfrac{[B]}{[A]} \] \[ K_{eq} = \dfrac{k_{-1}}{k_1} \] The recombination of bromine atoms in an organic solvent, like carbon tetrafloride, is considered as a diffusion-controlled process \[Br + Br \rightarrow Br_2\] we have the viscosity of CF is 9.80 x 10 Nsm at 30 C, what is the rate of recombination at 30 C? Calculate the rate constant of the diffusion-controlled reaction \[ 2 I \rightarrow I_2 \] in dichloromethane at 15°C, which has a viscosity of 0.449 mPa·s at 15°C. 1. Use equation 9.50 to calculate the rate constant. \[ k_D = \dfrac{8}{3}\dfrac{RT}{\eta} \] \[ k_D = \dfrac{8}{3} \dfrac{8.314 \dfrac{J}{mol \ K} \times 288\ K \times \dfrac{N\ m}{J}}{0.449\ mPa \ s \times \dfrac{Pa}{1000\ mPa} \times\dfrac{N/m^2}{Pa}} \times \dfrac{1000\ L}{m^3} \] Answer: \[ k_D = 1.42 \times 10^{10} M^{-1} s^{-1} \] First, convert the rate mCi s 1 mCi=1.10X10 Ci 1 Ci=3.7 X 10 s The rate (\(r\) can be derived as such \[r= (9.7\; Ci) \left( \dfrac{1\;Ci}{1000\; mCi} \right) \left( \dfrac{3.7 \times 10^{10}\; s^{-1}}{1\; Ci} \right) = 3.59 \times 10^8 s^{-1}\] The accepted value for the half life of I-131 is 8.02 days. Using this information, the number of I-131 atoms can be calculated using the nuclear decay equation. Use the Nuclear Decay Equation: \[ \lambda N = \dfrac{\ln 2}{t_{1/2}} N\] or \[N = \dfrac{\ln 2}{t_{1/2}} r\] The half-life for the radioactive beta decay of iodine is 8.02 days \[\ce{^{131}_{53}I} \rightarrow \ce{^{131}_{54}Xe} + \beta + \bar{\nu_e} \] therefore, the rate is \[r= 3.59 \times 10^8 s^{-1}\] Calculate the rate law for the following acid-catalyzed reaction: \[CH_3COCH_3 + Br_2 \overset{H^+}{\longrightarrow} CH_3COCH_2Br + H^+ + Br^-\] 0.3 Determine the rate law for the following reaction: \[N_2O_2 + H_2 \rightarrow H_2O + N_2O\] In addition, determine which of the following actions would alter the value of \(k\)? Consider the mechanism for the association of iodine atoms to create molecular iodine. \[2I_{(g)} \rightleftharpoons I_{2(g)}^*\] \[I_2^*(g) + M(g) \rightarrow I_{2(g)} + M_{(g)}\] With the respect of the first step is at equilibrium, determine the expected rate law (d/dt)[I2(g)] in terms of k1, k-1, k2, [I], and [M]. Consider the following reaction: If propane (C3H8) is burning at a rate of 0.15 M/s^-1, calculate the rate of formation of CO2. First, express the reaction with the differential rate equation for the reactants and products involved. -(d/dt)[C3H8] = (1/3)(d/dt)[CO2] Then, use the given burning rate of propane and plug it into the differential equation. (d/dt)[CO2] = 3(0.15M/s^-1) If kept in a refrigerator, fresh fish will last for 3 days. If kept in a freezer, it will last for 6 months. Assuming that the temperature in the refrigerator is 5°C, and the temperature in the freezer is -10°C, calculate the activation energy for the bacterial spoiling of fish. Assume that the spoiling time is the 1/e lifetime instead of the half-life. The kinetics of the browning of juice from Golden Delicious apples was studied; at 20°C k=7.87×10 /week, and at 37°C k=0.139/week. What is the activation energy for the browning of Golden Delicious apple juice? | 12,493 | 1,727 |
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It was shown in that benzene is \(36\)-\(38 \: \text{kcal}\) more stable than the hypothetical molecule 1,3,5-cyclohexatriene on the basis of the differences between experimental heats of combustion, or hydrogenation, and heats calculated from bond energies. We call this energy difference the stabilization energy (SE) of benzene. We have associated most of this energy difference with \(\pi\)-electron delocalization, which is the (DE). The difference between SE and DE will be small only if our bond-energy tables are reliable and steric and strain effects are small. The problem with bond energies is that we use bond energies that neglect changes in bond strength caused by environment. and \(\ce{C_H}\) bonds are assumed to have equal energies; \(\ce{C-C}\) single bonds are assumed to be equal, regardless of whether the bonds to the carbon atoms in question are single or multiple; and differences in energy between double bonds that are mono-, di-, tri-, or tetra-substituted are neglected, as are changes in bond energies associated with steric strain. Bond energies are strictly applicable to molecules in which the bonds are of the normal lengths. In the case of benzene, which has \(\ce{C-C}\) bonds with lengths intermediate between normal single and double bonds, there seems to be no clear agreement as to how to take the bond distances into account in computing the delocalization energy. In spite of these uncertainties the stabilization energies seem to give a good qualitative idea of the importance of electron delocalization in organic molecules. Tables 21-1 and 21-2 give stabilization energies for several substances that are best represented as hybrid structures. and (1977) | 1,728 | 1,730 |
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Iron supplements are available as dietary supplements in food markets (where they are not controlled by the FDA) or by presciption, for treatment of TAG Heuer Replica, the most common blood disorder. Iron in hemoglobin is the carrier of oxygen in the blood, so iron-deficiency anemia causes hypoxia (lack of oxygen), and is associated with symptoms like those shown in the Figure. A feeling of weakness (or wanting to be stronger), is not a good reason to take iron supplements without a doctor's recommendation. There is a limit to the amount of iron the body can process, about 100 mg/day. Iron is toxic if taken in overload, and can cause vomiting, diabetes, damage to the liver, heart, and endochrine glands, and may cause premature death . Approximately 3 g is lethal for a 2 year old, so supplements must be kept inaccessible to children. Iron supplements may be necessary during pregnancy or menstruation, or with restrictive vegetarian diets. Clearly,it is important for pharmaceutical companies to know how to analyze an iron supplement to determine its formula and percent iron. How can we tell from a formula which supplement contains the most of the active ingredient, Fe ions? Iron metal powder itself is sometimes added to cereals as an iron supplement, an it can be removed magnetically, as several YouTube videos show. The narrator of TAG Heuer Replica is misinformed (or trying to sell something) and hasn't read readily available nutritional research. The iron granules are toxic or harmful in any way, and they bioavailable. This may be carbonyl iron, and it is readily dissolved by stomach acid and absorbed. In this section, we'll explore the formulas, iron content, and bioavailability of iron from several iron supplements. The iron particles in cereal are probably made by decomposing iron carbonyl: "Iron Carbonyl" → Fe + CO How do we know what the formula for iron carbonyl, so we can better understand the synthesis of "carbonyl iron particles"? We can determine it as follows: When 10.00 g of iron carbonyl is decomposed, it yields 2.85 g of iron and gaseous CO, which contains 3.06 g of carbon and the remainder oxygen. Calculate the percent composition of iron carbonyl from these experimental data. The percentage of iron is the mass of iron divided by the total mass of compound times 100 percent: The iron and carbon constitute 2.85 + 3.06 = 5.91 g. The remainder of the compound (10.0 g – 5.91 g = 4.09 g) is oxygen. Similarly, the percent C is 30.6. As a check, verify that the percentages add to 100: To obtain the formula from percent-composition data, we must find how many carbon and oxygen atoms there are per iron atom. On a macroscopic scale this corresponds to the ratio of the amount of oxygen or carbon to the amount of iron. If the formula is FeC O , it not only indicates that there are two carbon and two oxygen per iron , it also says that there are 2 of carbon and oxygen atoms for each 1 of iron atoms. That is, the of carbon or oxygen is twice the of iron. The numbers in the ratio of the amount of oxygen to the amount of iron (2:1) are the subscripts of oxygen and iron 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, 28.5 g (28.5%) is iron, 30.6 g is carbon, and 40.9 g is oxygen. Each mass can be converted to an amount of substance \(\frac{n_{\text{C}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.55 mol C}}{\text{0}\text{.510 mol Fe}}=\frac{\text{5}\text{.01 mol C}}{\text{1 mol Fe}}\) \(\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.55 mol O}}{\text{0}\text{.510 mol Fe}}=\frac{\text{5}\text{.01 mol O}}{\text{1 mol Fe}}\) So the formula is FeC O or Fe(CO) . The ratio 5.01 mol C and 5.01 mol O to 1 mol Fe also implies that there are 5.01 oxygen atoms and 5.01 carbon atoms per 1 Fe atom. If the atomic theory is correct, there is no such thing as 0.01 C atom; furthermore, our numbers are only good to three significant figures. Therefore we round 2.01 to 2 and write the formula as FeC O . This is an interesting compound of iron and carbon monoxide. It is a liquid at room temperature, and is quite toxic. The dietary supplement is thus prepared according to the equation: Fe(CO) → Fe + 5 CO Iron Sulfate (Feratab , MyKidz Iron 10 , etc.) is made by oxidizing pyrite ore: But how do we know what the product is, so we know how use it, and how to balance the equation? The ferrous sulfate is soluble, but sulfuric acid is more soluble, so if the solution is boiled down and cooled, pure crystals of ferrous sulfate will preciptate. Once a pure product has been obtained, it may be possible to identify the substance by means of its physical and chemical properties. Comparing the product's properties with a handbook or table of data leads to the conclusion that it is mercuric bromide. But suppose you were the person who ever prepared ferrous sulfate. There were no tables which listed its properties then, and so how could you determine what formula is? One answer involves —the determination of the percentage by mass of each element in the compound. Such data are usually reported as the . Ferrous sulfate has the composition 36.8% Fe, 21.1% S and 42.1% O. Find its formula. Again assume a 100-g sample and calculate the amount of each element: \(\frac{n_{\text{S}}}{n_{\text{Fe}}}=\frac{\text{0}\text{.657 mol S}}{\text{0}\text{.659 mol Fe}}=\frac{\text{1}\text{.00 mol S}}{\text{1 mol Fe}}\) \(\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.63 mol O}}{\text{0}\text{.659 mol Fe}}=\frac{\text{3}\text{.99 mol O}}{\text{1 mol Fe}}\) We would therefore assign the formula FeSO , and the balanced equation for the synthesis is: As we'll see in the next example, our method can only determine the of elements in a compound. Dextran, a polymer whose structure is shown below, is commonly used to complex Fe ion to make injectible iron supplements, like INFeD . Suppose a sample of an iron dextran complex has a molecular mass of 450, and contains 32.0% C, 4.00% H, 24.8 % Fe and 39.1 % O. Determine its empirical and molecular formulas. Dextran monomers \(\begin{align}
& n_{\text{C}}=\text{32}\text{.0 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.66 mol C} \\
& n_{\text{H}}=\text{4}\text{.00 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{4}\text{.00 mol H} \\
& n_{\text{O}}=\text{39}\text{.1 g}\times \frac{\text{1 mol O}}{\text{15}\text{.999 g}}=\text{2}\text{.44 mol O} \\
& n_{\text{Fe}}=\text{24}\text{.8 g}\times \frac{\text{1 mol Fe}}{\text{55}\text{.845 g}}=\text{0}\text{.444 mol Fe} \\
\end{align}\) \(\frac{n_{\text{H}}}{n_{\text{Fe}}}=\frac{\text{4}\text{.00 mol H}}{\text{0}\text{.444 mol Fe}}=\frac{\text{9}\text{.01 mol H}}{\text{1 mol Fe}}\) \(\frac{n_{\text{C}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.66 mol C}}{\text{0}\text{.444 mol Fe}}=\frac{\text{5}\text{.99 mol C}}{\text{1 mol Fe}}\) \(\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.44 mol O}}{\text{0}\text{.444 mol Fe}}=\frac{\text{5}\text{.5 mol O}}{\text{1 mol Fe}}\) The ratios C H O Fe are the same as the ratios C H O Fe . or C H O Fe . Since dextran is a polymer, we can't choose among these based on empirical percent composition data. The formula determined by this method is called the or : C H O Fe . The empirical formula sometimes 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, and the molecular weight is given above as 450. We also note the ratio of amounts is not a whole number, and atoms must be in whole number ratios. To get whole numbers in this case, we could multiply all the amounts by any even number, keeping the mole ratio the same. We'll just calculate the empirical formula mass, and see what the multiplier is: The molecular weight corresponding to the empirical formula is (6 × 12.01) + (9 × 1.008) + (5.5 × 16) + (1 × 55.846) = 225 Since the experimental molecular weight is twice as great, all subscripts must be doubled and the molecular formula is C H O Fe . It may be less obvious how to convert some mole ratios to whole numbers. Suppose the ratio of amounts of carbon to oxygen, for example, were 2.25:1, and we're sure enough of the analysis to know that it's not 2.33:1 or 2.50:1. We must convert 2.25 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .25 becomes ¼. Thus 2.25 = 2¼ = \(\tfrac{\text{9}}{\text{4}}\), and \(\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{2}\text{.25 mol C}}{\text{1 mol O}}=\frac{\text{9 mol C}}{\text{4 mol O}}\) Once someone has determined a formula–empirical or molecular—it is possible to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows. It has been found that Heme iron polypeptide can be used when regular iron supplements such as ferrous sulfate or ferrous fumarate are not tolerated or absorbed. A clinical study demonstrated that HIP increased serum iron levels 23 times greater than ferrous fumarate on a milligram-per-milligram basis.[3] HIP contains the central unit of hemoglobin in animal flesh, possibly with attached proteins. Since this molecule contains a much smaller percent iron than Ferrous sulfate, it is amazing that the iron is 23 time more bioavailable. Let's analyze these data. Elemental analysis: 66.24%C, 5.23%H, 9.09%N HIP, C H O FeN (Molecular weight: 616.498), Ferrous sulfate, FeSO , and Ferrous fumarate, C H FeO Molecular weight: 169.9, all are used as iron supplements. Which has the highest percent of iron? We will show the detailed calculation only for the case of C H FeO Iron Fumarate 1 mol C H FeO contains 4 mol C, 2 mol H, and 4 mol O, and 1 mol Fe. The molar mass is thus = (4 × 12.0 + 2 × 1.008 + 4 × 16.00 + 1 × 55.85) g mol = 169.9 g mol A 1-mol sample would weigh 169.9 g. The mass of 4 mol C it contains is \(m_{\text{C}}\text{ = 4 mol C }\times \text{ }\frac{\text{12}\text{.0 g}}{\text{1 mol C}}\text{ = 48}\text{.0 g}\) Therefore the percentage of C is \(\text{ }\!\!%\!\!\text{ C = }\frac{m_{\text{C}}}{m_{\text{C}_{\text{4}}\text{H}_{\text{2}}\text{O}_{\text{4}}\text{Fe}_{\text{1}}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{48}\text{.0 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 28}\text{.3 }\!\!%\!\!\text{ }\) The percentages of H, Fe, and O are easily calculated as \(\begin{align}
& m_{\text{H}}\text{ = 2 mol H }\times \text{ = }\frac{\text{1}\text{.008 g}}{\text{1 mol H}}\text{ = 2}\text{.016 g} \\
& \text{ }\!\!%\!\!\text{ H = }\frac{\text{2}\text{.016 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 1}\text{.19 }\!\!%\!\!\text{ } \\ #10; & m_{\text{O}}\text{ = 4 mol O }\times \text{ = }\frac{\text{16}\text{.00 g}}{\text{1 mol O}}\text{ = 64}\text{.00 g} \\
& \text{ }\!\!%\!\!\text{ O = }\frac{\text{64}\text{.00 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 37}\text{.7 }\!\!%\!\!\text{ } \\
& m_{\text{Fe}}\text{ = 1 mol O }\times \text{ = }\frac{\text{55}\text{.85 g}}{\text{1 mol Fe}}\text{ = 55}\text{.85 g} \\
& \text{ }\!\!%\!\!\text{ Fe = }\frac{\text{55}\text{.85 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 32}\text{.97 }\!\!%\!\!\text{ } \\
\end{align}\) Though not strictly needed to answer the problem, the percentages of C, H, and O provide a check of the results. The total 28.3 + 1.19% + 37.7% + 32.97% = 100.00% as it should. Similar calculations for FeSO , and HIP yield 36.8% and 9.1% iron, respectively. It's amazing that the HIP with only 9.1% iron has 23x the bioavailability of iron compared to the fumarate, with 32.97% iron. Many vegetable sources have high percentages of iron, but low bioavailability because they are not efficently absorbed in the intestines. | 12,088 | 1,731 |
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We have seen that, for a closed system, the Gibbs energy is related to pressure and temperature as follows: \[dG =VdP - SdT \label{16.15} \] For a constant temperature process, \[dG = VdP \label {16.16} \] Equation \(\ref{16.16}\) can be evaluated for an ideal gas \[ dG =\dfrac{RT}{P} dP \label{16.17} \] \[dG = RT\;d\ln P \;\;\; \text{ @ constant T } \label{16.18} \] This expression by itself is strictly applicable to ideal gases. However, Lewis, in 1905, suggested extending the applicability of this expression to all substances by defining a new thermodynamic property called fugacity, \(f\), such that: \[dG = RT\ln f \;\;\; \text{ @ constant T } \label {16.19} \] This definition implies that for ideal gases, ‘f’ must be equal to ‘P’. For mixtures, this expression is written as: \[d\bar{G} = RT\ln f_i \;\;\; \text{ @ constant T } \label {16.20} \] where and f are the partial molar Gibbs energy and fugacity of the i-th component, respectively. Fugacity can be readily related to chemical potential because of the one-to-one relationship of Gibbs energy to chemical potential, which we have discussed previously. Therefore, the definition of fugacity in terms of chemical potential becomes: For a pure substance, @ const T
(ideal gas limit)
For a component in a mixture,
@ const T = partial pressure (ideal gas limit) The fugacity coefficient (Φ ) is defined as the ratio of fugacity to its value at the ideal state. Hence, for pure substances: and for a component in a mixture, The fugacity coefficient takes a value of unity when the substance behaves like an ideal gas. Therefore, the fugacity coefficient is also regarded as a ; the closer the value of the fugacity coefficient is to unity, the closer we are to the ideal state. Fugacity turns out to be an auxiliary function to chemical potential. Even though the concept of thermodynamic equilibrium which we discussed in the previous section is given in terms of chemical potentials, above definitions allow us to restate the same principle in terms of fugacity. To do this, previous expressions can be integrated for the change of state from liquid to vapor at saturation conditions to obtain: For equilibrium, ,hence, Therefore: ; i = 1, 2, … n For equilibrium, fugacities must be the same as well! This is, for a system to be in equilibrium, both the fugacity and the chemical potential of each component in each of the phases must be equal. Conditions (16.14) and (16.25) are equivalent. Once one of them is satisfied, the other is satisfied immediately. Using or to describe equilibrium is a matter of choice, but generally the fugacity approach is preferred. | 2,718 | 1,732 |
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The potential energy curve linking the carboxyloids can be used as a guide to how these compounds can be interconverted. In general, it is possible to take a compound that is higher on the ski hill and convert it to a compound that is lower on the ski hill. The potential energy surface linking carboxloids. For example, acid chlorides are widely used to make other carboxyloids. By choosing the correct nucleophile, an acid chloride could be converted to any of the other derivatives. This is really the whole point of an acid chloride; it has no other function other than to provide an easy way to make other derivatives. Acid anhydrides, also high on the potential energy curve, are also used in the same way. They could be used to make any of the derivatives lower than they are on the ski hill. In turn, acid anhydrides could conceivably be made from acid chlorides. However, because an acid anhydride plays the same role as an acid chloride -- providing a source with which to make the other derivatives -- we wouldn't normally make an acid anhydride from an acid chloride. We would either make one of the other derivatives directly from the acid chloride, or else make it from an acid anhydride that was obtained in another way. Show all of the ways that you could make the following compounds by going down the ski hill. (Show the starting material, the reagent added and a reaction arrow going to the product.) For help with names, see the appendix. Converting a carboxylic acid into an amide is complicated by a side reaction, as a result of which amide formation becomes an uphill process. Sometimes, two carboxyloids are close enough on the ski hill that it may be possible to convert in either direction between them. In other words, there is an equilibrium between these two compounds, and the equilibrium constant is close enough to unity (K = 1) that the equilibrium can be pushed in either direction. For selected cases, we can choose which direction the equilibrium will go by changing the reaction conditions. To do so, we can use an important concept of equilibrium: le Chatelier's principle (luh sha-TELL-yay). According to le Chatelier, if conditions in the reaction cause the reaction to move away from equilibrium, the reaction will shift direction until it is back at equilibrium again. In other words, if the actual ratio of reactants to products strays from what it ought to be, the correct reaction will occur so that the ratio returns to normal. le Chatelier's principle should be partly intuitive. If a reaction is occurring in both reactions, there are really two reactions, one going in each direction. The ability of these reactions to occur depends partly on the amount of reactants available for the forward reaction or the reverse reaction. If extra starting materials are added (on the left side of the reaction), there is too much reactant as defined by the equilibrium constant. The denominator gets bigger and the ratio of products to reactants goes down. However, because there is extra starting material for the forward reaction, more product is quickly made, until the ratio returns to normal. The opposite situation applies if too much product is made. From the point of view of the reverse reaction, those products of the forward reaction are really the reactants needed to go in the opposite direction. The reverse reaction has more material to work with, and this material can quickly be converted into the stuff on the left hand side of the reaction. Draw the mechanism for the conversion of: In the interconversion of carboxyloids, equilibrium can be influenced in different ways. The conversion between esters and carboxylic acids can be influenced by the solvent used for the reaction. For example, an ester might be converted to a carboxylic acid under aqueous conditions, but a carboxylic acid might be converted to an ester using the appropriate alcohol as the solvent. In other words, because water and alcohol can be viewed as reactants in these cases, adding more can shift the reaction to one side. Solvent is usually present in concentrations many tens or hundreds or thousands of times higher than the reactants and products. Changing the solvent thus has a big impact on the direction of equilibrium. Another strategy used to influence the equilibrium involves removal of product. For example, if water is a product of the reaction, a drying agent can be added to absorb the water. Drying agents include compounds such as MgSO or zeolites (mixed aluminosilicates containing Al, Si, O and other metal ions such as Mg , Ca or Ti ). If a carboxylic acid is a product of a reaction, its concentration can be lowered by converting it to a carboxylate salt; this would happen easily in the presence of base. In either case, the disappearance of a product of the reaction (or a side product) would draw the reaction to the right in order to replace those products an re-establish the correct equilibrium ratio. Show, with curved arrows, how a magnesium ion could remove water from solution. Show, with curved arrows, how sodium hydroxide would remove butanoic acid from solution. , | 5,151 | 1,733 |
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The Wave Nature of the Electron Lecture Demonstrations Commercial Photoelectric Effect Demonstrators are excellent for this purpose, but if one is not available: Tape a piece of plastic diffraction grating to the lens of an overhead projector, as in Figure, and place an opaque cover on the Fresnel lens as shown, to project a continuous spectrum. Projecting a Spectrum On the wall, tape a piece of phosphorescent sheeting protected from room lights with a cover made from file folders (etc). Dim the room lights, and project the spectrum on the phosphorescent material as shown. After several seconds, turn off the projector, and observe that the greatest phosphorescent intensity is below the visible range. Use metal forks connected to the AC output of a Variac to excite the sodium wavelength in a dill pickle. | 830 | 1,735 |
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To illustrate how mechanisms may affect the concentration profile for a reaction, we can examine some simple mechanisms In this type of reaction, one substance is simply converting into another. An example of this type of reaction might be the isomerization of methylisocyanide to form acetonitrile (methylcyanide) (Redmon, Purvis, & Bartlett, 1978). If the reaction mechanism consists of a single unimolecular step, which is characterized by the rate constant \(k_1\): \[ A \xrightarrow{k_1} B \nonumber \] then rate of change of the concentrations of \(A\) and \(B\) may be written \[\dfrac{d[A]}{dt} = - k_1 [A] \nonumber \] and \[\dfrac{d[B]}{dt} = + k_1 [A] \nonumber \] A plot the concentrations as a function of time would look as follows: It can be easily seen that the concentration of the reactant (A) decreases as time moves forward, and that of the product (B) increases. This will continue until reactant A is depleted. When the system can establish equilibrium, the rate of change of the concentration of A and B will depend on both the forward and reverse reactions. If k is the rate constant that characterizes the forward reaction \[ A \xrightarrow{k_1} B \nonumber \] and k that which characterizes the reverse \[ B \xrightarrow{k_{-1}} A \nonumber \] the the rate of change on concentrations of A and B can be expressed \[\dfrac{d[A]}{dt} = - k_1 [A] + k_{-1} [B] \nonumber \] and \[\dfrac{d[B]}{dt} = + k_1 [A] - k_{-1} [B] \nonumber \] The concentration profile for this situation looks as follows: This profile is characterized by the fact that after a certain amount of time, the system achieves equilibrium and the concentrations stop changing (even though the forward and reverse reactions are still taking place. This is the nature of a about which we speak off of the time in chemistry. The final concentrations of [A] and [B] once equilibrium is established will depend on the ratio of k and k Since the rate of formation of \(A\) (from the reverse step) is equal to the rate of consumption of \(A\) (from the forward step, the overall rate of change of the concentration of A is zero once equilibrium has been established. So it should be clear that \[k_1[A] = k_{-1}[B] \nonumber \] or \[\dfrac{k_1}{k_{-1}} = \dfrac{[B]}{[A]} \nonumber \] and the ratio o f\(k_1\) to \(k_{-1}\) gives the value of the equilibrium constant! Some reactions require a to mediate the conversion of reactants in to products. The definition of a catalyst is a species that must be added (it is not formed as an intermediate) shows up in the mechanism (usually in a very early step) and this ends up as part of the rate law, but is reformed later on so that it does not appear in the overall stoichiometry. If the reaction \[A \rightarrow B \nonumber \] is aided by a catalyst \(C\), then one possible (single-step) reaction mechanism might be \[A + C \rightarrow B + C \nonumber \] In this case, \(C\) is acting as a catalyst to the reaction. The rate of change of the concentrations can be found by \[\dfrac{[A]}{dt} = -k [A,C] \nonumber \] \[\dfrac{[B]}{dt} = k [A,C] \nonumber \] \[\dfrac{[C]}{dt} = - k [A,C] + k [A,C] = 0 \nonumber \] This is a very simplified picture of a catalyzed reaction. Generally a catalyzed reaction will require at least two steps: \[A +C \rightarrow AC \nonumber \] \[AC \rightarrow B + C \nonumber \] Later, we will see how the steady-state approximation actually predicts the above depicted concentration profile for the two-step mechanism when \(AC\) is a short-lived species that can be treated as having a constant and small concentration. Another important (and very common) mechanistic feature is the formation of an intermediate. This is a species that is formed in at least one of the mechanism step, but does not appear in the overall stoichiometry for the reaction. This is different from a catalyst which must be added to speed the reaction. A simple example of a reaction mechanism involving the formation of a catalyst is \[ A \xrightarrow{k_1} B \nonumber \] \[ B \xrightarrow{k_2} C \nonumber \] In this case, \(C\) cannot form until an appreciable concentration of the intermediate \(B\) has been created by the first step of the mechanism. The rate of change of the concentrations of \(A\), \(B\), and \(C\) can be expressed \[\dfrac{[A]}{dt} = -k_1 [A] \nonumber \] \[\dfrac{[B]}{dt} = k_1 [A] - k_2 [B] \nonumber \] \[\dfrac{[C]}{dt} = k_2 [B] \nonumber \] The concentration profile is then shown below. Notice the delay in the formation of \(C\). In many cases, the formation of an intermediate involves a reversible step. This step is sometimes referred to as a step since it oftentimes will establish a near equilibrium while the reaction progresses. The result of combining a pre-equilibrium with an intermediate produces a profile that shows features of both of the simpler mechanisms. An example of such a mechanism is \[ A \xrightleftharpoons [k_1]{k_{-1}} B \nonumber \] \[ B \xrightarrow{k_2} C \nonumber \] In this case, the rate of change for the concentrations of \(A\), \(B\), and \(C\) can be expressed by \[\dfrac{[A]}{dt} = -k_1 [A] + -k_{-1} [B] \nonumber \] \[\dfrac{[B]}{dt} = k_1 [A] -k_{-1} [B] - k_2 [B] \nonumber \] \[\dfrac{[C]}{dt} = k_2 [B] \nonumber \] The concentration profile for this mechanism is shown below. Again, notice the delay in the production of the product \(C\), due to the requirement that the concentration of B be sufficiently high to allow the second step to occur with an appreciable rate. There are many cases where a reactant can follow pathways to different products (or sometimes even the same products!), and those pathways compete with one another. An example is the following simple mechanism: \[ A \xrightarrow{k_{-1}} B \nonumber \] \[ A \xrightarrow{k_2} C \nonumber \] In this case, the rate of change on concentrations can be expressed as \[\dfrac{[A]}{dt} = -k_1 [A] + -k_2 [A] \nonumber \] \[\dfrac{[B]}{dt} = + k_1 [A] \nonumber \] \[\dfrac{[C]}{dt} = + k_2 [A] \nonumber \] Overall, the profile looks like two first order decompositions occurring at the same time, with the final concentration of the product formed with the larger rate constant being favored. One of the goals of studying chemical kinetics is to understand how to alter reaction condition to favor the production of desirable reaction products. This can be accomplished by a number of means, such as alteration of concentrations, temperature, addition of catalysts, etc. Understanding the basics will (hopefully) lead to a better understanding of how concentration profiles can be altered by changing conditions. | 6,631 | 1,736 |
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When an atom or molecule is electronically excited by the absorption of a photon, there are a number of schemes by which that species can then return to the ground state. One mode involves the emission of a photon of light, which causes the substance to fluoresce or phosphorese. Alternatively, species can lose their excitation by nonradiative means: The first of these nonradiative schemes is called dynamic quenching, and will be our primary focus in the exercise below. Fluorescence or phosphorescence lifetime measurements are often used to investigate the various modes of de-excitation that occur following photo-excitation. In such an experiment, a pulsed laser or gas discharge lamp is used to excite a small population of excited state atoms or molecules. In solution, the lifetimes of excited states can be extremely short (on the order of nanoseconds). Therefore, the light source that is employed in solution studies must emit pulses that are generally 1 ns or shorter. Let us assume that we fire the laser once and generate a temporary concentration of excited state molecules \([A^*]\) at some moment in time. Let us further assume that \(A^*\) is strongly fluorescent, and that we can follow the intensity of this fluorescence at later times using a sensitive light detector like a photomultiplier tube. How do we expect the fluorescence to decay in time? Will it decay linearly, in an exponential fashion, or some other way? And what will the shape of this decay curve tell us about the various de-excitation schemes that are occurring in the system? If there are no quenching agents present in our system (i.e. there are no species in solution that can effectively quench the fluorescence through bimolecular collisions), then A* can return to the ground state by simple fluorescence \[A^* \ce{ ->[k_1]} A + h\nu \label{1}\] and by non-radiative decay \[A^* \ce{ ->[k_2]} A \label{2}\] where \(k_1\) and \(k_2\) are the rate constants for these two processes. With only these two paths to the ground state available, the rate equation for \([A^*]\) can be written as \[ \dfrac{[A^*]}{dt} = - k_1 [A^*] - k_2 [A^*] = -( k_1 + k_2)[A^*] \label{3}\] Rearrangment and integration of Equation \ref{3} with respect to initial conditions \(t = 0\) and \([A^*] = [A^*]_0\) gives \[ [A^*] = [A^*]_0 e^{-(k_1+k_2)/t}. \label{4}\] According to this result, the concentration of excited species (and the fluorescence) is expected to decay away in an exponential fashion. The rate constants \(k_1\) and \(k_2\) quantify how quickly the fluorescence decays. For convenience, we will define a ‘fluorescence lifetime in the absence of quencher’ \[ \tau_o = \dfrac{1}{k_1+k_2} \label{5}\] \(\tau_o\) is the amount of time that it take for the fluorescence intensity to decay to \(1/e\) its initial value. A typical fluorescence decay curve is shown below in Figure \(\Page {1}\) (the time axis is in units of nanoseconds). The fluorescence intensity spikes upward when the sample is irradiated by the light pulse (at 10 ns), and the intensity decays away in an exponential fashion (the lifetime associated with this particular curve is approximately 30 ns). If a quenching agent (\(Q\)) is present in solution, then a third process becomes available for returning \(A^*\) molecules to the ground state; \[A^* +Q \ce{ ->[k_q]} A \label{6}\] and the rate equation (Equation \ref{3}) for \([A^*]\) then becomes \[ \dfrac{[A^*]}{dt} = -( k_1 + k_2 + k_q[Q])[A^*] \label{7}\] \(k_q\) which appears in Equations \ref{6} and \ref{7} is called the . Assuming \([Q] \gg [A^*]\), which means \([Q]\) can be treated as a constant (the concentration of quencher is normally orders of magnitude larger than \(A^*\) in practice), then Equation \ref{7} can be integrated to give \[ [A^*] = [A^*]_0 e^{-(k_1+k_2+k_q[Q])/t}. \label{8}\] Consequently, the fluorescence decay is still exponential, but the fluorescence lifetime (with quencher) becomes \[ \tau_o = \dfrac{1}{k_1+k_2+k_q[Q]} \label{9}\] To isolate the effects of quenching, fluorescence lifetime measurements are carried out over a range of quenching agent concentrations (including \([Q] = 0\)). A fluorescence decay curve is recorded for each trial and each decay curve is fit to an exponential function, which yields a lifetime for each trial. Dividing Equation \ref{9} into Equation \ref{5} gives \[ \begin{align} \dfrac{\tau_o}{\tau} &= \dfrac{k_1+k_2+k_q[Q]}{k_1+k_2} \nonumber \\[5pt] &= 1 + k_q\tau_o[Q] \nonumber \end{align} \label{10}\] According to Equation \ref{10}, a plot of \(\tau_o/\tau\) versus \([Q]\) should be linear with an intercept equal to one, and the slope can be analyzed to obtain the bimolecular quenching rate constant, \(k_q\). Such a plot is called a . In the exercise below, you will analyze fluorescent lifetime data for two different systems (each with its own quencher) and will be prompted to determine kq in each case. The following data where obtained from L.K. Fraiji, D.M. Hayes, and T.C. Werner, Journal of Chemical Education, 69(5), 424 (1992). In the first experiment, seven aqueous solutions of the fluorescent agent N-methylacridinium iodide (N-MEAI) were prepared with increasing concentrations of a quencher, (guanosine 5'-monophosphate, disodium salt trihydrate), which is abbreviated GMP. Each sample was irradiated with a 358 nm pulsed light source and the fluorescence was recorded at 485 nm. The lifetime associated with each trial was obtained by fitting an exponential function to the measured decay curve. These lifetimes are indicated in the table below. In the second experiment, five solutions of the fluorescent agent 1-pyrenesulfonic acid, sodium salt (PSA) were prepared with increasing concentrations iodide ion (which is an effective quenching agent for many fluorescent species). For these solutions, each sample was irradiated with a 337 nm pulsed light source and the fluorescence was recorded at 395 nm. | 5,936 | 1,737 |
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There are two fundamental kinds of equilibrium problems: We saw previously the equilibrium constant for the decomposition of \(\ce{CaCO3(s)}\) to \(\ce{CaO(s)}\) and \(\ce{CO2(g)}\) is \(K = [\ce{CO2}]\). At 800°C, the concentration of \(\ce{CO2}\) in equilibrium with solid \(\ce{CaCO3}\) and \(\ce{CaO}\) is \(2.5 \times 10^{-3}\; M\). Thus \(K\) at 800°C is \(2.5 \times 10^{-3}\). (Remember that equilibrium constants are unitless.) A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane). This reaction can be written as follows: \[\ce{n-butane_{(g)} \rightleftharpoons isobutane_{(g)}} \label{Eq1}\] and the equilibrium constant \[K = \dfrac{[\text{isobutane}]}{[\text{n-butane}]}.\] At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression, \[\begin{align} K &=\dfrac{[isobutane]}{[n-butane]} \\[4pt] &=\dfrac{0.041\; M}{0.016\,M} \\[4pt] &= 2.6 \label{Eq2} \end{align}\] Thus the equilibrium constant for the reaction as written is 2.6. The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid: \[\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \nonumber\] A mixture of \(\ce{SO_2}\) and \(\ce{O_2}\) was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained Calculate \(K\) and \(K_p\) at this temperature. : balanced equilibrium equation and composition of equilibrium mixture : equilibrium constant Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain \(K\). Substituting the appropriate equilibrium concentrations into the equilibrium constant expression, \[\begin{align*} K &=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]} \\[4pt] &=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})} \\[4pt] &=7.9 \times 10^4 \end{align*}\] To solve for \(K_p\), we use the relationship between \(K\) and \(K_p\), where \(Δn = 2 − 3 = −1\): \[\begin{align*} K_p &=K(RT)^{Δn} \\[4pt] &=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1} \\[4pt] &=1.2 \times 10^3 \end{align*}\] Hydrogen gas and iodine react to form hydrogen iodide via the reaction \[\ce{H2(g) + I2(g) <=> 2HI(g)} \nonumber\] A mixture of \(\ce{H_2}\) and \(\ce{I_2}\) was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained Calculate \(K\) and \(K_p\) for this reaction. : Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example \(\Page {2}\) shows one way to do this. A 1.00 mol sample of \(NOCl\) was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of \(Cl_2\). Calculate \(K\) at this temperature. The equation for the decomposition of \(NOCl\) to \(NO\) and \(Cl_2\) is as follows: \[\ce{2 NOCl(g) <=> 2NO(g) + Cl2(g)} \nonumber\] : balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium : \(K\) The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows: \[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2} \nonumber\] To obtain the concentrations of \(\ce{NOCl}\), \(\ce{NO}\), and \(\ce{Cl_2}\) at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations. \[\ce{2 NOCl(g) <=> 2NO(g) + Cl2(g)} \nonumber\] Initially, the system contains 1.00 mol of \(NOCl\) in a 2.00 L container. Thus \([NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M\). The initial concentrations of \(NO\) and \(Cl_2\) are \(0\; M\) because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of \(Cl_2\) in a 2.00 L container, so \([Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M\). We insert these values into the following table: \[\ce{2 NOCl(g) <=> 2NO(g) + Cl2(g)} \nonumber\] We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of \(Cl_2\), the substance for which initial and final concentrations are known: \[Δ[Cl_2] = 0.028 \;M_{(final)} − 0.00\; M_{(initial)}] = +0.028\; M \nonumber\] According to the coefficients in the balanced chemical equation, 2 mol of \(NO\) are produced for every 1 mol of \(Cl_2\), so the change in the \(NO\) concentration is as follows: \[Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M\] Similarly, 2 mol of \(NOCl\) are consumed for every 1 mol of \(Cl_2\) produced, so the change in the \(NOCl\) concentration is as follows: \[Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M\] We insert these values into our table: \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber\] D We sum the numbers in the \([NOCl]\) and \([NO]\) columns to obtain the final concentrations of \(\ce{NO}\) and \(\ce{NOCl}\): \[[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M\] \[[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M\] We can now complete the table: \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\] We can now calculate the equilibrium constant for the reaction: \[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4}\] The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (\(\ce{NH3}\)) by reacting \(0.1248\; M \; \ce{H_2}\) and \(0.0416\; M \;\ce{N_2}\) at about 500°C. At equilibrium, the mixture contained 0.00272 M \(\ce{NH_3}\). What is \(K\) for the reaction \[\ce{N2 + 3H2 <=> 2NH_3}\] at this temperature? What is \(K_p\)? Answer: Using Q to Find Equilibrium Concentrations: To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation \(\ref{Eq1}\)), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example \(\Page {2}\). \[\text{n-butane}(g) \rightleftharpoons \text{isobutane}(g)\] The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as +x, then the change in the concentration of n-butane is Δ[n-butane] = −x. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone. \[\text{n-butane}(g) \rightleftharpoons \text{isobutane}(g)\] Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation, \[K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\dfrac{x}{1.00−x}=2.6\] Rearranging and solving for \(x\), \[\begin{align*} x &=2.6(1.00−x) \\[4pt] x &= 2.6−2.6x \\[4pt] x+2.6x &= 2.6 \\[4pt] x &=0.72 \end{align*}\] We obtain the final concentrations by substituting this \(x\) value into the expressions for the final concentrations of n-butane and isobutane listed in the table: \[\begin{align*}[\text{n-butane}]_f &= (1.00 − x) M \\[4pt] &= (1.00 − 0.72) M \\[4pt] &= 0.28\; M \\[5pt] [\text{isobutane}]_f &= (0.00 + x) M \\[4pt] &= (0.00 + 0.72) M \\[4pt] &= 0.72\; M \end{align*}\] We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same \(K\) that we used in the calculation: \[\begin{align*} K &=\dfrac{[\text{isobutane}]}{[\text{n-butane}]} \\[4pt] &=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) \\[4pt] &=2.6 \end{align*}\] This is the same \(K\) we were given, so we can be confident of our results. Example \(\Page {3}\) illustrates a common type of equilibrium problem that you are likely to encounter. The water–gas shift reaction is important in several chemical processes, such as the production of \(\ce{H2}\) for fuel cells. This reaction can be written as follows: \[\ce{H2(g) + CO2(g) <=> H2O(g) + CO(g)} \nonumber\] \(K = 0.106\) at 700 K. If a mixture of gases that initially contains 0.0150 M \(\ce{H_2}\) and 0.0150 M \(\ce{CO_2}\) is allowed to equilibrate at 700 K, what are the final concentrations of all substances present? : balanced equilibrium equation, \(K\), and initial concentrations Asked for: final concentrations : The initial concentrations of the reactants are \([\ce{H_2}]_i = [\ce{CO_2}]_i = 0.0150\; M\). Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of \(\ce{H_2O}\) as \(x\), then \(Δ[\ce{H_2O}] = +x\). We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of \(x\). For example, 1 mol of \(\ce{CO}\) is produced for every 1 mol of \(\ce{H_2O}\), so the change in the \(\ce{CO}\) concentration can be expressed as \(Δ[\ce{CO}] = +x\). Similarly, for every 1 mol of \(\ce{H_2O}\) produced, 1 mol each of \(\ce{H_2}\) and \(\ce{CO_2}\) are consumed, so the change in the concentration of the reactants is \(Δ[\ce{H_2}] = Δ[\ce{CO_2}] = −x\). We enter the values in the following table and calculate the final concentrations. \[H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\] B We can now use the equilibrium equation and the given \(K\) to solve for \(x\): \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x}=\dfrac{x^2}{(0.0150−x)^2}=0.106\] We could solve this equation with the quadratic formula, but it is far easier to solve for \(x\) by recognizing that the left side of the equation is a perfect square; that is, \[\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106\] Taking the square root of the middle and right terms, \[\begin{align*} \dfrac{x}{(0.0150−x)} &=(0.106)^{1/2} \\[4pt] &=0.326 \\[4pt] x &=(0.326)(0.0150)−0.326x \\[4pt] 1.326x &=0.00489 \\[4pt] x &=0.00369=3.69 \times 10^{−3} \end{align*}\] The final concentrations of all species in the reaction mixture are as follows: We can check our work by inserting the calculated values back into the equilibrium constant expression: \[\begin{align*} K &=\dfrac{[H_2O,CO]}{[H_2,CO_2]} \\[4pt] &=\dfrac{(0.00369)^2}{(0.0113)^2} \\[4pt] &=0.107 \end{align*}\] To two significant figures, this \(K\) is the same as the value given in the problem, so our answer is confirmed. Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation: \[\ce{H2(g) + I2(g) <=> 2HI(g)}\] \(K = 54\) at 425°C. If 0.172 M \(H_2\) and \(I_2\) are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture? : In Example \(\Page {3}\), the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example \(\Page {4}\). In the water–gas shift reaction shown in Example \(\Page {3}\), a sample containing 0.632 M \(\ce{CO2}\) and 0.570 M \(\ce{H_2}\) is allowed to equilibrate at 700 K. At this temperature, \(K = 0.106\). What is the composition of the reaction mixture at equilibrium? : balanced equilibrium equation, concentrations of reactants, and \(K\) : composition of reaction mixture at equilibrium : \([\ce{CO_2}]_i = 0.632\; M\) and \([\ce{H_2}]_i = 0.570\; M\). Again, x is defined as the change in the concentration of \(\ce{H_2O}\): \(Δ[\ce{H_2O}] = +x\). Because 1 mol of \(CO\) is produced for every 1 mol of \(H_2O\), the change in the concentration of \(\ce{CO}\) is the same as the change in the concentration of \(\ce{H2O}\), so Δ[\ce{CO}] = +x. Similarly, because 1 mol each of \(\ce{H_2}\) and \(\ce{CO_2}\) are consumed for every 1 mol of \(H_2O\) produced, \(Δ[H_2] = Δ[CO_2] = −x\). The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium. \[H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\] We can now use the equilibrium equation and the known \(K\) value to solve for \(x\): \[\begin{align*} K &=\dfrac{[\ce{H_2O},\ce{CO}]}{[\ce{H_2},\ce{CO_2}]} \\[4pt] &=\dfrac{x^2}{(0.570−x)(0.632−x)} \\[4pt] &=0.106 \end{align*}\] In contrast to Example \(\Page {3}\), however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator: \[x^2 = 0.106(0.360 − 1.20x + x^2) \nonumber\] Collecting terms on one side of the equation, \[0.894x^2 + 0.127x − 0.0382 = 0 \nonumber\] This equation can be solved using the quadratic formula: \[\begin{align*} x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\[4pt] &= \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \end{align*}\] so \[x =0.148 \text{ and } −0.290 \nonumber\] Only the answer with the positive value has any physical significance, so \(Δ[H_2O] = Δ[CO] = +0.148 M\), and \(Δ[H_2] = Δ[CO_2] = −0.148\; M\). The final concentrations of all species in the reaction mixture are as follows: We can check our work by substituting these values into the equilibrium constant expression: \[\begin{align*} K &=\dfrac{[H_2O,CO]}{[H_2,CO_2]} \\[4pt] &=\dfrac{(0.148)^2}{(0.422)(0.484)} \\[4pt] &=0.107 \end{align*}\] Because \(K\) is essentially the same as the value given in the problem, our calculations are confirmed. The exercise in Example \(\Page {1}\) showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which \(K = 54\) at 425°C. If a sample containing 0.200 M \(H_2\) and 0.0450 M \(I_2\) is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture? : In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small (\(K ≤ 10^{−3}\)) or very large (\(K ≥ 10^3\)), which means that the change in the concentration (defined as x) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example \(\Page {5}\). Atmospheric nitrogen and oxygen react to form nitric oxide: \[N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\] with \(K_p = 2.0 \times 10^{−31}\) at 25°C. What is the partial pressure of \(\ce{NO}\) in equilibrium with \(\ce{N_2}\) and \(\ce{O_2}\) in the atmosphere (at 1 atm, \(P_{\ce{N_2}} = 0.78\; atm\) and \(P_{\ce{O_2}} = 0.21\; atm\)? : balanced equilibrium equation and values of \(K_p\), \(P_{\ce{O_2}}\), and \(P_{\ce{N_2}}\) : partial pressure of NO : Because we are given \(K_p\) and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of \(\ce{O_2}\) is 0.21 atm and that of \(\ce{N_2}\) is 0.78 atm. If we define the change in the partial pressure of \(\ce{NO}\) as 2x, then the change in the partial pressure of \(\ce{O_2}\) and of \(\ce{N_2}\) is −x because 1 mol each of \(\ce{N_2}\) and of \(\ce{O_2}\) is consumed for every 2 mol of \(\ce{NO}\) produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium. \[\ce{N2(g) + O2(g) <=> 2NO(g)}\] Substituting these values into the equation for the equilibrium constant, \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \nonumber\] In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the \(x\) value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x) = 0.78 and (0.21 − x) = 0.21. Substituting these expressions into our original equation, \[\begin{align*} \dfrac{(2x)^2}{(0.78)(0.21)} &= 2.0 \times 10^{−31} \\[4pt] \dfrac{4x^2}{0.16} &=2.0 \times10^{−31} \\[4pt] x^2 &=\dfrac{0.33 \times 10^{−31}}{4} \\[4pt] x&=9.1 \times 10^{−17} \end{align*}\] Substituting this value of x into our expressions for the final partial pressures of the substances, From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, \(2.0 \times 10^{−16}\) is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or \(10^{−3} > K > 10^3\), then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic \(NO\), an ingredient of smog, does not form from atmospheric concentrations of \(N_2\) and \(O_2\) to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation: \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 \times 10^{−31} \nonumber\] The final \(K_p\) agrees with the value given at the beginning of this example. Under certain conditions, oxygen will react to form ozone, as shown in the following equation: \[\ce{3O2(g) <=> 2O3(g)} \nonumber\] with \(K_p = 2.5 \times 10^{−59}\) at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere (\(P_{O_2}=0.21\; atm\))? \(4.8 \times 10^{−31} \;atm\) Finding Equilibrium Concentrations for Reactions with a Small K Value: Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (K ≥ 103). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example \(\Page {6}\). The chemical equation for the reaction of hydrogen with ethylene (\(C_2H_4\)) to give ethane (\(C_2H_6\)) is as follows: \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}\] with \(K = 9.6 \times 10^{18}\) at 25°C. If a mixture of 0.200 M \(H_2\) and 0.155 M \(C_2H_4\) is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture? : balanced chemical equation, \(K\), and initial concentrations of reactants : equilibrium concentrations : : From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example \(\Page {5}\). If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions. \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \nonumber\] Substituting values into the equilibrium constant expression, \[K=\dfrac{[C_2H_6]}{[H_2,C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18}\] Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x) = 0.045 and (0.155 − x) = 0.155] as follows: \[K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18}\] \[x=3.6 \times 10^{−19}\] The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table: We can verify our calculations by substituting the final concentrations into the equilibrium constant expression: \[K=\dfrac{[C_2H_6]}{[H_2,C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18}\] This \(K\) value agrees with our initial value at the beginning of the example Hydrogen reacts with chlorine gas to form hydrogen chloride: \[\ce{ H2(g) + Cl 2(g) <=> 2HCl(g)} \nonumber\] with \(K_p = 4.0 \times 10^{31}\) at 47°C. If a mixture of 0.257 M \(H_2\) and 0.392 M \(Cl_2\) is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture? : When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium. Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. | 24,042 | 1,738 |
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: pH measurements are presented for a series of HCl solutions of increasing concentration. The data are analyzed to determine the activities and activity coefficients of the hydrogen-ion. The results are compared to theoretical predictions from the extended Debye-Hückel equation. : A knowledge of activities, activity coefficients and the Debye-Huckel equation for mean activity coefficients. : The calculations and graphing associated with this exercise can be carried out within any quantitative analysis software. Students of chemistry are well acquainted with the relationship between solution pH and hydrogen-ion concentration: \[pH = -\log_{10} [\ce{H^{+}}] .\label{1}\] However, this expression is only accurate in the dilute limit. The exact relationship depends on the activity of the hydrogen ion: \[pH = -\log_{10} a_{\ce{H^{+}}} \label{2}\] Consequently, activities for hydrogen-ion in a given solution can be determined through simple pH measurements, and the activity coefficient (\(γ\)) can be evaluated using the relationship \[a = \gamma C \label{3}\] where \(C\) is concentration. Activities can be defined in terms of molar concentrations (\(M\)) or molal concentrations (\(m\)). Activities calculated through the use of Equation \ref{2} will be 'molar-based', which in turn yields a molar-based activity coefficient. This distinction is made clear here by rewriting Equation \ref{3} as \[a_M = \gamma_M [\ce{H^{+}}] \label{4}\] The corresponding expression for a 'molality-based' activity is given by \[a_m = \gamma_m m[\ce{H^{+}}] \label{5}\] where the two types of activity coefficients are related by the expression \[ \gamma_M = \dfrac{\gamma_m [ \ce{H^{+}}]D}{m_{\ce{H^{+}}}} \label{6}\] and \(D\) is the solution density. For moderately dilute aqueous solutions at ambient conditions, the density of the solution is equal to that of pure water (\(D = 1\, g/mL\)), meaning the molar and molal concentration are essentially equal (within experimental error). Consequently, the molar and molal activity coefficients are equal to one another. In electrolytic solutions, a convention has been adopted whereby one assumes that the non-ideality of the solution is shared by the cation and anion. This is accomplished by defining a mean-activity coefficient according to the expression \[\gamma_\pm = (\gamma_+^{\nu_+}\gamma_-^{\nu_-})^{1/(\nu_++\nu_-)} \label{7}\] where \(\nu_+\) and \(\nu_-\) are the number of cations and anions, respectively, in one formula unit of the electrolyte. A number of theoretical approaches are available for estimating mean activity coefficients. One expression that is valid up to moderate concentrations is an empirical modification of the Debye-Huckel limiting law (the ‘extended’ Debye Huckel equation) given by \[\log \gamma_\pm = -A | z_+z_-| \left(\dfrac{\sqrt{I}}{1 + B \sqrt{I}} \right) \label{8}\] where \(A\) is a constant the depends on properties of the solvent (0.5085 for water at 25 °C), \(z_+\) and \(z_-\) are the ionic charges, \(I\) is the , and \(B\) is an empirical constant. Activity coefficients calculated using Equation \ref{8} are molal-based. The molar ionic strength, \(I\), of a solution is a function of the concentration of all ions present in that solution \[I = \dfrac{1}{2} \sum_{i=1}^{n} c_i z_i^2\] where one half is because we are including both cations and anions, \(c_i\) is the molar concentration of ion \(i\) (M, mol/L), \(z_i\) is the charge number of that ion, and the sum is taken over all ions in the solution.
