Chapter 16, Problem 66A

Interpretation Introduction

Interpretation:

To determine the experimental rate law of the reaction ${\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right){\text{+N}}_{\text{2}}\left(\text{g}\right)$.

Concept introduction:

The rate law of the general reaction $\text{aA}\to \text{Products}$ is

$\text{rate}=k{\left[\text{A}\right]}^m$

Here, $\left[\text{A}\right]$ is the concentration of reactant A, k represents rate and m represents order of corresponding reactants.

Answer to Problem 66A

The rate law of the reaction is $k\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]{}^1$. It is a first order reaction. Its concentration of $\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$ doubled therefore rate also doubled.

Explanation of Solution

The given reaction is as follows:

${\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right){\text{+N}}_{\text{2}}\left(\text{g}\right)$

Rate law of the reaction is:

$\text{Rate}=k\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]{}^1$

Two different experiments are given with different initial concentration of reactant in the table.

Experiment Number	Initial $\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]\text{M}$	Initial reaction rate $\left(\text{mol/L}\text{.s}\right)$
1	$0.012$	$2.5\times {10}^{-6}$
2	$0.024$	$5.00\times {10}^{-6}$

From the table it is clear that concentration of ${\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}$ is doubled in second experiment than the first. It is a first order reaction.

Conclusion

Rate of the reaction increased with increase in the reactant concentration.