Chapter 16, Problem 68A

Interpretation Introduction

Interpretation:

To predict the reaction rate when initial concentration of $\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$ is $0.048\text{M}$.

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 68A

The reaction rate is $1.0\times {10}^{-5}\times \frac{\text{mol}}{\text{L}\text{.s}}$ when initial concentration of $\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$ is $0.048\text{M}$.

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	$0.024$	$5.00\times {10}^{-6}$

Substitute the value of experiment 1 in above rate expression.

$\begin{array}{c}\text{Rate}=k\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]{}^1\\ 2.5\times {10}^{-6}\text{mol/L}\text{.s=}k\left(0.012\text{M}\right)\\ k=\frac{2.5\times {10}^{-6}\text{mol/L}\text{.s}}{0.012\text{mol/L}}\\ =208.33\times {10}^{-6}{\text{s}}^{-1}\end{array}$

$k=2.1\times {10}^{-4}{\text{s}}^{-1}$

Therefore, k value is $2.1\times {10}^{-4}{\text{s}}^{-1}$.

Predict the reaction rate when initial concentration of $\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$ is $0.048\text{M}$ as follows:

$\begin{array}{c}\text{Rate}=k\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]{}^1\\ =2.1\times {10}^{-4}{\text{s}}^{-1}\times {\left(0.048\text{M}\right)}^1\\ =0.1008\times {10}^{-4}\times \frac{\text{mol}}{\text{L}\text{.s}}\\ =1.008\times {10}^{-5}\times \frac{\text{mol}}{\text{L}\text{.s}}\end{array}$

Therefore, reaction rate is $1.0\times {10}^{-5}\times \frac{\text{mol}}{\text{L}\text{.s}}$.

Conclusion

The reaction rate is $1.0\times {10}^{-5}\times \frac{\text{mol}}{\text{L}\text{.s}}$ when initial concentration of $\left[{\text{CH}}_{\text{3}}{\text{N}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$ is $0.048\text{M}$.