Suppose the formation of tert-butanol proceeds by the following mechanism:step elementary reaction rate constant1 →CH33CBraq+CH33C+aqBr−aq k12 →+CH33C+aqOH−aqCH33COHaq k2Suppose also k1 ≪ k2. 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. =ratek Note: your answer should not contain the concentrations of any intermediates.
Question
Suppose the formation of tert-butanol proceeds by the following mechanism:step elementary reaction rate constant1 →CH33CBraq+CH33C+aqBr−aq k12 →+CH33C+aqOH−aqCH33COHaq k2Suppose also k1 ≪ k2. 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. =ratek Note: your answer should not contain the concentrations of any intermediates.
Solution
The overall balanced chemical equation for the reaction is obtained by adding the elementary reactions together:
CH3-C(CH3)3Br + OH- → CH3-C(CH3)3OH + Br-
The rate law for the overall reaction is determined by the slowest (rate-determining) step. Since the first step is much slower than the second, it is the rate-determining step. The rate law for the overall reaction is therefore:
rate = k1[CH3-C(CH3)3Br]
Note that the rate law does not include the concentration of any intermediates, as requested.
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