Question

A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction. Consider the following multistep reaction:
A+B ----- AB(slow)
A+AB-----A2B(fast
....................................................
2A+B ----- A2B(overall)
Based on this mechanism, determine the rate law for the overall reaction.
a) rate = kA2BAB
b) rate = kAB
c) rate = kAAB
d) rate = kA2B

Answer 1

Answer:

b) rate = kAB.

Explanation:

Hello,

In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:

$A+B \rightarrow AB\ \ (slow)$

In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:

$rate=k[A][B]$

Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.

Best regards.