Suppose the formation of tert-butanol proceeds by the following mechanism: step elementary reaction rate constant (CH), СBr (aq) —(CH),С (aq) + Br (aq) c* 1 2 | (CH), С"(аq) + он (аа) — (СH;), СоН(а) (CH3), COH(aq) Suppose also k,«k,. That is, the first step is much slower than the second. Write the balanced chemical equation for the 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.

Physical Chemistry
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Author:Ball, David W. (david Warren), BAER, Tomas
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Chapter22: Surfaces
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Problem 22.44E: Are the following processes examples of homogeneous or heterogeneous catalysis? a Hydrolysis of...
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Suppose the formation of tert-butanol proceeds by the following mechanism:
step
elementary reaction
rate constant
CBr(aq)
(CH), С"(аg) + Br (ag)
c*
k
1
2 (CH3),C"(aq) + OH (aq) → (CH,),COH(aq)
k2
3
3
Suppose also k, «k,. That is, the first step is much slower than the second.
Write the balanced
chemical equation for the
overall chemical reaction:
Write the experimentally-
observable rate law for the
overall chemical reaction.
rate = k U
Note: your answer should
not contain the
concentrations of any
intermediates.
Express the rate constant
k for the overall chemical
reaction in terms of k1, k2,
and (if necessary) the rate
constants k.1 and k-2 for
0
k =
the reverse of the two
elementary reactions in
the mechanism.
Transcribed Image Text:Suppose the formation of tert-butanol proceeds by the following mechanism: step elementary reaction rate constant CBr(aq) (CH), С"(аg) + Br (ag) c* k 1 2 (CH3),C"(aq) + OH (aq) → (CH,),COH(aq) k2 3 3 Suppose also k, «k,. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. rate = k U Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k.1 and k-2 for 0 k = the reverse of the two elementary reactions in the mechanism.
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