The exercise outlined below will involve using pH data to experimentally determine activity coefficients for hydrogen-ion and subsequently using this data to test the validity of 'extended' Debye Huckel law. The following table gives experimentally determined pH values for a series of HCl solutions of increasing concentration at 25 °C. Data based upon information contained in Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752 (2006) and G.N. Lewis, M. Randall, K. Pitzer, D.F. Brewer, Thermodynamics (McGraw-Hill: New York, 1961; pp. 233-34). | 4,102 | 1,739 |
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We can measure the standard potentials for a wide variety of chemical substances, some of which are listed in Table P2. These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table \(\Page {1}\). All reactants that lie below the SHE in the table are stronger oxidants than H , and all those that lie above the SHE are weaker. The strongest oxidant in the table is F , with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element. Not all oxidizers and reducers are created equal. The standard reduction potentials in Table \(\Page {1}\) can be interpreted as a ranking of substances according to their oxidizing and reducing power. Strong oxidizing agents are typically compounds with elements in high oxidation states or with high electronegativity, which gain electrons in the redox reaction (Figure \(\Page {1}\)). Examples of strong oxidizers include hydrogen peroxide, permanganate, and osmium tetroxide. Reducing agents are typically electropositive elements such as hydrogen, lithium, sodium, iron, and aluminum, which lose electrons in redox reactions. Hydrides (compounds that contain hydrogen in the formal -1 oxidation state), such as sodium hydride, sodium borohydride and lithium aluminum hydride, are often used as reducing agents in organic and organometallic reactions. Similarly, all species in Table \(\Page {1}\) that lie above H reductants H H reductant ΔH reductant The standard reduction potentials can be interpreted as a ranking of substances according to their oxidizing and reducing power. Species in Table \(\Page {1}\) that lie above H are stronger reducing agents (more easily oxidized) than H . Species that lie below \(\ce{H2}\) are stronger oxidizing agents. Because the half-reactions shown in Table \(\Page {1}\) are arranged in order of their E° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions (E° > 0), as illustrated in Example \(\Page {1}\). The black tarnish that forms on silver objects is primarily \(\ce{Ag2S}\). The half-reaction for reversing the tarnishing process is as follows: \[\ce{Ag2S(s) + 2e^{−} → 2Ag(s) + S^{2−} (aq)} \quad E°=−0.69\, V \nonumber\] reduction half-reaction, standard electrode potential, and list of possible reductants reductants for \(\ce{Ag2S}\), strongest reductant, and potential reducing agent for removing tarnish From their positions in Table \(\Page {1}\), decide which species can reduce \(\ce{Ag2S}\). Determine which species is the strongest reductant. Use Table \(\Page {1}\) to identify a reductant for \(\ce{Ag2S}\) that is a common household product. We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag S/Ag couple in Table \(\Page {1}\) or (2) compare E° for each species with E° for the Ag S/Ag couple (−0.69 V). Use the data in Table \(\Page {1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: redox reaction and list of standard electrode potentials (Table \(\Page {1}\)) reaction spontaneity Adding the two half-reactions gives the overall reaction: \[\begin{align*}\textrm{cathode} &: \ce{Be^{2+}(aq)} +\ce{2e^-} \rightarrow \ce{Be(s)} \\ \textrm{anode} &: \ce{Sn(s)} \rightarrow \ce{Sn^{2+}(s)} +\ce{2e^-} \\ \hline \textrm{overall} &: \ce{Sn(s)} + \ce{Be^{2+}(aq)} \rightarrow \ce{Sn^{2+}(aq)} + \ce{Be(s)} \end{align*}\] with \[\begin{align*} E^\circ_{\textrm{cathode}} &=\textrm{-1.99 V} \\[4pt] E^\circ_{\textrm{anode}} &=\textrm{-0.14 V} \\[4pt] E^\circ_{\textrm{cell}} &=E^\circ_{\textrm{cathode}}-E^\circ_{\textrm{anode}} \\[4pt] &=-\textrm{1.85 V} \end{align*}\] The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot reduce Be to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn ) by metallic beryllium, which has a positive value of E° , will occur spontaneously. The two half-reactions and their corresponding potentials are as follows: \[\begin{align*}\textrm{cathode} &: \ce{MnO_2(s)}+\ce{4H^+(aq)}+\ce{2e^-} \rightarrow\ce{Mn^{2+}(aq)}+\ce{2H_2O(l)} \\ \ce{anode} &: \ce{H_2O_2(aq)}\rightarrow\ce{O_2(g)}+\ce{2H^+(aq)}+\ce{2e^-} \\\hline \textrm{overall} &: \ce{MnO_2(s)}+\ce{H_2O_2(aq)}+\ce{2H^+(aq)}\rightarrow\ce{O_2(g)}+\ce{Mn^{2+}(aq)}+\ce{2H_2O(l)} \end{align*}\] with \[\begin{align*} E^\circ_{\textrm{cathode}} &=\textrm{1.23 V} \\[4pt] E^\circ_{\textrm{anode}} &=\textrm{0.70 V} \\[4pt] E^\circ_{\textrm{cell}} &=E^\circ_{\textrm{cathode}}-E^\circ_{\textrm{anode}} \\[4pt] &=+\textrm{0.52 V} \end{align*}\] The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce \(\ce{MnO2}\), and oxygen gas will evolve from the solution. Use the data in Table \(\Page {1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: spontaneous with \(E^o_{cell} = 1.61\, V - 1.396\, V = 0.214\, V\) nonspontaneous with \(E^°_{cell} = −0.20\, V\) Although the sign of \(E^o_{cell}\) tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics. The relative strengths of various oxidants and reductants can be predicted using \(E^o\) values. The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase. 5. | 6,745 | 1,740 |
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The Electronic Structure of Atoms Lecture Demonstrations Observe aquarium lights (long, thin, vertically mounted tungsten bulbs) and similar shaped socket mounted fluorescent lamps through "Rainbow Glasses" or "Spectrum Glasses" to see the difference between and . Typical "Spectrum Tubes" available from scientific suppliers, and a variety of colored lamps may also be used. Sources: American Paper Optics, Inc. Bartlett, TN 901.381.1515; 800.767.8427 ; Rainbow Symphony, Inc.,6860 Canby Ave. Suite 120 • Reseda · CA · 91335; (818) 708-8400 (818) 708-8400 • (818) 708-8470 (818) 708-8470 Fax Show that atoms are the transducers of energy from all forms to light energy; 1. heat energy in, light out: Flame Tests 2. UV in, light out: phosphorescence 3. chemical energy in, light out: chemiluminescence (glow sticks, luminol, etc.) 4. electrical energy in: neon signs, lights (as above) | 905 | 1,741 |
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One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes. CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands. We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure \(\Page {1a}\)). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure \(\Page {1b}\), the d and d orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the e orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (d , d , and d , collectively called the t orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge. The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δ ), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δ .) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two e orbitals increase in energy by 0.6Δ , whereas the three t orbitals decrease in energy by 0.4Δ . Thus the total change in energy is \[2(0.6Δ_o) + 3(−0.4Δ_o) = 0.\] Crystal field splitting does not change the total energy of the d orbitals. Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure \(\Page {1a}\)). We can use the d-orbital energy-level diagram in Figure \(\Page {1}\) to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure \(\Page {2}\), for d –d systems—such as [Ti(H O) ] , [V(H O) ] , and [Cr(H O) ] , respectively—the electrons successively occupy the three degenerate t orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H O) ] , for example, by saying that the chromium ion has a d electron configuration or, more succinctly, Cr is a d ion. When we reach the d configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty e orbitals or one of the singly occupied t orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the . If Δ is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty e orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H O) ] ion, is called a high-spin complex. Conversely, if Δ is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN) ] ion, is called a low-spin complex. Similarly, metal ions with the d , d , or d electron configurations can be either high spin or low spin, depending on the magnitude of Δ . In contrast, only one arrangement of d electrons is possible for metal ions with d –d electron configurations. For example, the [Ni(H O) ] ion is d with two unpaired electrons, the [Cu(H O) ] ion is d with one unpaired electron, and the [Zn(H O) ] ion is d with no unpaired electrons. If Δ is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δ is greater, a low-spin configuration forms. The magnitude of Δ dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δ (i.e., Δ > P) yield a low-spin complex, whereas small values of Δ (i.e., Δ < P) produce a high-spin complex. As we noted, the magnitude of Δ depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δ for some representative transition-metal complexes are given in Table \(\Page {1}\). Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994). Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δ increases as the charge on the metal ion increases. Typically, Δ for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H O) ] , Δ = 11,800 cm ; for [V(H O) ] , Δ = 17,850 cm . For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δ increases with increasing principal quantum number: Δ (3d) < Δ (4d) < Δ (5d). The data for hexaammine complexes of the trivalent metals illustrate this point: [Rh(NH ) ] : Δ = 34,100 cm The increase in Δ with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions. Experimentally, it is found that the Δ observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δ decreases as the size of the donor atom increases. For example, Δ values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F , interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH , results in significantly larger Δ values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F . The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δ : The values of Δ listed in Table \(\Page {1}\) illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand. The largest Δ splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. : complexes structure, high spin versus low spin, and the number of unpaired electrons B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d electron configuration. C Because of the weak-field ligands, we expect a relatively small Δ , making the compound high spin. D In a high-spin octahedral d complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t orbitals, giving four unpaired electrons. B C Because rhodium is a second-row transition metal ion with a d electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δ , making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order d d , and d (these are degenerate); d ; and d . D The eight electrons occupy the first four of these orbitals, leaving the d . orbital empty. Thus there are no unpaired electrons. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. octahedral; high spin; five square planar; low spin; no unpaired electrons Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d complex such as [Ti(H O) ] is located in one of the t orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δ . The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t orbitals by the energy of those orbitals (−0.4Δ ), multiplying the number of electrons in e orbitals by the energy of those orbitals (+0.6Δ ), and summing the two. Table \(\Page {2}\) gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d configuration. CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences. Octahedral d and d complexes and low-spin d , d , d , and d complexes exhibit large CFSEs. Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δ ), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δ is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δ is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δ , whereas weak-field ligands interact more weakly and give a smaller Δ . The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT. | 15,184 | 1,742 |
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In an acid–base titration, a is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration. Figure \(\Page {1a}\) shows a plot of the pH as 0.20 M \(\ce{HCl}\) is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as \(\ce{HCl}\) is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of \(\ce{HCl}\) (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M \(\ce{NaOH}\) is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of \(\ce{NaOH}\) as shown in Figure \(\Page {1b}\). As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid. Suppose that we now add 0.20 M \(\ce{NaOH}\) to 50.0 mL of a 0.10 M solution of \(\ce{HCl}\). Because \(\ce{HCl}\) is a strong acid that is completely ionized in water, the initial \([H^+]\) is 0.10 M, and the initial pH is 1.00. Adding \(\ce{NaOH}\) decreases the concentration of H+ because of the neutralization reaction (Figure \(\Page {2a}\)): \[\ce{OH^{−} + H^{+} <=> H_2O}. \nonumber \] Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the \(\ce{H^{+}}\) ions originally present have been consumed. For the titration of a monoprotic strong acid (\(\ce{HCl}\)) with a monobasic strong base (\(\ce{NaOH}\)), we can calculate the volume of base needed to reach the equivalence point from the following relationship: \[moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \label{Eq1} \] If 0.20 M \(\ce{NaOH}\) is added to 50.0 mL of a 0.10 M solution of \(\ce{HCl}\), we solve for \(V_b\): \[V_b(0.20 Me)=0.025 L=25 mL \nonumber \] At the equivalence point (when 25.0 mL of \(\ce{NaOH}\) solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more \(\ce{NaOH}\) produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M \(NaOH\). As shown in Figure \(\Page {2b}\), the titration of 50.0 mL of a 0.10 M solution of \(\ce{NaOH}\) with 0.20 M \(\ce{HCl}\) produces a titration curve that is nearly the mirror image of the titration curve in Figure \(\Page {2a}\). The pH is initially 13.00, and it slowly decreases as \(\ce{HCl}\) is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M \(\ce{HCl}\). The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities. The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities. Calculate the pH of the solution after 24.90 mL of 0.200 M \(\ce{NaOH}\) has been added to 50.00 mL of 0.100 M \(\ce{HCl}\). : volumes and concentrations of strong base and acid : pH : Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of \(\ce{H^{+}}\) in 50.00 mL of 0.100 M \(\ce{HCl}\) can be calculated as follows: \[ 50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \nonumber \] The number of millimoles of \(\ce{NaOH}\) added is as follows: \[ 24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \nonumber \] Thus \(\ce{H^{+}}\) is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of \(\ce{OH^{-}}\) to the \(\ce{HCl}\) solution. Because only 4.98 mmol of \(OH^-\) has been added, the amount of excess \(\ce{H^{+}}\) is 5.00 mmol − 4.98 mmol = 0.02 mmol of \(H^+\). The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of \(\ce{H^{+}}\) is as follows: \[ \left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{-4} \; M \nonumber \] Hence, \[pH \approx −\log[\ce{H^{+}}] = −\log(3 \times 10^{-4}) = 3.5 \nonumber \] This is significantly less than the pH of 7.00 for a neutral solution. Calculate the pH of a solution prepared by adding \(40.00\; mL\) of \(0.237\; M\) \(HCl\) to \(75.00\; mL\) of a \(0.133 M\) solution of \(NaOH\). 11.6 pH after the addition of 10 ml of Strong Base to a Strong Acid: pH at the Equivalence Point in a Strong Acid/Strong Base Titration: In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding \(K_a\) or \(K_b\). As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, \([\ce{H^{+}}]\) of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its \(pK_a\) and its concentration. Because only a fraction of a weak acid dissociates, \([\(\ce{H^{+}}]\) is less than \([\ce{HA}]\). Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Figure \(\Page {3a}\) shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M \(\ce{NaOH}\) superimposed on the curve for the titration of 0.100 M \(\ce{HCl}\) shown in part (a) in Figure \(\Page {2}\). Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the \(\ce{HCl}\) solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the \(\ce{HCl}\) titration; the magnitude of the pH change at the equivalence point depends on the \(pK_a\) of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess \(\ce{NaOH}\) present, regardless of whether the acid is weak or strong. The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the \(K_a\) or \(K_b\). The titration curve in Figure \(\Page {3a}\) was created by calculating the starting pH of the acetic acid solution before any \(\ce{NaOH}\) is added and then calculating the pH of the solution after adding increasing volumes of \(NaOH\). The procedure is illustrated in the following subsection and Example \(\Page {2}\) for three points on the titration curve, using the \(pK_a\) of acetic acid (4.76 at 25°C; \(K_a = 1.7 \times 10^{-5}\). As explained discussed, if we know \(K_a\) or \(K_b\) and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define \(x\) as \([\ce{H^{+}}]\) due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows: \[\ce{CH3CO2H(aq) <=> H^{+}(aq) + CH3CO2^{−}} \nonumber \] In this and all subsequent examples, we will ignore \([H^+]\) and \([OH^-]\) due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified. Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations), \[ \begin{align*} K_a &=\dfrac{[H^+,CH_3CO_2^-]}{[CH_3CO_2H]} \\[4pt] &=\dfrac{(x)(x)}{0.100 - x} \\[4pt] &\approx \dfrac{x^2}{0.100} \\[4pt] &\approx 1.74 \times 10^{-5} \end{align*} \nonumber \] Solving this equation gives \(x = [H^+] = 1.32 \times 10^{-3}\; M\). Thus the pH of a 0.100 M solution of acetic acid is as follows: \[pH = −\log(1.32 \times 10^{-3}) = 2.879 \nonumber \] pH at the Start of a Weak Acid/Strong Base Titration: Now consider what happens when we add 5.00 mL of 0.200 M \(\ce{NaOH}\) to 50.00 mL of 0.100 M \(CH_3CO_2H\) (part (a) in Figure \(\Page {3}\)). Because the neutralization reaction proceeds to completion, all of the \(OH^-\) ions added will react with the acetic acid to generate acetate ion and water: \[ CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2} \] All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine \([\ce{H^{+}}]\) of the resulting solution. To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of \(\ce{CH_3CO_2H}\) in the original solution and the amount of \(\ce{OH^{-}}\) in the \(\ce{NaOH}\) solution that was added. The acetic acid solution contained \[ 50.00 \; \cancel{mL} (0.100 \;mmol (\ce{CH_3CO_2H})/\cancel{mL} )=5.00\; mmol (\ce{CH_3CO_2H}) \nonumber \] The \(\ce{NaOH}\) solution contained 5.00 mL=1.00 mmol \(NaOH\) Comparing the amounts shows that \(CH_3CO_2H\) is in excess. Because \(OH^-\) reacts with \(CH_3CO_2H\) in a 1:1 stoichiometry, the amount of excess \(CH_3CO_2H\) is as follows: 5.00 mmol \(CH_3CO_2H\) − 1.00 mmol \(OH^-\) = 4.00 mmol \(CH_3CO_2H\) Each 1 mmol of \(OH^-\) reacts to produce 1 mmol of acetate ion, so the final amount of \(CH_3CO_2^−\) is 1.00 mmol. The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations. \[\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)} \nonumber \] This ICE table gives the initial amount of acetate and the final amount of \(OH^-\) ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of \(CH_3CO_2^−\) in equilibrium is insignificant compared to the amount of \(OH^-\) added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of \(OH^-\), but the amount of \(OH^-\) due to the autoionization of water is insignificant compared to the amount of \(OH^-\) added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem. To calculate \([\ce{H^{+}}]\) at equilibrium following the addition of \(NaOH\), we must first calculate [\(\ce{CH_3CO_2H}\)] and \([\ce{CH3CO2^{−}}]\) using the number of millimoles of each and the total volume of the solution at this point in the titration: \[ final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \nonumber \] \[ \left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \nonumber \] \[ \left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber \] Knowing the concentrations of acetic acid and acetate ion at equilibrium and \(K_a\) for acetic acid (\(1.74 \times 10^{-5}\)), we can calculate \([H^+]\) at equilibrium: \[ K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \nonumber \] \[ \left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \nonumber \] Calculating \(−\log[\ce{H^{+}}]\) gives \[pH = −\log(6.95 \times 10^{−5}) = 4.158. \nonumber \] Comparing the titration curves for \(\ce{HCl}\) and acetic acid in Figure \(\Page {3a}\), we see that adding the same amount (5.00 mL) of 0.200 M \(\ce{NaOH}\) to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for \(\ce{HCl}\) (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example \(\Page {2}\), we calculate another point for constructing the titration curve of acetic acid. pH Before the Equivalence Point of a Weak Acid/Strong Base Titration: What is the pH of the solution after 25.00 mL of 0.200 M \(\ce{NaOH}\) is added to 50.00 mL of 0.100 M acetic acid? Given: volume and molarity of base and acid : pH Ignoring the spectator ion (\(Na^+\)), the equation for this reaction is as follows: \[CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \nonumber \] The initial numbers of millimoles of \(OH^-\) and \(CH_3CO_2H\) are as follows: 25.00 mL(0.200 mmol OH−mL=5.00 mmol \(OH-\) \[50.00\; mL (0.100 CH_3CO_2 HL=5.00 mmol \; CH_3CO_2H \nonumber \] The number of millimoles of \(OH^-\) equals the number of millimoles of \(CH_3CO_2H\), so neither species is present in excess. Because the number of millimoles of \(OH^-\) added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form. \[CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l) \nonumber \] C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction: \[ [\ce{CH_3CO_2}]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \nonumber \] The equilibrium reaction of acetate with water is as follows: \[\ce{CH_3CO^{-}2(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-} (aq)} \nonumber \] The equilibrium constant for this reaction is \[K_b = \dfrac{K_w}{K_a} \label{16.18} \] where \(K_a\) is the acid ionization constant of acetic acid. We therefore define x as \([\ce{OH^{−}}]\) produced by the reaction of acetate with water. Here is the completed table of concentrations: \[H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)} \nonumber \] We can obtain \(K_b\) by substituting the known values into Equation \ref{16.18}: \[ K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23} \] Substituting the expressions for the final values from the ICE table into Equation \ref{16.23} and solving for \(x\): \[ \begin{align*} \dfrac{x^{2}}{0.0667} &= 5.80 \times 10^{-10} \\[4pt] x &= \sqrt{(5.80 \times 10^{-10})(0.0667)} \\[4pt] &= 6.22 \times 10^{-6}\end{align*} \nonumber \] Thus \([OH^{−}] = 6.22 \times 10^{−6}\, M\) and the pH of the final solution is 8.794 (Figure \(\Page {3a}\)). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce \(\ce{OH^{-}}\). Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M \(\ce{HCl}\) solution to 125.0 mL of a 0.150 M solution of ammonia. The \(pK_b\) of ammonia is 4.75 at 25°C. 9.23 As shown in part (b) in Figure \(\Page {3}\), the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid. The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure \(\Page {4}\) illustrates the shape of titration curves as a function of the \(pK_a\) or the \(pK_b\). As the acid or the base being titrated becomes weaker (its \(pK_a\) or \(pK_b\) becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point. One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures \(\Page {4a}\) and \(\Page {4b}\) for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows: \[K_a=\dfrac{[H_3O^+,A^−]}{[HA]} \nonumber \] If \([HA] = [A^−]\), this reduces to \(K_a = [H_3O^+]\). Taking the negative logarithm of both sides, \[−\log K_a = −\log[H_3O+] \nonumber \] From the definitions of \(pK_a\) and pH, we see that this is identical to \[pK_a = pH \label{16.52} \] Thus the pH at the midpoint of the titration of a weak acid is equal to the \(pK_a\) of the weak acid, as indicated in part (a) in Figure \(\Page {4}\) for the weakest acid where we see that the midpoint for \(pK_a\) = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the \(pK_a\) (or the \(pK_b\)) of a weak acid (or a weak base). The pH at the midpoint of the titration of a weak acid is equal to the \(pK_a\) of the weak acid. When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the \(pK_a\) values are separated by at least three \(pK_a\) units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid \(H_3PO_4\) with \(\ce{NaOH}\) is illustrated in Figure \(\Page {5}\) and shows two well-defined steps: the first midpoint corresponds to \(pK_a\)1, and the second midpoint corresponds to \(pK_a\) . Because HPO is such a weak acid, \(pK_a\) has such a high value that the third step cannot be resolved using 0.100 M \(\ce{NaOH}\) as the titrant. The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure \(\Page {5}\). The initial pH is high, but as acid is added, the pH decreases in steps if the successive \(pK_b\) values are well separated. Table E1 lists the ionization constants and \(pK_a\) values for some common polyprotic acids and bases. Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M \(\ce{NaOH}\) solution to 100.0 mL of a 0.0510 M solution of oxalic acid (\(\ce{HO_2CCO_2H}\)), a diprotic acid (abbreviated as \(\ce{H2ox}\)). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion (\(\ce{O2CCO2^{2−}}\), abbreviated \(\ce{ox^{2-}}\)).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as \(\ce{Ca^{2+}}\) are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids. : volume and concentration of acid and base : pH : : gives the \(pK_a\) values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present: \[ 100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \nonumber \] \[ 55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \nonumber \] The strongest acid (\(H_2ox\)) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of \(OH^-\) to react with Hox−, forming ox and H O. The reactions can be written as follows: \[ \underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \nonumber \] \[ \underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \nonumber \] In tabular form, B The equilibrium between the weak acid (\(\ce{Hox^{-}}\)) and its conjugate base (\(\ce{ox^{2-}}\)) in the final solution is determined by the magnitude of the second ionization constant, \(K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}\). To calculate the pH of the solution, we need to know \(\ce{[H^{+}]}\), which is determined using exactly the same method as in the acetic acid titration in Example \(\Page {2}\): \[\text{final volume of solution} = 100.0\, mL + 55.0\, mL = 155.0 \,mL \nonumber \] Thus the concentrations of \(\ce{Hox^{-}}\) and \(\ce{ox^{2-}}\) are as follows: \[ \left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \nonumber \] \[ \left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \nonumber \] We can now calculate [H+] at equilibrium using the following equation: \[ K_{a2} =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \nonumber \] Rearranging this equation and substituting the values for the concentrations of \(\ce{Hox^{−}}\) and \(\ce{ox^{2−}}\), \[ \left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \nonumber \] So \[ pH = -\log\left [ H^{+} \right ]= -\log\left ( 3.7 \times 10^{-4} \right )= 3.43 \nonumber \] This answer makes chemical sense because the pH is between the first and second \(pK_a\) values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than \(pK_{a1}\)), but we added only enough to titrate less than half of the second, less acidic proton, with \(pK_{a2}\). If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to \(pK_{a2}\). Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine (\(pK_{b1}\) = 4.27, \(pK_{b2}\) = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M \(\ce{HCl}\) (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine. pH=4.9 In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful. We can describe the chemistry of indicators by the following general equation: \[ \ce{ HIn (aq) <=> H^{+}(aq) + In^{-}(aq)} \nonumber \] where the protonated form is designated by \(\ce{HIn}\) and the conjugate base by \(\ce{In^{−}}\). The ionization constant for the deprotonation of indicator \(\ce{HIn}\) is as follows: \[ K_{In} =\dfrac{ [\ce{H^{+}} , \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3} \] The \(pK_{in}\) (its \(pK_a\)) determines the pH at which the indicator changes color. Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure \(\Page {6}\)). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers. Irrespective of the origins, a good indicator must have the following properties: Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure \(\Page {7}\) shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values. It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units. We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure \(\Page {8}\). This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M \(NaOH\). The pH ranges over which two common indicators (methyl red, \(pK_{in} = 5.0\), and phenolphthalein, \(pK_{in} = 9.5\)) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the \(\ce{HCl}\) titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of \(\ce{NaOH}\) has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the \(\ce{HCl}\) titration, the phenolphthalein indicator will turn pink when about 50 mL of \(\ce{NaOH}\) has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of \(\ce{NaOH}\) will therefore cause the methyl red indicator to change color, resulting in a huge error. The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M \(NaOH\). Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the \(pK_a\) of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point. In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used. The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure \(\Page {9}\)). pH Indicators: Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the \(pK_a\), and the \(pK_b\) of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the \(pK_a\) of the weak acid or the \(pK_b\) of the weak base. Thus titration methods can be used to determine both the concentration and the \(pK_a\) (or the \(pK_b\)) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | 32,404 | 1,743 |
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We normally experience carbon dioxide (\(\ce{CO2}\)) as a gas, but if it were cooled down to about −78°C, it would become a solid. The everyday term for solid carbon dioxide is dry ice. Why “dry” ice? Solid carbon dioxide is called dry ice because it converts from a solid to a gas directly, without going through the liquid phase, in a process called sublimation. Thus, there is no messy liquid phase to worry about. Although it is a novelty, dry ice has some potential dangers. Because it is so cold, it can freeze living tissues very quickly, so people handling dry ice should wear special protective gloves. The cold carbon dioxide gas is also heavier than air (because it is cold and more dense), so people in the presence of dry ice should be in a well-ventilated area. Dry ice has several common uses. Because it is so cold, it is used as a refrigerant to keep other things cold or frozen (e.g., meats or ice cream). In the medical field, dry ice is used to preserve medical specimens, blood products, and drugs. It also has dermatological applications (e.g., freezing off warts). Organs for transplant are kept cool with dry ice until the recipient of the new organ is ready for surgery. In this respect, carbon dioxide is much like water—more than one phase of the same substance has significant uses in the real world. | 1,340 | 1,744 |
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The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at with respect to a specific unit.This article explains how to write equilibrium constant expressions, and introduces the calculations involved with both the concentration and the partial pressure equilibrium constant. A homogeneous reaction is one where the of the products and reactions are all the same (the word "homo" means "same"). In most cases, the solvent determines the state of matter for the overall reaction. For example, the synthesis of methanol from a carbon monoxide-hydrogen mixture is a mixture, which contains two or more substances: \[ CO (g)+ 2H_2 (g) \rightleftharpoons CH_3OH (g)\] At equilibrium, the rate of the forward and reverse reaction are equal, which is demonstrated by the arrows. The equilibrium constant, however, gives the ratio of the units (pressure or concentration) of the products to the reactants when the reaction is at equilibrium. The synthesis of ammonia is another example of a mixture: \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] A heterogeneous reaction is one in which one or more states within the reaction differ (the Greek word "heteros" means "different"). For example, the formation of an aqueous solution of lead(II) iodide creates a mixture dealing with particles in both the and states: \[PbI_{2 (s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2I^-_{(aq)}\] The decomposition of sodium hydrogen carbonate (baking soda) at high elevations is another example of a mixture, this reaction deals with molecules in both the and states: \[ 2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + H_2O_{ (g)} + CO_{2 (g)} \] \[ C_{(s)} + O_{2 (g)} \rightleftharpoons CO_{2 (g)} \] This difference between homogeneous and heterogeneous reactions is emphasized so that students remember that solids, pure liquids, and solvents are treated differently than gases and solutes when approximating the activities of the substances in equilibrium constant expressions. The numerical value of an equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. The ratio of the product concentrations to reactant concentrations is calculated. Because the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations. This knowledge allowed scientists to derive a model expression that can serve as a "template" for any reaction. This basic "template" form of an equilibrium constant expression is examined here. The thermodynamically correct equilibrium constant expression relates the activities of of the species present in the reaction. Although the concept of activity is too advanced for a typical General Chemistry course, it is essential that the explanation of the derivation of the equilibrium constant expression starts with activities so that no misconceptions occur. For the hypothetical reaction: \[bB + cC \rightleftharpoons dD + eE \] the equilibrium constant expression is written as \[ K = \dfrac{a_D^d ·a_E^e}{a_B^b · a_C^c}\] *The lower case letters in the balanced equation represent the number of of each substance, the upper case letters represent the substance itself. To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentration approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).The equilibrium constant expression is written as \(K_c\), as in the expression for the reaction: \[HF_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + F^-_{(aq)} \] \[ K_c = \dfrac{a_{H_3O^+}· a_{F^-}}{a_{HF} · a_{H_2O}} ≈ \dfrac{[H_3O^+,F^-]}{[HF](1)} = \dfrac{[H_3O^+,F^-]}{[HF]} \] Here, the letters inside the brackets represent the concentration (in molarity) of each substance. Notice the mathematical product of the chemical products raised to the powers of their respective coefficients is the numerator of the ratio and the mathematical product of the reactants raised to the powers of their respective coefficients is the denominator. This is the case for every equilibrium constant. A ratio of molarities of products over reactants is usually used when most of the species involved are dissolved in water. A ratio of concentrations can also be used for reactions involving gases if the volume of the container is known. . Gaseous reaction equilibria are often expressed in terms of partial pressures. The equilibrium constant of pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium (again, the pressures of all species are raised to the powers of their respective coefficients). The equilibrium constant is written as \(K_p\), as shown for the reaction: \[aA_{(g)} + bB_{(g)} \rightleftharpoons gG_{(g)} + hH_{(g)} \] \[ K_p= \dfrac{p^g_G \, p^h_H}{ p^a_A \,p^b_B} \] To convert K to K , the following equation is used: \[K_p = K_c(RT)^{\Delta{n_{gas}}}\] where: Another quantity of interest is the reaction quotient, \(Q\), which is the numerical value of the ratio of products to reactants at any point in the reaction. The reaction quotient is calculated the same way as is \(K\), but is not necessarily equal to \(K\). It is used to determine which way the reaction will proceed at any given point in time. \[Q = \dfrac{[G]^g[H]^h}{[A]^a[B]^b}\] The same process is employed whether calculating \(Q_c\) or \(Q_p\). The most important consideration for a heterogeneous mixture is that . From a mathematical perspective, with the activities of solids and liquids and solvents equal one, these substances do not affect the overall K or Q value. This convention is extremely important to remember, especially in dealing with heterogeneous solutions. \[ aA_{(s)} + bB_{(l)} \rightleftharpoons gG_{(aq)} + hH_{(aq)} \] \[K_c = \dfrac{[G]^g[H]^h}{1 \times 1} = [G]^g[H]^h\] In this case, since solids and liquids have a fixed value of 1, the numerical value of the expression is independent of the amounts of A and B. If the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction: \[ H^+_{(aq)} + OH^–_{(aq)} \rightarrow H_2O_{ (l)}\] The equilibrium constant expression would be: \[ K_c= \dfrac{1}{ [H^+,OH^-]}\] which is the reciprocal of the of water (\(K_w\)) \[ K_c = \dfrac{1}{K_w}=1 \times 10^{14}\] The equilibrium constant expression must be manipulated if a reaction is reversed or split into elementary steps. When the reaction is reversed, the equilibrium constant expression is inverted. The new expression would be written as: \[K'= \dfrac{1}{\dfrac{[G]^g[H]^h}{[A]^a[B]^b}} = \dfrac{[A]^a[B]^b}{[G]^g[H]^h}\] When there are multiple steps in the reaction, each with its own K (in a scenario similar to problems), then the successive K values for each step are together to calculate the overall K. Because the concentration of reactants and products are not dimensionless (i.e. they have units) in a reaction, the actual quantities used in an equilibrium constant expression are . is expressed by the dimensionless ratio \(\frac{[X]}{c^{\circ}}\) where \([X]\) signifies the molarity of the molecule and c is the chosen reference state: \[ a_b=\dfrac{[B]}{c^{\circ}} \] For gases that do not follow the ideal gas laws, using activities will accurately determine the equilibrium constant that changes when concentration or pressure varies. Thus, the units are canceled and \(K\) becomes unitless. What is \(K_c\) for the Reaction 1) Kc: 24.5 Kp: 1.002 Atm 2) Q = 83.33 > K therefore the reaction shifts to the left | 7,914 | 1,745 |
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The atoms in all substances that contain more than one atom are held together by electrostatic interactions —interactions between electrically charged particles such as protons and electrons. Electrostatic attraction between oppositely charged species (positive and negative) results in a force that causes them to move toward each other, like the attraction between opposite poles of two magnets. In contrast, electrostatic repulsion between two species with the same charge (either both positive or both negative) results in a force that causes them to repel each other, as do the same poles of two magnets. Atoms form chemical compounds when the attractive electrostatic interactions between them are stronger than the repulsive interactions. Collectively, we refer to the attractive interactions between atoms as chemical bonds . Chemical bonds are generally divided into two fundamentally different kinds: ionic and covalent. In reality, however, the bonds in most substances are neither purely ionic nor purely covalent, but they are closer to one of these extremes. Although purely ionic and purely covalent bonds represent extreme cases that are seldom encountered in anything but very simple substances, a brief discussion of these two extremes helps us understand why substances that have different kinds of chemical bonds have very different properties. Ionic compounds consist of positively and negatively charged ions held together by strong electrostatic forces, whereas covalent compounds generally consist of molecules , which are groups of atoms in which one or more pairs of electrons are shared between bonded atoms. In a covalent bond , the atoms are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share. We begin our review of structures and formulas by describing covalent compounds. Ionic compounds consist of ions of opposite charges held together by strong electrostatic forces, whereas pairs of electrons are shared between bonded atoms in covalent compounds. Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the (“two atoms”) H , N , O , F , Cl , Br , and I (part (a) in ). Similarly, a few pure elements are polyatomic (“many atoms”) , such as elemental phosphorus and sulfur, which occur as P and S (part (b) in ). Each covalent compound is represented by a molecular formula , which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as H O. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as CO . Elements That Exist as Covalent Molecules Covalent compounds that contain predominantly carbon and hydrogen are called organic compounds . One convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH O is methyl alcohol, a fuel). Another convention better represents the molecular structure as a structural formula, as, for example, writing the formula for methyl alcohol asCH OH, where CH is the methyl group and OH the hydroxyl. Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compounds ; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table, such as we see in CO or SF . Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13–15, the hydrogen is usually listed last in the formula. Examples are ammonia (NH ) and silane (SiH ). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H O, not OH . The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H SO ), are described in . For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up. Write the molecular formula of each compound. identity of elements present and number of atoms of each molecular formula Identify the symbol for each element in the molecule. Then identify the substance as either an organic compound or an inorganic compound. If the substance is an organic compound, arrange the elements in order beginning with carbon and hydrogen and then list the other elements alphabetically. If it is an inorganic compound, list the elements beginning with the one farthest left in the periodic table. List elements in the same group starting with the lower element and working up. From the information given, add a subscript for each kind of atom to write the molecular formula. According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl F. We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl F. Exercise Write the molecular formula for each compound. Molecular formulas give only the elemental composition of molecules. In contrast, structural formulas show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. These are the Lewis structures we learned about. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically. The structural formula for H can be drawn as H–H and that for I as I–I, where the line indicates a single pair of shared electrons, a single bond . Two pairs of electrons are shared in a double bond , which is indicated by two lines— for example, O is O=O. Three electron pairs are shared in a triple bond , which is indicated by three lines—for example, N is N≡N (see ). Carbon is unique in the extent to which it forms single, double, and triple bonds to itself and other elements. The number of bonds formed by an atom in its covalent compounds is arbitrary. As we have learned in , hydrogen, oxygen, nitrogen, and carbon have a very strong tendency to form substances in which they have one, two, three, and four bonds to other atoms, respectively ( ). Molecules That Contain Single, Double, and Triple Bonds The Number of Bonds That Selected Atoms Commonly Form to Other Atoms The structural formula for water can be drawn as follows, but including the two lone pairs on the oxygen provides more information: Because as we learned from VSEPR, the latter approximates the experimentally determined shape of the water molecule, it is more informative. Similarly, ammonia (NH ) and methane (CH ) are often written as planar molecules but at least for ammonia, it would be more informative to include the lone pair on the nitrogen atom: As shown in , however, we know that the the actual three-dimensional structure of NH looks like a pyramid with a triangular base of three hydrogen atoms. Again using VSEPR, or what we know about sp3 orbitals on the central atom in all three molecules the structure of CH , with four hydrogen atoms arranged around a central carbon atom as shown in , is . That is, the hydrogen atoms are positioned at every other vertex of a cube. Many compounds—carbon compounds, in particular—have four bonded atoms arranged around a central atom to form a tetrahedron. The Three-Dimensional Structures of Water, Ammonia, and Methane , and illustrate different ways to represent the structures of molecules. It should be clear that there is no single “best” way to draw the structure of a molecule; the method you use depends on which aspect of the structure you want to emphasize and how much time and effort you want to spend. shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol (part (a) in ) gives only the number of each kind of atom; writing methanol as CH O tells nothing about its structure. In contrast, the structural formula (part (b) in ) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in ) and the perspective drawing (part (d) in ) show the three-dimensional structure of the molecule. The latter (also called a representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in ) illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. Also, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.” Different Ways of Representing the Structure of a Molecule Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a (part (f) in ), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH OH, which tells us that the molecule contains a CH unit that looks like a fragment of methane (CH ). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point. Write the molecular formula for each compound. The condensed structural formula is given. condensed structural formula molecular formula Identify every element in the condensed structural formula and then determine whether the compound is organic or inorganic. As appropriate, use either organic or inorganic convention to list the elements. Then add appropriate subscripts to indicate the number of atoms of each element present in the molecular formula. The molecular formula lists the elements in the molecule and the number of atoms of each. Exercise Write the molecular formula for each molecule. The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively charged protons in the nucleus is equal to the number of negatively charged electrons. In contrast, are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations . Conversely, ions that contain more electrons than protons have a net negative charge and are called anions . contain both cations and anions in a ratio that results in no net electrical charge. Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge. In covalent compounds, electrons are shared between bonded atoms and are simultaneously attracted to more than one nucleus. In contrast, ionic compounds contain cations and anions rather than discrete neutral molecules. Ionic compounds are held together by the attractive electrostatic interactions between cations and anions. In an ionic compound, the cations and anions are arranged in space to form an extended three-dimensional array that maximizes the number of attractive electrostatic interactions and minimizes the number of repulsive electrostatic interactions ( ). As shown in , the electrostatic energy of the interaction between two charged particles is proportional to the product of the charges on the particles and inversely proportional to the distance between them:
