CHEMISTRY 112 Question Set IV: Answers

CHEMISTRY 112 School of Pharmacy University of California, San Francisco Question Set IV: Answers 1. A value of the entropy of activation that is posi...

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CHEMISTRY 112 School of Pharmacy University of California, San Francisco Question Set IV: Answers 1. A value of the entropy of activation that is positive indicates that the entropy of the transition state is higher than the entropy of the reactants. Thus, the transition state is less ordered than the reactants. An entropy of activation of zero implies that the transition state has the same amount of order as the reactants. Since k = 2 x 1010 Te(∆S‡ /R) e-∆H‡ /RT, we can see that if ∆S‡ <0, than k is smaller than it would be if ∆S‡ = 0. Thus the reaction would be slower if ∆S‡ <0 than it would be if ∆S‡ = 0, all other things being the same. The entropy of reaction is defined as Sproducts - Sreactants while the entropy of activation refers to the difference Stransition state - Sreactants. 2. Enthalpies of activation are almost always positive for single step reactions. The reason is that almost all reactions have to surmount an energy barrier in order for rearrangement of reacting species to product species to occur. Free energies of activation for single step reactions are always positive. If this were not the case, then the transition state would be more stable than the reactants; if this were true, then the transition state would be more likely to be an isolatable species than the reactants. This is contrary to the assumptions of the transition state theory, which implies that the transition state does not have a finite lifetime. Note, however, in multistep reactions the overall enthalpy of activation and overall free energy of activation can conceivably be negative; the reasons for this are similar to those developed for the overall Ea for a reaction sequence, as developed in Problem 2 in the current Problem Set. 3. Primary kinetic salt effects are those exerted on the slowest step in the forward direction of a reaction sequence. Secondary kinetic salt effects are those exerted on a step preceding the slowest step in a reaction sequence. Both of these effects are described quantitatively by the Brønsted-Bjerrum equation and have their physical origin in the effects of ionic strength on the activities of ionic reacting species in solution. The Brønsted-Bjerrum equation was derived under the assumption that these ionic strength effects could be quantitatively represented by Debye-Huckel theory, a theory of electrolyte solutions that assumes dilute ionic solutions. 4. The difference is in the form of the rate law describing the catalytic pathway. A catalytic pathway is said to be specific acid catalyzed if the rate law describing that path has the form k[X]...[H+], where [X]... stands for a product of reactant concentrations. It is general acid catalyzed if the rate law describing that the catalytic pathway has the form k[X]...[HA], where HA is an undissociated acid. See Problem 6 of the current problem set for a mechanism that, under certain conditions, can lead to either specific or general acid catalysis. Problem Set IV: Solutions 1. For the forward reaction at T1 , we can write k11 = 2 x 1010 T1 e(∆S1 ‡ /R) e-∆H‡ /RT1 and at T2 , k12 = 2 x 1010 T2 e∆S‡ e-∆H‡ /RT2 . Then, we obtain k12 /k11 = (T2 /T1 )e-(∆H‡ /R)(1/T2 - 1/T1 ). Taking natural logs of both sides, we find ln (k12 /k11 ) = ln (T2 /T1 ) - (∆H‡ /R)(1/T2 - 1/T1 ). Putting in numbers and letting T2 = 298, we find ln 12 = 2.48 = ln (298/273) - (∆H‡ /R){(1/298) (1/273)}. Solving, we find ∆H‡ = 64.90 kJ. Substituting this number into the expression for k12 , we find that e(∆S‡ /R) = 3.81 x 10-4 and ∆S‡ = -63.4 J/K. Now, for the reversible reaction at 298 K, we can write k1 /k-1 = e{(∆S1 ‡ - ∆S-1 ‡ )/R}e-(∆H1 ‡ - ∆H-1 ‡ )/RT. Now the ratio of k1 /k-1 at 298 K is the equilibrium constant at 298 K and thus we have one equation in two unknowns. Since the problem stated that the structures of the reactant and product are quite similar, it is reasonable to assume that, having the same transition state, the entropy of activation of the forward and reverse reactions are the same; thus ∆S1 ‡ - ∆S-1 ‡ is zero. Then we can solve for the quantity (∆H1 ‡ - ∆H-1 ‡ ) at 298K. Substituting the appropriate numbers, we find ln (k1 /k-1 ) = ln 50 = 3.91 = -(∆H1 ‡ - ∆H-1 ‡ )/RT = -(64.78 kJ - ∆H-1 ‡ )/298R. Solving, we find ∆H-1 ‡ = 74.59 kJ.

