Chapter 16, Problem 64A

Interpretation Introduction

Interpretation:

To explain how the instantaneous rate of the reaction ${\text{H}}_2\left(\text{g}\right)\text{+2NO}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right){\text{+N}}_{\text{2}}\text{O}\left(\text{g}\right)$ changes as the initial concentration of ${\text{H}}_2$ changed.

Concept introduction:

The given reaction is as follows:

${\text{H}}_2\left(\text{g}\right)\text{+2NO}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right){\text{+N}}_{\text{2}}\text{O}\left(\text{g}\right)$

Rate law of the reaction is-

$\text{Rate}=k{\left[\text{NO}\right]}^{\text{2}}\left[{\text{H}}_{\text{2}}\right]$

The rate law was calculated from experimental data suggesting that the rate depends on the reactant concentration as follows: If [NO] doubles, the rate quadruples; if ${\text{H}}_2$ doubles, the rate doubles.

Answer to Problem 64A

Instantaneous rate of the reaction increases by $1.8$ times with change in initial concentration of ${\text{H}}_2$. And it is the part of the rate law.

Explanation of Solution

The given reaction is as follows:

${\text{H}}_2\left(\text{g}\right)\text{+2NO}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right){\text{+N}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concentration of ${\text{H}}_2$ measured at three successive points during the experiment is given in table.

$\left[{\text{H}}_{\text{2}}\right]\left(\text{mol/L}\right)$	Instantaneous rate $\left(\text{mol/L}\text{.s}\right)$
$0.18$	$6.00\times {10}^{-3}$
$0.32$	$1.07\times {10}^{-2}$
$0.58$	$1.93\times {10}^{-2}$

From the table it is clear that concentration of ${\text{H}}_2$ increased by $1.8$ times from first line to the second and also rate is increased by about $1.8$ times.

Again, see the changes of concentration of ${\text{H}}_2$ from lines 2 to 3. Concentration of ${\text{H}}_2$ increased by $1.8$ times and also rate is increased by about $1.8$ times. Hence, ${\text{H}}_2$ is a part of the rate law and it is first order reaction.

Conclusion

Rate of the reaction increased with increase in the reactant concentration. The reactant concentration ${\text{H}}_2$ is increased in each step by $1.8$. Therefore, it is a part of the rate law.