\( electrostatic\; energy=\frac{Q_{1}Q_{2}}{r} \tag{6.1.1} \) where and are the electrical charges on particles 1 and 2, and is the distance between them. When and are both positive, corresponding to the charges on cations, the cations repel each other and the electrostatic energy is positive. When and are both negative, corresponding to the charges on anions, the anions repel each other and the electrostatic energy is again positive. The electrostatic energy is negative only when the charges have opposite signs; that is, positively charged species are attracted to negatively charged species and vice versa. As shown in , the strength of the interaction is proportional to the of the charges and decreases as the between the particles increases as we have seen previously If the electrostatic energy is positive, the particles repel each other; if the electrostatic energy is negative, the particles are attracted to each other. Covalent and Ionic Bonding The Effect of Charge and Distance on the Strength of Electrostatic Interactions One example we have studied of an ionic compound is sodium chloride (NaCl; ), formed from sodium and chlorine. In forming chemical compounds, many elements have a tendency to gain or lose enough electrons to attain the same number of electrons as the noble gas closest to them in the periodic table. When sodium and chlorine come into contact, each sodium atom gives up an electron to become a Na ion, with 11 protons in its nucleus but only 10 electrons (like neon), and each chlorine atom gains an electron to become a Cl ion, with 17 protons in its nucleus and 18 electrons (like argon), as shown in part (b) in . Solid sodium chloride contains equal numbers of cations (Na ) and anions (Cl ), thus maintaining electrical neutrality. Each Na ion is surrounded by 6 Cl ions, and each Cl ion is surrounded by 6 Na ions. Because of the large number of attractive Na Cl interactions, the total attractive electrostatic energy in NaCl is great. Sodium Chloride: an Ionic Solid Consistent with a tendency to have the same number of electrons as the nearest noble gas, when forming ions, elements in groups 1, 2, and 3 tend to one, two, and three electrons, respectively, to form cations, such as Na and Mg . They then have the same number of electrons as the nearest noble gas: neon. Similarly, K , Ca , and Sc have 18 electrons each, like the nearest noble gas: argon. In addition, the elements in group 13 three electrons to form cations, such as Al , again attaining the same number of electrons as the noble gas closest to them in the periodic table. Because the lanthanides and actinides formally belong to group 3, the most common ion formed by these elements is M , where M represents the metal. Conversely, elements in groups 17, 16, and 15 often react to one, two, and three electrons, respectively, to form ions such as Cl , S , and P . Ions such as these, which contain only a single atom, are called monatomic ions . The names of the single atom cations are simply the name of the metal from which they are derived. The names of the single atom anions add the suffix -ide to the first syllable of the atom, for example oxide, chloride, nitride, etc. You can predict the charges of most monatomic ions derived from the main group elements by simply looking at the periodic table and counting how many columns an element lies from the extreme left or right. For example, you can predict that barium (in group 2) will form Ba to have the same number of electrons as its nearest noble gas, xenon, that oxygen (in group 16) will form O to have the same number of electrons as neon, and cesium (in group 1) will form Cs to also have the same number of electrons as xenon. This method does not usually work for most of the transition metals. Some common monatomic ions are in . Elements in groups 1, 2, and 3 tend to form 1+, 2+, and 3+ ions, respectively; elements in groups 15, 16, and 17 tend to form 3−, 2−, and 1− ions, respectively. Some Common Monatomic Ions and Their Names Li lithium Be beryllium N nitride (azide) O oxide F fluoride Na sodium Mg magnesium Al aluminum P phosphide S sulfide Cl chloride K potassium Ca calcium Sc scandium Ga gallium As arsenide Se selenide Br bromide Rb rubidium Sr strontium Y yttrium In indium Te telluride I iodide Cs cesium Ba barium La lanthanum Predict the charge on the most common monatomic ion formed by each element. element ionic charge Identify the group in the periodic table to which the element belongs. Based on its location in the periodic table, decide whether the element is a metal, which tends to lose electrons; a nonmetal, which tends to gain electrons; or a semimetal, which can do either. After locating the noble gas that is closest to the element, determine the number of electrons the element must gain or lose to have the same number of electrons as the nearest noble gas. Exercise Predict the charge on the most common monatomic ion formed by each element. In general, ionic and covalent compounds have different physical properties. As we have discussed, ionic compounds usually form hard crystalline solids that melt at rather high temperatures and are resistant to evaporation. These properties stem from the characteristic internal structure of an ionic solid, illustrated schematically in part (a) in , which shows the three-dimensional array of alternating positive and negative ions held together by strong electrostatic attractions. In contrast, as shown in part (b) in , most covalent compounds consist of discrete molecules held together by comparatively weak forces (the forces between molecules), even though the atoms within each molecule are held together by strong covalent bonds (the forces within the molecule). Covalent substances can be gases, liquids, or solids at room temperature and pressure, depending on the strength of the intermolecular interactions. Covalent molecular solids tend to form soft crystals that melt at rather low temperatures and evaporate relatively easily. The covalent bonds that hold the atoms together in the molecules are unaffected when covalent substances melt or evaporate, so a liquid or vapor of discrete, independent molecules is formed. For example, at room temperature, methane, the major constituent of natural gas, is a gas that is composed of discrete CH molecules. A comparison of the different physical properties of ionic compounds and covalent molecular substances is given in . The Physical Properties of Typical Ionic Compounds and Covalent Molecular Substances Interactions in Ionic and Covalent Solids (a) The positively and negatively charged ions in an ionic solid such as sodium chloride (NaCl) are held together by strong electrostatic interactions. (b) In this representation of the packing of methane (CH ) molecules in solid methane, a prototypical molecular solid, the methane molecules are held together in the solid only by relatively weak intermolecular forces, even though the atoms within each methane molecule are held together by strong covalent bonds. The atoms in chemical compounds are held together by attractive electrostatic interactions known as . contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most consist of , groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called , whereas compounds that consist primarily of elements other than carbon and hydrogen are . contain two atoms, and contain more than two. A indicates the composition and approximate structure and shape of a molecule. , and are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ; they can have either a positive charge ( ) or a negative charge ( ). Ions can consist of one atom ( ) or several ( ). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure. Ionic and covalent compounds are held together by electrostatic attractions between oppositely charged particles. Describe the differences in the nature of the attractions in ionic and covalent compounds. Which class of compounds contains pairs of electrons shared between bonded atoms? Which contains fewer electrons than the neutral atom—the corresponding cation or the anion? What is the difference between an organic compound and an inorganic compound? What is the advantage of writing a structural formula as a condensed formula? The majority of elements that exist as diatomic molecules are found in one group of the periodic table. Identify the group. Discuss the differences between covalent and ionic compounds with regard to Why do covalent compounds generally tend to have lower melting points than ionic compounds? Covalent compounds generally melt at lower temperatures than ionic compounds because the intermolecular interactions that hold the molecules together in a molecular solid are weaker than the electrostatic attractions that hold oppositely charged ions together in an ionic solid. The structural formula for chloroform (CHCl ) was shown in Example 2. Based on this information, draw the structural formula of dichloromethane (CH Cl ). What is the total number of electrons present in each ion? What is the total number of electrons present in each ion? Predict how many electrons are in each ion. Predict how many electrons are in each ion. Predict the charge on the most common monatomic ion formed by each element. Predict the charge on the most common monatomic ion formed by each element. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. | 24,395 | 1,747 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Aluminosilicates/Aluminosilicates |
While you are admiring this beautiful picture of faujasite, remember that the oxygen atoms have two unshared electron pairs in addition to the (Al,Si)-O-Si(or Al) bonds. Thus the oxygen atoms are sites to interact with positive site of molecules that passes by these structures. At present over 150 synthetic zeolites & zeotypes and 40 natural zeolites are known. Synthesis of zeolite is a very active field of study. Aluminosilicates have three major minerals: , and . Zeochem has been developing and manufacturing molecular sieve adsorbents since 1977. Simply put, their adsorbents are used to "screen" out impurities from a variety of applications by attracting and trapping the targeted contaminants. For example, in natural gas processing, molecular sieves are used to remove specific molecules from the gas stream to allow for more efficient downstream processing. Faujasite is a typical zeolite. As you have read above that there are many different kinds of zeolites, each with a definite structure and associate with it are unique properties. In terms of applications, we are assuming zeolites as porous aluminosilicates with large tunnels and cages for a fluid (gas and liquid) to pass through. The applications are based on the interactions between the fluid phase and the atoms or ions of the zeolites. In general terms, zeolites have many applications: = 494 g That 80 % of protons of the is used means that we require a little more than stoichiometric quantities.
Zeolites are aluminosilicates with open frames strcutures discussed above. Replacement of each Si atom by an Al atom in silicates results in having an extra negative charge on the frame. These charges must be balanced by trapping positive ions: H , Na K , Ca , Cu or Mg . Water molecules are also trapped in the frame work of zeolites. In this example, we assume that when we soak the zeolite in water containing Ca , and Mg ions, these ions are more attrative to the zeolite than the small, singly charged protons. We further assumed that 80 percent of the protons in zeolite are replaced by other ions.
How much salt is required if 60% of the sodium ions are effectively used to replace all the divalent ions? | 2,220 | 1,748 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Solubility_Product_Constant%2C_Ksp |
The solubility product constant, \(K_{sp}\) is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the \(K_{sp}\) value it has. Consider the general dissolution reaction below (in aqueous solutions): \[\ce{aA(s) <=> cC(aq) + dD(aq)} \nonumber \] To solve for the \(K_{sp}\) it is necessary to take the molarities or concentrations of the products (\(\ce{cC}\) and \(\ce{dD}\)) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power(and also multiply the concentration by that coefficient). This is shown below: \[ K_{sp} = [C]^c [D]^d \nonumber \] Note that the reactant, aA, is not included in the \(K_{sp}\) equation. Solids are not included when calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted. Hence, \(K_{sp}\) represents the maximum extent that a solid that can dissolved in solution. What is the solubility product constant expression for \(MgF_2\)? The relavant equilibrium is \[MgF_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2F^-_{(aq)} \nonumber \] so the associated equilibrium constant is \[K_{sp} = [Mg^{2+},F^-]^2\nonumber \] What is the solubility product constant expression for \(Ag_2CrO_4\)? The relavant equilibrium is \[Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4(aq)}\nonumber \] so the associated equilibrium constant is \[K_{sp} = [Ag^{+}]^2[CrO_4^{2-}]\nonumber \] | 1,682 | 1,750 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/22%3A_Biophysical_Reaction_Dynamics/22.04%3A_Analyzing_Trajectories |
Analyzing Trajectories Waiting‐Time Distributions, P τ : Waiting time between arriving and leaving a state or P(k,t) P : Probability of making k jumps during a time interval, t. → Survival probability P : Probability of waiting a time τ between jumps? Waiting time distribution → FPT distribution Let’s relate these... Assume independent events. No memory of history – where it was in trajectory. Flux: \(\dfrac{dP_R}{dt}=J\) J: Probability of jump during ∆t. ∆t is small enough that J\(\ll\) 1, but long enough to lose memory of earlier configurations. The probability of seeing k jumps during a time interval t, where t is divided into N intervals of width Δt (t = N∆t) is given by the binomial distribution \[P(k,N)=\dfrac{N!}{k!(N-k)!}J^k(1-J)^{N-k} \] Here N≫k. Define rate λ in terms of the average number of jumps per unit time \[ \lambda = \dfrac{\langle k \rangle}{t}=\dfrac{1}{\langle \tau_W \rangle} \nonumber \] \[J=\lambda \Delta t \rightarrow J=\dfrac{\lambda t}{N} \nonumber \] Substituting this into eq. (22.4.1) Error! Reference source not found.. For N ≫ k, recognize \[ (1-J)^{N-k} \approx (1-J)^N = \left( 1-\dfrac{\lambda t}{N} \right)^N \approx e^{-\lambda t} \nonumber \] The last step is exact for lim N → ∞. Poisson distribution for the number of jumps in time t. \[ \langle P(k,t) \rangle = \langle \lambda t \rangle = \dfrac{\lambda t}{\langle P^2(k,t)\rangle^{1/2}}=(\lambda t)^{1/2} \nonumber \] Fluctuations: \( \sigma / \langle P(k,t) \rangle = (\lambda t)^{-1/2} \) OK, now what about P the waiting time distribution? Consider the probability of not jumping during time t: \[P_k(0,t) = e^{-\lambda t} \nonumber \] As you wait longer and longer, the probability that you stay in the initial state drops exponentially. Note that P (0, t) is related to P by integration over distribution of waiting times. \[ \int_{t}^{\infty}P_w(t')dt'=P(0,t)=e^{-\lambda t} \nonumber \] \[ \int_{t}^{\infty}P_wdt \rightarrow \text {probability of staying for t} \nonumber \] \[ \int_{0}^{t}P_wdt \rightarrow \text {probability of jumping within t} \nonumber \] Probability of jumping between t and t+∆t: Probability of no decay for time <t decay on last \[\begin{aligned}
P_{w}(t) \Delta t &=\overbrace{\left(1-\langle k\rangle \Delta t_{1}\right)} \overbrace{\left(1-\langle k\rangle \Delta t_{2}\right) \ldots\left(1-\langle k\rangle \Delta t_{N}\right)} \overbrace{k \Delta t} \\ &=(1-\langle k\rangle \Delta t)^{N} k \Delta t \approx k e^{-k t} \Delta t \end{aligned} \nonumber \] \[\begin{aligned} P_{w} &=\lambda e^{-\lambda t} \\ \langle\tau\rangle &=\int_{0}^{\infty} t p_{w}(t) t
\end{aligned} \nonumber \]
\[ \begin{aligned} \langle\tau_{w}\rangle &=1 / \lambda \rightarrow \text {the average waiting time is the lifetime} (1/\lambda)\\
\langle\tau_{w}^{2}\rangle-\langle\tau_{w}\rangle^{2} &= (1 / \lambda)^{2} \end{aligned} \nonumber \] Reduction of Complex Kinetics from Trajectories \[ \dfrac{dP_m}{dt}=\sum_{n}k_{n\rightarrow m}P_n-\sum_{n}k_{m \rightarrow n}P_m \nonumber \] \(k_{n\rightarrow m}\) is rate constant for transition from state n to state m. Units: probability/time. Or in matrix form: P= P where is the transition rate matrix. With detailed balance, conservation of population all initial conditions will converge on equilibrium | 3,323 | 1,751 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Fission_and_Fusion/Fission_and_Fusion |
The energy harnessed in nuclei is released in nuclear reactions. Fission is the splitting of a heavy nucleus into lighter nuclei and fusion is the combining of nuclei to form a bigger and heavier nucleus. The consequence of fission or fusion is the absorption or release of energy. Protons and neutrons make up a nucleus, which is the foundation of nuclear science. Fission and fusion involves the dispersal and combination of elemental nucleus and isotopes, and part of nuclear science is to understand the process behind this phenomenon. Adding up the individual masses of each of these subatomic particles of any given element will always give you a greater mass than the mass of the nucleus as a whole. The missing idea in this observation is the concept called nuclear binding energy. is the energy required to keep the protons and neutrons of a nucleus intact, and the energy that is released during a nuclear fission or fusion is nuclear power. There are some things to consider however. The mass of an element's nucleus as a whole is less than the total mass of its individual protons and neutrons. The difference in mass can be attributed to the nuclear binding energy. Basically, nuclear binding energy is considered as mass, and that mass becomes "missing". This missing mass is called , which is the nuclear energy, also known as the mass released from the reaction as neutrons, photons, or any other trajectories. In short, mass defect and nuclear binding energy are interchangeable terms. N is the splitting of a heavy nucleus into two lighter ones. Fission was discovered in 1938 by the German scientists Otto Hahn, Lise Meitner, and Fritz Strassmann, who bombarded a sample of uranium with neutrons in an attempt to produce new elements with Z > 92. They observed that lighter elements such as barium (Z = 56) were formed during the reaction, and they realized that such products had to originate from the neutron-induced fission of uranium-235: \[_{92}^{235}\textrm U+\,_0^1\textrm n \rightarrow \,_{56}^{141}\textrm{Ba}+\,_{36}^{92}\textrm{Kr}+3_0^1\textrm n \label{21.6.11}\] This hypothesis was confirmed by detecting the krypton-92 fission product. As discussed in , the nucleus usually divides asymmetrically rather than into two equal parts, and the fission of a given nuclide does not give the same products every time. In a typical nuclear fission reaction, more than one neutron is released by each dividing nucleus. When these neutrons collide with and induce fission in other neighboring nuclei, a self-sustaining series of nuclear fission reactions known as a can result ( ). For example, the fission of U releases two to three neutrons per fission event. If absorbed by other U nuclei, those neutrons induce additional fission events, and the rate of the fission reaction increases geometrically. Each series of events is called a generation. Experimentally, it is found that some minimum mass of a fissile isotope is required to sustain a nuclear chain reaction; if the mass is too low, too many neutrons are able to escape without being captured and inducing a fission reaction. The minimum mass capable of supporting sustained fission is called the . This amount depends on the purity of the material and the shape of the mass, which corresponds to the amount of surface area available from which neutrons can escape, and on the identity of the isotope. If the mass of the fissile isotope is greater than the critical mass, then under the right conditions, the resulting supercritical mass can release energy explosively. The enormous energy released from nuclear chain reactions is responsible for the massive destruction caused by the detonation of nuclear weapons such as fission bombs, but it also forms the basis of the nuclear power industry. , in which two light nuclei combine to produce a heavier, more stable nucleus, is the opposite of nuclear fission. As in the nuclear transmutation reactions discussed in , the positive charge on both nuclei results in a large electrostatic energy barrier to fusion. This barrier can be overcome if one or both particles have sufficient kinetic energy to overcome the electrostatic repulsions, allowing the two nuclei to approach close enough for a fusion reaction to occur. The principle is similar to adding heat to increase the rate of a chemical reaction. As shown in the plot of nuclear binding energy per nucleon versus atomic number in , fusion reactions are most exothermic for the lightest element. For example, in a typical fusion reaction, two deuterium atoms combine to produce helium-3, a process known as deuterium–deuterium fusion (D–D fusion): \[2_1^2\textrm H\rightarrow \,_2^3\textrm{He}+\,_0^1\textrm n \label{21.6.12}\] In another reaction, a deuterium atom and a tritium atom fuse to produce helium-4 (Figure \(\Page {1}\)), a process known as deuterium–tritium fusion (D–T fusion): \[_1^2\textrm H+\,_1^3\textrm H\rightarrow \,_2^4\textrm{He}+\,_0^1\textrm n \label{21.6.13}\] Initiating these reactions, however, requires a temperature comparable to that in the interior of the sun (approximately 1.5 × 10 K). Currently, the only method available on Earth to achieve such a temperature is the detonation of a fission bomb. For example, the so-called hydrogen bomb (or H bomb) is actually a deuterium–tritium bomb (a D–T bomb), which uses a nuclear fission reaction to create the very high temperatures needed to initiate fusion of solid lithium deuteride ( LiD), which releases neutrons that then react with Li, producing tritium. The deuterium-tritium reaction releases energy explosively. Example 21.6.3 and its corresponding exercise demonstrate the enormous amounts of energy produced by nuclear fission and fusion reactions. In fact, fusion reactions are the power sources for all stars, including our sun. To calculate the energy released during mass destruction in both nuclear fission and fusion, we use Einstein’s equation that equates energy and mass: \[ E=mc^2 \label{1} \] with Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when the neutron-induced fission of U produces Cs, Rb, and two neutrons: \(_{92}^{235}\textrm U+\,_0^1\textrm n\rightarrow \,_{55}^{144}\textrm{Cs}+\,_{37}^{90}\textrm{Rb}+2_0^1\textrm n\) balanced nuclear reaction energy released in electronvolts per atom and kilojoules per mole Following the method used in Example 21.6.1, calculate the change in mass that accompanies the reaction. Convert this value to the change in energy in electronvolts per atom. Calculate the change in mass per mole of U. Then use to calculate the change in energy in kilojoules per mole. The change in mass that accompanies the reaction is as follows: The change in energy in electronvolts per atom is as follows: The change in mass per mole of \(_{92}^{235}\textrm{U}\) is −0.188386 g = −1.88386 × 10 kg, so the change in energy in kilojoules per mole is as follows: \(\begin{align}\Delta E&=(\Delta m)c^2=(-1.88386\times10^{-4}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2
\\&=-1.693\times10^{13}\textrm{ J/mol}=-1.693\times10^{10}\textrm{ kJ/mol}\end{align}\) Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when deuterium and tritium fuse to give helium-4 and a neutron: \(_1^2\textrm H+\,_1^3\textrm H\rightarrow \,_2^4\textrm{He}+\,_0^1\textrm n\) ΔE = −17.6 MeV/atom = −1.697 × 10 kJ/mol Figure \(\Page {1}\): Binding energy per nucleon of common isotopes. Fission is the splitting of a nucleus that releases free neutrons and lighter nuclei. The fission of heavy elements is highly exothermic which releases about 200 million eV compared to burning coal which only gives a few eV. The amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only 0.1 percent of the original nuclei is converted to energy. Daughter nucleus, energy, and particles such as neutrons are released as a result of the reaction. The particles released can then react with other radioactive materials which in turn will release daughter nucleus and more particles as a result, and so on. The unique feature of nuclear fission reactions is that they can be harnessed and used in chain reactions. This chain reaction is the basis of nuclear weapons. One of the well known elements used in nuclear fission is \(\ce{^{235}U}\), which when is bombarded with a neutron, the atom turns into \(\ce{^{236}U}\) which is even more unstable and splits into daughter nuclei such as Krypton-92 and Barium-141 and free neutrons. The resulting fission products are highly radioactive, commonly undergoing \(\beta^-\) decay. Nuclear fission is the splitting of the nucleus of an atom into nuclei of lighter atoms, accompanied by the release of energy, brought on by a neutron bombardment. The original concept of this nuclei splitting was discovered by Enrico Femi in 1934—who believed might be produced by bombarding uranium with neutrons, because the loss of Beta particles would increase the atomic number. However, the products that formed did not correlate with the properties of elements with higher atomic numbers than uranium (Ra, Ac, Th, and Pa). Instead, they were radioisotopes of much lighter elements such as Sr and Ba. The amount of mass lost in the fission process is equivalent to an energy of \(3.20 \times 10^{-11}\; J\). Consider the neutron bonbardment \[ \ce{_{92}^{235}U + _{1}^{0}n \rightarrow _{92}^{236}U} \rightarrow \; \text{fission products} \] which releases \(3.20 \times 10^{-11}\; J\) per \(\ce{^{235}U}\) atom. How much energy would be released if \(1.00\;g\) of \(\ce{^{235}U}\) were to undergo fission? \[ (1.00\;\rm{g}\; \ce{^{235}U} ) \times \left(\dfrac{1\; mol\; \ce{^{235}U}}{235\; g\; \ce{^{235}U}}\right) \times \left(\dfrac{ 6.022 \times 10^{23}\; \text{atoms}\; \ce{^{235}U}}{1\; mol\; \ce{^{235}U}} \right) \times \left(\dfrac{3.20 \times 10^{-11}\; J}{1\; atom \; \ce{^{235}U}}\right) = 8.20 \times 10^{10}\; J\] Clearly, the fission of a small amount of atoms can produce an enormous amount of energy, in the form of warmth and radiation (gamma waves). When an atom splits, each of the two new particles contains roughly half the neutrons and protons of the original nucleus, and in some cases a 2:3 ratio. The explosion of a bomb only occurs if the chain reaction exceeds its critical mass. The critical mass is the point at which a chain reaction becomes self-sustaining. If the neutrons are lost at a faster rate than they are formed by fission, the reaction will not be self-sustaining. The spontaneous nuclear fission rate is the probability per second that a given atom will fission spontaneously--that is, without any external intervention. In nuclear power plants, nuclear fission is controlled by a medium such as water in the nuclear reactor. The water acts as a heat transfer medium to cool down the reactor and to slow down neutron particles. This way, the neutron emission and usage is a controlled. If nuclear reaction is not controlled because of lack of cooling water for example, then a meltdown will occur. Nuclear fusion is the joining of two nuclei to form a heavier nuclei. The reaction is followed either by a release or absorption of energy. Fusion of nuclei with lower mass than iron releases energy while fusion of nuclei heavier than iron generally absorbs energy. This phenomenon is known as . The opposite occurs with nuclear fission. The power of the energy in a fusion reaction is what drives the energy that is released from the sun and a lot of stars in the universe. Nuclear fusion is also applied in nuclear weapons, specifically, a hydrogen bomb. Nuclear fusion is the energy supplying process that occurs at extremely high temperatures like in stars such as the sun, where smaller nuclei are joined to make a larger nucleus, a process that gives off great amounts of heat and radiation. When uncontrolled, this process can provide almost unlimited sources of energy and an uncontrolled chain provides the basis for a hydrogen bond, since most commonly hydrogen is fused. Also, the combination of deuterium atoms to form helium atoms fuel this thermonuclear process. For example: \[ \ce{^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n} + \text{energy}\] However, a controlled fusion reaction has yet to be fully demonstrated due to many problems that present themselves including the difficulty of forcing deuterium and tritium nuclei within a close proximity, achieving high enough thermal energies, and completely ionizing gases into plasma. A necessary part in nuclear fusion is , which is a mixture of atomic nuclei and electrons that are required to initiate a self-sustaining reaction which requires a temperature of more than 40,000,000 K. Why does it take so much heat to achieve nuclear fusion even for light elements such as hydrogen? The reason is because the nucleus contain protons, and in order to overcome electrostatic repulsion by the protons of both the hydrogen atoms, both of the hydrogen nucleus needs to accelerate at a super high speed and get close enough in order for the nuclear force to start fusion. The result of nuclear fusion releases more energy than it takes to start the fusion so ΔG of the system is negative which means that the reaction is exothermic. And because it is exothermic, the fusion of light elements is self-sustaining given that there is enough energy to start fusion in the first place. Figure \(\Page {1}\): Scientists have yet to find a method for controlling fusion reactions. Fission reactions on the other hand is the type used in nuclear power plants and can be controlled. Atomic bombs and hydrogen bombs are examples of uncontrolled nuclear reactions. | 13,776 | 1,752 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Potential_Energy |
Potential Energy is the energy due to position, composition, or arrangement. Also, it is the energy associated with forces of attraction and repulsion between objects. Any object that is lifted from its resting position has stored energy therefore it is called potential energy because it has a potential to do work when released. For example, when a ball is released from a certain height, it is pulled by gravity and the potential energy is converted to kinetic energy during the fall. As this energy converts from potential to kinetic, it is important to take into consideration that energy cannot be created nor destroyed (law of conservation of energy). This potential energy becomes kinetic energy as the ball accelerates towards the ground. The object's total energy can be found through the sum of these to energies. In an exothermic chemical reaction, potential energy is the source of energy. During an exothermic reaction bonds break and new bonds form and protons and electrons go from a structure of higher potential energy to lower potential energy. During this change, potential energy is converted to kinetic energy, which is the heat released in reactions. In an endothermic reaction the opposite occurs. The protons and electrons move from an area of low potential energy to an area of high. This takes in energy. Energy stored in bonds and static interactions are: \[PE= Fx\] where \(F\) is the opposing force and \(x\) is the distance moved. To calculate the potential energy of an object on Earth or within any other force field the formula \[PE=mgh \label{pe1}\] with If the units above are used for the \(m\), \(g\), and \(h\), then the final answer should be given in Joules. A 15 gram ball sits on top of a 2 m high refrigerator. What is the potential energy of the ball at the top of the refrigerator? Use Equation \ref{pe1} with \(m =15\, grams\). This mass however has to be in kilograms. The conversion to grams to kilograms is: 1,000 grams per 1 kg \[PE=(0.015 \, kg)(9.8 \, m/s^2)(2\,m)=0.294\, J \nonumber\] What is the mass of a cart full of groceries that is sitting on top of a 2 m hill if its gravitational potential energy is 0.3 J? Use Equation \ref{pe1} \[0.3\,J=(m)(9.8\, m/s^2)(2\,m) \nonumber\] and solve for mass \[m=0.015 \,kg=15\, g. \nonumber\] A 200 gram weight is placed on top of a shelf with a potential energy of 5 J. How high is the weight resting? \[5\,J=\left(\dfrac{200\,g}{1000\,g/kg}\right)(9.8 m/s^2)(h) \nonumber\] and solve for height \[h=2.55\, m \nonumber\] The potential energy of two charged particles at a distance can be found through the equation: \[E= \dfrac{q_1 q_2}{4π \epsilon_o r} \label{Coulomb}\] where For charges with the same sign, \(E\) has a + sign and tends to get smaller as \(r\) increases. This can explain why like charges repel each other. Systems prefer a low potential energy and thus repel each other which increases the distance between them and lowers the potential energy. For charges with different charges, the opposite of what is stated above is true. E has a - sign which becomes even more negative as the opposite charged particles attract, or come closer together. Calculate the potential energy associated with two particles with charges of \(3 \times 10^{-6}\, C\) and \(3.9 \times 10^{-6}\, C\) are separated by a distance of \(1\, m\) Using Equation \ref{Coulomb} \[\begin{align*} E &=\dfrac{(3\times 10^{-6}\,C)(3.9 \times 10^{-6}\,C)}{4π \,8.85 \times 10^{-12} \,C^2/Jm} \\[4pt] &=0.105 \,J \end{align*}\] Find the distance between two particles that have a potential energy of \(0.2\, J\) and charges of \(2.5 \times 10^{-6}\, C\) and \(3.1 \times 10^{-6}\, C\). \[\begin{align*} 0.2 &=\dfrac{(2.5 \times 10^{-6}\,C)(3.1 \times 10^{-6} \,C)}{4\pi (8.85 \times 10^{-12} \,C^2/Jm) r} \\[4pt] &=\dfrac{(8.99 \times 10^9)(7.75 \times 10^{-11})}{r} \\[4pt] &=\dfrac{0.6967}{r} \end{align*}\] cross multiply and solve for \(r\) \[r=3.5\, m \nonumber\] Includes all interactions in the system such as: in nucleus of atoms; in atoms; between atoms in a molecule (intra-molecular forces); and between different molecules (inter-molecular forces). | 4,159 | 1,753 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.E%3A_Solids_Liquids_and_Gases_(Exercises) |
1. What types of intermolecular interactions can exist in compounds? 2. What is the difference between covalent network and covalent molecular compounds? 1. ionic bonding, network covalent, dispersion forces, dipole-dipole interactions, and hydrogen bonding. 2. Covalent network compounds contain atoms that are covalently bonded to other individual atoms in a giant 3-dimensional network. Covalent molecular compounds contain individual molecules that are attracted to one another through dispersion, dipole-dipole or hydrogen bonding. List the three common phases in the order you are likely to find them—from lowest temperature to highest temperature. List the three common phases in the order they exist from lowest energy to highest energy. List these intermolecular interactions from weakest to strongest: London forces, hydrogen bonding, and ionic interactions. List these intermolecular interactions from weakest to strongest: covalent network bonding, dipole-dipole interactions, and dispersion forces. What type of intermolecular interaction is predominate in each substance? What type of intermolecular interaction is predominate in each substance? Explain how a molecule like carbon dioxide (CO ) can have polar covalent bonds but be nonpolar overall. Sulfur dioxide (SO ) has a formula similar to that of carbon dioxide (see Exercise 7) but is a polar molecule overall. What can you conclude about the shape of the SO molecule? What are some of the physical properties of substances that experience covalent network bonding? What are some of the physical properties of substances that experience only dispersion forces? solid, liquid, and gas How do the strengths of intermolecular interactions in solids and liquids differ? Solids have stronger intermolecular interactions than liquids do. What are the general properties of solids? What are the general properties of liquids What are the general properties of gases? What phase or phases have a definite volume? What phase or phases do not have a definite volume? Name a common substance that forms a crystal in its solid state. Name a common substance that forms an amorphous solid in its solid state. Are substances with strong intermolecular interactions likely to be solids at higher or lower temperatures? Explain. Are substances with weak intermolecular interactions likely to be liquids at higher or lower temperatures? Explain. State two similarities between the solid and liquid states. State two differences between the solid and liquid states. If individual particles are moving around with respect to each other, a substance may be in either the _______ or ________ state but probably not in the _______ state. If individual particles are in contact with each other, a substance may be in either the ______ or _______ state but probably not in the ______ state. hard, specific volume and shape, high density, cannot be compressed What is pressure, and what units do we use to express it? Pressure is the force per unit area; its units can be pascals, torr, millimeters of mercury, or atmospheres. What is the kinetic theory of gases? According to the kinetic theory of gases, the individual gas particles are (always, frequently, never) moving. Why does a gas exert pressure? Why does the kinetic theory of gases allow us to presume that all gases will show similar behavior? Arrange the following pressure quantities in order from smallest to largest: 1 mmHg, 1 Pa, and 1 atm. Which unit of pressure is larger—the torr or the atmosphere? How many torr are there in 1.56 atm? Convert 760 torr into pascals. Blood pressures are expressed in millimeters of mercury. What would be the blood pressure in atmospheres if a patient’s systolic blood pressure is 120 mmHg and the diastolic blood pressure is 82 mmHg? (In medicine, such a blood pressure would be reported as “120/82,” spoken as “one hundred twenty over eighty-two.”) In weather forecasting, barometric pressure is expressed in inches of mercury (in. Hg), where there are exactly 25.4 mmHg in every 1 in. Hg. What is the barometric pressure in millimeters of mercury if the barometric pressure is reported as 30.21 in. Hg? What conditions of a gas sample should remain constant for Boyle’s law to be used? What conditions of a gas sample should remain constant for Charles’s law to be used? Does the identity of a gas matter when using Boyle’s law? Why or why not? Does the identity of a gas matter when using Charles’s law? Why or why not? A sample of nitrogen gas is confined to a balloon that has a volume of 1.88 L and a pressure of 1.334 atm. What will be the volume of the balloon if the pressure is changed to 0.662 atm? Assume that the temperature and the amount of the gas remain constant. A sample of helium gas in a piston has a volume of 86.4 mL under a pressure of 447 torr. What will be the volume of the helium if the pressure on the piston is increased to 1,240 torr? Assume that the temperature and the amount of the gas remain constant. If a gas has an initial pressure of 24,650 Pa and an initial volume of 376 mL, what is the final volume if the pressure of the gas is changed to 775 torr? Assume that the amount and the temperature of the gas remain constant. A gas sample has an initial volume of 0.9550 L and an initial pressure of 564.5 torr. What would the final pressure of the gas be if the volume is changed to 587.0 mL? Assume that the amount and the temperature of the gas remain constant. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is 18°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and amount of the gas remain constant. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is −10°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and the amount of the gas remain constant. An air/gas vapor mix in an automobile cylinder has an initial temperature of 450 K and a volume of 12.7 cm . The gas mix is heated to 565°C. If pressure and amount are held constant, what is the final volume of the gas in cubic centimeters? Given the following conditions for a gas: = 0.665 L, = 23.6°C, = 1.034 L. What is in degrees Celsius and kelvins? Assuming the amount remains the same, what must be the final volume of a gas that has an initial volume of 387 mL, an initial pressure of 456 torr, an initial temperature of 65.0°C, a final pressure of 1.00 atm, and a final temperature of 300 K? When the nozzle of a spray can is depressed, 0.15 mL of gas expands to 0.44 mL, and its pressure drops from 788 torr to 1.00 atm. If the initial temperature of the gas is 22.0°C, what is the final temperature of the gas? Use the ideal gas law to show that 1 mol of a gas at STP has a volume of about 22.4 L. Use a standard conversion factor to determine a value of the ideal gas law constant that has units of L•torr/mol•K. How many moles of gas are there in a 27.6 L sample at 298 K and a pressure of 1.44 atm? How many moles of gas are there in a 0.066 L sample at 298 K and a pressure of 0.154 atm? A 0.334 mol sample of carbon dioxide gas is confined to a volume of 20.0 L and has a pressure of 0.555 atm. What is the temperature of the carbon dioxide in kelvins and degrees Celsius? What must be for a gas sample if = 4.55 mol, = 7.32 atm, and = 285 K? 21. What is the pressure of 0.0456 mol of Ne gas contained in a 7.50 L volume at 29°C? 22. What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of −86.0°C? 23. A mixture of the gases \(N_2\), \(O_2\), and \(Ar\) has a total pressure of 760 mm Hg. If the partial pressure of \(N_2\) is 220 mm Hg and of \(O_2\) is 470 mm Hg, What is the partial pressure of \(Ar\)? 24. What percent of the gas above is Ar? 25. Apply Henry’s Law to the diagram below to explain: why oxygen diffuses from the alveoli of the lungs into the blood and from the blood into the tissues of the body. why carbon dioxide diffuses from the tissues into the blood and from the blood into the alveoli and then finally out into the atmosphere. \(\dfrac{760\: torr}{1\: atm}\) How many grams of oxygen gas are needed to fill a 25.0 L container at 0.966 atm and 22°C? A breath of air is about 1.00 L in volume. If the pressure is 1.00 atm and the temperature is 37°C, what mass of air is contained in each breath? Use an average molar mass of 28.8 g/mol for air. The balanced chemical equation for the combustion of propane is as follows: \[C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(ℓ)}\] The equation for the formation of ammonia gas (NH ) is as follows: \[N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}\] At 500°C and 1.00 atm, 10.0 L of N gas are reacted to make ammonia. At 20°C, 1 g of liquid H O has a volume of 1.002 mL. What volume will 1 g of water vapor occupy at 20°C if its pressure is 17.54 mmHg? By what factor has the water expanded in going from the liquid phase to the gas phase? At 100°C, 1 g of liquid H O has a volume of 1.043 mL. What volume will 1 g of steam occupy at 100°C if its pressure is 760.0 mmHg? By what factor has the water expanded in going from the liquid phase to the gas phase? Predict whether NaCl or NaI will have the higher melting point. Explain. (Hint: consider the relative strengths of the intermolecular interactions of the two compounds.) Predict whether CH or CH OH will have the lower boiling point. Explain. (Hint: consider the relative strengths of the intermolecular interactions of the two compounds.) A standard automobile tire has a volume of about 3.2 ft (where 1 ft equals 28.32 L). Tires are typically inflated to an absolute pressure of 45.0 pounds per square inch (psi), where 1 atm equals 14.7 psi. Using this information with the ideal gas law, determine the number of moles of air needed to fill a tire if the air temperature is 18.0°C. Another gas law, Amontons’s law, relates pressure and temperature under conditions of constant amount and volume: \(\mathrm{\dfrac{P_i}{T_i}=\dfrac{P_f}{T_f}}\) If an automobile tire (see Exercise 9) is inflated to 45.0 psi at 18.0°C, what will be its pressure if the operating temperature (i.e., the temperature the tire reaches when the automobile is on the road) is 45.0°C? Assume that the volume and the amount of the gas remain constant. 31.9 g 3. 57.75 L; an expansion of 57,600 times NaCl; with smaller anions, NaCl likely experiences stronger ionic bonding. 9. 11.6 mol | 10,484 | 1,754 |
