Question

Consider the mechanism. Step 1: A + B − ⇀ ↽ − C equilibrium Step 2: C + A ⟶ D slow Overall: 2 A + B ⟶ D Determine the rate law for the overall reaction, where the overall rate constant is represented as

Answer 1

Answer : The rate law for the following mechanism is $Rate=k'[A]^2[B]$

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 : $A+B\rightleftharpoons C$    (fast)

Step 2 : $C+A\rightarrow D$    (slow)

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

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,

$C+A\rightarrow D$

The expression of rate law for this reaction will be,

$Rate=k[C][A]$         ............(1)

Now applying steady state approximation for $C$, we get:

$K=\frac{[C]}{[A][B]}$

$[C]=K\times [A][B]$     ...........(2)

Now substituting equation 2 in 1, we get:

$Rate=k[A]\times K\times [A][B]$

$Rate=K\times k[A]^2[B]$

$Rate=k'[A]^2[B]$

Hence, the rate law for the following mechanism is $Rate=k'[A]^2[B]$