Question

For the net reaction: 2AB + 2C → A2 + 2BC, the following slow first steps have been proposed. AB → A + B 2C + AB → AC + BC 2AB → A2 + 2B C + AB → BC + A

Answer 1

This is an incomplete question, here is a complete question.

For the net reaction: 2AB + 2C → A2 + 2BC, the following slow first steps have been proposed.

(1) AB → A + B

(2) 2C + AB → AC + BC

(3) 2AB → A₂ + 2B

(4) C + AB → BC + A

What rate law is predicted by each of these steps?

Answer : The rate law expression for the following reactions are:

(1) $\text{Rate}=k[AB]$

(2) $\text{Rate}=k[C]^2[AB]$

(3) $\text{Rate}=k[AB]^2$

(4) $\text{Rate}=k[C][AB]$

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.

The general reaction is:

$A+B\rightarrow C+D$

The general rate law expression for the reaction is:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

R = rate  law

k = rate constant

$[A]$ and $[B]$ = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

(1) The balanced equations will be:

$AB\rightarrow A+B$

In this reaction, $AB$ is the reactant.

The rate law expression for the reaction is:

$\text{Rate}=k[AB]$

(2) The balanced equations will be:

$2C+AB\rightarrow AC+BC$

In this reaction, $C$ and $AB$ are the reactants.

The rate law expression for the reaction is:

$\text{Rate}=k[C]^2[AB]$

(3) The balanced equations will be:

$2AB\rightarrow A_2+2B$

In this reaction, $AB$ is the reactant.

The rate law expression for the reaction is:

$\text{Rate}=k[AB]^2$

(4) The balanced equations will be:

$C+AB\rightarrow BC+A$

In this reaction, $C$ and $AB$ are the reactants.

The rate law expression for the reaction is:

$\text{Rate}=k[C][AB]$