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Water is composed of two hydrogen atoms and an oxygen atom. It exhibits polarity and is naturally found in the liquid, solid, and vapor states. Its polarity makes it a good solvent and is commonly known as the universal solvent. Because of its abundance on earth, it is important to note that it is involved in many chemical reactions. Many of these chemical reactions behave in trends that can be categorized using the periodic table. A common characteristic of most is their ability to displace H (g) from water. This is represented by their large, negative electrode potentials. In this event, the Group 1 metal is oxidized to its metal ion and water is reduced to form hydrogen gas and hydroxide ions. The general reaction of an alkali metal (M) with H O (l) is given in the following equation: \[\ce{ 2M(s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-}(aq) + H2 (g)} \] From this reaction it is apparent that OH is produced, creating a basic or alkaline environment. Group 1 elements are called alkali metals because of their ability to displace H (g) from water and create a basic solution. Alkali metals are also known to react violently and explosively with water. This is because enough heat is given off during the exothermic reaction to ignite the H (g). elements also react with water to create basic solutions. Alkali metals react with oxygen to form monoxides, peroxides, or superoxides. These species react with water in different ways: \[\ce{M2O(s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-} (aq)} \label{1}\] \[\ce{M2O2(s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-} (aq) + H2O2(aq)} \label{2}\] \[\ce{2MO2 (s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-} (aq) + H2O2 (aq) + O2(g)} \label{3}\] Similarly to the Group 1 oxides, the hydrides of the Group 1 elements react with water to form a basic solution. In this case, however, hydrogen gas is produced with the metal hydroxide. The general reaction for alkali metal hydrides and water is given below: \[MH_{(s)} + H_2O_{(l)} \longrightarrow M^+_{(aq)}+OH^-_{(aq)}+H_{2(g)} \label{4}\] This reaction can be generalized to all alkali metal hydrides. The majority of also produce hydroxides when reacted with water. The hydroxides of calcium, strontium, and barium are only slightly soluble in water; however, enough hydroxide ions are produced to make a basic environment. The general reaction of calcium, strontium, and barium with water is represented below, where M represents calcium, strontium, or barium: \[M_{(s)} + 2H_2O_{(l)} \longrightarrow M(OH)_{2(aq)}+H_{2\;(g)} \label{5}\] Magnesium (Mg) reacts with water vapor to form magnesium hydroxide and hydrogen gas. (Be) is the only alkaline earth metal that does not react with water. This is due to its small size and high ionization energy in relation to the other elements in the group. Similarly to the alkali metal oxides, alkaline earth metal monoxides combine with water to form metal hydroxide salts (as illustrated in the equation below). The exception to this general assumption is beryllium, whose oxide (BeO) does not react with water. \[MO_{(s)}+H_2O_{(l)} \longrightarrow M(OH)_{2(s)} \label{6}\] One of the most familiar alkaline earth metal oxides is CaO or quicklime. This substance is often used to treat water and to remove harmful \(SO_{2(g)}\) from industrial smokestacks. With the exception of beryllium (Be), the alkaline metal hydrides react with water to produce the metal hydroxide and hydrogen gas. The reaction of these metal hydrides can be described below: \[MH_{2(s)}+2H_2O_{(l)} \longrightarrow M(OH)_{2(aq)}+2H_{2(g)} \label{7} \] The two types of include temporary hard water and permanent hard water. Temporary hard water contains bicarbonate (HCO ) which forms CO (aq), CO (g), and H O when heated. The bicarbonate ions react with alkaline earth cations and precipitate out of solution, causing boiler scale and problems in water heaters and plumbing. Common cations in the water include Mg and Ca . In order to soften the water, water treatment plants add an alkaline earth metal hydroxide, such as [Ca(OH) ]. This solid dissolves in the water producing a metal ion (M ) and hydroxide ions (OH ). The hydroxide ions combine with the bicarbonate ions in the water to produce water and a carbonate ion. The carbonate ion then precipitates out with the metal ion to form MCO (s). Water treatment plants are able to remove the precipitated metal carbonate and thus soften the water. The other type of hard water is permanent hard water. Permanent hard water contains bicarbonate ions (HCO ) as well as other anions such as sulfate ions (SO ). The hardening species often cannot be boiled off. To soften permanent water, sodium carbonate (Na CO ) is added. Sodium carbonate precipitates out the Mg and Ca ions out as the respective metal carbonates and introduces Na ions into the solutions. Group 13 elements are not very reactive with water. In fact, boron (B) does not react at with water. One notable reaction within this group is aluminum's (Al) reaction with water. Aluminum does not appear to react with water because an outer layer of aluminum oxide (Al O ) solid forms and protects the rest of the metal. For the most part, Group 14 elements do not react with water. One interesting consequence of this is that tin (Sn) is often sprayed as a protective layer on iron cans to prevent the can from corroding. The pure elements in this family do not tend to react with water. Compounds of nitrogen (nitrates and nitrites) as well as nitrogen gas (N ) dissolve in water but do not react. As mentioned earlier, many Group 1 and Group 2 oxides react with water to form metal hydroxides. The nonmetal oxides react with water to form oxoacids. Examples include phosphoric acid and sulfuric acid. Generally halogens react with water to give their halides and hypohalides. The halogen gases vary in their reactions with water due to their different electronegativities. Because fluorine (\(\ce{F2}\)) is so electronegative, it can displace oxygen gas from water. The products of this reaction include oxygen gas and hydrogen fluoride. The hydrogen halides react with water to form acids (\(\ce{HX}\)). With the exception of \(\ce{HF}\), the hydrohalic acids are strong acids in water. Hydrochloric acid (\(\ce{HCl}\)), a strong acid, is an example. \[\ce{Cl2(g) + 2H2O(l) → HCl(aq) + HOCl(aq)}\] Hypochlorous (\(\ce{HOCl}\)) acid is a strong bleaching agent and is not very stable in solution and readily decomcomposes, especially when exposed to sunlight, yielding oxygen. \[\ce{ 2 HClO -> 2 HCl + O2}\] Bromine liquid dissolves slowly in water to form a yellowish-brown solution. \[\ce{Br2(g) + 2H2O(l) → HBr(aq) + HOBr(aq)}\] Hypobromous (\(\ce{HOBr}\)) acid is a weak bleaching agent. \[\ce{I2(g) + 2H2O(l) → HI(aq) + HOI(aq)}\] Only a little iodine dissolves in water to form a yellowish solution and hypoiodous (\(\ce{HOI}\)) acid has very weak bleaching characteristic. The noble gases do not react with water. \(NaH{(s)}+2H_{2}O_{(l)} \longrightarrow\) \(NaH{(s)}+2H_{2}O_{(l)} \longrightarrow Na^+_{(aq)}+OH^-_{(aq)}+H_2 \; {(g)}\) | 7,143 | 1,755 |
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This page describes how simple acid-base indicators work, and how to choose the right one for a particular titration. Litmus is a weak acid and is one of the oldest forms of a pH indicator and is used to test materials for acidity. It has a seriously complicated molecule which we will simplify to HLit. The "H" is the proton which can be given away to something else. The "Lit" is the rest of the weak acid molecule. There will be an equilibrium established when this acid dissolves in water. Taking the simplified version of this equilibrium: The un-ionised litmus is red, whereas the ion is blue. Now use Le Chatelier's Principle to work out what would happen if you added hydroxide ions or some more hydrogen ions to this equilibrium. Adding hydroxide ions: Adding hydrogen ions: At some point during the movement of the position of equilibrium, the concentrations of the two colors will become equal. The color you see will be a mixture of the two. The reason for the inverted commas around "neutral" is that there is no reason why the two concentrations should become equal at pH 7. For litmus, it so happens that the 50 / 50 color does occur at close to pH 7 - that's why litmus is commonly used to test for acids and alkalis. As you will see below, that isn't true for other indicators. Methyl orange is one of the indicators commonly used in titrations. In an alkaline solution, methyl orange is yellow and the structure is: Now, you might think that when you add an acid, the hydrogen ion would be picked up by the negatively charged oxygen. That's the obvious place for it to go. Not so! In fact, the hydrogen ion attaches to one of the nitrogens in the nitrogen-nitrogen double bond to give a structure which might be drawn like this: You have the same sort of equilibrium between the two forms of methyl orange as in the litmus case - but the colors are different. You should be able to work out for yourself why the color changes when you add an acid or an alkali. The explanation is identical to the litmus case - all that differs are the colors. In the methyl orange case, the half-way stage where the mixture of red and yellow produces an orange color happens at pH 3.7 - nowhere near neutral. This will be explored further down this page. Phenolphthalein is another commonly used indicator for titrations, and is another weak acid. In this case, the weak acid is colorless and its ion is bright pink. Adding extra hydrogen ions shifts the position of equilibrium to the left, and turns the indicator colorless. Adding hydroxide ions removes the hydrogen ions from the equilibrium which tips to the right to replace them - turning the indicator pink. The half-way stage happens at pH 9.3. Since a mixture of pink and colorless is simply a paler pink, this is difficult to detect with any accuracy! Think about a general indicator, HInd - where "Ind" is all the rest of the indicator apart from the hydrogen ion which is given away: \[HInd_{(aq)} \rightleftharpoons H^+_{(aq)} + Ind^-_{(aq)} \tag{1.6.1}\] Because this is just like any other weak acid, you can write an expression for \(K_a\) for it. We will call it \(K_{ind}\) to stress that we are talking about the indicator. \[ K_{ind} = \dfrac{[H^+,Ind^-]}{[HInd]} \tag{1.6.2}\] Think of what happens half-way through the color change. At this point the concentrations of the acid and its ion are equal. \[ [Ind^-] = [HInd] \tag{1.6.3}\] In that case, they will cancel out of the K expression. \[ K_{ind} = \dfrac{[H^+]\cancel{[Ind^-]}}{\cancel{[HInd]}} \tag{1.6.4}\] \[ K_{ind} = [H^+] \tag{1.6.5}\] You can use this to work out what the pH is at this half-way point. If you re-arrange the last equation so that the hydrogen ion concentration is on the left-hand side, and then convert to pH and pK , you get: \[ [H^+] = K_{ind} \tag{1.6.5}\] \[ pH = -\log_{10} [H^+] = -\log_{10} K_{ind} = pK_{ind} \tag{1.6.6}\] That means that the end point for the indicator depends entirely on what its pK value is. For the indicators we've looked at above, these are: Indicators don't change color sharply at one particular pH (given by their pK ). Instead, they change over a narrow range of pH. Assume the equilibrium is firmly to one side, but now you add something to start to shift it. As the equilibrium shifts, you will start to get more and more of the second color formed, and at some point the eye will start to detect it. For example, suppose you had methyl orange in an alkaline solution so that the dominant color was yellow. Now start to add acid so that the equilibrium begins to shift. At some point there will be enough of the red form of the methyl orange present that the solution will begin to take on an orange tint. As you go on adding more acid, the red will eventually become so dominant that you can no longer see any yellow. There is a gradual smooth change from one color to the other, taking place over a range of pH. As a rough "rule of thumb", the visible change takes place about 1 pH unit either side of the pK value. The exact values for the three indicators we've looked at are: The litmus color change happens over an unusually wide range, but it is useful for detecting acids and alkalis in the lab because it changes color around pH 7. Methyl orange or phenolphthalein would be less useful. This is more easily seen diagramatically. For example, methyl orange would be yellow in any solution with a pH greater than 4.4. It couldn't distinguish between a weak acid with a pH of 5 or a strong alkali with a pH of 14. Remember that the equivalence point of a titration is where you have mixed the two substances in exactly equation proportions. You obviously need to choose an indicator which changes color as close as possible to that equivalence point. That varies from titration to titration. The next diagram shows the pH curve for adding a strong acid to a strong base. Superimposed on it are the pH ranges for methyl orange and phenolphthalein. You can see that neither indicator changes color at the equivalence point. However, the graph is so steep at that point that there will be virtually no difference in the volume of acid added whichever indicator you choose. However, it would make sense to titrate to the best possible color with each indicator. If you use phenolphthalein, you would titrate until it just becomes colorless (at pH 8.3) because that is as close as you can get to the equivalence point. On the other hand, using methyl orange, you would titrate until there is the very first trace of orange in the solution. If the solution becomes red, you are getting further from the equivalence point. This time it is obvious that phenolphthalein would be completely useless. However, methyl orange starts to change from yellow towards orange very close to the equivalence point. You have to choose an indicator which changes color on the steep bit of the curve. This time, the methyl orange is hopeless! However, the phenolphthalein changes color exactly where you want it to. The curve is for a case where the acid and base are both equally weak - for example, ethanoic acid and ammonia solution. In other cases, the equivalence point will be at some other pH. You can see that neither indicator is any use. Phenolphthalein will have finished changing well before the equivalence point, and methyl orange falls off the graph altogether. It may be possible to find an indicator which starts to change or finishes changing at the equivalence point, but because the pH of the equivalence point will be different from case to case, you can't generalise. On the whole, you would never titrate a weak acid and a weak base in the presence of an indicator. This is an interesting special case. If you use phenolphthalein or methyl orange, both will give a valid titration result - but the value with phenolphthalein will be exactly half the methyl orange one. It so happens that the phenolphthalein has finished its color change at exactly the pH of the equivalence point of the first half of the reaction in which sodium hydrogencarbonate is produced. \[ Na_2CO_{3(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + NaHCO_{3(aq)} \tag{1.6.1}\] The methyl orange changes color at exactly the pH of the equivalence point of the second stage of the reaction. \[ NaHCO_{3(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + CO_{2(g)} + H_2O_{(l)} \tag{1.6.2}\] | 8,366 | 1,756 |
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The quantitative relationship of temperature effect on chemical reaction rates is discussed in form of activation energy, depicted as E in the diagram. For reactions having large E values, high temperatures are required to have a measurable reaction rate. The reason behind this is due to the small number of molecules having sufficient energy to overcome E , the energy barrier of forming an activated reaction intermediate. At a certain temperature, not all gas molecules are moving with the same speed, some fast and some slow. The kinetic energies of molecules are not all equal. The fast moving ones have high kinetic energy, and they may have enough energy to overcome the reaction energy barrier E . A plot of number of gas molecules at certain speed versus speed gives a distribution curve. Careful studies of gases show that the distribution is not a bell or normal distribution, but a Maxwell- Boltzmann distribution. A sketch of a Maxwell distribution is given here. The peaks are not symmetrical. There are more molecules at higher speed than at lower speeds. When the temperature increases, the peak shifts to the right. A more carefully plotted diagram is shown below. You will learn the theory and the statistics at a higher year. For the moment, you may notice how the distribution curve shifts as temperature increases.
Name the various distributions.
Explain how molecular masses affect the average speed.
The root-mean-square speed is \(\sqrt{\dfrac{3 R T}{M}}\), where = gas constant, = temperature, = molecular mass.
The formulas for the three speeds mentioned here can be derived by mathematical techniques. The most probable speed is \(\sqrt{\dfrac{2 R T}{M}}\), where = gas constant, = temperature, = molecular mass.
It is easy to calculate the most probable speed if you know the formula to use. Note that 500 C is higher than 700 F. Do you know what the temperature scales measure? | 1,964 | 1,758 |
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Gibbs energy is the energy of a chemical reaction that can be used to do non-mechanical work. Gibbs Energy is described as \[G = H - TS \label{1}\] Where H is , T is temperature, and S is . ΔG is used to predict spontaneity within a system by At a constant temperature and pressure, the Gibbs Energy of a system can be described as \[\Delta G_{sys} = \Delta H_{sys} – T \Delta S_{sys} \label{3}\] This equation can be used to determine the spontaneity of the process. Gibbs Energy is a useful tool to describe in what manner the reaction is conducted. If ΔH >> TΔS, the reaction is enthalpy driven. However, if TΔS >> ΔH, the reaction is driven by entropy. The Clausius-Clapeyron Equation is an application derived from Gibb's energy: \[\ln \left(\dfrac{P_2}{P_1}\right) = \dfrac{Δ_{vap}H}{R} \left(\dfrac{T_2 - T_1}{T_2T_1} \right) \label{4}\] Another important application of Gibb's energy is the Maxwell relations (also available in a link at the end of the wiki page.) When the pressure & temperature of a reaction are not held constant, \[G = H - TS \label{5}\] For an infinitesimal process, \[ΔG = ΔH – Δ(TS) \label{6}\] \[ΔG = ΔH – TΔS –SΔT \label{7}\] For a reaction where temperature is held constant, \[ΔG = ΔH – TΔS \label{8}\] From the , we know \[H= U +PV \label{9}\] \[ΔH = ΔU + PΔV+VΔP \label{10}\] Since \[ΔU = TΔS – PΔV \label{11}\] We find that \[ΔG = VΔP-SΔT \label{12}\] showing the obvious dependence of ΔG on temperature and pressure. To observe the change in Gibbs energy due to (pressure held constant) the equation becomes \[ ΔG=-SΔT \label{13}\] Solving for S and plugging it into Eq. (1), the Gibbs-Helmholtz Equation is found: \[ \left(\dfrac{∂(ΔG/T)}{∂T}\right)_p = -\dfrac{ΔH}{T^2} \label{14}\] The Gibbs-Helmholtz Equation is very important because it relates the change in Gibbs energy to its temperature dependence, and the position of equilibrium to change in enthalpy. To observe the change in Gibbs energy due to (temperature held constant) the equation becomes \[ΔG=VΔP \label{15}\] If the gas is assumed to be then \[ΔG=nRT \ln \left(\dfrac{P_2}{P_1}\right) \label{16}\] for an initial and final pressure (P1 and P2) at a constant T. Gibbs Energy is defined as a (a property that depends only on conditions describing the system, not how the change occurs as in a .) This is because each component of the equation (H,T, and S) are all state functions. Therefore, we can know the change in Gibbs energy without knowing every detail of the process. In a process that takes place at constant temperature and pressure (298 K, 1 atm) the standard molar free energy of formation can be determined by the change in free energy from the reactants and products. Using predetermined values, Eqn. (17) can be used \[ΔG^o)f = \sum (Coefficient_{products} G^°_{products}) - \sum (Coefficient_{reactants} G^°_{reactants}) \label{17}\] Standard-State Free Energy of Reaction Gibbs Energy can be found at standard-state conditions using \[ΔG° = ΔH° - TΔS° \label{18}\] ΔH° and ΔS° values can be found in the appendix of any general chemistry textbook, or using this . Gibbs energy can be found at any conditions by relating it to the standard-state free energy of reaction, using \[ΔG = ΔG° +RT \ln Q \label{19}\] Where Q is the . Very rarely does chemistry actually occur at the given "standard-state" conditions. Using the above equation and standard-state values, chemists can determine the overall Gibb's energy for the system, regardless of the conditions. | 3,529 | 1,759 |
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In previous chapters, we used the principles of chemical bonding, thermodynamics, and kinetics to provide a conceptual framework for understanding the chemistry of the elements. Beginning in , we use the periodic table to guide our discussion of the properties and reactions of the elements and the synthesis and uses of some of their commercially important compounds. We begin this chapter with a review of periodic trends as an introduction, and then we describe the chemistry of hydrogen and the other -block elements. In , we consider the chemistry of the -block elements; presents the transition metals, in which the -subshell is being filled. In this chapter, you will learn why potassium chloride is used as a substitute for sodium chloride in a low-sodium diet, why cesium is used as a photosensor, why the heating elements in electric ranges are coated with magnesium oxide, and why exposure to a radioactive isotope of strontium is more dangerous for children than for adults. . Heating a compound in a very hot flame results in the formation of its component atoms in electronically excited states. When an excited atom decays to the ground state, it emits light. Each element emits light at characteristic frequencies. Flame tests are used to identify many elements based on the color of light emitted in the visible region of the electromagnetic spectrum. As shown here, sodium compounds produce an intense yellow light, whereas potassium compounds produce a crimson color. by Sully Science from YouTube | 1,542 | 1,760 |
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In a crystal lattice, the ionic radius is a measure of the size of the atom's ion. When formed, ionic atoms change in size with respect to their orginal atom. Cation radii will decrease and the anion radii will increase in size compared to their neutral atoms. Questions such as: "What methodology is used by chemists to measure ionic radii?" and "Are there any non-experimental ways to estimate the size of ionic radii?" will be answered in this module. Accordingly, there are many ways to determine ionic radii. In the past, after an atom is ionized, X-ray diffraction is used to measure how much the radius of the atom increased or decreased. However, scientists wanted to use another technique, due to the fact, that X-ray diffraction is difficult to distinguish a boundary between two ions. As a result, the can be used. The Hard-Sphere model are impenetrable spheres that do not overlap in space. The Hard-Sphere model has been tested by well-known scientists; Lande', Pauling and Goldsmidt. The ion radii measured under crystal state of ionic compound which cations and anions are stacking in pattern as shown below. The Hard-Sphere model can be applied to metallic and ionic compounds such as NaCl, which is shown below. In general, scientists uses formula of Internuclear distance to test out the radii of ion then compared with the ion radii had done on X-ray diffraction: Internuclear distance (d) = r + r *To calculate ion radii, Lande used ionic compound under solid state (ex: NaCl). This will minize the distribution of electrons. As described earlier, cations are smaller in size compared to their neutral atoms while anions are larger in size.Cations are smaller than its neutral atoms because the positive nuclear charge, which holds the electrons in closer, exceeds the negative charge when a metal atom loses an electron. On the contrary, anions are larger because the electrons are not held as tightly, repulsions of electrons increase, and the electrons spread out more due to nonmetal atoms gaining an electron. Refer to the outside link to learn more about the periodic trends for ionic radii 1. What is the most general formula that used to determine the ion radii for hard sphere model? 2. Find radius for Cacium ion in Calcium Chloride (CaCl List out all the steps (numbers are not necessary) 3. Determine which is larger: a) K or Cs ? b) La or Lu ? c) Ca or Zn ? 3. a.) Cs b.) La c.) Ca | 2,501 | 1,761 |
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Ernest Rutherford had proposed a model of atoms based on the \(\alpha\)-particle scattering experiments of Hans Geiger and Ernest Marsden. In these experiments helium nuclei (\(\alpha\)-particles) were shot at thin gold metal foils. Most of the particles were not scattered; they passed unchanged through the thin metal foil. Some of the few that were scattered were scattered in the backward direction; i.e. they recoiled. This backward scattering requires that the foil contain heavy particles. When an \(\alpha\)-particle hits one of these heavy particles it simply recoils backward, just like a ball thrown at a brick wall. Since most of the α-particles don’t get scattered, the heavy particles (the nuclei of the atoms) must occupy only a very small region of the total space of the atom. Most of the space must be empty or occupied by very low-mass particles. These low-mass particles are the electrons that surround the nucleus. There are some basic problems with the Rutherford model. The Coulomb force that exists between oppositely charge particles means that a positive nucleus and negative electrons should attract each other, and the atom should collapse. To prevent the collapse, the electron was postulated to be orbiting the positive nucleus. The Coulomb force (discussed below) is used to change the direction of the velocity, just as a string pulls a ball in a circular orbit around your head or the gravitational force holds the moon in orbit around the Earth. The origin for this hypothesis that suggests this perspective is plausible is the similarity of gravity and Coulombic interactions. The expression for the force of gravity between two masses ( ) is \[F_{gravity} \propto \dfrac{m_1m_2}{r^2}\label{1.8.1} \] with \(m_1\) and \(m_2\) representing the of object 1 and 2, respectively and \(r\) representing the distance between the objects centers The expression for the Coulomb force between two charged species is \[F_{Coulomb} \propto \dfrac{Q_1Q_2}{r^2}\label{1.8.2} \] with \(Q_1\) and \(Q_2\) representing the of object 1 and 2, respectively and \(r\) representing the distance between the objects centers. However, this analogy has a problem too. An electron going around in a circle is constantly being accelerated because its velocity vector is changing. A charged particle that is being accelerated emits radiation. This property is essentially how a radio transmitter works. A power supply drives electrons up and down a wire and thus transmits energy (electromagnetic radiation) that your radio receiver picks up. The radio then plays the music for you that is encoded in the waveform of the radiated energy. If the orbiting electron is generating radiation, it is losing energy. If an orbiting particle loses energy, the radius of the orbit decreases. To conserve angular momentum, the frequency of the orbiting electron increases. The frequency increases continuously as the electron collapses toward the nucleus. Since the frequency of the rotating electron and the frequency of the radiation that is emitted are the same, both change continuously to produce a continuous spectrum and not the observed discrete lines. Furthermore, if one calculates how long it takes for this collapse to occur, one finds that it takes about \(10^{‑11}\) seconds. This means that nothing in the world based on the structure of atoms could exist for longer than about \(10^{-11}\) seconds. Clearly something is terribly wrong with this classical picture, which means that something was missing at that time from the known laws of physics. A conservative force is dependent only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point. When an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken. The potential can be constructed as simple derivatives for 1-D forces: \[F = -\dfrac{dV}{dx} \nonumber \] or as gradients in 3-D forces \[F = -\nabla V \nonumber \] where \(\nabla\) is the vector of partial derivatives \[\nabla = \left ( \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right) \nonumber \] The most familiar conservative forces are gravity and Coloumbic forces. The Coulomb force law (Equation \(\ref{1.8.2}\)) comes from the corresponding Coulomb potential (sometimes call electrostatic potential) \[V(r)=\dfrac{kQ_1 Q_2}{r} \label{1.8.5} \] and it can be easily verified that the Coulombic force from this interaction (\(F(r)\)) is \[F(r)=-\dfrac{dV}{dr} \label{1.8.6} \] \(r\) \(V(r)\) (Figure \(\Page {2; left}\)) If \(Q_1\) and \(Q_2\) are the same sign, then the curve which is a purely , i.e., the energy increases monotonically as the charges are brought together and decreases monotonically as they are separated. From this, it is easy to see that like charges (charges of the same sign) repel each other. If the charges are of opposite sign, then the curve appears roughly Figure \(\Page {2; right}\) and this is a purely . Thus, the energy decreases as the charges are brought together, implying that opposite charges attract It is observed that line spectra discussed in the previous sections show that hydrogen atoms absorb and emit light at only discrete wavelengths. Quantum mechanics postulates that, in contrast to classical mechanics, the energy of a system can only take on certain discrete values. This leaves us with the question: How do we determine what these allowed discrete energy values are? After all, it seems that Planck's formula for the allowed energies came out of nowhere. The model we will describe here, due to Niels Bohr in 1913, is an early attempt to predict the allowed energies for single-electron atoms such as \(\ce{H}\), \(\ce{He^{+}}\), \(\ce{Li^{2+}}\), \(\ce{Be^{3+}}\), etc. Although Bohr's reasoning relies on classical concepts and hence, is not a correct explanation, the reasoning is interesting, and so we examine this model for its historical significance. Consider a nucleus with charge \(+Ze\) and one electron orbiting the nucleus. In this analysis, we will use another representation of the constant \(k\) in Coulomb's law (Equation \(\ref{1.8.5}\)), which is more commonly represented in the form: \[k=\dfrac{1}{4\pi \epsilon_0} \label{1.8.7} \] \(\epsilon_0\) with \(\epsilon_0 = 8.8541878\times 10^{-12} \ C^2 J^{-1} m^{-1}\) \[E_{total}=\underset{\text{kinetic energy}}{\dfrac{p^2}{2m_e}} - \underset{\text{potential energy}}{\dfrac{Ze^2}{4\pi \epsilon_0 r}} \nonumber \] \[\vec{F}=-\dfrac{Ze^2}{4\pi \epsilon_0 r^3}r \nonumber \] and its magnitude is \[F=|\vec{F}|=\dfrac{Ze^2}{4\pi \epsilon_0 r^3}|r|=\dfrac{Ze^2}{4\pi \epsilon_0 r^2} \nonumber \] \(\vec{F}=m_e \vec{a}\) \(|\vec{F}|=m_e |\vec{a}|\) \[|a|=\dfrac{v^2}{r} \nonumber \] \(v\) \(|F|\) \(m_e |a|\) \[\dfrac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\dfrac{v^2}{r} \nonumber \] \[\dfrac{Ze^2}{4\pi \epsilon_0}=m_e v^2 r \nonumber \] \[\dfrac{Ze^2 m_e r}{4\pi \epsilon_0}=(m_e vr)^2 \label{1.8.14} \] \(m_e vr\) 1.8.1
\(m_e vr\) \[m_e vr=n\hbar\label{1.8.15} \] with \(n=1,2,3,...\). Note that the electron must be in motion, so \(n=0\) is not allowed. Substituting Equation \(\ref{1.8.15}\) into the Equation \(\ref{1.8.14}\), we find \[\dfrac{Ze^2 m_e r}{4\pi \epsilon_0}=n^2 (\hbar)^2 \label{1.8.16} \] \[\begin{align}r_n &= \dfrac{4\pi \epsilon_0 \hbar^2}{Ze^2 m_e}n^2 \\ &=\dfrac{a_0}{Z}n^2 \label{1.8.16B} \end{align} \] with \(n=1,2,3,...\). T \(a_0\) \[a_0=\dfrac{4\pi \epsilon_0 \hbar^2}{e^2 m_e} \label{1.8.17} \] e We can also calculate the allowed momenta since \(m_e vr=n\hbar\), and \(p=m_e v\). Thus, \[\begin{align}p_n r_n &=n\hbar\\[4pt] p_n &=\dfrac{n\hbar}{r_n}\\[4pt] &=\dfrac{\hbar Z}{a_0 n} \\[4pt] &= \dfrac{Ze^2 m_e}{4\pi \epsilon_0 \hbar n}\end{align} \label{1.8.18} \] \(p_n\) \(r_n\) \[E_n=\dfrac{p^2_n}{2m_e}-\dfrac{Ze^2}{4\pi \epsilon_0 r_n} \label{1.8.19} \] \(p_n\) \(r_n\) \[E_n=-\dfrac{Z^2 e^4 m_e}{32\pi^2 \epsilon_{0}^{2}\hbar^2}\dfrac{1}{n^2}=-\dfrac{e^4 m_e}{8 \epsilon_{0}^{2}h^2}\dfrac{Z^2}{n^2} \label{1.8.20} \] W \[1 \ Ry = \dfrac{e^4 m_e}{8\epsilon_{0}^{2} h^2} =2.18\times 10^{-18} \ J. \nonumber \] \[E_n=-(2.18\times 10^{-18})\dfrac{Z^2}{n^2} \ J=-\dfrac{Z^2}{n^2} \ R_y \label{1.8.21} \] Hence, the energy of the electron in an atom also is quantized. Equation \(\ref{1.8.21}\) gives the energies of the electronic states of the hydrogen atom. It is very useful in analyzing spectra to represent these energies graphically in an energy-level diagram. An energy-level diagram has energy plotted on the vertical axis with a horizontal line drawn to locate each energy level 1.8.4
. These turn out to be the correct energy levels, apart from small corrections that cannot be accounted for in this pseudo-classical treatment. Despite the fact that the energies are essentially correct, the Bohr model masks the true quantum nature of the electron, which only emerges from a fully quantum mechanical analysis. Calculate a value for the Bohr radius using Equation \(\ref{1.8.16}\) to check that this equation is consistent with the value 52.9 pm. What would the radius be for \(n = 1\) in the \(\ce{Li^{2+}}\) ion? Starting from Equation \ref{1.8.16} and solving for \(r\): \[ \begin{align*} \dfrac{Ze^2m_er}{4πϵ_0} &=n^2ℏ^2 \\ r &=\dfrac{4 n^2 \hbar^2 πϵ_0}{Z e^2 m_e} \end{align*} \nonumber \] with For the ground-state of the hydrogen atom: \(Z=1\) and \(n=1\). \[ \begin{align*} r &=\dfrac{4 \hbar^2 πϵ_0}{e^2m_e} \\ &= \dfrac{4 (1.0546 \times 10^{-34}m^2kg/s)^2 \times π \times 8.854 \times 10^{-12}C^2N^{-1}m^{-2}}{(1.60217662 \times 10^{-19}C)^2(9.10938356 \times 10^{-31}kg)} \\ &=5.29 \times 10^{-11}m = 52.9\, pm\end{align*} \nonumber \] For the ground-state of the lithium +2 ion: \(Z=3\) and \(n=1\) \[ \begin{align*} r &=\dfrac{4 \hbar^2 πϵ_0}{3 e^2m_e} \\ &= \dfrac{4 (1.0546 \times 10^{-34}m^2kg/s)^2 \times π \times 8.854\times10^{-12}C^2N^{-1}m^{-2}}{3(1.60217662 \times 10^{-19}C)^2(9.10938356 \times 10^{-31}kg)} \\ &=1.76 \times 10^{-11}m = 17.6 \,pm\end{align*} \nonumber \] As expected, the \(\ce{Li^{2+}}\) has a smaller radius than the \(\ce{H}\) atoms because of the increased nuclear charge. How do the radii of the hydrogen orbits vary with \(n\)? Prepare a graph showing \(r\) as a function of \(n\). States of hydrogen atoms with \(n = 200\) have been prepared (called Rydberg states). What is the diameter of the atoms in these states? This is a straightforward application of Equation of \ref{1.8.16B}. The hydrogen atom has only certain allowable radii and these radii can be predicted from the equation that relates them with each \(n\). Note that the electron must be in motion so \(n = 0\) is not allowed. \(4 \pi \epsilon_{0}=1.113 \times 10^{-10} \mathrm{C}^{2} \mathrm{J}^{-1} \mathrm{m}^{-1}\) and \(\hbar=1.054 \times 10^{-34} \mathrm{J} \mathrm{s},\) also knowing \[\begin{aligned}
e &=1.602 \times 10^{-19} \mathrm{C} \text { with } \\
m_{e} &=9.109 \times 10^{-31} \mathrm{kg}
\end{aligned} \nonumber \] and \(Z\) is the nuclear charge, we use this equation directly. A simplification can be made by taking advantage of the fact that \[a_{0}=\frac{4 \pi \epsilon_{0} \hbar^{2}}{e^{2} m_{e}} \nonumber \] resulting in \[r_{n}=\frac{a_{0}}{Z} n^{2} \nonumber \] where \(a_{0}=5.292 \times 10^{-11} \mathrm{m}\) which is the Bohr Radius. Suppose we want to find the radius where \(n=200 . n^{2}=40000\) so plugging in directly we have \[ \begin{align*} r_{n} &=\frac{\left(5.292 \times 10^{-11}\right)}{(1)}(40000) \\[4pt] &=2.117 \times 10^{-6} m \end{align*} \nonumber \] for the radius of a hydrogen atom with an electron excited to the \(\mathrm{n}=200\) state. The diameter is then \(4.234 \times 10^{-6} \mathrm{m}\). The above discussion is based off of a classical picture of an orbiting electron with the quantization from the angular momentum (Equation \(\ref{1.8.15}\)) requirement lifted from Planck's quantization arguments. Hence, only allows certain trajectories are stable (with differing radii). However, as discussed previously, the electron will have a wavelike property also with a de Broglie wavelength \(\lambda\) \[\lambda = \dfrac{h}{p} \nonumber \] Hence, a larger momentum \(p\) implies a shorter wavelength. That means as \(n\) increases (Equation \(\ref{1.8.21}\)), the wavelength must also increase; this is a common feature in quantum mechanics and will be often observed. In the Bohr atom, the circular symmetry and the wave property of the electron requires that the electron waves have an integer number of wavelengths (Figure \(\Page {1A}\)). If not, then the waves will overlap imperfectly and cancel out (i.e., the electron will cease to exist) as demonstrated in Figure \(\Page {1B}\). A more detailed discussion of the effect of electron waves in atoms will be discuss in the following chapters. Given a prediction of the allowed energies of a system, how could we go about verifying them? The general experimental technique known as permits us to probe the various differences between the allowed energies. Thus, if the prediction of the actual energies, themselves, is correct, we should also be able to predict these differences. Let us assume that we are able to place the electron in Bohr's hydrogen atom into an energy state \(E_n\) for \(n>1\), i.e. one of its so-called . The electron will rapidly return to its lowest energy state, known as the and, in doing so, emit light. The energy carried away by the light is determined by the condition that the total energy is conserved (Figure 1.8.6
). Thus, if \(n_i\) is the integer that characterizes the initial (excited) state of the electron, and \(n_f\) is the final state (here we imagine that \(n_f =1\), but is applicable in cases that \(n_f <n_i\), i.e., emission) \[E_{nf}=E_{ni}-h\nu \label{1.8.22} \] or \[\nu=\dfrac{E_{ni}-E_{nf}}{h}=\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2} h^3}\left ( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right ) \label{1.8.23} \] We can now identify the Rydberg constant \(R_H\) with the ratio of constants on the right hand side of Equation \(\ref{1.8.23}\) \[ R_H = \dfrac {m_ee^4}{8 \epsilon ^2_0 h^3 } \label {2-22} \] Evaluating \(R_H\) from the fundamental constants in this formula gives a value within 0.5% of that obtained experimentally from the hydrogen atom spectrum. Thus, by observing the emitted light, we can determine the energy difference between the initial and final energy levels, which results in the emission spectra discussed in Sections 1.4 and 1.5. Different values of \(n_f\) determine which emission spectrum is observed, and the examples shown in the figure are named after the individuals who first observed them. The figure below shows some of the transitions possible for different \(n_f\) and \(n_i\) values discussed previously. If the atom absorbs light it ends up in an excited state as a result of the absorption. The absorption is only possible for light of certain frequencies, and again, conservation of energy determines what these frequencies are. If light is absorbed, then the final energy \(E_{nf}\) will be related to the initial energy \(E_{ni}\) with \(n_f >n_i\) by \[E_{nf}=E_{ni}+h\nu \label{1.8.24} \] \[\nu=\dfrac{E_{nf}-E_{ni}}{h}=\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2}h^3}\left ( \dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}}\right ) \label{1.8.25} \] a: \[\begin{align*}
E_{\text{nf}} &= E_{ni} - h\nu \\
E_{photon} = h\nu &= E_{nf} - E_{ni}\\
&= \frac{Z^2e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\\
&=\frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{1^2} - \frac{1}{4^2}\right)\\
&=2.18 \times 10^{-18}\left(1 - \frac{1}{16} \right)\\
&=2.04 \times 10^{-18} J
\end{align*} \nonumber \] b: As \(n_i \rightarrow \infty\) \[\begin{align*}
E_{photon} &= \frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\\
\frac{1}{n_i^2} &\rightarrow 0\\
E_{photon} &\rightarrow \frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2}\right)\\
\end{align*} \nonumber \] Bohr’s proposal explained the hydrogen atom spectrum, the origin of the Rydberg formula, and the value of the Rydberg constant. Specifically it demonstrated that the integers in the Rydberg formula are a manifestation of quantization. The energy, the angular momentum, and the radius of the orbiting electron all are quantized. This quantization also parallels the concept of stable orbits in the Bohr model. Only certain values of \(E\), \(M\), and \(r\) are possible, and therefore the electron cannot collapse onto the nucleus by continuously radiating energy because it can only have certain energies, and it cannot be in certain regions of space. The electron can only jump from one orbit (quantum state) to another. The quantization means that the orbits are stable, and the electron cannot spiral into the nucleus in spite of the attractive Coulomb force. Although Bohr’s ideas successfully explained the hydrogen spectrum, they failed when applied to the spectra of other atoms. In addition a profound question remained. Why is angular momentum quantized in units of \(\hbar\)? As we shall see, de Broglie had an answer to this question, and this answer led Schrödinger to a general postulate that produces the quantization of angular momentum as a consequence. This quantization is not quite as simple as proposed by Bohr, and we will see that it is not possible to determine the distance of the electron from the nucleus as precisely as Bohr thought. In fact, since the position of the electron in the hydrogen atom is not at all as well defined as a classical orbit (such as the moon orbiting the earth) it is called an orbital. An electron orbital represents or describes the position of the electron around the nucleus in terms of a mathematical function called a wavefunction that yields the of positions of the electron. ( ) ") | 18,009 | 1,762 |
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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 | 1,763 |
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Circular Dichroism, an absorption spectroscopy, uses circularly polarized light to investigate structural aspects of optically active chiral media. It is mostly used to study biological molecules, their structure, and interactions with metals and other molecules. Circular Dichroism (CD) is an absorption spectroscopy method based on the differential absorption of left and right circularly polarized light. Optically active chiral molecules will preferentially absorb one direction of the circularly polarized light. The difference in absorption of the left and right circularly polarized light can be measured and quantified. UV CD is used to determine aspects of protein secondary structure. Vibrational CD, IR CD, is used to study the structure of small organic molecules, proteins and DNA. UV/Vis CD investigates charge transfer transitions in metal-protein complexes. Electromagnetic radiation consists of oscillating electric and magnetic fields perpendicular to each other and the direction of propagation. Most light sources emit waves where these fields oscillate in all directions perpendicular to the propagation vector. Linear polarized light occurs when the electric field vector oscillates in only one plane. In circularly polarized light, the electric field vector rotates around the propagation axis maintaining a constant magnitude. When looked at down the axis of propagation the vector appears to trace a circle over the period of one wave frequency (one full rotation occurs in the distance equal to the wavelength). In linear polarized light the direction of the vector stays constant and the magnitude oscillates. In circularly polarized light the magnitude stays constant while the direction oscillates. As the radiation propagates the electric field vector traces out a helix. The magnetic field vector is out of phase with the electric field vector by a quarter turn. When traced together the vectors form a double helix.
