The reaction A rightarrow D has the following mechanism It is safe to assume tha

Question

The reaction A rightarrow D has the following mechanism It is safe to assume that the formation of the intermediate C is the very slow, and thus the concentration of C remains small and constant throughout the reaction. In addition. assume that the equilibrium between A and B is step (1) is very fast compared to all other rates in the problem. What are the two approximations that you can make given the informal ion above? (?) Use these approximation to derive a rate expression In terms of only the rate constant and the concentration of (A). In the limit where [A] is very large. for example at the beginning of the reaction, what is the order of the overall reaction? In this same limit. give the observed rate constant. In terms of the rate constants of the mechanism (k_1, k_3, K_2 and k_3).

Explanation / Answer

a) Since the formation of C is very slow and step 1 is very fast compared to the other rates, then one can assume step 1 would be the rate determining step

b) d[D]/dt = k3[B][C]

From step 1:

k1[A] = k-1[B]

[B] = k1*[A]/k-1

From step 2:

k2[A] = k-2[C]

[C] = k2*[A]/k-2

substituting for B and C,

d[D]/dt = k3* k1* k2*[A]2 / k-1*k-2

c) when A is large, order of reaction is 2

kobs = k3k1k2/k-1k-2

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