Question

Given the following proposed mechanism, predict the rate law for the overall reaction. A2 + 2 B → 2 AB (overall reaction) Mechanism A2 ⇌ 2 A fast A + B → AB slow

Answer 1

Answer:

The rate of the over all reaction is ;

$R=K[A_2]^{1/2}[B]$

Explanation:

Step 1 : $A_2\rightleftharpoons 2 A$ fast

Step 2  : $A + B\rightarrow AB$ slow

Equilibrium constant of the reaction in step 1:

$K_1=\frac{[A]^2}{[A_2]}$....[1]

Overall reaction:

$A_2 + 2 B \rightarrow 2 AB$

When there is a chemical reaction which taking place in more than 1 step than the rate of the over all reaction is determined by the slowest step occurring during that process;

Here step 2 is slow step, so the rate of the reaction will be;

$R=k[A][B]$..[2]

Putting value of [A] from [1] in [2]:

$R=k\times \sqrt{K_1\times [A_2]}\times [B]$

$K=k\times (K_1)^{1/2}$

K = rate constant of the reaction

The rate of the over all reaction is ;

$R=K[A_2]^{1/2}[B]$