Light can be circularly polarized in two directions: left and right. If the vector rotates counterclockwise when the observer looks down the axis of propagation, the light is left circularly polarized (LCP). If it rotates clockwise, it is right circularly polarized (RCP). If LCP and RCP of the same amplitude, they are superimposed on one another and the resulting wave will be linearly polarized. As with linear polarized light, circularly polarized light can be absorbed by a medium. An optically active chiral compound will absorb the two directions of circularly polarized light by different amounts \[ \Delta A = A_l - A_r \] The CD spectrum is often reported in degrees of ellipticity, \(\theta\), which is a measure of the ellipticity of the polarization given by:
\[tan \theta = \frac{E_l-E_r}{E_l+E_r}\] where E is the magnitude of the electric field vector. The change in polarization is usually small and the signal is often measured in radians where \(\theta = \frac{2.303}{4}(A_l - A_r)\) and is a function of wavelength. \(\theta\) can be converted to degrees by multiplying by \(\frac{180}{\pi}\) which gives \(\theta = 32.98 \Delta A\)
The historical reported unit of CD experiments is molar ellipticity, \([\theta]\), which removes the dependence on concentration and path length \[[\theta] = 3298 \Delta \varepsilon\] where the 3298 converts from the units of molar absorptivity to the historical units of degrees\(\cdot\) cm \(\cdot\)dmol . Most commercial CD instruments are based on the modulation techniques introduced by Grosjean and Legrand. Light is linearly polarized and passed through a monochromator. The single wavelength light is then passed through a modulating device, usually a photoelastic modulator (PEM), which transforms the linear light to circular polarized light. The incident light on the sample switches between LCP and RCP light. As the incident light swtches direction of polarization the absorption changes and the differention molar absorptivity can be calculated. The most widely used application of CD spectroscopy is identifying structural aspects of proteins and DNA. The peptide bonds in proteins are optically active and the ellipticity they exhibit changes based on the local conformation of the molecule. Secondary structures of proteins can be analyzed using the far-UV (190-250 nm) region of light. The ordered \(\alpha\)-helices, \(\beta\)-sheets, \(\beta\)-turn, and random coil conformations all have characteristic spectra. These unique spectra form the basis for protein secondary structure analysis. It should be noted that in CD only the relative fractions of residues in each conformation can be determined but not specifically where each structural feature lies in the molecule. Some information about the tertiary structure of proteins can be determined using near-UV spectroscopy. Absorptions between 250-300 nm are due to the dipole orientation and surrounding environment of the aromatic amino acids, phenylalanine, tyrosine, and tryptophan, and cysteine residues which can form disulfide bonds. Near-UV techniques can also be used to provide structural information about the binding of prosthetic groups in proteins. Vibrational CD (VCD) spectroscopy uses IR light to determine 3D structures of short peptides, nucleic acids, and carbohydrates. VCD has been used to show the shape and number of helices in A-, B-, and Z-DNA. VCD is still a relatively new technique and has the potential to be a very powerful tool. Resolving the spectra requires extensive calculations, as well as, high concentrations and must be performed in water, which may force the molecule into a nonnative conformation. | 5,641 | 1,764 |
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According to the scale of oxidation levels established for carbon (see Table 11-1), primary alcohols \(\left( \ce{RCH_2OH} \right)\) are at a lower oxidation level than either aldehydes \(\left( \ce{RCHO} \right)\) or carboxylic acids \(\left( \ce{RCO_2H} \right)\). With suitable oxidizing agents, primary alcohols in fact can be oxidized first to aldehydes and then to carboxylic acids. Unlike the reactions discussed previously in this chapter, oxidation of alcohols involves the alkyl portion of the molecule, or more specifically, the \(\ce{C-H}\) bonds of the hydroxyl-bearing carbon (the \(\alpha\) carbon). Secondary alcohols, which have only one such \(\ce{C-H}\) bond, are oxidized to ketones, whereas tertiary alcohols, which have no \(\ce{C-H}\) bonds to the hydroxylic carbon, are oxidized only with accompanying degradation into smaller fragments by cleavage of carbon-carbon bonds. Conversion of ethanol to ethanal is carried out on a commercial scale by passing gaseous ethanol over a copper catalyst at \(300^\text{o}\): At room temperature this reaction is endothermic with an equilibrium constant of about \(10^{22}\). At \(300^\text{o}\) conversions of \(20\%\)-\(50\%\) per pass can be realized and, by recycling the unreacted alcohol, the yield can be greater than \(90\%\). Another commercial process uses a silver catalyst and oxygen to combine with the hydrogen, which makes the net reaction substantially exothermic: In effect, this is a partial combustion reaction and requires very careful control to prevent overoxidation. In fact, by modifying the reaction conditions (alcohol-to-oxygen ratio, temperature, pressure, and reaction time), the oxidation proceeds smoothly to ethanoic acid: Reactions of this type are particularly suitable as industrial processes because they generally can be run in continuous-flow reactors, and can utilize a cheap oxidizing agent, usually supplied directly as air. Laboratory oxidation of alcohols most often is carried out with chromic acid \(\left( \ce{H_2CrO_4} \right)\), which usually is prepared as required from chromic oxide \(\left( \ce{CrO_3} \right)\) or from sodium dichromate \(\left( \ce{Na_2Cr_2O_7} \right)\) in combination with sulfuric acid. Ethanoic (acetic) acid is a useful solvent for such reactions: The mechanism of the chromic acid oxidation of 2-propanol to 2-propanone (acetone) has been investigated very thoroughly. It is a highly interesting reaction in that it reveals how changes of oxidation state can occur in a reaction involving a typical inorganic and a typical organic compound. The initial step is reversible formation of an isopropyl ester of chromic acid. This ester is unstable and is not isolated: The subsequent step is the slowest in the sequence and appears to involve attack of a base (water) at the alpha hydrogen of the chromate ester concurrent with elimination of the \(\ce{HCrO_3^-}\) group. There is an obvious analogy between this step and an \(E2\) reaction ( ): The transformation of chromic acid \(\left( \ce{H_2CrO_4} \right)\) to \(\ce{H_2CrO_3}\) amounts to the reduction of chromium from an oxidation state of \(+6\) to \(+4\). Disproportionation of Cr(IV) occurs rapidly to give compounds of Cr(III) and Cr(VI): or The \(E2\) character of the ketone-forming step has been demonstrated in two ways. First, the rate of decomposition of isopropyl hydrogen chromate to 2-propanone and \(\ce{H_2CrO_3}\) is strongly accelerated by efficient proton-removing substances. Second, the hydrogen on the \(\alpha\) carbon clearly is removed in a slow reaction because . No significant slowing of oxidation is noted for 2-propanol having deuterium in the methyl groups: Carbon-deuterium bonds normally are broken more slowly than carbon-hydrogen bonds. This so-called provides a general method for determining whether particular carbon-hydrogen bonds are broken in slow reaction steps. Primary alcohols are oxidized by chromic acid in sulfuric acid solution to aldehydes, but to stop the reaction at the aldehyde stage, it usually is necessary to remove the aldehyde from the reaction mixture as it forms. This can be done by distillation if the aldehyde is reasonably volatile: Unsaturated alcohols can be oxidized to unsaturated ketones by chromic acid, because chromic acid usually attacks double bonds relatively slowly: However, complications are to be expected when the double bond of an unsaturated alcohol is particularly reactive or when the alcohol rearranges readily under strongly acidic conditions. It is possible to avoid the use of strong acid through the combination of chromic oxide with the weak base azabenzene (pyridine). A crystalline solid of composition \(\ce{(C_5H_5N)_2} \cdot \ce{CrO_3}\) is formed when \(\ce{CrO_3}\) is added to excess pyridine at low temperatures. (Addition of pyridine to \(\ce{CrO_3}\) is likely to give an uncontrollable reaction resulting in a fire.) The pyridine-\(\ce{CrO_3}\) reagent is soluble in chlorinated solvents such as dichloromethane, and the resulting solutions rapidly oxidize to at ordinary temperatures: The yields usually are good, partly because the absence of strong acid minimizes degradation and rearrangement, and partly because the product can be isolated easily. The inorganic products are insoluble and can be separated by filtration, thereby leaving the oxidized product in dichloromethane from which it can be easily recovered. Permanganate ion, \(\ce{MnO_4^-}\), oxidizes both primary and secondary alcohols in either basic or acidic solution. With primary alcohols the product normally is the carboxylic acid because the intermediate aldehyde is oxidized rapidly by permanganate: Oxidation under conditions evidently involves the alkoxide ion rather than the neutral alcohol. The oxidizing agent, \(\ce{MnO_4^-}\), abstracts the alpha hydrogen from the alkoxide ion either as an atom (one-electron transfer) or as hydride, \(\ce{H}^\ominus\) (two-electron transfer). The steps for the two-electron sequence are: In the second step, permanganate ion is reduced from Mn(VII) to Mn(V). However, the stable oxidation states of manganese are \(+2\), \(+4\), and \(+7\); thus the Mn(V) ion formed disproportionates to Mn(VII) and Mn(IV). The normal manganese end product from oxidations in basic solution is manganese dioxide, \(\ce{MnO_2}\), in which \(\ce{Mn}\) has an oxidation state of \(+4\) corresponding to Mn(IV). In we described the use of permanganate for the oxidation of alkenes to 1,2-diols. How is it possible to control this reaction so that it will stop at the diol stage when permanganate also can oxidize to ? Overoxidation with permanganate is always a problem, but the relative reaction rates are very much a function of the pH of the reaction mixture and, in basic solution, potassium permanganate oxidizes unsaturated groups more rapidly than it oxidizes alcohols: There are many biological oxidations that convert a primary or secondary alcohol to a carbonyl compound. These reactions cannot possibly involve the extreme pH conditions and vigorous inorganic oxidants used in typical laboratory oxidations. Rather, they occur at nearly neutral pH values and they all require enzymes as catalysts, which for these reactions usually are called . An important group of biological oxidizing agents includes the pyridine nucleotides, of which nicotinamide adenine dinucleotide (\(\ce{NAD}^\oplus\), \(13\)) is an example: This very complex molecule functions to accept hydride \(\left( \ce{H}^\ominus \right)\) or the equivalent \(\left( \ce{H}^\oplus + 2 \ce{e^-} \right)\) from the \(\alpha\) carbon of an alcohol: The reduced form of \(\ce{NAD}^\oplus\) is abbreviated as \(\ce{NADH}\) and the \(\ce{H}^\ominus\) is added at the 4-position of the pyridine ring: Some examples follow that illustrate the remarkable specificity of this kind of redox system. One of the last steps in the metabolic breakdown of glucose (glycolysis; ) is the reduction of 2-oxopropanoic (pyruvic) acid to \(L\)-2-hydroxypropanoic (lactic) acid. The reverse process is oxidation of \(L\)-lactic acid. The enzyme catalyses this reaction, and it functions only with the \(L\)-enantiomer of lactic acid: A second example, provided by one of the steps in metabolism by way of the Krebs citric acid cycle (see ), is the oxidation of \(L\) -2-hydroxy-butanedioic (\(L\)-malic) acid to 2-oxobutanedioic (oxaloacetic) acid. This enzyme functions only with \(L\)-malic acid: All of these reactions release energy. In biological oxidations much of the energy is utilized to form ATP from ADP and inorganic phosphate ( ). That is to say, electron-transfer reactions are coupled with ATP formation. The overall process is called . Another important oxidizing agent in biological systems is flavin adenine dinucleotide, \(\ce{FAD}\). Like \(\ce{NAD}^\oplus\), it is a two-electron acceptor, but unlike \(\ce{NAD}^\oplus\), it accepts two electrons as \(2 \ce{H} \cdot\) rather than as \(\ce{H}^\ominus\). The reduced form, \(\ce{FADH_2}\), has the hydrogens at ring nitrogens: and (1977) | 9,095 | 1,766 |
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As has been seen in previous chapters, may important thermochemical quantities can be expressed in terms of partial derivatives. Two important examples are the molar heat capacities \(C_p\) and \(C_V\) which can be expressed as \[ C_p = \left(\dfrac{\partial H}{\partial T}\right)_p \nonumber \] and \[ C_V = \left(\dfrac{\partial U}{\partial T}\right)_V \nonumber \] These are properties that can be measured experimentally and tabulated for many substances. These quantities can be used to calculate changes in quantities since they represent the slope of a surface (\(H\) or \(U\)) in the direction of the specified path (constant \(p\) or \(V\)). This allows us to use the following kinds of relationships: \[ dH = \left(\dfrac{\partial H}{\partial T}\right)_p dT \nonumber \] and \[ \Delta H = \int \left(\dfrac{\partial H}{\partial T}\right)_p dT \nonumber \] Because thermodynamics is kind enough to deal in a number of , the functions that define how those variable change must behave according to some very well determined mathematics. This is the true power of thermodynamics! | 1,100 | 1,769 |
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Because a crystalline solid consists of repeating patterns of its components in three dimensions (a crystal lattice), we can represent the entire crystal by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cell. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. In this section, we describe the arrangements of atoms in various unit cells. Unit cells are easiest to visualize in two dimensions. In many cases, more than one unit cell can be used to represent a given structure, as shown for the Escher drawing in the chapter opener and for a two-dimensional crystal lattice in Figure 12.2. Usually the smallest unit cell that completely describes the order is chosen. The only requirement for a valid unit cell is that repeating it in space must produce the regular lattice. Thus the unit cell in part (d) in Figure 12.2 is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in Figure 12.3. There are seven fundamentally different kinds of unit cells, which differ in the relative lengths of the edges and the angles between them (Figure 12.4). Each unit cell has six sides, and each side is a parallelogram. We focus primarily on the cubic unit cells, in which all sides have the same length and all angles are 90°, but the concepts that we introduce also apply to substances whose unit cells are not cubic. If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called simple cubic (part (a) in Figure 12.5). If the unit cell also contains an identical component in the center of the cube, then it is body-centered cubic (bcc) (part (b) in Figure 12.5). If there are components in the center of each face in addition to those at the corners of the cube, then the unit cell is face-centered cubic (fcc) (part (c) in Figure 12.5). As indicated in Figure 12.5, a solid consists of a large number of unit cells arrayed in three dimensions. Any intensive property of the bulk material, such as its density, must therefore also be related to its unit cell. Because density is the mass of substance per unit volume, we can calculate the density of the bulk material from the density of a single unit cell. To do this, we need to know the size of the unit cell (to obtain its volume), the molar mass of its components, and the number of components per unit cell. When we count atoms or ions in a unit cell, however, those lying on a face, an edge, or a corner contribute to more than one unit cell, as shown in Figure 12.5. For example, an atom that lies on a face of a unit cell is shared by two adjacent unit cells and is therefore counted as 12 atom per unit cell. Similarly, an atom that lies on the edge of a unit cell is shared by four adjacent unit cells, so it contributes 14 atom to each. An atom at a corner of a unit cell is shared by all eight adjacent unit cells and therefore contributes 18 atom to each.The statement that atoms lying on an edge or a corner of a unit cell count as 14 or 18 atom per unit cell, respectively, is true for all unit cells except the hexagonal one, in which three unit cells share each vertical edge and six share each corner (Figure 12.4), leading to values of 13 and 16 atom per unit cell, respectively, for atoms in these positions. In contrast, atoms that lie entirely within a unit cell, such as the atom in the center of a body-centered cubic unit cell, belong to only that one unit cell. Metallic gold has a face-centered cubic unit cell (part (c) in Figure 12.5). How many Au atoms are in each unit cell? : unit cell : number of atoms per unit cell : Using Figure 12.5, identify the positions of the Au atoms in a face-centered cubic unit cell and then determine how much each Au atom contributes to the unit cell. Add the contributions of all the Au atoms to obtain the total number of Au atoms in a unit cell. : As shown in Figure 12.5, a face-centered cubic unit cell has eight atoms at the corners of the cube and six atoms on the faces. Because atoms on a face are shared by two unit cells, each counts as \({1 \over 2}\) atom per unit cell, giving 6×\({1 \over 2}\)=3 Au atoms per unit cell. Atoms on a corner are shared by eight unit cells and hence contribute only \({1 \over 8}\) atom per unit cell, giving 8×\({1 \over 8}\) =1 Au atom per unit cell. The total number of Au atoms in each unit cell is thus 3 + 1 = 4. Metallic iron has a body-centered cubic unit cell (part (b) in Figure 12.5). How many Fe atoms are in each unit cell? : two Now that we know how to count atoms in unit cells, we can use unit cells to calculate the densities of simple compounds. Note, however, that we are assuming a solid consists of a perfect regular array of unit cells, whereas real substances contain impurities and defects that affect many of their bulk properties, including density. Consequently, the results of our calculations will be close but not necessarily identical to the experimentally obtained values. Calculate the density of metallic iron, which has a body-centered cubic unit cell (part (b) in Figure 12.5) with an edge length of 286.6 pm. : unit cell and edge length : density : : We know from Example 1 that each unit cell of metallic iron contains two Fe atoms. The molar mass of iron is 55.85 g/mol. Because density is mass per unit volume, we need to calculate the mass of the iron atoms in the unit cell from the molar mass and Avogadro’s number and then divide the mass by the volume of the cell (making sure to use suitable units to get density in g/cm ): \[ mass \; of \; Fe=\left ( 2 \; \cancel{atoms} \; Fe \right )\left ( \dfrac{ 1 \; \cancel{mol}}{6.022\times 10^{23} \; \cancel{atoms}} \right )\left ( \dfrac{55.85 \; g}{\cancel{mol}} \right ) =1.855\times 10^{-22} \; g \] \[ volume=\left [ \left ( 286.6 \; pm \right )\left ( \dfrac{10^{-12 }\; \cancel{m}}{\cancel{pm}} \right )\left ( \dfrac{10^{2} \; cm}{\cancel{m}} \right ) \right ] =2.345\times 10^{-23} \; cm^{3} \] \[ density = \dfrac{1.855\times 10^{-22} \; g}{2.345\times 10^{-23} \; cm^{3}} = 7.880 g/cm^{3} \] Calculate the density of gold, which has a face-centered cubic unit cell (part (c) in Figure 12.5) with an edge length of 407.8 pm. : 19.29 g/cm Our discussion of the three-dimensional structures of solids has considered only substances in which all the components are identical. As we shall see, such substances can be viewed as consisting of identical spheres packed together in space; the way the components are packed together produces the different unit cells. Most of the substances with structures of this type are metals. The arrangement of the atoms in a solid that has a simple cubic unit cell was shown in part (a) in Figure 12.5. Each atom in the lattice has only six nearest neighbors in an octahedral arrangement. Consequently, the simple cubic lattice is an inefficient way to pack atoms together in space: only 52% of the total space is filled by the atoms. The only element that crystallizes in a simple cubic unit cell is polonium. Simple cubic unit cells are, however, common among binary ionic compounds, where each cation is surrounded by six anions and vice versa. The body-centered cubic unit cell is a more efficient way to pack spheres together and is much more common among pure elements. Each atom has eight nearest neighbors in the unit cell, and 68% of the volume is occupied by the atoms. As shown in part (b) in Figure 12.5, the body-centered cubic structure consists of a single layer of spheres in contact with each other and aligned so that their centers are at the corners of a square; a second layer of spheres occupies the square-shaped “holes” above the spheres in the first layer. The third layer of spheres occupies the square holes formed by the second layer, so that each lies directly above a sphere in the first layer, and so forth. All the alkali metals, barium, radium, and several of the transition metals have body-centered cubic structures. The most efficient way to pack spheres is the close-packed arrangement, which has two variants. A single layer of close-packed spheres is shown in part (a) in Figure 12.6. Each sphere is surrounded by six others in the same plane to produce a hexagonal arrangement. Above any set of seven spheres are six depressions arranged in a hexagon. In principle, all six sites are the same, and any one of them could be occupied by an atom in the next layer. Actually, however, these six sites can be divided into two sets, labeled B and C in part (a) in Figure 12.6. Sites B and C differ because as soon as we place a sphere at a B position, we can no longer place a sphere in any of the three C positions adjacent to A and vice versa. If we place the second layer of spheres at the B positions in part (a) in Figure 12.6, we obtain the two-layered structure shown in part (b) in Figure 12.6. There are now two alternatives for placing the first atom of the third layer: we can place it directly over one of the atoms in the first layer (an A position) or at one of the C positions, corresponding to the positions that we did not use for the atoms in the first or second layers (part (c) in Figure 12.6). If we choose the first arrangement and repeat the pattern in succeeding layers, the positions of the atoms alternate from layer to layer in the pattern ABABAB…, resulting in a hexagonal close-packed (hcp) structure (part (a) in Figure 12.7). If we choose the second arrangement and repeat the pattern indefinitely, the positions of the atoms alternate as ABCABC…, giving a cubic close-packed (ccp) structure (part (b) in Figure 12.7). Because the ccp structure contains hexagonally packed layers, it does not look particularly cubic. As shown in part (b) in Figure 12.7, however, simply rotating the structure reveals its cubic nature, which is identical to a fcc structure. The hcp and ccp structures differ only in the way their layers are stacked. Both structures have an overall packing efficiency of 74%, and in both each atom has 12 nearest neighbors (6 in the same plane plus 3 in each of the planes immediately above and below). Table 12.1 compares the packing efficiency and the number of nearest neighbors for the different cubic and close-packed structures; the number of nearest neighbors is called the coordination number. Most metals have hcp, ccp, or bcc structures, although several metals exhibit both hcp and ccp structures, depending on temperature and pressure. Properties of the Common Structures of Metals The smallest repeating unit of a crystal lattice is the unit cell. The simple cubic unit cell contains only eight atoms, molecules, or ions at the corners of a cube. A body-centered cubic (bcc) unit cell contains one additional component in the center of the cube. A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. Simple cubic and bcc arrangements fill only 52% and 68% of the available space with atoms, respectively. The hexagonal close-packed (hcp) structure has an ABABAB… repeating arrangement, and the cubic close-packed (ccp) structure has an ABCABC… repeating pattern; the latter is identical to an fcc lattice. The hcp and ccp arrangements fill 74% of the available space and have a coordination number of 12 for each atom in the lattice, the number of nearest neighbors. The simple cubic and bcc lattices have coordination numbers of 6 and 8, respectively. A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure. 1. Why is it valid to represent the structure of a crystalline solid by the structure of its unit cell? What are the most important constraints in selecting a unit cell? 2. All unit cell structures have six sides. Can crystals of a solid have more than six sides? Explain your answer. 3. Explain how the intensive properties of a material are reflected in the unit cell. Are all the properties of a bulk material the same as those of its unit cell? Explain your answer. 4. The experimentally measured density of a bulk material is slightly higher than expected based on the structure of the pure material. Propose two explanations for this observation. 5. The experimentally determined density of a material is lower than expected based on the arrangement of the atoms in the unit cell, the formula mass, and the size of the atoms. What conclusion(s) can you draw about the material? 6. Only one element (polonium) crystallizes with a simple cubic unit cell. Why is polonium the only example of an element with this structure? 7. What is meant by the term coordination number in the structure of a solid? How does the coordination number depend on the structure of the metal? 8. Arrange the three types of cubic unit cells in order of increasing packing efficiency. What is the difference in packing efficiency between the hcp structure and the ccp structure? 9. The structures of many metals depend on pressure and temperature. Which structure—bcc or hcp—would be more likely in a given metal at very high pressures? Explain your reasoning. 10. A metal has two crystalline phases. The transition temperature, the temperature at which one phase is converted to the other, is 95°C at 1 atm and 135°C at 1000 atm. Sketch a phase diagram for this substance. The metal is known to have either a ccp structure or a simple cubic structure. Label the regions in your diagram appropriately and justify your selection for the structure of each phase. 1. Metallic rhodium has an fcc unit cell. How many atoms of rhodium does each unit cell contain? 2. Chromium has a structure with two atoms per unit cell. Is the structure of this metal simple cubic, bcc, fcc, or hcp? 3. The density of nickel is 8.908 g/cm . If the metallic radius of nickel is 125 pm, what is the structure of metallic nickel? 4. The density of tungsten is 19.3 g/cm . If the metallic radius of tungsten is 139 pm, what is the structure of metallic tungsten? 5. An element has a density of 10.25 g/cm and a metallic radius of 136.3 pm. The metal crystallizes in a bcc lattice. Identify the element. 6. A 21.64 g sample of a nonreactive metal is placed in a flask containing 12.00 mL of water; the final volume is 13.81 mL. If the length of the edge of the unit cell is 387 pm and the metallic radius is 137 pm, determine the packing arrangement and identify the element. 7. A sample of an alkali metal that has a bcc unit cell is found to have a mass of 1.000 g and a volume of 1.0298 cm . When the metal reacts with excess water, the reaction produces 539.29 mL of hydrogen gas at 0.980 atm and 23°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers. 8. A sample of an alkaline earth metal that has a bcc unit cell is found to have a mass 5.000 g and a volume of 1.392 cm . Complete reaction with chlorine gas requires 848.3 mL of chlorine gas at 1.050 atm and 25°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers. 9. Lithium crystallizes in a bcc structure with an edge length of 3.509 Å. Calculate its density. What is the approximate metallic radius of lithium in picometers? 10. Vanadium is used in the manufacture of rust-resistant vanadium steel. It forms bcc crystals with a density of 6.11 g/cm at 18.7°C. What is the length of the edge of the unit cell? What is the approximate metallic radius of the vanadium in picometers? 11. A simple cubic cell contains one metal atom with a metallic radius of 100 pm. a. Determine the volume of the atom(s) contained in one unit cell [the volume of a sphere = (\({4 \over 3} \))πr ]. b. What is the length of one edge of the unit cell? (Hint: there is no empty space between atoms.) c. Calculate the volume of the unit cell. d. Determine the packing efficiency for this structure. e. Use the steps in Problem 11 to calculate the packing efficiency for a bcc unit cell with a metallic radius of 1.00 Å. 1. four 3. fcc 5. molybdenum 7. sodium, unit cell edge = 428 pm, r = 185 pm 9. d = 0.5335 g/cm , r =151.9 pm | 16,471 | 1,770 |
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According to the scale of oxidation levels established for carbon (see Table 11-1), primary alcohols \(\left( \ce{RCH_2OH} \right)\) are at a lower oxidation level than either aldehydes \(\left( \ce{RCHO} \right)\) or carboxylic acids \(\left( \ce{RCO_2H} \right)\). With suitable oxidizing agents, primary alcohols in fact can be oxidized first to aldehydes and then to carboxylic acids. Unlike the reactions discussed previously in this chapter, oxidation of alcohols involves the alkyl portion of the molecule, or more specifically, the \(\ce{C-H}\) bonds of the hydroxyl-bearing carbon (the \(\alpha\) carbon). Secondary alcohols, which have only one such \(\ce{C-H}\) bond, are oxidized to ketones, whereas tertiary alcohols, which have no \(\ce{C-H}\) bonds to the hydroxylic carbon, are oxidized only with accompanying degradation into smaller fragments by cleavage of carbon-carbon bonds. Conversion of ethanol to ethanal is carried out on a commercial scale by passing gaseous ethanol over a copper catalyst at \(300^\text{o}\): At room temperature this reaction is endothermic with an equilibrium constant of about \(10^{22}\). At \(300^\text{o}\) conversions of \(20\%\)-\(50\%\) per pass can be realized and, by recycling the unreacted alcohol, the yield can be greater than \(90\%\). Another commercial process uses a silver catalyst and oxygen to combine with the hydrogen, which makes the net reaction substantially exothermic: In effect, this is a partial combustion reaction and requires very careful control to prevent overoxidation. In fact, by modifying the reaction conditions (alcohol-to-oxygen ratio, temperature, pressure, and reaction time), the oxidation proceeds smoothly to ethanoic acid: Reactions of this type are particularly suitable as industrial processes because they generally can be run in continuous-flow reactors, and can utilize a cheap oxidizing agent, usually supplied directly as air. Laboratory oxidation of alcohols most often is carried out with chromic acid \(\left( \ce{H_2CrO_4} \right)\), which usually is prepared as required from chromic oxide \(\left( \ce{CrO_3} \right)\) or from sodium dichromate \(\left( \ce{Na_2Cr_2O_7} \right)\) in combination with sulfuric acid. Ethanoic (acetic) acid is a useful solvent for such reactions: The mechanism of the chromic acid oxidation of 2-propanol to 2-propanone (acetone) has been investigated very thoroughly. It is a highly interesting reaction in that it reveals how changes of oxidation state can occur in a reaction involving a typical inorganic and a typical organic compound. The initial step is reversible formation of an isopropyl ester of chromic acid. This ester is unstable and is not isolated: The subsequent step is the slowest in the sequence and appears to involve attack of a base (water) at the alpha hydrogen of the chromate ester concurrent with elimination of the \(\ce{HCrO_3^-}\) group. There is an obvious analogy between this step and an \(E2\) reaction ( ): The transformation of chromic acid \(\left( \ce{H_2CrO_4} \right)\) to \(\ce{H_2CrO_3}\) amounts to the reduction of chromium from an oxidation state of \(+6\) to \(+4\). Disproportionation of Cr(IV) occurs rapidly to give compounds of Cr(III) and Cr(VI): or The \(E2\) character of the ketone-forming step has been demonstrated in two ways. First, the rate of decomposition of isopropyl hydrogen chromate to 2-propanone and \(\ce{H_2CrO_3}\) is strongly accelerated by efficient proton-removing substances. Second, the hydrogen on the \(\alpha\) carbon clearly is removed in a slow reaction because . No significant slowing of oxidation is noted for 2-propanol having deuterium in the methyl groups: Carbon-deuterium bonds normally are broken more slowly than carbon-hydrogen bonds. This so-called provides a general method for determining whether particular carbon-hydrogen bonds are broken in slow reaction steps. Primary alcohols are oxidized by chromic acid in sulfuric acid solution to aldehydes, but to stop the reaction at the aldehyde stage, it usually is necessary to remove the aldehyde from the reaction mixture as it forms. This can be done by distillation if the aldehyde is reasonably volatile: Unsaturated alcohols can be oxidized to unsaturated ketones by chromic acid, because chromic acid usually attacks double bonds relatively slowly: However, complications are to be expected when the double bond of an unsaturated alcohol is particularly reactive or when the alcohol rearranges readily under strongly acidic conditions. It is possible to avoid the use of strong acid through the combination of chromic oxide with the weak base azabenzene (pyridine). A crystalline solid of composition \(\ce{(C_5H_5N)_2} \cdot \ce{CrO_3}\) is formed when \(\ce{CrO_3}\) is added to excess pyridine at low temperatures. (Addition of pyridine to \(\ce{CrO_3}\) is likely to give an uncontrollable reaction resulting in a fire.) The pyridine-\(\ce{CrO_3}\) reagent is soluble in chlorinated solvents such as dichloromethane, and the resulting solutions rapidly oxidize to at ordinary temperatures: The yields usually are good, partly because the absence of strong acid minimizes degradation and rearrangement, and partly because the product can be isolated easily. The inorganic products are insoluble and can be separated by filtration, thereby leaving the oxidized product in dichloromethane from which it can be easily recovered. Permanganate ion, \(\ce{MnO_4^-}\), oxidizes both primary and secondary alcohols in either basic or acidic solution. With primary alcohols the product normally is the carboxylic acid because the intermediate aldehyde is oxidized rapidly by permanganate: Oxidation under conditions evidently involves the alkoxide ion rather than the neutral alcohol. The oxidizing agent, \(\ce{MnO_4^-}\), abstracts the alpha hydrogen from the alkoxide ion either as an atom (one-electron transfer) or as hydride, \(\ce{H}^\ominus\) (two-electron transfer). The steps for the two-electron sequence are: In the second step, permanganate ion is reduced from Mn(VII) to Mn(V). However, the stable oxidation states of manganese are \(+2\), \(+4\), and \(+7\); thus the Mn(V) ion formed disproportionates to Mn(VII) and Mn(IV). The normal manganese end product from oxidations in basic solution is manganese dioxide, \(\ce{MnO_2}\), in which \(\ce{Mn}\) has an oxidation state of \(+4\) corresponding to Mn(IV). In we described the use of permanganate for the oxidation of alkenes to 1,2-diols. How is it possible to control this reaction so that it will stop at the diol stage when permanganate also can oxidize to ? Overoxidation with permanganate is always a problem, but the relative reaction rates are very much a function of the pH of the reaction mixture and, in basic solution, potassium permanganate oxidizes unsaturated groups more rapidly than it oxidizes alcohols: There are many biological oxidations that convert a primary or secondary alcohol to a carbonyl compound. These reactions cannot possibly involve the extreme pH conditions and vigorous inorganic oxidants used in typical laboratory oxidations. Rather, they occur at nearly neutral pH values and they all require enzymes as catalysts, which for these reactions usually are called . An important group of biological oxidizing agents includes the pyridine nucleotides, of which nicotinamide adenine dinucleotide (\(\ce{NAD}^\oplus\), \(13\)) is an example: This very complex molecule functions to accept hydride \(\left( \ce{H}^\ominus \right)\) or the equivalent \(\left( \ce{H}^\oplus + 2 \ce{e^-} \right)\) from the \(\alpha\) carbon of an alcohol: The reduced form of \(\ce{NAD}^\oplus\) is abbreviated as \(\ce{NADH}\) and the \(\ce{H}^\ominus\) is added at the 4-position of the pyridine ring: Some examples follow that illustrate the remarkable specificity of this kind of redox system. One of the last steps in the metabolic breakdown of glucose (glycolysis; ) is the reduction of 2-oxopropanoic (pyruvic) acid to \(L\)-2-hydroxypropanoic (lactic) acid. The reverse process is oxidation of \(L\)-lactic acid. The enzyme catalyses this reaction, and it functions only with the \(L\)-enantiomer of lactic acid: A second example, provided by one of the steps in metabolism by way of the Krebs citric acid cycle (see ), is the oxidation of \(L\) -2-hydroxy-butanedioic (\(L\)-malic) acid to 2-oxobutanedioic (oxaloacetic) acid. This enzyme functions only with \(L\)-malic acid: All of these reactions release energy. In biological oxidations much of the energy is utilized to form ATP from ADP and inorganic phosphate ( ). That is to say, electron-transfer reactions are coupled with ATP formation. The overall process is called . Another important oxidizing agent in biological systems is flavin adenine dinucleotide, \(\ce{FAD}\). Like \(\ce{NAD}^\oplus\), it is a two-electron acceptor, but unlike \(\ce{NAD}^\oplus\), it accepts two electrons as \(2 \ce{H} \cdot\) rather than as \(\ce{H}^\ominus\). The reduced form, \(\ce{FADH_2}\), has the hydrogens at ring nitrogens: and (1977) | 9,095 | 1,771 |
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Chemists use the elemental composition of compounds to determine their formulas (and vice-versa). We'll look at some of the hyperbole surrounding salt and salt additives to explore the meaning of formulas. Here are a few of the claims by salt manufacturers that we can examine scientifically: FDA requires that food grade salt be at least 97.5% NaCl, but it's usually much purer. Sea salt is usually about 99% NaCl, because it is the salt in highest concentration, and precipitates first in nearly pure form when sea water is condensed. Sea salt is not a significant source of any nutrients except NaCl. We can tell from the formula that salt must be 39.34% Na and 60.66% Cl (see below), so there can be little debate about one kind of salt being different or "better" than another, except if there are significant amounts of additives. Some "sea salts" have no additives, but most salt contains potassium iodide (KI) or cuprous iodide (CuI) as additives which provide the essential mineral nutrient iodine. This additive prevents thyroid disease. We'll see below that the additive with the least iodine per gram is often used, and explain why. If KI is added to the salt, an additional additive must be added to protect it from oxidation by air to give elemental iodine (I ), which has no nutritive value (and is actually toxic). One reducing sugar is explored below in an example. Salts also contain anti-caking agents to prevent them from clumping in damp air, but those additives are usually insoluble and innocuous, and present at very low levels. Morton's table salt contains 0.2 to 0.7% calcium silicate , which explains why solutions of table salt are cloudy. Morton's Coarse Kosher Salt contains sodium ferrocyanide [Na Fe(CN) ]as an anticaking agent , which can decompose in acid to give cyanide, but the concentration is 0.0013%, so low that it can't be a problem. Kosher salt is made for coating raw meat or poultry to purify it, so it's crystals are irregular and large which may create "bursts of saltiness" superior to regular table salt when used to salt food. This is the only reason why freshly ground salt is superior (it has no aroma to release as freshly ground pepper does). Boutique salts from around the world may be gray to black , pinkish to red, or other hues, and do have different flavors due to the impurities they contain. We have presented a microscopic view of the chemical reaction between potassium (a reactive metal) and iodine (a poisonous purple solid) to form white, nutritious KI. The equation \[\ce{K(s) + I2(s) → KI(s)} \label{1} \] represent the same event in terms of chemical symbols and formulas. We also noted that the reaction between Cu and I , or between Cu and I gives CuI, not CuI : \[\ce{2 Cu^{2+} + 4 I^{-} → 2 CuI + I2} \label{2} \] But how does a practicing chemist what is occurring on the microscopic scale? In the case of reaction (2) above, we would expect CuI so how do we know the product is CuI? When a reaction is carried out 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 low solubility of CuI in water would permit purification by . The product could be dissolved in a small quantity of hot water and filtered to remove undissolved impurities. Upon cooling and partial evaporation of the water, crystals of relatively pure CuI would form. Comparing its properties, like color and melting point, with a handbook or table of data leads to the conclusion that it is CuI. But suppose you were the person who ever prepared CuI. There were no tables which listed its properties then, and so how could you determine that the formula should be CuI? 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 copper reacts with sufficient iodine, 29.97 g of a pure compound is formed. Calculate the percent composition from these experimental data. The percentage of mercury is the mass of mercury divided by the total mass of compound times 100 percent: \(\text{ }\!\!%\!\!\text{ Cu = }\frac{m_{\text{Cu}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{10}\text{.0 g}}{\text{29}\text{.97 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 33}\text{.4 }\!\!%\!\!\text{ }\) The remainder of the compound (29.97 g – 10 g = 19.97 g) is iodine: \(\text{ }\!\!%\!\!\text{ I = }\frac{m_{\text{I}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{19}\text{.97 g}}{\text{29}\text{.97 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 66}\text{.6 }\!\!%\!\!\text{ }\) As a check, verify that the percentages add to 100: 66.6% + 33.4% = 100% To obtain the formula from percent-composition data, we must find how many iodine atoms there are per copper atom. On a macroscopic scale this corresponds to the ratio of the amount of iodine to the amount of copper. If the formula is CuI, it not only indicates that there is one iodine per copper , it also says that there is 1 of iodine atoms for each 1 of copper atoms. That is, the of iodine is the same as the of copper. The numbers in the ratio of the amount of iodine to the amount of copper (1:1) are the subscripts of iodine and copper in the formula, although when they are 1, the subscripts are omitted (Cu I = CuI). 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, 66.6 g (66.6%) is iodine and 33.4 g is copper. Each mass can be converted to an amount of substance \(\begin{align}
& n_{\text{I}}=\text{66}\text{.6 g}\times \frac{\text{1 mol I}}{\text{126}\text{.9 g}}=\text{0}\text{.525 mol I} \\
& n_{\text{Cu}}=\text{34}\text{.4 g}\times \frac{\text{1 mol Cu}}{\text{63}\text{.55 g}}=\text{0}\text{.541 mol Cu} \\
\end{align}\) Dividing the larger amount by the smaller, we have \(\frac{n_{\text{Cu}}}{n_{\text{I}}}=\frac{\text{0}\text{.541 mol Cu}}{\text{0}\text{.525 mol Hg}}=\frac{\text{1}\text{.03 mol Cu}}{\text{1 mol I}}\) The ratio 1.03 mol Cu to 1 mol I also implies that there is 1.03 Cu atom per 1 I atom. If the atomic theory is correct, there is no such thing as 0.03 Cu atom; furthermore, our numbers are only good to three significant figures. Therefore we round 1.03 to 1 and write the formula as CuI. An unstable iodide of copper is isolated at low temperature and is found to have the composition 20.0% Cu, 80.0% I. Find its formula. Again assume a 100-g sample and calculate the amount of each element: \(\begin{align}
& n_{\text{Cu}}=\text{20}\text{.0 g}\times \frac{\text{1 mol Cu}}{\text{63}\text{.55 g}}=\text{0}\text{.315 mol Cu} \\
& n_{\text{I}}=\text{80}\text{.0 g}\times \frac{\text{1 mol I}}{\text{126}\text{.90 g}}=\text{0}\text{.630 mol Br} \\
\end{align}\) The ratio is \(\frac{n_{\text{I}}}{n_{\text{Cu}}}=\frac{\text{0}\text{.630 mol I}}{\text{0}\text{.315 mol Cu}}=\frac{\text{2}\text{.00 mol I}}{\text{1 mol Cu}}\) We would therefore assign the formula CuI . The formula determined by this method is called the or . Occasionally, the empirical formula differs from the actual molecular composition, or the , because the ratio 1:2 is the same as 2:4. For example, a compound of N and O with the empirical formula NO may actually be N O . Experimental determination of the molecular weight in addition to percent composition permits calculation of the molecular formula. If KI is added to salt, a "reducing sugar" is added to protect the KI from oxidation by air to give iodine (which is violet, but vaporizes easily and is lost by the salt), especially in the presence of moisture and acid, which supplies H : 4 H + 4 KI + O → 2 H O + I The reducing sugar reacts with the oxygen (or other oxidizing agents which may be present) before this reaction can occur. An antioxidant in Morton's table salt is found to have a composition of 40.00% C, 53.28% O, and 6.713% H, and its molecular weight is 180.16 by freezing point depression measurements. Determine its empirical and molecular formulas. \(\begin{align}
& n_{\text{C}}=\text{40}\text{.00 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{3}\text{.33 mol C} \\
& n_{\text{H}}=\text{6}\text{.713 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{6}\text{.660 mol H} \\
& n_{\text{O}}=\text{53}\text{.28 g}\times \frac{\text{1 mol O}}{\text{15}\text{.999 g}}=\text{3}\text{.330 mol O} \\
\end{align}\) Dividing each by the smallest: \(\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{6}\text{.66 mol H}}{\text{3}\text{.33 mol C}}=\frac{\text{2}\text{.0 mol H}}{\text{1 mol C}}\) Dividing each by the smallest: \(\frac{n_{\text{O}}}{n_{\text{C}}}=\frac{\text{3}\text{.33 mol O}}{\text{3}\text{.33 mol C}}=\frac{\text{2}\text{.0 mol O}}{\text{1 mol C}}\) The empirical formula is therefore CH O. The molecular weight corresponding to the empirical formula is 12.01 + 2 × 1.008 + 15.9994= 30.03 Since the experimental molecular weight is 180.16, this represents 180.16/30.03 or 6.0 x as great, all subscripts must be multiplied by 6, and the molecular formula is C H O . Dextrose is a reducing sugar, due to the C=O group of the open chain structure. Occasionally the ratio of amounts is not a whole number. A common anticaking agent is calcium silicate, CaSiO , but zeolites and calcium minerals like "bone ash" are commonly used. A sample of "bone ash" anticaking agent contains 38.76% Ca, 19.971% P, and 41.26% O. What is its empirical formula? \(\begin{align}
& n_{\text{P}}=\text{18}\text{.97 g}\times \frac{\text{1 mol P}}{\text{30}\text{.974 g}}=\text{0}\text{.6448 mol P} \\
& n_{\text{Ca}}=\text{38}\text{.764 g}\times \frac{\text{1 mol Ca}}{\text{40}\text{.078 g}}=\text{0}\text{.9672 mol Ca} \\
& n_{\text{O}}=\text{41}\text{.264 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{2}\text{.579 mol O} \\
\end{align}\) Divide all three by the smallest amount of substance \(\begin{align}
& \frac{n_{\text{Ca}}}{n_{\text{P}}}=\frac{\text{0}\text{.9672 mol Ca}}{\text{0}\text{.6448 mol P}}=\frac{\text{1}\text{.50 mol Ca}}{\text{1 mol P}} \\
& \frac{n_{\text{O}}}{n_{\text{P}}}=\frac{\text{2}\text{.579 mol O}}{\text{0}\text{.6448 mol P}}=\frac{\text{4}\text{.00 mol O}}{\text{1 mol P}} \\
\end{align}\) Clearly there are four times as many O atoms as P atoms, but the ratio of Ca to P is less obvious. We must convert 1.5 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .5 becomes 1/2. Thus 1.5 = 2/2 + 1/2 = \(\tfrac{\text{3}}{\text{2}}\). [in a more complicated case, like 2.25, .25 becomes ¼. Thus 2.25 = 2¼ = \(\tfrac{\text{9}}{\text{4}}\)]. \(\frac{n_{\text{Ca}}}{n_{\text{P}}}=\frac{\text{1}\text{.5 mol Ca}}{\text{1 mol P}}=\frac{\text{3 mol Ca}}{\text{2 mol P}}\) We can also write \(\frac{n_{\text{O}}}{n_{\text{P}}}=\frac{\text{4 mol O}}{\text{1 mol P}}=\frac{\text{8 mol O}}{\text{2 mol P}}\) Thus the empirical formula is Ca P O , which is tricalcium phosphate (Ca (PO ) ), an important nutritional supplement and mineral. 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. a. Prove that all NaCl is 39.34% Na and 60.66% Cl, as claimed above. b. Which of the nutritional supplements, KI or CuI, has the highest percent I? a. 1 mol NaCl contains 1 mol Na and 1 mol Cl. The molar mass is thus = 22.99 + 35.45 = 58.44 g mol A 1-mol sample of NaCl would weigh 58.44 g. The mass of 1 mol Na it contains is \(m_{\text{I}}\text{ = 1 mol Na }\times \text{ }\frac{\text{22}\text{.99 g}}{\text{1 mol Na}}\text{ = 22}\text{.99 g}\) Therefore the percentage of Na is \(\text{ }\!\!%\!\!\text{ Na = }\frac{m_{\text{Na}}}{m_{NaCl}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{22}\text{.99 g}}{\text{58}\text{.44 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 39}\text{.34 }\!\!%\!\!\text{ }\) The percentage of Cl must be 100% - 39.34% = 60.66%, but we can check: \(m_{\text{Cl}}\text{ = 1 mol Cl }\times \text{ }\frac{\text{35}\text{.45 g}}{\text{1 mol Cl}}\text{ = 35}\text{.45 g}\) Therefore the percentage of Cl is \(\text{ }\!\!%\!\!\text{ Cl = }\frac{m_{\text{Cl}}}{NaCl}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{35}\text{.45 g}}{\text{58}\text{.44 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 35}\text{.45 }\!\!%\!\!\text{ }\) b. 1 mol KI contains 1 mol K and 1 mol I. The molar mass is thus = 39.098 + 126.9 = 166.0 g mol Similarly, the molar mass of CuI is 190.45. A 1-mol sample of KI would weigh 166.0 g. The mass of 1 mol I it contains is \(m_{\text{I}}\text{ = 1 mol K }\times \text{ }\frac{\text{126}\text{.9 g}}{\text{1 mol K}}\text{ = 126}\text{.9 g}\) Therefore the percentage of I is \(\text{ }\!\!%\!\!\text{ I = }\frac{m_{\text{I}}}{m_{KI}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{126}\text{.9 g}}{\text{166}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 76}\text{.5 }\!\!%\!\!\text{ }\) The percentage of K must be 100% - 76.5% = 23.5%, but we can check: \(m_{\text{K}}\text{ = 1 mol K }\times \text{ }\frac{\text{39}\text{.098 g}}{\text{1 mol K}}\text{ = 39}\text{.98 g}\) Therefore the percentage of K is \(\text{ }\!\!%\!\!\text{ K = }\frac{m_{\text{K}}}{KI}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{39}\text{.098 g}}{\text{166}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 23}\text{.6 }\!\!%\!\!\text{ }\) The percentages of I in CuI is \(\text{ }\!\!%\!\!\text{ I = }\frac{m_{\text{I}}}{m_{CuI}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{126}\text{.9 g}}{\text{190}\text{.45 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 66}\text{.6 }\!\!%\!\!\text{ }\) Even though CuI has a smaller %I, (KI provides more I per gram), it is often used because it is less subject to the oxidation of the iodide than KI. | 14,239 | 1,773 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Colinergic_Drugs_II_-_Anticholinesterase_Agents_and_Acetylcholine_Antagonists |
Acetylcholine is inactivated by the enzyme acetylcholinesterase (enlarged), which is located at cholinergic synapses and breaks down the acetylcholine molecule into choline and acetate. Three particularly well-known drugs, neostigmine, physostigmine, and diisopropyl fluorophosphate, inactivate acetylcholinesterase so that it cannot hydrolyze the acetylcholine released at the nerve ending. As a result, acetylcholine increases in quantity with successive nerve impulses so that large amounts of acetylcholine can accumulate and repetitively stimulate receptors. In view of the widespread distribution of cholinergic neurons, it is not surprising that the anticholinesterase agents as a group have received more extensive application as toxic agents, in the form of agricultural insecticides and potential chemical-warfare "nerve gases," than as therapeutic agents. Nevertheless, several members of this class of compounds are clinically useful. The active center of acetylcholinesterase consists of a negative subsite, which attracts the quaternary group of choline through both coulombic and hydrophobic forces, and an esteratic subsite, where nucleophilic attack occurs on the acyl carbon of the substrate. The catalytic mechanism resembles that of other serine esterases, where a serine hydroxyl group is rendered highly nucleophilic through a charge-relay system involving the close apposition of an imidazole group and , presumably, a carboxyl group on the enzyme. During enzymatic attack on the ester, a tetrahedral intermediate between enzyme and ester is formed that collapses to an acetyl enzyme conjugate with the concomitant release of choline. The acetyl enzyme is labile to hydrolysis, which results in the formation of acetate and active enzyme. Acetylcholinesterase is one of the most efficient enzymes known and has the capacity to hydrolyze \(3 \times 10^5\) acetylcholine molecules per molecule of enzyme per minute: this is equivalent to a turnover time of 150 microseconds. Drugs such as neostigmine and pydrostigmine that have a carbamyl ester linkage are hydrolyzed by acetylcholinesterase, but much more slowly than acetylcholine because they from more stable intermediates. Therefore, these drugs inactivate acetylcholinesterase for up to several hours, after which they are displaced so that acetylcholinesterase. These drugs are used for their effects on skeletal muscle and the eye (pupillary constriction and decreased intraocular pressure) and for the treatment of atropine poisoning. On the other hand, diisopropyl fluorophosphate, which has military potential as a powerful nerve gas poison, inactivates acetylcholinesterase for weeks, which makes this a particularly lethal poison. Generally the pharmacological properties of anticholinesterase agents can be predicted merely by knowing those loci where acetylcholine is released physiologically by nerve impulses. and the responses of the corresponding effector organs to acetylcholine. While this is true of the main, the diverse locations of cholinergic synapses increase the complexity of the response. Potentially, the antcholiesterase agents can produce all of the following effects 1) stimulation of muscarinic receptor responses at autonomic effector organs 2) stimulation, followed by paralysis, of all autonomic ganglia and skeletal muscle and 3) stimulation, with occasional subsequent depression, of cholinergic receptor sites in the CNS. However, with smaller doses, particularly those employed therapeutically, several modifying factors are significant. Neostigmine and pydrostigmine are among the principal anticholinesterases. These drugs have only a few clinical uses, mainly in augmenting gastric and intestinal contractions (in treatment of obstructions of the digestive tract), in generally augmenting muscular contractions (in the treatment of myasthenia gravis), and in constricting the eye pupils (in the treatment of glaucoma). Other anticholinesterases in larger doses, however, are widely used as toxins that achieve their effects by causing a continual stimulation of the parasympathetic nervous system. Parathion and malathion are thus highly effective agricultural insecticides, while tabun and serin are nerve gases used in chemical warfare to induce nausea, vomiting, convulsions, and death in humans. This action produces a decrease in the rate of destruction of acetylcholine in the synaptic cleft and hence an increase in the amount of transmitter available to interact with the receptors.
