Chapter 16.3, Problem 22PP

Interpretation Introduction

Interpretation:

The mixing initial rate for the given reaction and rate law needs to be determined.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}\stackrel{}{\to }{\text{CH}}_{\text{4(g)}}{\text{+ CO}}_{\text{(g)}}\\ {\text{Rate = k[CH}}_{\text{3}}{\text{CHO]}}^{\text{2}}\end{array}$

Concept introduction:

Chemical kinetic is the branch of chemistry that deals with kinetic or rate of a chemical reaction. The rate of reaction can be defined as the change in the concentration of reactant and product with time. The rate of reaction mainly depends on the activation energy.

Answer to Problem 22PP

$\text{Rate = 4}{\text{.32×10}}^{\text{-10}}\text{M}\text{.sec}$

Explanation of Solution

Given information:

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}\stackrel{}{\to }{\text{CH}}_{\text{4(g)}}{\text{+ CO}}_{\text{(g)}}\\ {\text{Rate = k[CH}}_{\text{3}}{\text{CHO]}}^{\text{2}}\end{array}$

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}\stackrel{}{\to }{\text{CH}}_{\text{4(g)}}{\text{+ CO}}_{\text{(g)}}\\ {\text{Rate = k[CH}}_{\text{3}}{\text{CHO]}}^{\text{2}}\end{array}$

Substitute the value of concentration of CH₃CHO and rate from trial 1 or 2 to calculate rate constant k:

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}\stackrel{}{\to }{\text{CH}}_{\text{4(g)}}{\text{+ CO}}_{\text{(g)}}\\ {\text{Rate = k[CH}}_{\text{3}}{\text{CHO]}}^{\text{2}}\\ \text{10}{\text{.8×10}}^{\text{-11}}\text{=k[4}{\text{.00×10}}^{\text{-3}}{\text{]}}^{\text{2}}\\ \text{k =}\frac{\text{10}{\text{.8×10}}^{\text{-11}}}{{\text{[4}{\text{.00×10}}^{\text{-3}}\text{]}}^{\text{2}}}=6.75\times {\text{10}}^{\text{-6}}\end{array}$

Calculate initial rate for trial (3):

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}\stackrel{}{\to }{\text{CH}}_{\text{4(g)}}{\text{+ CO}}_{\text{(g)}}\\ {\text{Rate = k[CH}}_{\text{3}}{\text{CHO]}}^{\text{2}}\\ \text{Rate = 6}{\text{.75×10}}^{\text{-6}}{\text{[8}}^{{\text{.00×10}}^{\text{-3}}}\\ \text{Rate = 4}{\text{.32×10}}^{\text{-10}}\text{M}\text{.sec}\end{array}$

Conclusion

Thus,

$\text{Rate = 4}{\text{.32×10}}^{\text{-10}}\text{M}\text{.sec}$