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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.