Atropine, a naturally occurring alkaloid of " ", the deadly nightshade, inhibits the actions of acetylcholine on autonomic effectors innervated by postganglionic cholinergic nerves as well as on smooth muscles that lack cholinergic innervation. Since atropine antagonizes the muscarinic actions of acetylcholine, it is known as an antimuscarinic agent. Evidence shows that atropine and related compounds compete with muscarinic agonists for identical binding sites on muscarinic receptors. In general, antimuscarinic agents have only a moderate effect on the actions of acetylcholine at nicotinic receptor sites. Thus, at autonomic ganglia, where transmission primarily involves an action of acetylcholine on nicotinic receptors, atropine produces only partial block. At the neuomuscular junction, where the receptors are exclusively nicotinic, extremely high doses of atropine or related drugs are required to cause any degree of blockade. In the central nervous system, cholinergic transmission appears to be predominantly nicotinic in the spinal chord and both nicotinic and muscarinic in the brain. Many of the CNS effects of atropine-like drugs are probably attributable to their central antimuscarinic actions.
When used as premedication for anaesthesia, atropine decreases bronchial and salivary secretions, blocks the bradycardia associated with some drugs used in anaesthesia such as halothane, suxamethonium and neostigmine, and also helps prevent bradycardia from excessive vagal stimulation. There is usually an increase in heart rate and sometimes a tachycardia as well as inhibition of secretions (causing a dry mouth) and relaxation of smooth muscle in the gut, urinary tract and biliary tree. Since atropine crosses the blood brain barrier CNS effects in the elderly may include amnesia, confusion and excitation. Pupillary dilatation and paralysis of accommodation occur, with an increase in intraocular pressure especially in patients with glaucoma. Occasionally small intravenous doses may be accompanied by slowing of the heart rate due to a central effect - this resolves with an extra increment of intravenous atropine Being a sympathetic cholinergic blocking agent, signs of parasympathetic block may occur such as dryness of the mouth, blurred vision, increased intraocular tension and urinary retention. Sarin is a nerve agent in the organophosphate family. It is dispersed in a droplet or mist form. Upon inhalation, for instance, the symptoms (in order of occurrence) include: runny nose, bronchial secretions, tightness in the chest, dimming of vision, pin-point pupils, drooling, excessive perspiration, nausea, vomiting, involuntary defecationand urination, muscle tremors, convulsions, coma, and death. Primary treatment for nerve agents is atropine sulfate. It is commonly carried in auto-injectors by military personnel in dosages of 1-2 mgs. However, in many cases, massive doses may be necessary to reverse the effects of the anticholinesterase agents. Frequently, 20-40 mgs. of atropine may be necessary. In 1799 the famous Prussian explorer and scientist Baron Von Humboldt discovered a potent drug called curare. On an expedition into the jungles of Venezuela, he watched an Indian hunter bring down a large animal with a single shot from his bow and arrow. The arrow had been poisoned with curare, a potion with two curious properties, derived from the jungle plants. Curare injected into the bloodstream, as it was when hunting animals, was deadly. It immobilized the body, attacked the vital organs, and caused death almost instantaneously. Humboldt discovered the second property of curare in a more dramatic fashion. He became sick, and a native witch doctor forced Humboldt to drink some curare that had been diluted with water. Terrified that he was going to die, Humboldt was surprised to find that after drinking the curare, he felt significantly better. Curare, when it was diluted and taken orally, he discovered, could have a positive medicinal value without causing any damage to vital organs. Curare is a generic term for various South American arrow poisons. The main active ingredient in curare is d-tubocurarine, which has the chemical structure shown below. In brief, d-tubocurarine is an antagonist of the cholinergic receptor sites at the post junctional membrane and thereby blocks competitively the transmitter action of acetylcholine. When the drug is applied directly to the end-plate of a single isolated muscle fiber under microscopic control, the muscle cell becomes insensitive to motor-nerve impulses and to direct applied acetylcholine; however, the end -plate region and the remainder of the muscle fiber membrane retain their normal sensitivity to the application of potassium ions, and the muscle fiber still responds to direct electrical stimulation. Because acetylcholine release into the neuromuscular junction muscle is what initiates contraction, curare causes muscle relaxation and paralysis. There are several clinical application for neuromuscular blockage. The most important by far is the induction of muscle relaxation during anesthesia for effective surgery. Without such drugs deeper anesthesia, requiring more anesthetic, would be needed to achieve the same degree of muscle relaxation: tracheal intubation would also be impossible because of strong reflex response to tube insertion. It is the decreased need for anesthetics, however, that represents increased surgical safety. Neuromuscular blockers also find limited utility in convulsive situations such as those precipitated by tetanus infections and to minimize injury to patients undergoing electroconvulsive therapy. Manipulation of fractured or dislocated bones may also be aided by such drugs. * is a bacterial poison that prevents the release of acetylcholine by all types of cholinergic nerve terminals. Apparently, the toxin blocks release of vesicular acetylcholine at the preterminal portion of the axon, but why this is confined to cholinergic fibers is not known. | 10,550 | 1,774 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.03%3A_Molar_Reaction_Enthalpy |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Recall that \(\Del H\m\rxn\) is a molar integral reaction enthalpy equal to \(\Del H\rxn/\Del\xi\), and that \(\Delsub{r}H\) is a molar differential reaction enthalpy defined by \(\sum_i\!\nu_i H_i\) and equal to \(\pd{H}{\xi}{T,p}\). During a process in a closed system at constant pressure with expansion work only, the enthalpy change equals the energy transferred across the boundary in the form of heat: \(\dif H=\dq\) (Eq. 5.3.7). Thus for the molar reaction enthalpy \(\Delsub{r}H = \pd{H}{\xi}{T,p}\), which refers to a process not just at constant pressure but also at constant temperature, we can write \begin{gather} \s{ \Delsub{r}H = \frac{\dq}{\dif\xi} } \tag{11.3.1} \cond{(constant \(T\) and \(p\), \(\dw'{=}0\))} \end{gather} Note that when there is nonexpansion work (\(w'\)), such as electrical work, the enthalpy change is not equal to the heat. For example, if we compare a reaction taking place in a galvanic cell with the same reaction in a reaction vessel, the heats at constant \(T\) and \(p\) for a given change of \(\xi\) are different, and may even have opposite signs. The value of \(\Delsub{r}H\) is the same in both systems, but the ratio of heat to advancement, \(\dq/\dif\xi\), is different. An reaction is one for which \(\Delsub{r}H\) is negative, and an reaction is one for which \(\Delsub{r}H\) is positive. Thus in a reaction at constant temperature and pressure with expansion work only, heat is transferred out of the system during an exothermic process and into the system during an endothermic process. If the process takes place at constant pressure in a system with thermally-insulated walls, the temperature increases during an exothermic process and decreases during an endothermic process. These comments apply not just to chemical reactions, but to the other chemical processes at constant temperature and pressure discussed in this chapter. A , \(\Delsub{r}H\st\), is the same as the molar integral reaction enthalpy \(\Del H\m\rxn\) for the reaction taking place under standard state conditions (each reactant and product at unit activity) at constant temperature. At constant temperature, partial molar enthalpies depend only mildly on pressure. It is therefore usually safe to assume that unless the experimental pressure is much greater than \(p\st\), the reaction is exothermic if \(\Delsub{r}H\st\) is negative and endothermic if \(\Delsub{r}H\st\) is positive. The of a substance is the reaction in which the substance, at a given temperature and in a given physical state, is formed from the constituent elements in their reference states at the same temperature. The is usually chosen to be the standard state of the element in the allotropic form and physical state that is stable at the given temperature and the standard pressure. For instance, at \(298.15\K\) and \(1\br\) the stable allotrope of carbon is crystalline graphite rather than diamond. Phosphorus is an exception to the rule regarding reference states of elements. Although red phosphorus is the stable allotrope at \(298.15\K\), it is not well characterized. Instead, the reference state is white phosphorus (crystalline P\(_4\)) at \(1\br\). At \(298.15\K\), the reference states of the elements are the following: A principle called can be used to calculate the standard molar enthalpy of formation of a substance at a given temperature from standard molar reaction enthalpies at the same temperature, and to calculate a standard molar reaction enthalpy from tabulated values of standard molar enthalpies of formation. The principle is an application of the fact that enthalpy is a state function. Therefore, \(\Del H\) for a given change of the state of the system is independent of the path and is equal to the sum of \(\Del H\) values for any sequence of changes whose net result is the given change. (We may apply the same principle to a change of state function.) This value is one of the many standard molar enthalpies of formation to be found in compilations of thermodynamic properties of individual substances, such as the table in Appendix H. We may use the tabulated values to evaluate the standard molar reaction enthalpy \(\Delsub{r}H\st\) of a reaction using a formula based on Hess’s law. Imagine the reaction to take place in two steps: First each reactant in its standard state changes to the constituent elements in their reference states (the reverse of a formation reaction), and then these elements form the products in their standard states. The resulting formula is \begin{gather} \s{ \Delsub{r}H\st = \sum_i\nu_i \Delsub{f}H\st(i) } \tag{11.3.3} \cond{(Hess’s law)} \end{gather} where \(\Delsub{f}H\st(i)\) is the standard molar enthalpy of formation of substance \(i\). Recall that the stoichiometric number \(\nu_i\) of each reactant is negative and that of each product is positive, so according to Hess’s law the standard molar reaction enthalpy is the sum of the standard molar enthalpies of formation of the products minus the sum of the standard molar enthalpies of formation of the reactants. Each term is multiplied by the appropriate stoichiometric coefficient from the reaction equation. A standard molar enthalpy of formation can be defined for a to use in Eq. 11.3.3. For instance, the formation reaction of aqueous sucrose is \[ \textstyle \tx{12 C(s, graphite)} + \tx{11 H\(_2\)(g)} + \frac{11}{2}\tx{O\(_2\)(g)} \arrow \tx{C\(_{12}\)H\(_{22}\)O\(_{11}\)(aq)} \] and \(\Delsub{f}H\st\) for C\(_{12}\)H\(_{22}\)O\(_{11}\)(aq) is the enthalpy change per amount of sucrose formed when the reactants and product are in their standard states. Note that this formation reaction does include the formation of the solvent H\(_2\)O from H\(_2\) and O\(_2\). Instead, the solute once formed combines with the amount of pure liquid water needed to form the solution. If the aqueous solute is formed in its standard state, the amount of water needed is very large so as to have the solute exhibit infinite-dilution behavior. There is no ordinary reaction that would produce an individual from its element or elements without producing other species as well. We can, however, prepare a consistent set of standard molar enthalpies of formation of ions by assigning a value to a single reference ion. ({This procedure is similar to that described in Sec. 9.2.4 for partial molar volumes of ions.) We can use these values for ions in Eq. 11.3.3 just like values of \(\Delsub{f}H\st\) for substances and nonionic solutes. Aqueous hydrogen ion is the usual reference ion, to which is assigned the arbitrary value \begin{equation} \Delsub{f}H\st\tx{(H\(^+\), aq)} = 0 \qquad \tx{(at all temperatures)} \tag{11.3.4} \end{equation} To see how we can use this reference value, consider the reaction for the formation of aqueous HCl (hydrochloric acid): \begin{equation*} \ce{1/2H2}\tx{(g)} + \ce{1/2Cl2}\tx{(g)} \arrow \ce{H+}\tx{(aq)} + \ce{Cl-}\tx{(aq)} \end{equation*} The standard molar reaction enthalpy at \(298.15\K\) for this reaction is known, from reaction calorimetry, to have the value \(\Delsub{r}H\st = -167.08\units{kJ mol\(^{-1}\)}\). The standard states of the gaseous H\(_2\) and Cl\(_2\) are, of course, the pure gases acting ideally at pressure \(p\st\), and the standard state of each of the aqueous ions is the ion at the standard molality and standard pressure, acting as if its activity coefficient on a molality basis were \(1\). From Eq. 11.3.3, we equate the value of \(\Delsub{r}H\st\) to the sum \[ -\onehalf\Delsub{f}H\st\tx{(H\(_2\), g)} -\onehalf\Delsub{f}H\st\tx{(Cl\(_2\), g)} + \Delsub{f}H\st\tx{(H\(^+\), aq)} + \Delsub{f}H\st\tx{(Cl\(^-\), aq)} \] But the first three terms of this sum are zero. Therefore, the value of \(\Delsub{f}H\st\)(Cl\(^-\), aq) is \(-167.08\units{kJ mol\(^{-1}\)}\). Next we can combine this value of \(\Delsub{f}H\st\)(Cl\(^-\), aq) with the measured standard molar enthalpy of formation of aqueous sodium chloride \[ \ce{Na}\tx{(s)} + \ce{1/2Cl2}\tx{(g)} \arrow \ce{Na+}\tx{(aq)} + \ce{Cl-}\tx{(aq)} \] to evaluate the standard molar enthalpy of formation of aqueous sodium ion. By continuing this procedure with other reactions, we can build up a consistent set of \(\Delsub{f}H\st\) values of various ions in aqueous solution. The molar reaction enthalpy \(\Delsub{r}H\) is in general a function of \(T\), \(p\), and \(\xi\). Using the relations \(\Delsub{r}H=\sum_i\!\nu_i H_i\) (from Eq. 11.2.15) and \(C_{p,i}=\pd{H_i}{T}{p, \xi}\) (Eq. 9.2.52), we can write \begin{equation} \Pd{\Delsub{r}H}{T}{p, \xi} = \Pd{\sum_i\nu_i H_i}{T}{p, \xi} = \sum_i\nu_i C_{p,i} = \Delsub{r}C_p \tag{11.3.5} \end{equation} where \(\Delsub{r}C_p\) is the molar reaction heat capacity at constant pressure, equal to the rate at which the heat capacity \(C_p\) changes with \(\xi\) at constant \(T\) and \(p\). Under standard state conditions, Eq. 11.3.5 becomes \begin{equation} \dif\Delsub{r}H\st/\dif T = \Delsub{r}C_p\st \tag{11.3.6} \end{equation} Consider a reaction occurring with a certain finite change of the advancement in a closed system at temperature \(T'\) and at constant pressure. The reaction is characterized by a change of the advancement from \(\xi_1\) to \(\xi_2\), and the integral reaction enthalpy at this temperature is denoted \(\Del H\tx{(rxn, \(T'\))}\). We wish to find an expression for the reaction enthalpy \(\Del H\tx{(rxn, \(T''\))}\) for the same values of \(\xi_1\) and \(\xi_2\) at the same pressure but at a different temperature, \(T''\). The heat capacity of the system at constant pressure is related to the enthalpy by Eq. 5.6.3: \(C_p=\pd{H}{T}{p, \xi}\). We integrate \(\dif H=C_p\dif T\) from \(T'\) to \(T''\) at constant \(p\) and \(\xi\), for both the final and initial values of the advancement: \begin{equation} H(\xi_2, T'') = H(\xi_2, T') + \int_{T'}^{T''}\!\!C_p(\xi_2)\dif T \tag{11.3.7} \end{equation} \begin{equation} H(\xi_1, T'') = H(\xi_1, T') + \int_{T'}^{T''}\!\!C_p(\xi_1)\dif T \tag{11.3.8} \end{equation} Subtracting Eq. 11.3.8 from Eq. 11.3.7, we obtain \begin{equation} \Del H\tx{(rxn, \(T''\))} = \Del H\tx{(rxn, \(T'\))} + \int_{T'}^{T''}\!\!\!\Del C_p\dif T \tag{11.3.9} \end{equation} where \(\Del C_p\) is the difference between the heat capacities of the system at the final and initial values of \(\xi\), a function of \(T\): \(\Del C_p = C_p(\xi_2)-C_p(\xi_1)\). Equation 11.3.9 is the . When \(\Del C_p\) is essentially constant in the temperature range from \(T'\) to \(T''\), the Kirchhoff equation becomes \begin{equation} \Del H\tx{(rxn, \(T''\))} = \Del H\tx{(rxn, \(T'\))} + \Del C_p(T''-T') \tag{11.3.10} \end{equation} Figure 11.7 illustrates the principle of the Kirchhoff equation as expressed by Eq. 11.3.10. \(\Del C_p\) equals the difference in the slopes of the two dashed lines in the figure, and the product of \(\Del C_p\) and the temperature difference \(T''-T'\) equals the change in the value of \(\Del H\rxn\). The figure illustrates an exothermic reaction with negative \(\Del C_p\), resulting in a more negative value of \(\Del H\rxn\) at the higher temperature. We can also find the effect of temperature on the molar differential reaction enthalpy \(\Delsub{r}H\). From Eq. 11.3.5, we have \(\pd{\Delsub{r}H}{T}{p, \xi} = \Delsub{r}C_p\). Integration from temperature \(T'\) to temperature \(T''\) yields the relation \begin{equation} \Delsub{r}H(T''\!,\xi)=\Delsub{r}H(T'\!,\xi) + \int_{T'}^{T''}\!\!\Delsub{r}C_p(T,\xi)\dif T \tag{11.3.11} \end{equation} This relation is analogous to Eq. 11.3.9, using molar differential reaction quantities in place of integral reaction quantities. | 19,048 | 1,775 |
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This page looks at elimination from unsymmetric halogenoalkanes such as 2-bromobutane. 2-bromobutane is an unsymmetric halogenoalkane in the sense that it has a CH group one side of the C-Br bond and a CH CH group the other. The basic facts and mechanisms for these reactions are exactly the same as with simple halogenoalkanes like 2-bromopropane. This page only deals with the extra problems created by the possibility of more than one elimination product. You will remember that elimination happens when a hydroxide ion (from, for example, sodium hydroxide) acts as a base and removes a hydrogen as a hydrogen ion from the halogenoalkane. For example, in the simple case of elimination from 2-bromopropane: The hydroxide ion removes a hydrogen from one of the carbon atoms next door to the carbon-bromine bond, and the various electron shifts then lead to the formation of the alkene - in this case, propene. With an unsymmetric halogenoalkane like 2-bromobutane, there are several hydrogens which might possibly get removed. You need to think about each of these possibilities. The hydrogen has to be removed from a carbon atom adjacent to the carbon-bromine bond. If an OH ion hit one of the hydrogens on the right-hand CH group in the 2-bromobutane (as we've drawn it), there's nowhere for the reaction to go. To make room for the electron pair to form a double bond between the carbons, you would have to expel a hydrogen from the CH group as a hydride ion, H . That is energetically much too difficult, and so this reaction doesn't happen. That still leaves the possibility of removing a hydrogen either from the left-hand CH or from the CH group. The product is but-1-ene, CH =CHCH CH This time the product is but-2-ene, CH CH=CHCH . In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene. Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below. Which isomer gets formed is just a matter of chance. Elimination from 2-bromobutane leads to a mixture containing: | 2,257 | 1,776 |
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Given the relationship \[ \left( - \dfrac{\partial U}{\partial V} \right)_T= T\left( - \dfrac{\partial p}{\partial T} \right)_V-p \] show that \[\left( - \dfrac{\partial U}{\partial V} \right)_T =0 \] for an ideal gas. Determine if the following differential is exact, and if so, find the function \(z(x, y)\) that satisfies the expression. \[ dz = 4xy\,dz + 2x^2 dy\] For a van der Waals gas, \[\left(\dfrac{\partial U}{\partial V}\right)_T = \left(\dfrac{an^2}{V^2}\right) \] Find an expression in terms of \(a\), \(n\), \(V\), and \(R\) for \[\left(\dfrac{\partial T}{\partial V}\right)_U\] if \(C_V = 3/2 R\). Use the expression to calculate the temperature change for 1.00 mol of Xe ( = 4.19 atm L mol ) expanding against a vacuum from 10.0 L to 20.0 L. Given the following data, calculate the change in volume for 50.0 cm of due to an increase in pressure from 1.00 atm to 0.750 atm at 298 K. Consider a gas that follows the equation of state \[ p =\dfrac{nRT}{V-nb}\] derive an expression for \[\mu_{JT} = \dfrac{V}{C_p} (T \alpha -1)\] Given \[ \left(\dfrac{\partial H}{\partial p}\right)_T = -T \left(\dfrac{\partial V}{\partial T}\right)_p +V \] derive an expression for \[\left(\dfrac{\partial U}{\partial p}\right)_T \] in terms of measurable properties. Use your result to calculate the change in the internal energy of 18.0 g of water when the pressure is increased from 1.00 atm to 20.0 atm at 298 K. Derive an expression for \[\left(\dfrac{\partial U}{\partial T}\right)_p \] Begin with the definition of enthalpy, in order to determine \[ dH = dU + pdV + Vdp\] Finish by dividing by dT and constraining to constant pressure. Make substitutions for the measurable quantities, and solve for \[\left(\dfrac{\partial U}{\partial T}\right)_p .\] Derive an expression for the difference between \(C_p\) and \(C_V\) in terms of the internal pressure, \(\alpha\), \(p\) and \(V\). Using the definition for \(H\) as a starting point, show that \[\left(\dfrac{\partial H}{\partial T}\right)_p = \left(\dfrac{\partial U}{\partial T}\right)_p + p \left(\dfrac{\partial V}{\partial T}\right)_p \] Now, find an expression for by starting with \(U(V,T)\) and writing an expression for the total differential \(dU(V,T)\). Divide this expression by \(dp\) and constrain to constant \(T\). Substitute this into the previous expressions and solve for \[\left(\dfrac{\partial G}{\partial T}\right)_p - \left(\dfrac{\partial U}{\partial T}\right)_V .\] Evaluate the expression you derived in problem 8 for an ideal, assuming that the internal pressure of an ideal gas is zero.
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One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the . In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:
\( 2C_{8}H_{18}\left ( l \right ) + 25O_{2}\left ( g \right ) \rightarrow 16CO_{2}\left ( g \right ) + 18H_{2}O\left ( g \right ) \tag{14.6.1}\) For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction , involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.
\( NO_{2}\left ( g \right ) + CO\left ( g \right ) \rightarrow NO\left ( g \right ) + CO_{2} \left ( g \right ) \tag{14.6.2}\) From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO with a molecule of CO that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:
\( rate = k\left [ NO_{2} \right ]^{2} \tag{14.6.3} \) The fact that the reaction is second order in [NO ] and independent of [CO] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be rate = [NO ,CO]. The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2: According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The NO molecule is an intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step. The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction. The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as ; if there are two reactant molecules, it is ; and if there are three reactant molecules (a relatively rare situation), it is . Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!) Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order is the same as its molecularity ( ). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is rate = [A]. For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in For a bimolecular elementary reaction of the form A + B → products, the general rate law is rate = [A,B]. Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining step , that must give the experimentally determined rate law for the overall reaction. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions. The phenomenon of a rate-determining step can be compared to a succession of funnels. The smallest-diameter funnel controls the rate at which the bottle is filled, whether it is the first or the last in the series. Pouring liquid into the first funnel faster than it can drain through the smallest results in an overflow. Look at the rate laws for each elementary reaction in our example as well as for the overall reaction. The experimentally determined rate law for the reaction of NO with CO is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so for the overall reaction must equal . That is, NO is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2. Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect. In an alternative mechanism for the reaction of NO with CO, N O appears as an intermediate. Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = [NO ] )? elementary reactions rate law for each elementary reaction and overall rate law Determine the rate law for each elementary reaction in the reaction. Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step. The rate law for step 1 is rate = [NO ] ; for step 2, it is rate = [N O ,CO]. If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = [NO ] . This is the same as the experimentally determined rate law. Hence this mechanism, with N O as an intermediate, and the one described previously, with NO as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO and N O , directly. Exercise A Iodine monochloride (ICl) reacts with H as follows: The experimentally determined rate law is rate = [ICl,H ]. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: HI is an intermediate.) This mechanism is consistent with the experimental rate law the first step is the rate-determining step. Exercise B The reaction between NO and H occurs via a three-step process: Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: rate = [NO] [H ] ? Many reaction mechanisms, like those discussed so far, consist of only two or three elementary reactions. Many others consist of long series of elementary reactions. The most common mechanisms are chain reactions , in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. Chain reactions occur in fuel combustion, explosions, the formation of many polymers, and the tissue changes associated with aging. They are also important in the chemistry of the atmosphere. Chain reactions are described as having three stages. The first is , a step that produces one or more reactive intermediates. Often these intermediates are radicals , species that have an unpaired valence electron. In the second stage, , reactive intermediates are continuously consumed and regenerated while products are formed. Intermediates are also consumed but not regenerated in the final stage of a chain reaction, , usually by forming stable products. Let us look at the reaction of methane with chlorine at elevated temperatures (400°C–450°C), a chain reaction used in industry to manufacture methyl chloride (CH Cl), dichloromethane (CH Cl ), chloroform (CHCl ), and carbon tetrachloride (CCl ): \( CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl \) \( CH_{3}Cl + Cl_{2} \rightarrow CH_{2}Cl_{2} + HCl \) \(CH_{2}Cl_{2} + Cl_{2} \rightarrow CHCl_{3} + HCl \) \(CHCl_{3} + Cl_{2} \rightarrow CCl_{4} + HCl \) Direct chlorination generally produces a mixture of all four carbon-containing products, which must then be separated by distillation. In our discussion, we will examine only the chain reactions that lead to the preparation of CH Cl. In the initiation stage of this reaction, the relatively weak Cl–Cl bond cleaves at temperatures of about 400°C to produce chlorine atoms (Cl·): During propagation, a chlorine atom removes a hydrogen atom from a methane molecule to give HCl and CH ·, the methyl radical: The methyl radical then reacts with a chlorine molecule to form methyl chloride and another chlorine atom, Cl·: The sum of the propagation reactions is the same as the overall balanced chemical equation for the reaction: Without a chain-terminating reaction, propagation reactions would continue until either the methane or the chlorine was consumed. Because radical species react rapidly with almost anything, however, including each other, they eventually form neutral compounds, thus terminating the chain reaction in any of three ways: Here is the overall chain reaction, with the desired product (CH Cl) in bold: The chain reactions responsible for explosions generally have an additional feature: the existence of one or more , in which one radical reacts to produce two or more radicals, each of which can then go on to start a new chain reaction. Repetition of the branching step has a cascade effect such that a single initiation step generates large numbers of chain reactions. The result is a very rapid reaction or an explosion. The reaction of H and O , used to propel rockets, is an example of a chain branching reaction: Termination reactions occur when the extraordinarily reactive H· or OH· radicals react with a third species. The complexity of a chain reaction makes it unfeasible to write a rate law for the overall reaction. A is the microscopic path by which reactants are transformed into products. Each step is an . Species that are formed in one step and consumed in another are . Each elementary reaction can be described in terms of its , the number of molecules that collide in that step. The slowest step in a reaction mechanism is the . consist of three kinds of reactions: initiation, propagation, and termination. Intermediates in chain reactions are often , species that have an unpaired valence electron. How does the term relate to elementary reactions? How does it relate to the overall balanced chemical equation? What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation? When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction? If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step? Give the rate-determining step for each case. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health? Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows: where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction? Above approximately 500 K, the reaction between NO and CO to produce CO and NO follows the second-order rate law Δ[CO ]/Δ = [NO ,CO]. At lower temperatures, however, the rate law is Δ[CO ]/Δ = k′[NO ] , for which it is known that NO is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest? Nitramide (O NNH ) decomposes in aqueous solution to N O and H O. What is the experimental rate law (Δ[N O]/Δ ) for the decomposition of nitramide if the mechanism for the decomposition is as follows? Assume that the rates of the forward and reverse reactions in the first equation are equal. The following reactions are given: What is the relationship between the relative magnitudes of and if these reactions have the rate law Δ[F]/Δ = [A,B,E]/[C]? How does the magnitude of compare to that of ? Under what conditions would you expect the rate law to be Δ[F]/Δ = ′[A,B]? Assume that the rates of the forward and reverse reactions in the first equation are equal. The step is likely to be rate limiting; the rate cannot proceed any faster than the second step. Thumbnail from | 14,786 | 1,778 |
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It is difficult to do a quantitative determination of a formula as a lecture demonstration. This demonstration is at best semi-quantitative, but it's interesting and suggests methods that might be more quantitative, and points to variables that must be controlled in a quantitative experiment. A. Steel Wool: Weigh ~0.5 g of 000 or 0000 steel wool in an evaporating dish. With tongs, dip it in petroleum ether or hexane to remove surface oil which is present to prevent oxidation. Shake off most of the hexane, then hold the steel wool over the evaporating dish and ignite it with a Bunsen burner. Catch the product in the evaporating dish and reweigh. Calculate the formula from data, or from "optimal" data supplied by calculation. Discuss: (1) The product may be a mixture of FeO, Fe O , and Fe O . (2) some of the product may not have been collected. (3) the steel wool isn't really pure Fe. B. Pyrophoric Iron: Prepare ~0.5 g of pyrophoric iron by decomposition of FeC O under methane in a weighed 15 x 150 mm test tube with a two holed rubber stopper to allow inlet of methane product gases. Insert a long pipette in one hole, and ignite the product gases after allowing time for methane to fill the apparatus. Heat the iron oxalate gently so that it turns completly black, but avoid further heating. Weigh the tube plus product. Weigh an evaporating dish, and pour the product into it, then reweigh. Reweigh the empty tube. From the masses of the iron and product, calculate it's formula. 1. It's more fun to dump the product through several feet of air and watch the combustion reaction, but product will be lost. This may be inconsequential, because the calculations may be done with "optimal" data, noting the difficulties of making the demonstration as presented quantitative. 2. The product of the reaction is almost certainly not pure. It probably contains several oxides, as well as sintered (nonpyrophoric) iron that results from overheating the product. | 1,984 | 1,779 |
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The interactions that determine the solubility of a substance in a liquid depend largely on the chemical nature of the solute (such as whether it is ionic or molecular) rather than on its physical state (solid, liquid, or gas). We will first describe the general case of forming a solution of a molecular species in a liquid solvent and then describe the formation of a solution of an ionic compound. Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called , or hydration when the solvent is water. Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process. Because enthalpy is a , we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by \(ΔH_1\), \(ΔH_2\), and \(ΔH_3\) in Figure \(\Page {1}\). The overall enthalpy change in the formation of the solution (\( \Delta H_{soln}\)) is the sum of the enthalpy changes in the three steps: \[ \Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.3.2}\] When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent (\(\Delta H_1\)) and the solute (\(\Delta H_2\)). Because \(ΔH\) is positive for both steps 1 and 2, the solute–solvent interactions (\(\Delta H_3\)) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic (\(\Delta H_{soln} < 0\)). When the solute is an ionic solid, \(ΔH_2\) corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as \(\ce{MgO}\) (−3791 kJ/mol), are generally insoluble in all solvents. A positive value for \(ΔH_{soln}\) does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high \(ΔH_{soln}\) is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive \(ΔH_{soln}\) of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure \(\Page {2}\)). The London dispersion forces, dipole–dipole interactions, and hydrogen bonds that hold molecules to other molecules are generally weak. Even so, energy is required to disrupt these interactions. Unless some of that energy is recovered in the formation of new, favorable solute–solvent interactions, the increase in entropy on solution formation is not enough for a solution to form. For solutions of gases in liquids, we can safely ignore the energy required to separate the solute molecules (\(ΔH_2 = 0\)) because the molecules are already separated. Thus we need to consider only the energy required to separate the solvent molecules (\(ΔH_1\)) and the energy released by new solute–solvent interactions (\(ΔH_3\)). Nonpolar gases such as \(N_2\), \(O_2\), and \(Ar\) have no dipole moment and cannot engage in dipole–dipole interactions or hydrogen bonding. Consequently, the only way they can interact with a solvent is by means of London dispersion forces, which may be weaker than the solvent–solvent interactions in a polar solvent. It is not surprising, then, that nonpolar gases are most soluble in nonpolar solvents. In this case, \(ΔH_1\) and \(ΔH_3\) are both small and of similar magnitude. In contrast, for a solution of a nonpolar gas in a polar solvent, \(ΔH_1\) is far greater than \(ΔH_3\). As a result, nonpolar gases are less soluble in polar solvents than in nonpolar solvents. For example, the concentration of \(N_2\) in a saturated solution of \(N_2\) in water, a polar solvent, is only \(7.07 \times 10^{-4}\; M\) compared with \(4.5 \times 10^{-3}\; M\) for a saturated solution of \(N_2\) in benzene, a nonpolar solvent. The solubilities of nonpolar gases in water generally increase as the molecular mass of the gas increases, as shown in Table \(\Page {1}\). This is precisely the trend expected: as the gas molecules become larger, the strength of the solvent–solute interactions due to London dispersion forces increases, approaching the strength of the solvent–solvent interactions. Virtually all common organic liquids, whether polar or not, are miscible. The strengths of the intermolecular attractions are comparable; thus the enthalpy of solution is expected to be small (\(ΔH_{soln} \approx 0\)), and the increase in entropy drives the formation of a solution. If the predominant intermolecular interactions in two liquids are very different from one another, however, they may be immiscible. For example, organic liquids such as benzene, hexane, \(CCl_4\), and \(CS_2\) (S=C=S) are nonpolar and have no ability to act as hydrogen bond donors or acceptors with hydrogen-bonding solvents such as \(H_2O\), \(HF\), and \(NH_3\); hence they are immiscible in these solvents. When shaken with water, they form separate phases or layers separated by an interface (Figure \(\Page {3}\)), the region between the two layers. Just because two liquids are immiscible, however, does not mean that they are completely insoluble in each other. For example, 188 mg of benzene dissolves in 100 mL of water at 23.5°C. Adding more benzene results in the separation of an upper layer consisting of benzene with a small amount of dissolved water (the solubility of water in benzene is only 178 mg/100 mL of benzene). The solubilities of simple alcohols in water are given in Table \(\Page {2}\). Only the three lightest alcohols (methanol, ethanol, and n-propanol) are completely miscible with water. As the molecular mass of the alcohol increases, so does the proportion of hydrocarbon in the molecule. Correspondingly, the importance of hydrogen bonding and dipole–dipole interactions in the pure alcohol decreases, while the importance of London dispersion forces increases, which leads to progressively fewer favorable electrostatic interactions with water. Organic liquids such as acetone, ethanol, and tetrahydrofuran are sufficiently polar to be completely miscible with water yet sufficiently nonpolar to be completely miscible with all organic solvents. The same principles govern the solubilities of molecular solids in liquids. For example, elemental sulfur is a solid consisting of cyclic \(S_8\) molecules that have no dipole moment. Because the \(S_8\) rings in solid sulfur are held to other rings by London dispersion forces, elemental sulfur is insoluble in water. It is, however, soluble in nonpolar solvents that have comparable London dispersion forces, such as \(CS_2\) (23 g/100 mL). In contrast, glucose contains five –OH groups that can form hydrogen bonds. Consequently, glucose is very soluble in water (91 g/120 mL of water) but essentially insoluble in nonpolar solvents such as benzene. The structure of one isomer of glucose is shown here. Low-molecular-mass hydrocarbons with highly electronegative and polarizable halogen atoms, such as chloroform (\(CHCl_3\)) and methylene chloride (\(CH_2Cl_2\)), have both significant dipole moments and relatively strong London dispersion forces. These hydrocarbons are therefore powerful solvents for a wide range of polar and nonpolar compounds. Naphthalene, which is nonpolar, and phenol (\(C_6H_5OH\)), which is polar, are very soluble in chloroform. In contrast, the solubility of ionic compounds is largely determined not by the polarity of the solvent but rather by its dielectric constant, a measure of its ability to separate ions in solution, as you will soon see. Identify the most important solute–solvent interactions in each solution. : components of solutions : predominant solute–solvent interactions : Identify all possible intermolecular interactions for both the solute and the solvent: London dispersion forces, dipole–dipole interactions, or hydrogen bonding. Determine which is likely to be the most important factor in solution formation. Identify the most important interactions in each solution: : A solute can be classified as hydrophilic (literally, “water loving”), meaning that it has an electrostatic attraction to water, or hydrophobic (“water fearing”), meaning that it repels water. A hydrophilic substance is polar and often contains O–H or N–H groups that can form hydrogen bonds to water. For example, glucose with its five O–H groups is hydrophilic. In contrast, a hydrophobic substance may be polar but usually contains C–H bonds that do not interact favorably with water, as is the case with naphthalene and n-octane. Hydrophilic substances tend to be very soluble in water and other strongly polar solvents, whereas hydrophobic substances are essentially insoluble in water and soluble in nonpolar solvents such as benzene and cyclohexane. The difference between hydrophilic and hydrophobic substances has substantial consequences in biological systems. For example, vitamins can be classified as either fat soluble or water soluble. Fat-soluble vitamins, such as vitamin A, are mostly nonpolar, hydrophobic molecules. As a result, they tend to be absorbed into fatty tissues and stored there. In contrast, water-soluble vitamins, such as vitamin C, are polar, hydrophilic molecules that circulate in the blood and intracellular fluids, which are primarily aqueous. Water-soluble vitamins are therefore excreted much more rapidly from the body and must be replenished in our daily diet. A comparison of the chemical structures of vitamin A and vitamin C quickly reveals why one is hydrophobic and the other hydrophilic. Because water-soluble vitamins are rapidly excreted, the risk of consuming them in excess is relatively small. Eating a dozen oranges a day is likely to make you tired of oranges long before you suffer any ill effects due to their high vitamin C content. In contrast, fat-soluble vitamins constitute a significant health hazard when consumed in large amounts. For example, the livers of polar bears and other large animals that live in cold climates contain large amounts of vitamin A, which have occasionally proven fatal to humans who have eaten them. The following substances are essential components of the human diet: Using what you know of hydrophilic and hydrophobic solutes, classify each as water soluble or fat soluble and predict which are likely to be required in the diet on a daily basis. : chemical structures classification as water soluble or fat soluble; dietary requirement : Based on the structure of each compound, decide whether it is hydrophilic or hydrophobic. If it is hydrophilic, it is likely to be required on a daily basis. : These compounds are consumed by humans: caffeine, acetaminophen, and vitamin D. Identify each as primarily hydrophilic (water soluble) or hydrophobic (fat soluble), and predict whether each is likely to be excreted from the body rapidly or slowly. : Caffeine and acetaminophen are water soluble and rapidly excreted, whereas vitamin D is fat soluble and slowly excreted Solutions are not limited to gases and liquids; solid solutions also exist. For example, amalgams, which are usually solids, are solutions of metals in liquid mercury. Because most metals are soluble in mercury, amalgams are used in gold mining, dentistry, and many other applications. A major difficulty when mining gold is separating very small particles of pure gold from tons of crushed rock. One way to accomplish this is to agitate a suspension of the crushed rock with liquid mercury, which dissolves the gold (as well as any metallic silver that might be present). The very dense liquid gold–mercury amalgam is then isolated and the mercury distilled away. An alloy is a solid or liquid solution that consists of one or more elements in a metallic matrix. A solid alloy has a single homogeneous phase in which the crystal structure of the solvent remains unchanged by the presence of the solute. Thus the microstructure of the alloy is uniform throughout the sample. Examples are substitutional and interstitial alloys such as brass or solder. Liquid alloys include sodium/potassium and gold/mercury. In contrast, a partial alloy solution has two or more phases that can be homogeneous in the distribution of the components, but the microstructures of the two phases are not the same. As a liquid solution of lead and tin is cooled, for example, different crystalline phases form at different cooling temperatures. Alloys usually have properties that differ from those of the component elements. Network solids such as diamond, graphite, and \(ce{SiO_2}\) are insoluble in all solvents with which they do not react chemically. The covalent bonds that hold the network or lattice together are simply too strong to be broken under normal conditions. They are certainly much stronger than any conceivable combination of intermolecular interactions that might occur in solution. Most metals are insoluble in virtually all solvents for the same reason: the delocalized metallic bonding is much stronger than any favorable metal atom–solvent interactions. Many metals react with solutions such as aqueous acids or bases to produce a solution. However, in these instances the metal undergoes a chemical transformation that cannot be reversed by simply removing the solvent. Solids with very strong intermolecular bonding tend to be insoluble. Ionic substances are generally most soluble in polar solvents; the higher the lattice energy, the more polar the solvent must be to overcome the lattice energy and dissolve the substance. Because of its high polarity, water is the most common solvent for ionic compounds. Many ionic compounds are soluble in other polar solvents, however, such as liquid ammonia, liquid hydrogen fluoride, and methanol. Because all these solvents consist of molecules that have relatively large dipole moments, they can interact favorably with the dissolved ions. The interaction of water with Na+ and Cl− ions in an aqueous solution of \(\ce{NaCl}\) was illustrated in Figure 4.3. The ion–dipole interactions between Li+ ions and acetone molecules in a solution of LiCl in acetone are shown in Figure \(\Page {4}\). The energetically favorable Li+–acetone interactions make \(ΔH_3\) in Figure 13.1 sufficiently negative to overcome the positive \(ΔH_1\) and \(ΔH_2\). Because the dipole moment of acetone (2.88 D), and thus its polarity, is actually larger than that of water (1.85 D), one might even expect that LiCl would be more soluble in acetone than in water. In fact, the opposite is true: 83 g of LiCl dissolve in 100 mL of water at 20°C, but only about 4.1 g of LiCl dissolve in 100 mL of acetone. This apparent contradiction arises from the fact that the dipole moment is a property of a single molecule in the gas phase. A more useful measure of the ability of a solvent to dissolve ionic compounds is its dielectric constant (ε), which is the ability of a bulk substance to decrease the electrostatic forces between two charged particles. By definition, the dielectric constant of a vacuum is 1. In essence, a solvent with a high dielectric constant causes the charged particles to behave as if they have been moved farther apart. At 25°C, the dielectric constant of water is 80.1, one of the highest known, and that of acetone is only 21.0. Hence water is better able to decrease the electrostatic attraction between Li+ and Cl− ions, so LiCl is more soluble in water than in acetone. This behavior is in contrast to that of molecular substances, for which polarity is the dominant factor governing solubility. A solvent’s dielectric constant is the most useful measure of its ability to dissolve ionic compounds. A solvent’s polarity is the dominant factor in dissolving molecular substances. The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. The solubility of a substance in a liquid is determined by intermolecular interactions, which also determine whether two liquids are miscible. Solutes can be classified as hydrophilic (water loving) or hydrophobic (water fearing). Vitamins with hydrophilic structures are water soluble, whereas those with hydrophobic structures are fat soluble. Many metals dissolve in liquid mercury to form amalgams. Covalent network solids and most metals are insoluble in nearly all solvents. The solubility of ionic compounds is largely determined by the dielectric constant (ε) of the solvent, a measure of its ability to decrease the electrostatic forces between charged particles. Solutions of many ionic compounds in organic solvents can be dissolved using crown ethers, cyclic polyethers large enough to accommodate a metal ion in the center, or cryptands, compounds that completely surround a cation. | 17,389 | 1,781 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Thermodynamics_and_Structural_Consequences_of_d-Orbital_Splitting |