Returning to the relation ln (k1 /k-1 ) = -(∆H1 ‡ - ∆H-1 ‡ )/RT, we can calculate that ∆H1 ‡ = - 9.69 kJ and, hence, Keq (273) = k1 /k-1 = e-(∆H1 ‡ - ∆H-1 ‡ )/RT = e9690/273R = 71.4.

- ∆H-1 ‡

2. The overall rate constant k for the rate law describing the mechanism is given by k = {k1 k3 k4 /k5 }1/2 . The Arrhenius equation states that k = Ae-Ea\RT. However, each of the rate constants in the composite expression also has the same form (eg. k1 = A1 e-E1 \RT). Thus, we can write k = Ae-Ea\RT = {(A1 A3 A4 /A5 )e-(E1 + E3 + E4 - E5 )/RT}1/2 . Using the laws of exponents, we obtain k = Ae-Ea\RT (A1 A3 A4 /A5 )1/2 e-(E1 + E3 + E4 - E5 )/2RT. Equating like terms in these two expressions, we find that Ea = (E1 + E 3 + E 4 - E5 )/2. Substituting the values for the various activation energies given in the problem, we find that E3 = 198 kJ/mole. 3. (a) For this set of parallel reactions at T1 , we can write (k2 /k1 )T1 = ([C]/[B])T1. We can also write (k2 /k1 )T1 = (2 x 1010 T1 e(∆S2 ‡ /R) e-∆H2 ‡ /RT1 )/(2 x 1010 T1 e(∆S1 ‡ /R) e-∆H1 ‡ /RT1 ) = e∆∆ S‡ /R e-∆∆ H‡ /RT1 , where ∆∆S‡ = ∆S2 ‡ - ∆S1 ‡ and ∆∆H‡ = ∆H2 ‡ - ∆H1 ‡ . Similarly, at T2 , we can write (k2 /k1 )T2 = (2 x 1010 T2 e(∆S2 ‡ /R) e-∆H2 ‡ /RT2 )/(2 x 1010 T2 e(∆S1 ‡ /R) e-∆H1 ‡ /RT2 ) = e∆∆ S‡ /R e-∆∆ H‡ /RT2 . Taking the ratio of (k2 /k1 )T2 to (k2 /k1 )T1, we obtain (k2 /k1 )T2/(k2 /k1 )T1 = e-∆∆ H‡ /R[1/T2 - 1/T1 ]. The problem states that (k2 /k1 )T1 = 2 and (k2 /k1 )T2 = 1, where T1 = 300 K and T2 = 350 K. Thus, we can write 2 = e-∆∆ H‡ /R[1/350 - 1/300]. Taking the natural log of both sides, 0.693 = (-∆∆H‡ /R)(-0.000476). Solving for ∆∆H‡ , we find its value to be -12,104 J. Now returning to the equation (k2 /k1 )T1 = e∆∆ S‡ /R e-∆∆ H‡ /RT1 , we can evaluate ∆∆S‡ to be -34.60 J/K. (b) Using the relation (k2 /k1 )T = e∆∆ S‡ /R e-∆∆ H‡ /RT and the values for ∆∆H‡ and ∆∆S‡ determined above, we find that (k2 /k1 )400 = 0.59 = ([C]/[B])400 . Thus ([B]/[C])400 = 1.69. 4. (a) At least one uncharged species enters into the transition state of the rate determining step of the reaction. (b) The ions entering the activated complex corresponding to the slowest step have opposite charges. 5. (a) For the reaction Hg+ + Hg+ → Hg2 ++, the reaction would speed up, as the ions have the same charge. (b) For the reaction CH3 Br + OH- → CH3 OH + Br -, the addition of salt will have no effect, as one of the species is uncharged. (c) For the reaction Co(NH3 )5 Br + + + NO 2 - → Co(NH 3 )5 NO 2 + + + Br -, the reaction will be slower in the presence of inert salt, as the charges on the ions have the opposite sign. Dissolution of one mole of Li2 SO 4 gives three moles of ions, while dissolving one mole of LiNO3 yields two moles of ions. Thus, the Li2 SO 4 solution will have the highest ionic strength for a given molar concentration of salt and, therefore, have the largest effect on reaction rates. 6. We are interested in the rate of product formation, which is given by d[P1 ]/dt (or alternatively d[P2 ]/dt. We apply the steady state approximation to [SH+], as SH+ is the unstable intermediate in this mechanism. Then d[SH+]/dt = k2 [HA][S] - k3 [SH +][H2 O] - k-2 [[SH +][A-] ≈ 0. From this equation, we obtain [SH+] = k2 [HA][S]/{k3 [H2 O] + k-2 [A-]}. Then d[P1 ]/dt = k3 k2 [HA][S][H2 O]/{k3 [H2 O] + k -2 [A-]}. If k3 [H2 O] >>k-2 [A-], then d[P1 ]/dt = k[HA][S], which is a rate law of the form that describes general acid catalysis. If k3 [H2 O] << k-2 [A-], then d[P1 ]/dt = k2 k3 [HA][S][H2 O]/k-2 [A-]. Now [HA]/[A-] = [H+]/Ka, where Ka is the dissociation constant for the acid HA. Then d[P1 ]/dt = k2 k3 [S][H 2 O][H +]/Ka; this rate law is of a form that describes specific acid catalysis. 7. The relation of interest is the Brønsted-Bjerrum equation: log k = log k0 + 1.018I1/2 ZAZB, where I is the ionic strength. From the data given, we can prepare a graph of log k vs I1/2 , as shown below. From this graph, we can calculate the slope m = 1.9 = 1.018 ZAZB or ZAZB = 1.87. Since ionic charges can only take integer values, ZAZB can only take integer values. Since the calculated value of ZAZB is nearer to 2 than 1, we conclude that ZAZB = 2. Since I- has a -1 charge, (-1)ZB = 2. Therefore ZB = -2.

