6.Which letter in the above diagram represents the activation energy for the forward reaction, i.e for the uncatalyzed conversion of reactants into products? (a)A (b)B (c)D (d)E (e)None of the above. Answer: A Solution: The energy between the reactants and the peak of the larger curve is the activation energy for the uncatalyzed reaction. 7.Which letter represents the change in potential energy for the overall reaction? (a)A (b)B (c)D (d)E (e)None of the above. Answer: D Solution: The letter E is the change in potential energy for the overall reaction. This is the difference between the potential energy of the products and the reactants. E Reaction coordinatePotential Energy A B D C

solutions 1528.The rate law for a reaction is rate = k[A]2[B]. Which of the following mechanisms supports this rate law? i) A + B E (fast) E + B ⎯→C + D (slow) ii) A + B E (fast) E + A ⎯→C + D (slow) iii) A + A ⎯→E (slow) E + B ⎯→C + D (fast) (a)i (b)ii (c)iii (d)two of these (e)none of these. Answer: B Solution: It cannot be iii as the first step is the rate determining step. For ii: slow step rate = k[E][A] and fast equilibrium rate = k1[A][B] = k-1[E] [E] = k[A][B], therefore rate = [A]2[B] 9.If the reaction 2 HI (g) ⎯→H2(g) + I2(g) is second order, which of the following plots would be linear? (a)1/[HI] vs. time (b)[HI]2vs. time (c)ln[HI] vs. time (d)−[]dtdHIvs. time. (e)None of the above would be linear. Answer: A Solution: ktAAo+=][1][1, therefore from y = b + mx a plot of 1/[HI] vs. time would be linear. ⎯→←⎯⎯→←⎯

solutions 153Short Answer: 10.Suppose for an uncatalyzed reaction, it is observed that the rate constant, k, at T= 300 K and T = 1000 K are 3.88×10−5s−1and 5.98×107s−1, respectively. a)Assuming the Arrhenius law holds for the rates, what are the activation energy and pre-exponential factor for the reaction? 11lnlnRTEAKa−=and 22lnlnRTEAKa−=Hence, ⎟⎟⎠⎞⎜⎜⎝⎛−−=121211lnTTREKKaEa= ()()00233333.06.28314.81000130011088.31098.5ln314.811ln572112=⎟⎠⎞⎜⎝⎛−××=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−TTKKR= 100 kJ. 11/1/1RTERTEaaeKAAeK=⇒=−= (3.88×10−5) ()()300314.8000,100e= 1.00×1013sec−1. −b)Suppose that the reaction is catalyzed at T = 500 K and it is found that kcat(500) = 4.40×107s−1. Assuming the catalytic mechanism is entirely due to lowering of activation energy, how much has the transition state energy been lowered? Solution: ln K= ln A −Ea/RTln 4.40×107= ln 1.00×1013- )500)(314.8('EaSolve this equation to get Ea’ = 51.3 kJ Therefore ∆Ea = Ea – Ea’ = 48.7 kJ c)At what temperature does the uncatalyzed reaction proceed at the same rate as the catalyzed reaction at T = 500 K? Solution: Kcat= ARXERTEacaAeKe//−−==caaacaEETXRXERTE=⇒=. Ea= 100 kJ and caE= 51.3kJ ()⎟⎠⎞⎜⎝⎛=kJ51.3kJ100K500oX= 975K. .9 K.

solutions 15411.For the reaction 2 A + B ⎯→Products The following mechanism is proposed: A + B M Forward rate: k1Backward rate: k−1A + M ⎯→Products Forward rate: k2(a)Assuming that the second step is the rate-determining step and the first step is a fast equilibrium step, determine the rate law. Represent the rate constant in terms of k1, k−1, and k2.