The energy level splitting of the d-orbitals due to their interaction with the ligands in a complex has important structural and thermodynamic effects on the chemistry of transition-metal complexes. Although these two types of effects are interrelated, they are considered separately here. There are two major kinds of structural effects: effects on the ionic radii of metal ions with regular octahedral or tetrahedral geometries, and structural distortions observed for specific electron configurations. Figure \(\Page {1}\) is a plot of the ionic radii of the divalent fourth-period metal ions versus atomic number. Z 3d Cr Cu A similar effect is observed for the V ion, which has a d configuration. Because the three electrons in the t orbitals provide essentially no shielding of the ligands from the metal, the ligands experience the full increase of +1 in nuclear charge that occurs from Ti to V . Consequently, the observed ionic radius of the V ion is significantly smaller than that of the Ti ion. Skipping the Cr ion for the moment, consider the d Mn ion. Because the nuclear charge increases by +2 from V to Mn , Mn might be expected to be smaller than V . The two electrons added from V to Mn occupy the e orbitals, however, which are oriented directly toward the six ligands. Because these electrons are localized directly between the metal ion and the ligands, they are effective at screening the ligands from the increased nuclear charge. As a result, the ionic radius actually increases significantly from V to Mn , despite the higher nuclear charge of the latter. The same effects are observed in the second half of the first-row transition metals. In the Fe , Co , and Ni ions, the extra electrons are added successively to the t orbitals, resulting in increasingly poor shielding of the ligands from the nuclei and in abnormally small ionic radii. Skipping over Cu , adding the last two electrons causes a significant increase in the ionic radius of Zn , despite its greater nuclear charge. Because simple octahedral complexes are not observed for the Cr and Cu ions, only estimated values for their radii are shown in Figure \(\Page {1}\). Since both Cr and Cu ions have electron configurations with an odd number of electrons in the e orbitals. Because the single electron (in the case of Cr ) or the third electron (in the case of Cu ) can occupy either one of two degenerate e orbitals, both systems have degenerate ground states. The Jahn–Teller theorem states that such non-linear systems are not stable; they undergo a distortion that makes the complex less symmetrical and splits the degenerate states, which decreases the energy of the system. The distortion and resulting decrease in energy are collectively referred to as the Jahn–Teller effect. Neither the nature of the distortion nor its magnitude is specified, and in fact, they are difficult to predict. In principle, Jahn–Teller distortions are possible for many transition-metal ions; in practice, however, they are observed only for systems with an odd number of electrons in the e orbitals, such as the Cr and Cu ions. Consider an octahedral Cu complex, [Cu(H O) ] , which is elongated along the z axis. As indicated in Figure \(\Page {2}\), this kind of distortion splits both the e and t sets of orbitals. Because the axial ligands interact most strongly with the d orbital, the splitting of the e set (δ ) is significantly larger than the splitting of the t set (δ ), but both δ and δ are much, much smaller than the Δ . This splitting does not change the centerpoint of the energy within each set, so a Jahn–Teller distortion results in no net change in energy for a filled or half-filled set of orbitals. If, however, the e set contains one electron (as in the d ions, Cr and Mn ) or three electrons (as in the d ion, Cu ), the distortion decreases the energy of the system. For Cu , for example, the change in energy after distortion is 2(−δ /2) + 1(δ /2) = −δ /2. For Cu complexes, the observed distortion is always an elongation along the z axis by as much as 50 pm; in fact, many Cu complexes are distorted to the extent that they are effectively square planar. In contrast, the distortion observed for most Cr complexes is a compression along the z axis. In both cases, however, the net effect is the same: the distorted system is more stable than the undistorted system. are most important for d and high-spin d complexes; the distorted system is more stable than the undistorted one. Increasing the axial metal–ligand distances in an octahedral d complex is an example of a Jahn–Teller distortion, which causes the degenerate pair of e orbitals to split in energy by an amount δ ; δ and δ are much smaller than Δ . As a result, the distorted system is more stable (lower in energy) than the undistorted complex by δ /2. As previously noted, crystal field splitting energies (CFSEs) can be as large as several hundred kilojoules per mole, which is the same magnitude as the strength of many chemical bonds or the energy change in most chemical reactions. Consequently, CFSEs are important factors in determining the magnitude of hydration energies, lattice energies, and other thermodynamic properties of the transition metals. The hydration energy of a metal ion is defined as the change in enthalpy for the following reaction: \[M^{2+}_{(g)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} \label{1.1.1}\] Although hydration energies cannot be measured directly, they can be calculated from experimentally measured quantities using thermochemical cycles. In Figure \(\Page {3a}\), a plot of the hydration energies of the fourth-period metal dications versus atomic number forms a curve with two valleys. Note the relationship between the plot in Figure \(\Page {3a}\) and the plot of ionic radii in Figure \(\Page {1}\) the overall shapes are essentially identical, and only the three cations with spherically symmetrical distributions of d electrons (Ca , Mn , and Zn ) lie on the dashed lines. In Figure \(\Page {3a}\), the dashed line corresponds to hydration energies calculated based solely on electrostatic interactions. Subtracting the CFSE values for the [M(H O) ] ions from the experimentally determined hydration energies gives the points shown as open circles, which lie very near the calculated curve. Therefore, CFSEs are primarily responsible for the differences between the measured and calculated values of hydration energies. Values of the lattice energies for the fourth-period metal dichlorides are plotted against atomic number in part (b) of Figure \(\Page {3}\). Recall that the is defined as the negative of the enthalpy change for the reaction below. Like hydration energies, lattice energies are determined indirectly from a thermochemical cycle. \[M^{2+} (g) + 2Cl^− (g) \rightarrow MCl_2 (s) \label{1.1.2}\] The shape of the lattice-energy curve is essentially the mirror image of the hydration-energy curve in part (a) of Figure \(\Page {3}\), with only Ca , Mn , and Zn lying on the smooth curve. It is not surprising that the explanation for the deviations from the curve is exactly the same as for the hydration energy data: all the transition-metal dichlorides, except MnCl and ZnCl , are more stable than expected due to CFSE. Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δ . The crystal field stabilization energy (CFSE) is the additional stabilization of a complex due to placing electrons in the lower-energy set of d orbitals. CFSE explains the unusual curves seen in plots of ionic radii, hydration energies, and lattice energies versus atomic number. The Jahn–Teller theorem states that a non-linear molecule with a spatially degenerate electronic ground state undergoes a geometrical distortion to remove the degeneracy and lower the overall energy of the system. | 8,180 | 1,782 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Elimination_by_the_E2_mechanism |
At the beginning of this module, we saw the following reaction between tert-butyl bromide and cyanide. Clearly, this is not a substitution reaction. Below is a mechanistic diagram of an elimination reaction by the E2 pathway: . To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanim of this reaction is single-step, and is referred to as the E2 mechanism. In the methanol solvent used here, methanethiolate has greater nucleophilicity than methoxide by a factor of 100. Methoxide, on the other hand is roughly 10 times more basic than methanethiolate. As a result, we see a clear-cut difference in the reaction products, which reflects (bonding to an electrophilic carbon) versus (bonding to a proton). Kinetic studies of these reactions show that they are both second order (first order in R–Br and first order in Nu: ), suggesting a bimolecular mechanism for each. The substitution reaction is clearly . The corresponding designation for the elimination reaction is . An energy diagram for the single-step bimolecular E2 mechanism is shown on the right. We should be aware that the E2 transition state is less well defined than is that of S 2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the R–groups on the beta-carbon enhance the acidity of that hydrogen, then substantial breaking of C–H may occur before the other bonds begin to be affected. Similarly, groups that favor ionization of the halogen may generate a transition state with substantial positive charge on the alpha-carbon and only a small degree of C–H breaking. For most simple alkyl halides, however, it is proper to envision a balanced transition state, in which there has been an equal and synchronous change in all the bonds. Such a model helps to explain an important regioselectivity displayed by these elimination reactions. If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below. By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the . The main factor contributing to Zaitsev Rule behavior is the stability of the alkene. We noted earlier that carbon-carbon double bonds are stabilized (thermodynamically) by alkyl substituents, and that this stabilization could be evaluated by appropriate measurements. Since the E2 transition state has significant carbon-carbon double bond character, alkene stability differences will be reflected in the transition states of elimination reactions, and therefore in the activation energy of the rate-determining steps. From this consideration we anticipate that if two or more alkenes may be generated by an E2 elimination, the more stable alkene will be formed more rapidly and will therefore be the predominant product. This is illustrated for 2-bromobutane by the energy diagram on the right. The propensity of E2 eliminations to give the more stable alkene product also influences the distribution of product stereoisomers. In the elimination of 2-bromobutane, for example, we find that trans-2-butene is produced in a 6:1 ratio with its cis-isomer. The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used. Thus hydroxide, methoxide and ethoxide bases give comparable results. Bulky bases such as tert-butoxide tend to give higher yields of the less substituted double bond isomers, a characteristic that has been attributed to steric hindrance. In the case of 2-bromo-2,3-dimethylbutane, described above, tert-butoxide gave a 4:1 ratio of 2,3-dimethyl-1-butene to 2,3-dimethyl-2-butene ( essentially the opposite result to that obtained with hydroxide or methoxide). This point will be discussed further once we know more about the the structure of the E2 transition state. The importance of maintaining a planar configuration of the trigonal double-bond carbon components must never be overlooked. For optimum pi-bonding to occur, the p-orbitals on these carbons must be parallel, and the resulting doubly-bonded planar configuration is more stable than a twisted alternative by over 60 kcal/mole. This structural constraint is responsible for the existence of alkene stereoisomers when substitution patterns permit. It also prohibits certain elimination reactions of bicyclic alkyl halides, that might be favorable in simpler cases. For example, the bicyclooctyl 3º-chloride shown below appears to be similar to tert-butyl chloride, but it does not undergo elimination, even when treated with a strong base (e.g. KOH or KOC H ). There are six equivalent beta-hydrogens that might be attacked by base (two of these are colored blue as a reference), so an E2 reaction seems plausible. The problem with this elimination is that the resulting double bond would be constrained in a severely twisted (non-planar) configuration by the bridged structure of the carbon skeleton. The carbon atoms of this twisted double-bond are colored red and blue respectively, and a Newman projection looking down the twisted bond is drawn on the right. Because a pi-bond cannot be formed, the hypothetical alkene does not exist. Structural prohibitions such as this are often encountered in small bridged ring systems, and are referred to as . Bredt's Rule should not be applied blindly to all bridged ring systems. If large rings are present their conformational flexibility may permit good overlap of the p-orbitals of a double bond at a bridgehead. This is similar to recognizing that trans-cycloalkenes cannot be prepared if the ring is small (3 to 7-membered), but can be isolated for larger ring systems. The anti-tumor agent taxol has such a bridgehead double bond (colored red), as shown in the following illustration. The bicyclo[3.3.1]octane ring system is the smallest in which bridgehead double bonds have been observed. The drawing to the right of taxol shows this system. The bridgehead double bond (red) has a cis-orientation in the six-membered ring (colored blue), but a trans-orientation in the larger eight-membered ring. E2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram. Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state. The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates. A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the S 2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp to sp and the developing π-bond is drawn as dashed light blue lines. The symbol represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints -elimination has been observed. | 11,694 | 1,783 |
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In Gas Expansion, we assume Ideal behavior for the two types of expansions: This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant. The graphs clearly show work done (area under the curve) is greater in a reversible process. Adiabatic means when no heat exchange occurs during expansion between system and surrounding and the temperature is no longer held constant. This equation shows the relationship between PV and is useful only when it applies to ideal gas and reversible adiabatic change. The equation is very similar to Boyle's law except it has exponent (gamma) due to change in temperature. The work done by an adiabatic reversible process is given by the following equation: where T is less than T . The internal energy of the system decreases as the gas expands. The work can be calculated in two ways because the Internal energy (U) does not depend on path. The graph shows that less work is done in an adiabatic reversible process than an Isothermal reversible process. | 1,336 | 1,784 |
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Although the VSEPR model is a simple and useful method for qualitatively predicting the structures of a wide range of compounds, it is infallible. It predicts, for example, that H S and PH should have structures similar to those of \(\ce{H2O}\) and \(\ce{NH3}\), respectively. In fact, structural studies have shown that the H–S–H and H–P–H angles are more than 12° smaller than the corresponding bond angles in \(\ce{H2O}\) and \(\ce{NH3}\). More disturbing, the VSEPR model predicts that the simple group 2 halides (MX ), which have four valence electrons, should all have linear X–M–X geometries. Instead, many of these species, including \(\ce{SrF2}\) and \(\ce{BaF2}\), are significantly bent. A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called , which results in the formation of stronger bonds. As we have talked about using Lewis structures to depict the bonding in covalent compounds, we have been very vague in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together? The is introduced to describe bonding in covalent molecules. In this model, bonds are considered to form from the overlapping of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as H or HF, our present understanding of s and p atomic orbitals will suffice. To explain the bonding in organic molecules, however, we will need to introduce the concept of . The simplest case to consider is the hydrogen molecule, H . When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1 orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals. In simple terms, we can say that both electrons now spend more time the two nuclei and thus hold the atoms together. As we will see, the situation is not quite so simple as that, because the electron pair must still obey quantum mechanics - that is, the two electrons must now occupy a . This will be the essential principle of valence bond theory. How far apart are the two nuclei? That is a very important issue to consider. If they are too far apart, their respective 1 orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second. However, something else is happening at the same time: as the atoms get closer, the positive-positive interaction between the two nuclei also begins to increase. At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have a very unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the . For the H molecule, this distance is 74 x 10 meters, or 0.74 Å (Å means angstrom, or 10 meters). Likewise, the difference in potential energy between the lowest state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the . For the hydrogen molecule, this energy is equal to about 104 kcal/mol. Every covalent bond in a given molecule has a characteristic length and strength. In general, carbon-carbon single bonds are about 1.5 Å long (Å means angstrom, or 10 meters) while carbon-carbon double bonds are about 1.3 Å, carbon-oxygen double bonds are about 1.2 Å, and carbon-hydrogen bonds are in the range of 1.0 – 1.1 Å. Most covalent bonds in organic molecules range in strength from just under 100 kcal/mole (for a carbon-hydrogen bond in ethane, for example) up to nearly 200 kcal/mole. You can refer to tables in reference books such as the CRC Handbook of Chemistry and Physics for extensive lists of bond lengths, strengths, and many other data for specific organic compounds. Although we tend to talk about "bond length" as a specific distance, it is not accurate to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as which have a defined length when relaxed, but which can be compressed, extended, and bent. This ‘springy’ picture of covalent bonds will become very important, when we study the analytical technique known as infrared (IR) spectroscopy. You learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm (Figure \(\Page {1}\)). According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a , in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1 atomic orbital of one hydrogen atom overlaps with the singly occupied 1 atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a (Figure \(\Page {1}\)). Although Lewis and VSEPR structures also contain localized electron-pair bonds, neither description uses an atomic orbital approach to predict the stability of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theory, which is built on two assumptions: Figure \(\Page {2}\) shows an electron-pair bond formed by the overlap of two atomic orbitals, two atomic orbitals, and an and an orbital where = 2. Notice that bonding overlap occurs when the interacting atomic orbitals have the correct orientation (are "pointing at" each other) and are (represented by colors in Figure \(\Page {2}\) ). Maximum overlap occurs between orbitals with the same spatial orientation and similar energies. Let’s examine the bonds in BeH , for example. According to the VSEPR model, BeH is a linear compound with four valence electrons and two Be–H bonds. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1 2 electron configuration, and each H atom has a 1 electron configuration. Because the Be atom has a filled 2 subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1 orbitals on the H atoms. If a singly occupied 1 orbital on hydrogen were to overlap with a filled 2 orbital on beryllium, the resulting bonding orbital would contain electrons, but the maximum allowed by quantum mechanics is . How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2 electrons into an empty 2 orbital and reverse its spin, in a process called promotion: In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2 and one of the 2 orbitals), each of which could overlap with a singly occupied 1 orbital of an H atom to form an electron-pair bond. Although this would produce \(\ce{BeH2}\), the two Be–H bonds would not be equivalent: the 1 orbital of one hydrogen atom would overlap with a Be 2 orbital, and the 1 orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2 orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as \(\ce{BeH2}\) form, scientists developed the concept of hybridization. by (University of Minnesota, Morris) | 8,616 | 1,785 |
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Just as there were two mechanisms for nucleophilic substitution, there are two elimination mechanisms. The E1 mechanism is nearly identical to the S 1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the S 1 reaction, or a Brønsted acid, as in the E1 reaction. Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an S 1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific. (CH ) – + H [ (CH ) ] + + H (CH ) – H + (CH ) =CH + H + H To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes: Since the S 1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place. | 1,368 | 1,787 |
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One of the pioneers in the field of modern thermodynamics was James P. Joule (1818 - 1889). Among the experiments Joule carried out, was an attempt to measure the effect on the temperature of a sample of water that was caused by doing work on the water. Using a clever apparatus to perform work on water by using a falling weight to turn paddles within an insulated canister filled with water, Joule was able to measure a temperature increase in the water. Thus, Joule was able to show that work and heat can have the same effect on matter – a change in temperature! It would then be reasonable to conclude that heating, as well as doing work on a system will increase its energy content, and thus it’s ability to perform work in the surroundings. This leads to an important construct of the : The capacity of a system to do work is increased by heating the system or doing work on it. The (U) of a system is a measure of its capacity to supply energy that can do work within the surroundings, making U the ideal variable to keep track of the flow of heat and work energy into and out of a system. Changes in the internal energy of a system (\(\Delta U\)) can be calculated by \[\Delta U = U_f - U_i \label{FirstLaw} \] where the subscripts \(i\) and \(f\) indicate initial and final states of the system. \(U\) as it turns out, is a state variable. In other words, the amount of energy available in a system to be supplied to the surroundings is independent on how that energy came to be available. That’s important because the manner in which energy is transferred is path dependent. There are two main methods energy can be transferred to or from a system. These are suggested in the previous statement of the first law of thermodynamics. Mathematically, we can restate the first law as \[\Delta U = q + w \nonumber \] or \[dU = dq + dw \nonumber \] where q is defined as the amount of energy that flows into a system in the form of and w is the amount of energy lost due to the system doing on the surroundings. Heat is the kind of energy that in the absence of other changes would have the effect of changing the temperature of the system. A process in which heat flows into a system is from the standpoint of the system (\(q_{system} > 0\), \(q_{surroundings} < 0\)). Likewise, a process in which heat flows out of the system (into the surroundings) is called (\(q_{system} < 0\), \(q_{surroundings} > 0\)). In the absence of any energy flow in the form or work, the flow of heat into or out of a system can be measured by a change in temperature. In cases where it is difficult to measure temperature changes of the system directly, the amount of heat energy transferred in a process can be measured using a change in temperature of the soundings. (This concept will be used later in the discussion of calorimetry). An infinitesimal amount of heat flow into or out of a system can be related to a change in temperature by \[dq = C\, dT \nonumber \] where C is the and has the definition \[ C = \dfrac{dq}{\partial T} \nonumber \] Heat capacities generally have units of (J mol K ) and magnitudes equal to the number of J needed to raise the temperature of 1 mol of substance by 1 K. Similar to a heat capacity is a which is defined per unit mass rather than per mol. The specific heat of water, for example, has a value of 4.184 J g K (at constant pressure – a pathway distinction that will be discussed later.) How much energy is needed to raise the temperature of 5.0 g of water from 21.0 °C to 25.0 °C? \[\begin{align*} q &=mC \Delta T \\[4pt] &= (5.0 \,\cancel{g}) (4.184 \dfrac{J}{\cancel{g} \, \cancel{°C}}) (25.0 \cancel{°C} - 21.0 \cancel{°C}) \\[4pt] &= 84\, J \end{align*} \] A partial derivative, like a total derivative, is a slope. It gives a magnitude as to how quickly a function changes value when one of the dependent variables changes. Mathematically, a partial derivative is defined for a function \(f(x_1,x_2, \dots x_n)\) by \[\left( \dfrac{ \partial f}{\partial x_i} \right)_{x_j \neq i} = \lim_{\Delta _i \rightarrow 0} \left( \dfrac{f(x_1+ \Delta x_1 , x_2 + \Delta x_2, \dots, x_i +\Delta x_i, \dots x_n+\Delta x_n) - f(x_1,x_2, \dots x_i, \dots x_n) }{\Delta x_i} \right) \nonumber \] Because it measures how much a function changes for a change in a given dependent variable, infinitesimal changes in the in the function can be described by \[ df = \sum_i \left( \dfrac{\partial f}{\partial x_i} \right)_{x_j \neq i} \nonumber \] So that each contribution to the total change in the function \(f\) can be considered separately. For simplicity, consider an ideal gas. The pressure can be calculated for the gas using the ideal gas law. In this expression, pressure is a function of temperature and molar volume. \[ p(V,T) = \dfrac{RT}{V} \nonumber \] The partial derivatives of p can be expressed in terms of \(T\) and \(V\) as well. \[ \left( \dfrac{\partial p}{ \partial V} \right)_{T} = - \dfrac{RT}{V^2} \label{max1} \] and \[ \left( \dfrac{\partial p}{ \partial T} \right)_{V} = \dfrac{R}{V} \label{max2} \] So that the change in pressure can be expressed \[ dp = \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \label{eq3} \] or by substituting Equations \ref{max1} and \ref{max2} \[ dp = \left( - \dfrac{RT}{V^2} \right ) dV + \left( \dfrac{R}{V} \right) dT \nonumber \] Macroscopic changes can be expressed by integrating the individual pieces of Equation \ref{eq3} over appropriate intervals. \[ \Delta p = \int_{V_1}^{V_2} \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \int_{T_1}^{T_2} \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \nonumber \] This can be thought of as two consecutive changes. The first is an (constant temperature) expansion from \(V_1\) to \(V_2\) at \(T_1\) and the second is an (constant volume) temperature change from \(T_1\) to \(T_2\) at \(V_2\). For example, suppose one needs to calculate the change in pressure for an ideal gas expanding from 1.0 L/mol at 200 K to 3.0 L/mol at 400 K. The set up might look as follows. \[ \Delta p = \underbrace{ \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV}_{\text{isothermal expansion}} + \underbrace{ \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT}_{\text{isochoric heating}} \nonumber \] or \[ \begin{align*} \Delta p &= \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV + \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT \\[4pt] &= \left[ \dfrac{R(200\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} + \left[ \dfrac{RT}{3.0 \, L/mol} \right]_{ 200\,K}^{400\,K} \\[4pt] &= R \left[ \left( \dfrac{200\,K}{3.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{200\,K}{3.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \] Alternatively, one could calculate the change as an isochoric temperature change from \(T_1\) to \(T_2\) at \(V_1\) followed by an isothermal expansion from \(V_1\) to \(V_2\) at \(T_2\): \[ \Delta p = \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT + \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV \nonumber \] \[ \begin{align*} \Delta p &= \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT + \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV \\[4pt] &= \left[ \dfrac{RT}{1.0 \, L/mol} \right]_{ 200\,K}^{400\,K} + \left[ \dfrac{R(400\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} \\[4pt] &= R \left[ \left( \dfrac{400\,K}{1.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{400\,K}{1.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \] Work can take several forms, such as expansion against a resisting pressure, extending length against a resisting tension (like stretching a rubber band), stretching a surface against a surface tension (like stretching a balloon as it inflates) or pushing electrons through a circuit against a resistance. The key to defining the work that flows in a process is to start with an infinitesimal amount of work defined by what is changing in the system. The pattern followed is always an infinitesimal displacement multiplied by a resisting force. The total work can then be determined by integrating along the pathway the change follows. What is the work done by 1.00 mol an ideal gas expanding from a volume of 22.4 L to a volume of 44.8 L against a constant external pressure of 0.500 atm? \[dw = -p_{ext} dV \nonumber \] since the pressure is constant, we can integrate easily to get total work \[ \begin{align*} w &= -p_{exp} \int_{V_1}^{V_2} dV \\[4pt] &= -p_{exp} ( V_2-V_1) \\[4pt] &= -(0.500 \,am)(44.8 \,L - 22.4 \,L) \left(\dfrac{8.314 \,J}{0.08206 \,atm\,L}\right) \\[4pt] &= -1130 \,J = -1.14 \;kJ \end{align*} \] : The ratio of gas law constants can be used to convert between atm∙L and J quite conveniently! | 8,921 | 1,788 |
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Demo: Add 50.0 mL (about 39.5 g, 1.23 mol) of Methanol (CH OH) to one 50 mL volumetric flask and 50.0 mL (50 g, 2.77 mol) of water to another, then combine in a 100.0 mL volumetric flask . When solids and liquids are mixed, the total volume may be more than or less than the sum of the volumes (whether they're pure or solutions). This demonstration allows a graphic introduction to concentration terms: 1. What is the molar concentration? Note that additional solvent must be added to the 100 mL volumetric flask to allow accurate determination of the solution volume. 2. Which is the solvent? 3. What is the molal concentration? Note: the molal concentration must be calculated before the volume is brought up to 100 mL, or the mass of additional water must be determined by weighing the flask. 4. What is the percent by mass and Volume? | 851 | 1,789 |
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pH indicators are weak acids that exist as natural dyes and indicate the concentration of H (\(H_3O^+\)) ions in a solution via color change. A pH value is determined from the negative logarithm of this concentration and is used to indicate the acidic, basic, or neutral character of the substance you are testing. pH indicators exist as liquid dyes and dye-infused paper strips. They are added to various solutions to determine the pH values of those solutions. Whereas the liquid form of pH indicators is usually added directly to solutions, the paper form is dipped into solutions and then removed for comparison against a color/pH key. Very Acidic Acidic Neutral Basic Very Basic See Figure 1 and 2 to see a color range (1) of a universal indicator (2). Recall that the value of pH is related to the concentration of H (\(H_3O^+\)) of a substance. pH itself is approximated as the cologarithm or negative logarithm of the \(H^+\) ion concentration (Figure 3). \[pH \approx -log[H_3O^+] \tag{3} \] A pH of 7 indicates a neutral solution like water. A pH less than 7 indicates an acidic solution and a pH greater than 7 indicates a basic solution. Ultimately, the pH value indicates how much H has dissociated from molecules within a solution. The lower the pH value, the higher concentration of H ions in the solution and the stronger the acid. Likewise, the higher the pH value, the lower the concentration of H ions in the solution and the weaker the acid. The color change of a pH indicator is caused by the dissociation of the H ion from the indicator itself. Recall that pH indicators are not only natural dyes but also weak acids. The dissociation of the weak acid indicator causes the solution to change color. The equation for the dissociation of the H ion of the pH indicator is show below (Figure 4). \[HIn + H_2O \rightleftharpoons H_3O^+ + In^- \tag{4} \] with It is important here to note that the equation expressed in figure 4 is in equilibrium, meaning applies to it. Thus, as the concentration of \(H_3O^+\) (H ) increases or decreases, the equilibrium shifts to the left or right accordingly. An in the \(HIn\) acid concentration causes the equilibrium to shift to the (towards products), whereas an of the \(In^-\) base concentration causes the equilibrium to shift to the (towards reactants). pH indicators are specific to the range of pH values one wishes to observe. For example, common indicators such as phenolphthalein, methyl red, and bromothymol blue are used to indicate pH ranges of about 8 to 10, 4.5 to 6, and 6 to 7.5 accordingly. On these ranges, phenolphthalein goes from colorless to pink, methyl red goes from red to yellow, and bromothymol blue goes from yellow to blue. For universal indicators, however, the pH range is much broader and the number of color changes is much greater. See figures 1 and 2 in the introduction for visual representations. Usually, universal pH indicators are in the paper strip form. It is important to note that the is a : hence an increase of 1 pH unit corresponds to a ten times increase of \(H_3O^+\). For example, a solution with a pH of 3 will have an H (\(H_3O^+\)) concentration greater than that of a solution with a pH of 4. As pH is the negative logarithm of the H (\(H_3O^+\)) concentration of a foreign substance, the lower the pH value, the higher the concentration of H (\(H_3O^+\)) ions and the stronger the acid. Additionally, the higher the pH value, the lower the H (\(H_3O^+\)) concentration and the stronger the base. pH indicators can be used in a variety of ways, including measuring the pH of farm soil, shampoos, fruit juices, and bodies of water. Additionally, pH indicators can be found in nature, so therefore their presence in plants and flowers can indicate the pH of the soil from which they grow. Nature contains several natural pH indicators as well: for example, some flower petals (especially Roses and Hydrangeas), certain fruits (cherries, strawberries) and leaves can change color if the pH of the soil changes. See figure 7. (7)
In the lemon juice experiment, the pH paper turns from blue to vivid red, indicating the presence of \(H_3O^+\) ions: lemon juice is . Refer to the table of Universal Indicator Color change (figure 1 in the introduction) for clarification. The household detergent contained a concentrated solution of sodium bicarbonate, commonly known as baking soda. As shown, the pH paper turns a dark blue: baking soda (in solution) is .Refer to the table of Universal Indicator Color change (figure 1 in the introduction) for clarification. Here is a closer look of the pH papers before and after dipping them in the lemon juice and cleaning detergent (Figure 10):
neutral acidic neutral basic Figure 10: Here is a simple demonstration that you could try in the lab or at home to get a better sense of how indicator paper works. Make sure to always wear safety glasses and gloves when performing an experiment! Materials Procedure 1. A hair stylist walks into a store and wants to buy a shampoo with slightly acidic/neutral pH for her hair. She finds 5 brands that she really likes, but since she never took any introductory chemistry classes, she is unsure about which one to purchase. The first has a pH of , the second of , the third of , the fourth of and the fifth of . Which one should she buy? The brand that has a pH of 6.8 since it's under 7 (neutral) but very close to it, making it slightly acidic. 2. You decide to test the pH of your brand new swimming pool on your own. The instruction manual advises to keep it between 7.2-7.6. Shockingly, you realize it's set at 8.3! Horrified, you panic and are unsure whether you should add some basic or acidic chemicals in your pool (being mindful of the dose, of course. Those specific chemicals are included in the set, so no need to worry about which one you have to use and (eek!) if they are legal for public use). Which one should you add? Since the goal is to lower the pH to its ideal value, we must add to the pool. 3. Let's say the concentration of Hydronium ions in an aqueous solution is 0.033 mol/L. What is the corresponding pH of this solution, and based on your answer identify whether the solution is acidic, basic or neutral. Answer: Using the formula \(pH \approx -log[H_3O+]\) pH= -log[0.033]= : The solution is highly acidic! 4. Now let's do the inverse: Say you have a solution with a pH of 9.4. What is the H 0 ions concentration? [\(H_3O^+\)]= 10 = . Seem too low to be true? Think again, if the pH is >7, the solution will be , hence the hydronium ions will be low compared to the hydroxide (OH ions). 5. A more trickier one: 0.00026 moles of acetic acid are added to 2.5 L of water. What is the pH of the solution? M=n/L : M = 0.00026/2.5 = pH= -log[1.04E-4]= | 6,829 | 1,793 |
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Comparison of the physical properties of alcohols with those of hydrocarbons of comparable molecular weight shows several striking differences, especially for those with just a few carbons. Alcohols are substantially less volatile, have higher melting points, and greater water solubility than the corresponding hydrocarbons (see Table 15-1), although the differences become progressively smaller as molecular weight increases. The reason for these differences in physical properties is related to the high polarity of the hydroxyl group which, when substituted on a hydrocarbon chain, confers a measure of polar character to the molecule. As a result, there is a significant attraction of one molecule for another that is particularly pronounced in the solid and liquid states. This polar character leads to association of alcohol molecules through the rather positive hydrogen of one hydroxyl group with a correspondingly negative oxygen of another hydroxyl group: This type of association is called “hydrogen bonding,” and, although the strengths of such bonds are much less than those of most conventional chemical bonds, they are still significant (about \(5\) to \(10 \: \text{kcal}\) per bond). Clearly then, the reason alcohols have higher boiling points than corresponding alkyl halides, ethers, or hydrocarbons is because, for the molecules to vaporize, additional energy is required to break the hydrogen bonds. Alternatively, association through hydrogen bonds may be regarded as effectively raising the molecular weight, thereby reducing volatility (also see ). The water solubility of the lower-molecular-weight alcohols is pronounced and is understood readily as the result of hydrogen bonding with water molecules: and (1977) | 1,761 | 1,794 |
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Define and distinguish hazard and risk. Distinguish predictive tools (toxicity tests) and diagnostic tools (bioassays). What is the essence of diagnostic risk assessment? List and briefly describe bioassays at different levels of biological organisation. Name and briefly describe two approaches that aid to bridge policy goals and ecosystem responses to perturbation. Name advantages and disadvantages of effect-based and chemical-based monitoring strategies? Name at least three characteristics that make bioassays suitable for effect-based monitoring. Can the principle of the EROD assay also be used to develop a reporter gene assay? Explain your answer. In the benchmark approach (see text), toxicity profiles from sampling locations are compared to a reference profile. Should the reference profile always correspond to a clean situation? Motivate your answer. : Marja Lamoree : Timo Hamers, Jana Weiss You should be able to : extraction, bioassay testing, fractionation, identification, confirmation In general, the quality of the environment may be monitored by two complementary approaches: i) quantitative chemical analysis of selected (priority) pollutants and ii) effect-based monitoring using in vitro/vivo bioassays. Compared to the more classical chemical analytical approach that has been used for decades, effect-based monitoring is currently applied in an explorative manner and has not yet matured into a routinely implemented monitoring tool that is anchored in legislation. However, in an international framework, developments to formalize the role of effect-based monitoring and to standardize the use of bioassay testing for environmental quality assessment are underway. A weakness of the chemical approach is that because of the preselection of target compounds for quantitative analysis other compounds that are of relevance for the environmental quality may be missed. In comparison, inclusiveness is one of the advantages of effect-based monitoring: all compounds - and not only a few pre-defined ones - having a specific effect will contribute to the total, measured biological activity (see Section ). In turn, the effect-based approach strongly benefits from chemical analytical support to pinpoint which compounds are responsible for the observed activity and to be able to take measures for environmental protection, e.g. the reduction of the emission or discharge of a specific toxic compound into the environment. In Effect-Directed Analysis (EDA), the strengths of analytical chemical techniques and effect-based testing are combined with the aim to identify novel compounds that show activity in a biological analysis and that would have gone unnoticed using the chemical and the effect-based approach separately. A schematic representation of EDA is shown in Figure 1 and the various steps are described below in more detail. There is no limitation regarding the sample matrix: EDA has been applied to e.g. water, soil/sediment and biota samples. It is used for in-depth investigations at locations that are suspected to be contaminated but where the compounds responsible for the observed adverse effects are not known. In addition to environmental quality assessment, EDA is applied in the fields of food security analysis and drug discovery. In Table 1 examples of EDA studies are given. The first step is the preparation of an extract of the sample. For soil/sediment samples, a sieving step prior to the actual extraction may be necessary in order to remove large particles and obtain a sample that is well-defined in terms of particle size (e.g. <200 μm). Examples of biota samples are whole organism homogenates or parts of the organism, such as blood and liver. For the extraction of the samples, analytical techniques such as liquid/liquid or solid phase extraction are applied to concentrate the compounds of interest and to remove matrix constituents that may interfere with the later steps of the EDA. The choice of endpoint to include in an EDA study is very important, as it dictates the nature of the toxicity of the compounds that may be identified (see ). For application in EDA, typically bioassays that are carried out in multiwell (≥ 96 well) plates can be used, because of their low cost, high throughput and ease of use (see Section on ), although sometimes assays (see Section on ) are applied too. Endpoint Type of bioassay Sample matrix Type of compounds identified Estrogenicity Cell based reporter gene Sediment Endogenic hormones Anti-androgenicity Cell based reporter gene Sediment Plasticizers, organophosphorus flame retardants, synthetic fragrances idem Water Pharmaceuticals, pesticides, plasticizers, flame retardants, UV filters Mutagenicity Bacterial luminescence reporter strain Water Benzotriazoles Thyroid hormone disruption Radioligand binding Polar bear plasma Metabolites of PCBs, nonylphenols Photosystem II toxicity Pulse Amplitude Modulation fluorometry Water Pesticides Endocrine disruption Snail reproduction Sediment Phthalates, synthetic fragrances, alkylphenols Fractionation of the extract is achieved by the application of chromatography, resulting in the separation of the - in most cases - multitude of different compounds that are present in an extract of an environmental sample. Chromatographic separation is obtained after the migration of compounds through a sorbent bed. In most cases, the separation principle is based on the distribution of compounds between the liquid mobile phase and the solid stationary phase (liquid chromatography, or LC), but a chromatographic separation using the partitioning between the gas phase and a sorbent bed (gas chromatography, or GC) is also possible. At the end of the separation column, at specified time intervals fractions can be collected that are simpler in composition in comparison to the original extract: a reduction in the number of compounds per fraction is obtained. The collected fractions are tested in the bioassay and the responsive fractions are selected for further chemical analysis and identification (step 4). The time intervals for fraction collection vary between a few minutes in older applications and a few seconds in new applications of EDA, which enables fractionation directly into multiwell plates for high throughput bioassay testing. In cases where fractions are collected during time intervals in the order of minutes, the fractions are still so complex that a second round of fractionation to obtain fractions of reduced complexity is often necessary for the identification of compounds that are responsible for the observed effect (see Figure 2). Chemical analysis for the identification of the compounds that cause the effect in the bioassay is usually done by LC coupled to mass spectrometric (MS) detection. To obtain high mass accuracy that facilitates compound identification, high resolution mass spectrometry (HR-MS) is generally applied. Fractions obtained after one or two fractionation steps are injected into the LC-MS system. In studies where fractionation into multiwell plates is used (and thus small fractions in the order of microliters are collected), only one round of fractionation is applied. In these cases, identification and fraction collection can be done in parallel, using a splitter after the chromatographic column that directs part of the eluent from the column to the well plate and the other part to the MS (see Figure 3). This is called high throughput EDA (HT-EDA). The use of HR-MS is necessary to obtain mass information to establish the molecular weight with high accuracy (e.g. 119.12423 Dalton) to derive the molecular formula (e.g. C H N ) of the compound. Optimally, HR-MS instrumentation is equipped with an MS-MS mode, in which compound fragmentation is induced by collisions with other molecules, resulting in fragments that are specific for the original compound. Fragmentation spectra obtained using the MS-MS mode of HR-MS instruments help to elucidate the structure of the compounds eluting from the column, see for an example Figure 4. Other information such as log K may be calculated using dedicated software packages that use elemental composition and structure as input. To aid the identification process, compound and mass spectral libraries are used as well as the more novel databases containing toxicity information (e.g. PubChem Bioassay, Toxcast). Mass spectrometry instrumentation vendor software, public/web-based databases and databases compiled in-house enable suspect screening to identify compounds that are known, e.g. because they are applied in consumer products or construction materials. When MS signals cannot be attributed to known compounds or their metabolites/transformation products, the identification approach is called non-target screening, where additional identification techniques such as Nuclear Magnetic Resonance (NMR) may aid the identification. The identification process is complicated and often time consuming, and results in a suspect list that needs to be evaluated for further confirmation of the identification. For an unequivocal confirmation of the identity of a tentatively identified compound, it is necessary to obtain a standard of the compound to investigate whether its analytical chemical behaviour corresponds to that of the tentatively identified compound in the environmental sample. In addition, the biological activity of the standard should be measured and compared with the earlier obtained data. In case both the chemical analysis and bioassay testing results support the identification, confirmation of compound identity is achieved. In principle, the confirmation step of an EDA study is very straightforward, but in current practice the standards are mostly not commercially available. Dedicated synthesis is time consuming and costly, therefore the confirmation step often is a bottleneck in EDA studies. The application of EDA is suitable for samples collected at specific locations where comprehensive chemical analysis of priority pollutants and other chemicals of relevance has been conducted already, and where ecological quality assessment has revealed that the local conditions are compromised (see other Sections on ). Especially those samples that show a significant difference between the observed (in vitro) bioassay response and the activity that may be calculated according to the concept of Concentration Addition (see Section on ) by using the relative potencies and the concentrations of compounds active in that bioassay need a further in-depth investigation. EDA can be implemented at these 'hotspots' of environmental contamination to unravel the identity of compounds that have an effect, but that were not included in the chemical monitoring of the environmental quality. Knowledge with regard to the main drivers of toxicity at a specific location supports accurate decision making that is necessary for environmental quality protection. Draw a scheme of EDA and name the different steps. What is the aim of the fractionation of an extract? Describe the confirmation step of EDA. Give an example of an EDA study with regard to endpoint, bioassay, matrix and type of compounds identified. Explain why quantitative chemical analysis of known pollutants of e.g. a water sample is complementary to effect-based testing of that