0.15 0.1 0.05 log k 0 -0.05 -0.1 0

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I1 /2 8. A. (a) (competitive inhibition) v0 = Vm[S]/{K m(1 + (KI/[I])) + [S]} = (53 x 10-6 )(3.7 x 10-4 )/{2.8 x 10-5 (1 + (4.8 x 10-4 /1.7 x 10-5 ) + 3.7 x 10-4 } = 1.65 x 10-5 M/min (b) (noncompetitive inhibition) v0 = Vm[S]/{(K m + [S])(1 + ([I]/KI))} = (53 x 10 -6 )(3.7 x 10-4 )/ {(2.8 x 10-5 +3.7 x 10-4 )(1 + (4.8 x 10-4 /1.7 x 10-5 )}] = 1.69 x 10 -6 M min-1 . (assuming Ki = K i ') (c) (uncompetitive inhibition) v0 = Vm[S]/{K m + [S](1 + ([I]/KI))} = (53 x 10 -6 ) (3.7 x 10-4 )/(2.8 x 10-5 + 3.7 x 10-4 )(1 + (4.8 x 10-4 /1.7 x 10-5 )) = 1.8 x 10-6 M min-1 . B. We must first find the v0 for the situation with no inhibition. For this, we use the Michaelis-Menten equation v0 = Vm/(1 + (Km/[S]) = (53 x 10-6 )/(1 + (2.8 x 10 -5 /3.7 x 10-4 ) = 4.9 x 10-5 mole liter-1 min-1 . Then using the given relationship with the values obtained above, we find a = (a) 0.35 for competitive inhibition (b) 0.035 for noncompetitive inhibition and (c) 0.037 for uncompetitve inhibition. Thus i = (a) 66.3%, (b) 96.5% and (c) 96.3%. These values correspond to the percentage of the total enzyme that is inhibited (i.e. occupied by an inhibitor molecule). 9. (a) The Lineweaver-Burk plots of the uninhibited and inhibited oxidations are shown immediately below. This plot is characteristic of competitive inhibition. [I] = 0: y = 0.198x + 0.450 r2 = 0.995 [I] = 10 µM: y = 0.959x + 0.481 r 2 = 1.000

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(b) The X-intercept of the line for the uninhibited reaction has a value of -2.20 mM-1 . Since Km is given by the inverse of this intercept, we find Km = 0.454 mM. 1/Vm is given by the Y-intercept, which has a value of 0.45. Thus Vm = 2.22 mmoles/liter min. (c) The X-intercept for the pyrazole inhibited reaction, Xi , is -0.45 which can be equated to the quantity -(1/Km) x (KI/{KI + [I]}. Solving, we find KI = -K mXi [I]/(1+KmXi ). Substituting in values, we find KI = -0.454(-0.5)(10)/(1 + 0.454(-0.5)) = 2.58 µM. (d) From Problem 8B the degree of inhibition is given by i (%) = 100(1 - a), where the quantity a = (v0 )inhibition/v0 . The data in the Table in the statement of the problem gives a = 0.69/1.52 = 0.454. Thus i = 54.6 % (e) This corresponds to the situation where the degree of inhibition is 90% and, thus, a = 0.1. The appropriate relationship for the necessary calculation (from the Handout on Enzyme Kinetics) is 1/a = v0 /(v0 )inhibition = 10 = 1 + Km[I]/([S]KI + K mKI); employing the numerical values obtained in the analysis above gives the required value of [I] as 74.8 µM.