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.

Answer 1

Answer : The rate law for the overall reaction is, $Rate=k[A][B]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : $A+B\rightarrow AB$    (slow)

Step 2 : $A+AB\rightarrow A_2B$     (fast)

Overall reaction : $2A+B\rightarrow A_2B$

The rate law expression for overall reaction should be in terms of A and B.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow AB$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

Hence, the rate law for the overall reaction is $Rate=k[A][B]$