sample using e.g. an in vitro bioassay? (see sections on and on ) as test organisms for bioassays are generally the same as the ones selected for single species toxicity tests (see sections , , and n the Selection of ecotoxicity test organisms). Likewise, also the endpoints measured in bioassays are the same as those in single species ecotoxicity tests (see section on ). bioassays therefore have a relatively high ecological relevance, as they provide information on the survival, reproduction, growth, or behaviour of the species tested. A major difference between toxicity tests and bioassays is the selection of the controls. In laboratory toxicity experiments the controls consist of non-spiked 'clean' test medium (see section on ). In bioassays the choice of the controls is more complicated though. Non-treated test medium may be incorporated as a control in bioassays to check for the health and quality of the test organisms. But control media, like standard test water or artificial soil and sediment may differ in numerous aspects from natural environmental samples. Therefore, the control should preferably be a test medium that has exactly the same physicochemical properties as the contaminated sample, except for the chemical pollutants being present. This ideal situation, however, hardly ever exists. Hence, it is recommended to also incorporate environmental samples from less or non-contaminated reference sites into the bioassay and to compare the response of the organism to samples from contaminated sites with those from reference sites. Alternatively, controls can be selected as the least contaminated environmental samples from a gradient of pollution or as the dilution required to obtain no effect. As dilution medium artificial control medium can be used or medium from a reference site. Define bioassays and explain how bioassays are performed. Give examples of the most commonly used bioassays per environmental compartment. Motivate the necessity to incorporate several bioassays into a bioassay battery. List 5 advantages of an effect-based approach over a compound based approach for water quality assessment. Motivate the necessity of employing a bioassay battery in effect based monitoring approaches. Explain how bioassay responses are expressed in terms of toxicity equivalents of reference compounds? (you may wish to draw a figure). Translate the outcome of a bioassay battery into a ranking of contaminated sites based on ecotoxicological risk. (you may wish to draw a figure). Define biomonitoring. Explain the difference between passive and active biomonitoring. What can be measured in biomonitoring organisms after (re)collection? Explain the difference between passive and active biomonitoring. List the characteristics of suitable biomonitoring organisms. Name the advantage and disadvantage of bioassays Name the advantages and disadvantages of measuring contaminant concentrations in organisms. : Michiel Rutgers : Kees van Gestel, Michiel Kraak, Ad Ragas You should be able Triad, site-specific ecological risk assessment, weight of evidence Like the other diagnostic tools described in the previous sections (see sections on Effect-based monitoring and , , and and ), the TRIAD approach is a tool for site-specific ecological risk assessment of contaminated sites (Jensen et al., 2006; Rutgers and Jensen, 2011). Yet, it differs from the previous approaches by combining and integrating different techniques through a 'weight of evidence' approach. To this purpose, the TRIAD combines information on contaminant concentrations (environmental chemistry), the toxicity of the mixture of chemicals present at the site ((eco)toxicology), and observations of ecological effects (ecology) (Figure 1). The mere presence of contaminants is just an indication of potential ecological effects to occur. Additional data can help to better assess the ecological risks. For instance, information on actual toxicity of the contaminated site can be obtained from the exposure of test organisms to (extracts of) environmental samples (bioassays), while information on ecological effects can be obtained from an inventory of the community composition at the specific site. When these disciplines tend to converge to corresponding levels of ecological effects, a weight of evidence is established, making it possible to finalize the assessment and to support a decision for contaminated site management. The TRIAD approach thus combines the information obtained from three lines of evidence (LoE): The three lines of evidence form a weight of evidence when they are converging, meaning that when the independent lines of evidence are indicating a comparable risk level, there is sufficient evidence for providing advice to decision makers about the ecological risk at a contaminated site. When there is no convergence in risk information obtained from the three lines of evidence, uncertainty is large. Further investigations are then required to provide a unambiguous advice. The results of a site-specific ecological risk assessment (SS-ERA) applying the TRIAD approach are first organized basic tables for each sample and line of evidence separately. Table 1 shows an example. This table also collects supporting data, such as soil pH and organic matter content. Subsequently, these basic data are processed into ecological risk values by applying a risk scale running from zero (no effects) to one (maximum effect). An example of a metric used is the multi-substance Potentially Affected Fraction of species from the mixture of contaminants (see Section on ). These risk values are then collected in a TRIAD table (Table 2), for each endpoint separately, integrated per line of evidence individually, and finally integrated over the three lines of evidence. Also the level of agreement between the three lines of evidence is given a score. Weighting values are applied, e.g. equal weights for all ecological endpoints (depending on number of methods and endpoints), and equal weights for each line of evidence (33%). When differential weights are preferred, for instance when some data are judged as unreliable, or some endpoints are considered more important than others, the respective weight factors and the arguments to apply them must be provided in the same table and accompanying text. ISO (2017). ISO 19204: Soil quality -- Procedure for site-specific ecological risk assessment of soil contamination (soil quality TRIAD approach). International Standardization Organization, Geneva. . Jensen, J., Mesman, M. (Eds.) (2006). LIBERATION, Ecological risk assessment of contaminated land, decision support for site specific investigations. ISBN 90-6960-138-9, Report 711701047, RIVM, Bilthoven, The Netherlands. Rutgers, M., Bogte, J.J., Dirven-Van Breemen, E.M., Schouten, A.J. (2001) Locatiespecifieke ecologische risicobeoordeling - praktijkonderzoek met een Triade-benadering. RIVM-rapport 711701026, Bilthoven. Rutgers, M., Jensen, J. (2011). Site-specific ecological risk assessment. Chapter 15, in: F.A. Swartjes (Ed.), Dealing with Contaminated Sites - from Theory towards Practical Application, Springer, Dordrecht. pp. 693-720. Van der Waarde, J.J., Derksen, J.G.M, Peekel, A.F., Keidel, H., Bloem, J., Siepel, H. (2001) Risicobeoordeling van bodemverontreiniging met behulp van een triade benadering met chemische analyses, bioassays en biologische veldinventarisaties. Eindrapportage NOBIS 98-1-28, Gouda. A sediment sample was analyzed for Priority Hazardous Substances (PHSs), but none were detected. Yet, this sediment sample caused high mortality in laboratory bioassays with three sediment inhabiting species. At the site where the sample was taken biodiversity was very low. Explain these observations. In a sediment sample, the total concentration of Priority Hazardous Substances (PHSs) was shown to be very high. Yet, this sediment sample caused no mortality in laboratory bioassays with three sediment inhabiting species. Moreover, at the sample site in the field species rich communities were observed. Explain these observations. What is the added value of using bioassays and field observations over chemical analysis when assessing the potential risk of a contaminated site? What is the added value of performing an assessment along three independent Lines of Evidence (LoE)? : Leo Posthuma, Dick de Zwart : Allan Burton, Ad Ragas Learning objectives: You should be able to: : eco-epidemiology, mixture pollution, diagnosis, impact magnitude, probable causes, validation Approaches for environmental protection, assessment and management differ between 'classical' stressors (such as excess nutrients and pH) and chemical pollution. For the 'classical' environmental stress factors, ecologists use monitoring data to develop concepts and methods to prevent and reduce impacts. Although there are some clear-cut examples of chemical pollution impacts [e.g., the decline in vulture populations in South East Asia due to diclofenac (Oaks et al. 2004), and the suit of examples in the book 'Silent Spring' (Carson 1962)], ecotoxicologists commonly have assessed the stress from chemical pollution by evaluating exposures vis a vis laboratory toxicity data. Current pollution often consists of complex mixtures of chemicals, with highly variable patterns in space and time. This poses problems when one wants to evaluate whether observed impacts in ecosystems can be attributed to chemicals or their mixtures. Eco-epidemiological methods have been established to discern such pollution stress. These methods provide the diagnostic tools to identify the impact magnitude and key chemicals that cause impacts in ecosystems. The use of these methods is further relevant for validating the laboratory-based risk assessment approaches developed by ecotoxicology. Risk assessments of chemicals provide insights in expected exposures and impacts, commonly for separate chemicals. These are predictive outcomes with a high relevance for decision making on environmental protection and management. The validation of those risk assessments is key to avoid wrong protection and management decisions, but it is complex. It consists of comparing predicted risk levels to observed effects. This begs the question on how to discern effects of chemical pollution in the field. This question can be answered based on the principles of ecological bio-assessments combined with those of human epidemiology. A bio-assessment is a study of stressors and ecosystem attributes, made to delineate causes of impacts via (often) statistical associations between biotic responses and particular stressors. Epidemiology is defined as the study of the distribution and causation of health and disease conditions in specified populations. Applied epidemiology serves as a scientific basis to help counteracting the spreading of human health problems. Dr. John Snow is often referred to as the 'father of epidemiology'. Based on observations on the incidence, locations and timings of the 1854 cholera outbreak in London, he attributed the disease to contaminated water taken from the Broad Street pump well, counteracting the prevailing idea that the disease was caused by transmission via air. His proposals to control the disease were effective. Likewise, eco-epidemiology - in its ecotoxicological context - has been defined as the study of the distribution and causation of impacts of multiple stressor exposures in ecosystems. In its applied form, it supports the reduction of ecological impacts of chemical pollution. Human-health eco-epidemiology is concerned with environment-mediated disease. The first literature mention of eco-epidemiological analyses on chemical pollution stems from 1984 (Bro-Rasmussen and Løkke 1984). Those authors described eco-epidemiology as a discipline necessary to validate the risk assessment models and approaches of ecotoxicology. In its initial years, progress in eco-epidemiological research was slow due to practical constraints such as a lack of monitoring data, computational capacity and epidemiological techniques. Current eco-epidemiological studies in ecotoxicology aim to diagnose the impacts of chemical pollution in ecosystems, and utilize a combination of approaches in order to diagnose the role of chemical mixtures in causing ecological impacts in the field. The combination of approaches consists of: 1. Collection of monitoring data on abiotic characteristics and the occurrence and/or abundance of biotic species, for the environmental compartment under study; 2. If needed: data optimization, usually to align abiotic and biotic monitoring data, including the chemicals; 3. Statistical analysis of the data set using eco-epidemiological techniques to delineate impacts and probable causes, according to the approaches followed in 'classical' ecological bio-assessments; 4. Interpretation and use of the outcomes for either validation of ecotoxicological models and approaches, or for control of the impacts sensu Dr. Snow. Although impacts of chemicals in the environment were known before 1962, Rachel Carson's book Silent Spring (see Section on the ) can be seen as early and comprehensive eco-epidemiological study that synthesized the available information of impacts of chemicals in ecosystems. She considered effects of chemicals a novel force in natural selection when she wrote: "If Darwin were alive today the insect world would delight and astound him with its impressive verification of his theories of survival of the fittest. Under the stress of intensive chemical spraying the weaker members of the insect populations are being weeded out." Clear examples of chemical impacts on species are still reported. Amongst the best-known examples is a study on vultures. The population of Indian vultures declined more than 95% due to diclofenac exposure which was used intensively as a veterinary drug (Oaks et al. 2004). The analysis of chemical impacts in nature becomes however more complex over time. The diversity of chemicals produced and used has vastly increased, and environmental samples contain thousands of chemicals at often low concentrations. Hence, contemporary eco-epidemiology is complex. Nonetheless, various studies demonstrated that contemporary mixture exposures affect species assemblages. Starting from large-scale monitoring data and following the four steps mentioned above, De Zwart et al. (2006) were able to show that effects on fish species assemblages could be attributed to both habitat characteristics and chemical mixtures. Kapo and Burton Jr (2006) showed the impacts of multiple stressors and chemical mixtures in aquatic species assemblages with similar types of data, but slightly different techniques. Eco-epidemiological studies of the effects of chemicals and their mixtures currently represent different geographies, species groups, stressors and chemicals/mixtures that are considered. The potential utility eco-epidemiological studies was reviewed by Posthuma et al. (2016). The review showed that mixture impacts occur, and that they can be separated from natural variability and multiple-stressor impacts. That means that water managers can develop management plans to counteract stressor impacts. Thereby, the study outcomes are used to prioritize management to sites that are most affected, and to chemicals that contribute most to those effects. Based on sophisticated statistical analyses, Berger et al. (2016) suggested chemicals can induce effects in the environment at concentrations much lower than expected based on laboratory experiments. Schäfer et al. (2016) argued that eco-epidemiological studies that cover both mixtures and other stressors are essential for environmental quality assessment and management. In practice, however, the analysis of the potential impacts of chemical mixtures is often still separate from the analysis of impacts of other stressors. Various regulations require collection of monitoring data, followed by bio-assessment, such as the EU Water Framework Directive (see section on the ). Therefore, monitoring data sets are increasingly available. The data set is subsequently curated and/or optimized for the analyses. Data curation and management steps imply amongst others that taxonomic names of species are harmonized, and that metrics for abiotic and biotic variables represent the conditions for the same place and time as much as possible. Next, the data set is expanded with novel variables, e.g. a metric for the toxic pressure exerted by chemical mixtures. An example of such a metric is the multi-substance Potentially Affected Fraction of species (msPAF). This metric transfers measured or predicted concentrations into the Potentially Affected Fraction of species (PAF), the values of which are then aggregated for a total mixture (De Zwart and Posthuma 2005). This is crucial, as adding each chemical of interest as a separate variable implies an increasingly expanding number of required sampling sites to maintain statistical power to diagnose impacts and probable causation. The interpretation of the outcomes of the statistical analyses of the data set is the final step. Here, it must be acknowledged that statistical association is not equal to causation, and that care must be taken to explain the findings as indicative for mixture effects. Depending on the context of the study, this may then trigger a refined assessment, or alignment with other methods to collect evidence, or a direct use in an environmental management program. A very basic eco-epidemiological method is quantile regression. Whereas common regression methods explore the magnitude of the change of the mean of the response variable (e.g., biodiversity) in relation to a predictor variable (e.g., pollutant stress), the quantile regression looks at the tails of the distributions of the response variable. How this principle operates is illustrated in Figure 1. When a monitoring data set contains one stressor variable at different levels (i.e., a gradient of data), the observations typically take the shape of a common stressor-response relationship (see section on ). If the monitoring sites are affected by an extra stressor, the maximum-performance under the first stressor cannot be reached, so that the area under the curve contains the XY-points for this situation. Further addition of stressor variables and levels fills this space under the curve. When the raw data plotted as XY show an 'empty area' lacking XY-points, e.g. in the upper right corner, it is likely that the stressor variable can be identified as a stressor that limits the response variable, for example: chemicals limit biodiversity. The quantile regression calculates an upper percentiles (e.g., the 95th percentile) of the Y-values in assigned subgroups of X-values ("bins"). Such a procedure yields a picture such as Figure 1. More complex methods for analysis of (bio)monitoring data have been developed and applied. The methods are closely associated to those developed for, and utilized in, applied ecology. Well-known examples are 'species distribution models' (SDM), which are used to describe the abundance or presence of species as a function of multiple environmental variables. A well-known SDM is the bell-shaped curve relating species abundances to water pH: numbers of individuals of a species are commonly low at low and high pH, and the SDM is characterized as an optimum model for species abundance (Y) versus pH (X). Statistical models can also describe species abundance, presence or biodiversity, as a function of multiple stressors, for example via Generalized Linear Models. These have the general shape of: Log(Abundance)= (a. pH + a' pH ) + (b. OM + b' OM ) + …… + e, with a, a', b and b' being estimated from fitting the model to the data, whilst pH and OM are the abiotic stressor variables (acidity and Organic Matter, respectively); the quadratic terms are added to allow for optimum and minimum shaped relationships. When SSD models (see Section on ) are used to predict the multi-substance Potentially Affected Fraction of species, the resulting mixture stress proxy can be analysed together with the other stressor variables. Data analyses from monitoring data from the United States and the Netherlands have, for example, shown that the abundance of >60% of the taxa is co-affected by mixtures of chemicals. An example study is provided by Posthuma et al. (2016). In addition to the retrospective analysis of monitoring data, in search of chemical impacts, recent studies also show examples of prospective studies of effects of mixtures. Different land uses imply different chemical use patterns, summarized as 'signatures'. That is, agricultural land use will yield intermittent emissions of crop-specific plant protection products, aligning with the growing season. Emissions from populated areas will show continuous emission of household chemicals and discontinuous emissions of chemicals in street run-off associated to heavy rain events. The application of emission, fate and ecotoxicity models showed that aquatic ecosystems are subject to the 'signatures', with associated predicted impact magnitudes (Holmes et al. 2018; Posthuma et al. 2018). Although such prospective assessments did not yet prove ecological impacts, they can assist in avoiding impacts by preventing the emission 'signatures' that are identified as potentially most hazardous. Eco-epidemiological analysis outputs serve two purposes, closely related to prospective and retrospective risk assessment of chemical pollution: 1. Validation of ecotoxicological models and approaches; 2. Derivation of control measures, to reduce impacts of diagnosed probable causes of impacts. If needed, multiple lines of evidence can be combined, such as in the Triad approach (see section on ) or approaches that consider more than three lines of evidence (Chapman and Hollert, 2006). The higher the importance of a good diagnosis, the better the user may rely on multiple lines of evidence. Berger, E., Haase, P., Oetken, M., Sundermann, A. (2016). Field data reveal low critical chemical concentrations for river benthic invertebrates. Science of The Total Environment 544, 864-873. Bro-Rasmussen, F., Løkke, H. (1984). Ecoepidemiology - a casuistic discipline describing ecological disturbances and damages in relation to their specific causes; exemplified by chlorinated phenols and chlorophenoxy acids. Regulatory Toxicology and Pharmacology 4, 391-399. Carson, R. (1962). Silent spring. Boston, Houghton Mifflin. Chapman, P.M., Hollert, H. (2006). Should the sediment quality triad become a tetrad, a pentad, or possibly even a hexad? Journal of Soils and Sediments 6, 4-8. De Zwart, D., Dyer, S.D., Posthuma, L., Hawkins, C.P. (2006). Predictive models attribute effects on fish assemblages to toxicity and habitat alteration. Ecological Applications 16, 1295-1310. De Zwart, D., Dyer, S.D., Posthuma, L., Hawkins, C.P. (2006). Use of predictive models to attribute potential effects of mixture toxicity and habitat alteration on the biological condition of fish assemblages. Ecological Applications 16, 1295-1310. De Zwart, D., Posthuma, L. (2005). Complex mixture toxicity for single and multiple species: Proposed methodologies. Environmental Toxicology and Chemistry 24,: 2665-2676. Holmes, C.M., Brown, C.D., Hamer, M., Jones, R., Maltby, L., Posthuma, L., Silberhorn, E., Teeter, J.S., Warne, M.S.J., Weltje, L. (2018). Prospective aquatic risk assessment for chemical mixtures in agricultural landscapes. Environmental Toxicology and Chemistry 37, 674-689. Kapo, K.E., Burton Jr, G.A. (2006). A geographic information systems-based, weights-of-evidence approach for diagnosing aquatic ecosystem impairment. Environmental Toxicology and Chemistry 25, 2237-2249. Oaks, J.L., Gilbert, M., Virani, M.Z., Watson, R.T., Meteyer, C.U., Rideout, B.A., Shivaprasad, H.L., Ahmed, S., Chaudhry, M.J., Arshad, M., Mahmood, S., Ali, A., Khan, A.A. (2004). Diclofenac residues as the cause of vulture population decline in Pakistan. Nature 427(6975), 630-633. Posthuma, L., Brown, C.D., de Zwart, D., Diamond, J., Dyer, S.D., Holmes, C.M., Marshall, S., Burton, G.A. (2018). Prospective mixture risk assessment and management prioritizations for river catchments with diverse land uses. Environmental Toxicology and Chemistry 37, 715-728. Posthuma, L., De Zwart, D., Keijzers, R., Postma, J. (2016). Water systems analysis with the ecological key factor 'toxicity'. Part 2. Calibration. Toxic pressure and ecological effects on macrofauna in the Netherlands. Amersfoort, the Netherlands, STOWA. Posthuma, L., Dyer, S.D., de Zwart, D., Kapo, K., Holmes, C.M., Burton Jr, G.A. (2016). Eco-epidemiology of aquatic ecosystems: Separating chemicals from multiple stressors. Science of The Total Environment 573, 1303-1319. Posthuma, L., Suter, II, G.W., Traas, T.P. (Eds.) (2002). Species Sensitivity Distributions in Ecotoxicology. Boca Raton, FL, USA, Lewis Publishers. Schäfer, R.B., Kühn, B., Malaj, E., König, A., Gergs, R. (2016). Contribution of organic toxicants to multiple stress in river ecosystems. Freshwater Biology 61, 2116-2128 Which motivations do you know for executing eco-epidemiological analyses? Pollution in the 1950s and 1960s showed clear evidence for major impacts of chemicals in nature, whilst regulatory management actions taken since then have reduced clear recognition of chemical impacts in nature. Which ecotoxicological model has rejuvenated the development of eco-epidemiological methods? What is a simple approach, via which even raw-data plottings already show whether a stress factor (such as chemical mixture exposure) is limiting for ecology | 37,148 | 1,796 |
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under review : Jos Boesten, Theo Brock : Ad Ragas, Andreu Rico You should be able to: pesticides, exposure, scenarios, assessment goals, effects An exposure scenario describes the combination of circumstances needed to estimate exposure by means of models. For example, scenarios for modelling pesticides exposure can be defined as a combination of abiotic (e.g. properties and dimensions of the receiving environment and related soil, hydrological and climate characteristics) and agronomic (e.g. crops and related pesticide application) parameters that are thought to represent a realistic worst-case situation for the environmental context in which the exposure model is to be run. A scenario for exposure of aquatic organisms could be e.g. a ditch with a minimum water depth of 30 cm alongside a crop growing on a clay soil with annual applications of pesticide using a 20-year time series of weather data and including pesticide exposure via spray drift deposition and leaching from drainpipes. Such a scenario would require modelling of spray drift, leaching from drainpipes and exposure in surface water, ending up in a 20-year time series of the exposure concentration. In this chapter, we explain the use of exposure scenarios in prospective ERA by giving examples for the regulatory assessment of pesticides in particular. Between about 1995 and 2001 groundwater and surface water scenarios were developed for EU pesticide registration; also referred to as the FOCUS scenarios. The European Commission indicated that these should represent 'realistic worst-cases', a political concept which leaves considerable room for scientific interpretation. Risk assessors and managers agreed that the intention was to generate 90 percentile exposure concentrations. The concept of a 90 percentile exposure concentration assumes a statistical population of concentrations and 90% of these concentrations are lower than this 90 percentile (and thus 10% are higher). This 90 percentile approach has since then been followed for most environmental exposure assessments for pesticides at EU level. The selection of the FOCUS groundwater and surface water scenarios involved a considerable amount of expert judgement because this selection could not yet be based on well-defined GIS procedures and databases on properties of the receiving environment. The EFSA exposure assessment for soil organisms was the first environmental exposure assessment that could be based on a well-defined GIS procedure, using EU maps of parameters like soil organic matter, density of crops and weather. During the development of this exposure assessment, it became clear that the concept of a 90 percentile exposure concentration is too vague: it is essential to define also the statistical population of concentrations from which this 90 percentile is taken. Based on this insight, the EFSA Panel on Plant Protection Products and their Residues (PPR) developed the concept of the exposure assessment goals, which has become the standard within EFSA for developing regulatory exposure scenarios for pesticides. Figure 1 shows how an exposure assessment goal for the risk assessment of aquatic organisms can be defined following this EFSA procedure. The left part specifies the temporal dimensions and the right part the spatial dimensions. In box E1, the Ecotoxicologically Relevant type of Concentration (ERC) is defined, e.g. the freely dissolved pesticide concentration in water for pelagic organisms. In box E2, the temporal dimension of this concentration is defined, e.g. annual peak or time-weighted average concentration for a pre-defined period. Based on these elements, the multi-year temporal population of concentrations can be generated for one single water body (E5) which would consist of e.g. 20 peak concentrations in case of a time series of 20 years. The spatial part requires definition of the type of water body (e.g. ditch, stream or pond; box E3) and the spatial dimension of this body (e.g. having a minimum water depth of 30 cm; box E4). Based on these, the spatial population of water bodies can be defined (box E6), e.g. all ditches with a minimum water depth of 30 cm alongside fields treated with the pesticide. Finally, then in box E7 the percentile combination to be taken from the spatial-temporal population of concentrations is defined. Specification of the exposure assessment goals does not only involve scientific information, but also political choices because this specification influences the strictness of the exposure assessment. For instance, in case of exposure via spray drift a minimum water depth of 30 cm in box E4 leads to about a three times lower peak concentration in the water than a minimum water depth of 10 cm. The schematic approach of Figure 1 can easily be adapted to other exposure assessment goals. Nearly all the environmental protection goals for pesticides involve assessment of risk for organisms; only groundwater and drinking water from surface water are based on a concentration of 0.1 μg/L which is not related to possible ecotoxicological effects. The risk assessment for organisms is a combination of an exposure assessment and an effect assessment as is illustrated by Figure 2. Both the effect and the exposure assessment are tiered approaches with simple and conservative first tiers and less simple and more realistic higher tiers. A lower exposure tier may consist of a simple conservative scenario whereas a higher exposure tier may e.g. be based on a scenario selected using sophisticated spatial modelling. The top part of the scheme shows the link to the risk managers which are responsible for the overall level of protection. This overall level of protection is linked to the so-called Specific Protection Goals which will be explained in and form the basis for the definition of the effect and exposure assessment goals. So the exposure assessment goals and resulting exposure scenarios should be consistent with the Specific Protection Goals (e.g. algae and fish may require different scenarios). When linking the two assessments, it has to be ensured that the type of concentration delivered by the exposure assessment is consistent with that required by the effect assessment (e.g. do not use time-weighted average concentrations in acute effect assessment). Figure 2 shows that in the assessment procedure information flows always from the exposure assessment to the effect assessment because the risk assessment conclusion is based on the effect assessment. A relatively new development is to assess exposure and effects at the landscape level. This typically is a combination of higher-tier effect and exposure assessments. In such an approach, first the dynamics in exposure is assessed for the full landscape, and then combined with the dynamics of effects, for example based on spatially-explicit population models for species typical for that landscape. Such an approach makes a separate definition of the exposure and effect scenario redundant because this approach aims to deliver the exposure and effect assessment in an integrated way in space and time. Such an integrated approach requires the definition of "environmental scenarios". Environmental scenarios integrate both the parameters needed to define the exposure (exposure scenario) and those needed to calculate direct and indirect effects and recovery (ecological scenario) (see Figure 3). However, it will probably take at least a decade before landscape-level approaches, including agreed-upon environmental scenarios, will be implemented for regulatory use in prospective ERA. Boesten, J.J.T.I. (2017). Conceptual considerations on exposure assessment goals for aquatic pesticide risks at EU level. Pest Management Science 74, 264-274. Brock, T.C.M., Alix, A., Brown, C.D., et al. (2010). Linking aquatic exposure and effects: risk assessment of pesticides. SETAC Press & CRC Press, Taylor & Francis Group, Boca Raton, FL, 398 pp. Rico, A., Van den Brink, P.J., Gylstra, R., Focks, A., Brock, T.C.M. (2016). Developing ecological scenarios for the prospective aquatic risk assessment of pesticides. Integrated Environmental Assessment and Management 12, 510-521. Why is a detailed specification of exposure assessment goals needed ? Why does specification of the exposure assessment goals include political choices ? Why does the risk assessment of organisms consist of two parallel tiered schemes for effects and exposure ? in preparation : Els Smit, Eric Verbruggen : Alexandra Kroll, Inge Werner You should be able to: PNEC, quality standards, extrapolation, assessment factor The key question in environmental risk assessment is whether environmental exposure to chemicals leads to unacceptable risks for human and ecosystem health. This is done by comparing the measured or predicted concentrations in water, soil, sediment, or air, with a reference level. Reference levels represent a dose (intake rate) or concentration in water, soil, sediment or air below which unacceptable effects are not expected. The definition of 'no unacceptable effects' may differ between regulatory frameworks, depending on the protection goal. The focus of this section is the derivation of reference levels for aquatic ecosystems as well as for predators feeding on exposed aquatic species (secondary poisoning), but the derivation of reference values for other environmental compartments follows the same principles. Various technical terms are in use as reference values, e.g. the Predicted No Effect Concentration (PNEC) for ecosystems or the Acceptable Daily Intake (ADI) for humans (Section on ). The term "reference level" is a broad and generic term, which can be used independently of the regulatory context or protection goal. In contrast, the term "quality standard" is associated with some kind of legal status, e.g., inclusion in environmental legislation like the Water Framework Directive (WFD). Other terms exist, such as the terms 'guideline value' or 'screening level' which are used in different countries to indicate triggers for further action. While the scientific basis of these reference values may be similar, their implementation and the consequences of exceedance are not. It is therefore very important to clearly define the context of the derivation and the terminology used when deriving and publishing reference levels. A frequently used reference level for ecosystem protection is the Predicted No Effect Concentration (PNEC). The PNEC is the concentration below which adverse effects on the ecosystem are not expected to occur. PNECs are derived per compartment and apply to the organisms that are directly exposed. In addition, for chemicals that accumulate in prey, PNECs for secondary poisoning of predatory birds and mammals are derived. The PNEC for direct ecotoxicity is usually based on results from single species laboratory toxicity tests. In some case, data from field studies or mesocosms may be included. A basic PNEC derivation for the aquatic compartment is based on data from single species tests with algae, water fleas and fish. Effects on the level of a complex ecosystem are not fully represented by effects on isolated individuals or populations in a laboratory set-up. However, data from laboratory tests can be used to extrapolate to the ecosystem level if it is assumed that protection of ecosystem structure ensures protection of ecosystem functioning, and that effects on ecosystem structure can be predicted from species sensitivity. To account for the uncertainty in the extrapolation from single species laboratory tests to effects on real life ecosystems, the lowest available test result is divided by an assessment factor (AF). In establishing the size of the AF, a number of uncertainties must be addressed to extrapolate from single-species laboratory data to a multi-species ecosystem under field conditions. These uncertainties relate to intra- and inter-laboratory variation in toxicity data, variation within and between species (biological variance), test duration and differences between the controlled laboratory set-up and the variable field situation. The value of the AF depends on the number of studies, the diversity of species for which data are available, the type and duration of the experiments, and the purpose of the reference level. Different AFs are needed for reference levels for e.g. intermittent release, short-term concentration peaks or long-term (chronic) exposure. In particular, reference levels for intermittent release and short-term exposure may be derived on the basis of acute studies, but short-term tests are less predictive for a reference level for long-term exposure and larger AFs are needed to cover this. Table 1 shows the generic AF scheme that is used to derive PNECs for long-term exposure of freshwater organisms in the context of European regulatory framework for industrial chemicals (REACH; see Section on ). This scheme is also applied for the authorisation of biocidal products, pharmaceuticals and for derivation of long-term water quality standards for freshwater under the EU Water Framework Directive. Further details on the application of this scheme, e.g., how to compare acute and chronic data and how to deal with irregular datasets, are presented in guidance documents (see suggested reading: EC, 2018; ECHA, 2008). Similar schemes exist for marine waters, sediment, and soil. However, for the latter two environmental compartments often too little experimental information is available and risk limits have to be calculated by extrapolation from aquatic data using the Equilibrium Partitioning concept. The derivation of Regulatory Acceptable Concentrations (RAC) for plant protection products (PPPs) is also based on the extrapolation of laboratory data, but follows a different approach focussing on generating data for specific taxonomic groups, taking account of the mode of action of the PPP (see suggested reading: EFSA, 2013). At least one short-term L(E)C50 from each of three trophic levels (fish, invertebrates (preferred ) and algae) 1000 One long-term EC10 or NOEC (either fish or ) 100 Two long-term results (e.g. EC10 or NOECs) from species representing two trophic levels (fish and/or Daphnia and/or algae) 50 Long-term results (e.g. EC10 or NOECs) from at least three species (normally fish, and algae) representing three trophic levels 10 The AF approach was developed to account for the uncertainty arising from extrapolation from (potentially limited) experimental datasets. If enough data are available for other species than algae, daphnids and fish, statistical methods can be applied to derive a PNEC. Within the concept of species sensitivity distribution (SSD), the distribution of the sensitivity of the tested species is used to estimate the concentration at which 5% of all species in the ecosystem is affected (HC5; see section on ). When used for regulatory purposes in European regulatory frameworks, the dataset should meet certain requirements regarding the number of data points and the representation of taxa in the dataset, and an AF is applied to the HC5 to cover the remaining uncertainty from the extrapolation from lab to field. Where available, results from semi-field experiments (mesocosms, see section on ) can also be used, either on its own or to underpin the PNEC derived from the AF or SSD approach. SSDs and mesocosm-studies are also used in the context of authorisation of PPP. Substances might be toxic to wildlife because of bioaccumulation in prey or a high intrinsic toxicity to birds and mammals. If this is the case, a reference level for secondary poisoning is derived for a simple food chain: water è fish or mussel è predatory bird or mammal. The toxicity data from bird or mammal tests are transformed into safe concentrations in prey. This can be done by simply recalculating concentrations in laboratory feed into concentrations in fish using default conversion factors (see e.g., ECHA, 2008). For the derivation of water quality standards under the WFD, a more sophisticated method was introduced that uses knowledge on the energy demand of predators and energy content in their food to convert laboratory data to a field situation. Also, the inclusion of other, more complex and sometimes longer food chains is possible, for which field bioaccumulation factors are used rather than laboratory derived values. circabc.europa.eu/ui/group/9...7a2a6b/details EFSA (2013). Guidance on tiered risk assessment for plant protection products for aquatic organisms in edge-of-field surface waters. EFSA Journal 2013; 11(7): 3290 efsa.onlinelibrary.wiley.com...efsa.2013.3290 Traas, T.P., Van Leeuwen, C. (2007). Ecotoxicological effects. In: Van Leeuwen, C., Vermeire, T.C. (Eds.). Risk Assessment of Chemicals: an Introduction, Chapter 7. Springer. What is a PNEC? How is a basic PNEC commonly derived in Europe? Why are assessment factors applied? Which aspects are covered by the assessment factor? Within the EU REACH/WFD regulatory framework, which assessment factor may be applied to derive a PNEC for freshwater if you have one LC value for , one EC value for , one EC value for , and one NOEC value for ? : Leo Posthuma, Dick de Zwart : Ad Ragas, Keith Solomon You should be able to: : Species Sensitivity Distribution (SSD), benchmark concentration, Potentially Affected Fraction of species (PAF) The relationship between dose or concentration (X) and response (Y) is key in risk assessment of chemicals (see section on ). Such relationships are often determined in laboratory toxicity tests; a selected species is exposed under controlled conditions to a series of increasing concentrations to determine endpoints such as the No Observed Effect Concentration (NOEC), the EC50 (the Effect Concentration causing 50% effect on a studied endpoint such as growth or reproduction), or the LC50 (the Effect Concentration causing 50% lethal effects). For ecological risk assessment, multiple species are typically tested to characterise the (variation in) sensitivities across species or taxonomic groups within the ecosystem. In the mid-1980s it had been observed that-like many natural phenomena-a set of ecotoxicity endpoint data, representing effect concentrations for various species, follows a bell-shaped statistical distribution. The cumulative distribution of these data is a sigmoid (S-shaped) curve. It was recognized, that this distribution had particular utility for assessing, managing and protecting environmental quality regarding chemicals. The bell-shaped distribution was thereupon named a Species Sensitivity Distribution (SSD). Since then, the use of SSD models has grown steadily. Currently, the model is used for various purposes, providing important information for decision-making. Below, the dual utility of SSD models for environmental protection, assessment and management are shown first. Thereupon, the derivation and use of SSD models are elaborated in a stepwise sequence. A species sensitivity distribution (SSD) is a distribution describing the variance in sensitivity of multiple species exposed to a hazardous compound. The statistical distribution is often plotted using a log-scaled concentration axis (X), and a cumulative probability axis (Y, varying from 0 - 1; Figure 1). Figure 1 shows that different species (here the dots represent 3 test data for algal species, 2 data for invertebrate species and 2 data fish species) have different sensitivities to the studied chemical. First, the ecotoxicity data are collected, and log -transformed. Second, the data set can be visually inspected by plotting the bell-shaped distribution of the log-transformed data; deviations of the expected bell-shape can be visually identified in this step. They may originate from causes such as a low number of data points or be indicative for a selective mode of action of the toxicant, such as a high sensitivity of insects to insecticides. Third, common statistical software for deriving the two parameters of the log-normal model (the mean and the standard deviation of the ecotoxicity data) can be applied, or the SSD can be described with a dedicated software tool such as ET (see below), including a formal evaluation of the 'goodness of fit' of the model to the data. With the estimated parameters, the fitted model can be plotted, and this is often done in the intuitively attractive form of the S-shaped cumulative distribution. This curve then serves two purposes. First, the curve can be used to derive a so-called Hazardous Concentration on the X-axis: a benchmark concentration that can be used as regulatory criterion to protect the environment (YàX). That is, chemicals with different toxicities have different SSDs, with the more hazardous compounds plotted to the left of the less hazardous compounds. By selecting a protection level on the Y-axis-representing a certain fraction of species affected, e.g. 5%-one derives the compound-specific concentration standards. Second, one can derive the fraction of tested species probably affected at an ambient concentration (XàY), which can be measured or modelled. Both uses are popular in contemporary environmental protection, risk assessment, and management. The oldest use of the SSD model is the derivation of reference levels such as the PNEC (YàX). That is, given the policy goal to fully protect ecosystems against adverse effects of chemical exposures (see Section on ), the protective use is as follows. First, the user defines which ecotoxicity data are used. In the context of environmental protection, these have often been NOECs or low-effect levels (ECx, with low x, such as EC ) from chronic tests. This yields an SSD-NOEC or SSD-ECx. Then, the user selects a level of Y, that is: the maximum fraction of species for which the defined ecotoxicity endpoint (NOEC or ECx) may be exceeded, e.g., 0.05 (a fraction of 0.05 equals 5% of the species). Next, the user derives the Hazardous Concentration for 5% of the species (YàX). At the HC , 5% of the species are exposed to concentrations than their NOEC, but-which is the obverse-95% of the species are exposed to concentration than their NOEC. It is often assumed that the structural and functional integrity of ecosystems is sufficiently protected at the HC level if the SSD is based on NOECs. Therefore, many authorities use this level to derive regulatory PNECs (Predicted No Effect Concentration) or Environmental Quality Standards (EQS). The latter concepts are used as official reference levels in risk assessment, the first is the preferred abbreviation in the context of prospective chemical safety assessments, and the second is used in retrospective environmental quality assessment. Sometimes an extra assessment factor varying between 1 and 5 is applied to the HC5 to account for remaining uncertainties. Using SSDs for a set of compounds yields a set of HC values, which-in fact-represent a relative ranking of the chemicals by their potential to cause harm. The SSD model also can be used to explore how much damage is caused by environmental pollution. In this case, a predicted or measured ambient concentration is used to derive a Potentially Affected Fraction of species (PAF). The fraction ranges from 0-1 but, in practice, it is often expressed as a percentage (e.g., "24% of the species is likely affected"). According to this approach, users often have monitored or modelled exposure data from various water bodies, or soil or sediment samples, so that they can evaluate whether any of the studied samples contain a concentration higher than the regulatory reference level (previous section) and, if so how many species are affected. Evidently, the user must clearly express what type of damage is quantified, as damage estimates based on an SSD or an SSD quantify the fractions of species affected beyond the no effect level and at the 50% effect level, respectively. This use of SSDs for a set of environmental samples yields a set of PAF values, which, in fact, represent a relative ranking of the pollution levels at the different sites in their potential to cause harm. SSD model outcomes are used in various regulatory and practical contexts. Today, these three forms of use of SSD models have an important role in the practice of environmental protection, assessment and management on the global scale, which relates to their intuitive meaning, their ease of use, and the availability of a vast number of ecotoxicity data in the global databases. What is the basic concept underlying SSD models? What is the main assumption underlying SSD models? What is meant by "the dual utility of an SSD model" in environmental protection, assessment and management? Given that the SSD model is a statistical description of ecotoxicological differences in sensitivity between species for a chemical, what is a critical step in the derivation and use of SSD model outputs? Does an SSD describe or explain differences in species sensitivity for a chemical? under review | 25,108 | 1,797 |
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