Suppose the formation of tert-butanol proceeds by the following mechanism step e

Question

Suppose the formation of tert-butanol proceeds by the following mechanism step elementary reaction rate constant (CH3)3C. (aq) + OH (aq). + (CH),COH(aq)| Suppose also k «kz. That is, the first step is much slower than the second chermical equation for the Write the balanced overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. rate -k Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of ki, k 1. K2, and (if necessary) the rate- constants k.1 and k.2 for the reverse of the two elementary reactions in the mechanism.

Explanation / Answer

1)

Overall reaction is obtained by adding individual steps

That is

(CH3)3CBr (aq) + OH- (aq) —> (CH3)3COH (aq) + Br- (aq)

2)

Rate depends on the slowest step.

Here 1st step is slowest

So, rate law is:

rate = k1 [(CH3)3CBr]

let k1= k

rate = k [(CH3)3CBr]

3)

Since 1st step is slowest

k = k1

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