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

The rate of the over all reaction is ;

`R=K[A_2]^(1/2)[B]`

Step-by-step 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=([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* √(K_1* [A_2])* [B]`

`K=k* (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]`