CHEM344 Spring 2014 Problem set no. 2

Due: Fri, Jan 31 @ 2pm BEFORE CLASS!

(Use the boxed numbers 1-12 to mark your answers)

HW TO BE HANDED IN Atkins (9th ed.) Chapter 21: Exercises: 16(b), 18(b) : Problems: 2, 8, 12, 36, 38 Tinoco (5th ed.) Chapter 9 : 3, 12, 17, 33, 37 EXTRA-DO NOT HAND IN (good practice for exams) Atkins (9th ed.) Chapter 21: Exercises: 19(b) : Problems: 5, 9, 13, 15, 37

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Steady state most useful if first step (forward, k1) is slow compared to second step ( k2), so

that intermediate population does not build up, reacts away soon as forms. Note having

reverse step gives a denominator, but since both are 1st order steps, can divide out cleanly

and just get sum of constants.

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Recall: ln k = ln A – Ea/RT

So Ea can sum from:

ln k1k2 = ln k1 + ln k2

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Couple of points, this is quite compact version of solving the mechanism, but anaolysis is

simple rapid equilibrium problem. You need to focus on OI- as product, however it is

dependent on formation of HOI in mechanism, and second (last, K3) equilibrium is fast,

so rate controlled by middle step (K2), which also forms the other product Cl-

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EXTRA -DO NOT HAND IN (practice for exams)

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