I need help with problem number 1. Read: Chapter 14, p. 274 - 285 Problems: 14.1

Question

I need help with problem number 1.

Read: Chapter 14, p. 274 - 285 Problems: 14.1(5) and 1. The reaction mechanism for renaturation of a double helix from its strands A and B involves an unstable intermediate: k1 k2 A + B unstable helix-, stable helix k.1 Deduce the rate law for the reaction assuming pre-equilibrium. Deduce the rate law for the reaction using the steady-state approximation. a. b. c. d. Show that under certain conditions, the answers for a. and b. are equivalent. If the observed rate constant, kobs-0.25 M 's 1 and the reaction began with [Aj = 0.04 M and = 0.04 M, how long will it take until 95% of the strands have formed a stable helix? HINT: this is a special case where [A] -[B], so it simplifies to one of the common integrated rate laws. 2. Build a numerical kinetic model and use it to answer a few question (see other sheet)

Explanation / Answer

(a)

Assuming pre-equilibrium:

Equilibrium constant for step 1 = K = [unstable helix]/[A][B]

=> [unstable helix] = K[A][B]

Reaction rate = rate of step 2

= k2[unstable helix]

= k2K[A][B]

Thus the rate law is: Rate = k[A][B] where the rate constant k = k2K

(b)

Applying steady state approximation to [unstable helix]:

d[unstable helix]/dt = k1[A][B] - k-1[unstable helix] - k2[unstable helix] = 0

[unstable helix] = k1/(k-1 + k2)[A][B]

Reaction rate = rate of step 2

= k2[unstable helix]

= k1k2/(k-1 + k2)[A][B]

Since step 2 is slow and step 1 is fast: k2 << k-1 so (k-1 + k2) k-1

Reaction rate = k1k2/k-1[A][B]

Thus the rate law is: Rate = k[A][B] where the rate constant k = k1k2/k-1

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