Chapter DY, Problem 1PQ

Interpretation Introduction

Interpretation:

The gas-phase reaction, ${\text{A}}_{\text{2}}{\text{+B}}_{\text{2}}\stackrel{}{\to }\text{2AB}$ proceeds by bimolecular collisions between ${\text{A}}_{\text{2}}$ and ${\text{B}}_{\text{2}}$ molecules.  If the concentrations of both ${\text{A}}_{\text{2}}$ and ${\text{B}}_{\text{2}}$ are doubled, then by how much factor the reaction changes has to be given.

Concept Introduction:

Rate of a reaction represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how concentration of targets (products) formed in a unit of time is said to be rate of reaction.  Rate of a reaction is determined experimentally.

Rate law:

In rate law, the rate of a reaction is expressed as function of the reactants and products in the reaction.

For a reaction,

    $\begin{array}{l}\text{A+B}\stackrel{}{\to }\text{C}\\ \text{rate law υ = k[A][B]}\end{array}$

where k is called the rate constant.

Explanation of Solution

Reason for correct option:

Given equation is:

    ${\text{A}}_{\text{2}}{\text{+B}}_{\text{2}}\stackrel{}{\to }\text{2AB}$

Rate law is:

    ${\text{rate law υ = k[A}}_2{\text{][B}}_2\text{]}$.

When the concentration of both the reactants are doubled.  Then the rate law becomes,

    ${\text{rate law υ = k[2A}}_2{\text{][2B}}_2{\text{] = 4k[A}}_2{\text{][B}}_2\text{]}$

Thus the rate is increased by four times.  That is the rate is increased by a factor of 4.

Hence the correct option is (D).

Reason for incorrect option:

From the above explanation of rate law, it is clear that options (A), (B) and (C) are incorrect.