Chapter 16, Problem 73A

Interpretation Introduction

Interpretation:

The rate constant and concentration of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ after the reaction proceeds for 1.30 h needs to be determined.

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }{\text{4 NO}}_{\text{2}}{\text{ + O}}_{\text{2}}$

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.

In a chemical reaction to be occurred; two most important requirements are activation energy to reactant molecule and correct orientation of reactant molecules to colloid and form product.

Answer to Problem 73A

$\text{k = }6.2\times {\text{ 10}}^{\text{-4}}{\text{ min}}^{\text{-1}}$

Concentration after 1.30 hours = $\text{0}\text{.247M}$

Explanation of Solution

Given information:

Initial concentration of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ = 0.400 mol/L

Rate = 2.48 x 10-4 M/min

Time = 1.30 hour

For the given chemical reaction, the rate must be:

${\text{Rate = k [N}}_{\text{2}}{\text{O}}_{\text{5}}]$

Substitute the values of rate and initial concentration to calculate rate constant:

$\begin{array}{l}{\text{Rate = k [N}}_{\text{2}}{\text{O}}_{\text{5}}\text{] }\\ \text{2}\text{.48 }\times {\text{ 10}}^{\text{-4}}\text{M/min = k[0}\text{.400M] }\\ \text{k = }\frac{\text{2}\text{.48 }\times {\text{ 10}}^{\text{-4}}\text{M/min }}{\text{0}\text{.400M}}\\ \text{k = }6.2\times {\text{ 10}}^{\text{-4}}{\text{ min}}^{\text{-1}}\end{array}$

The integrated rate equation for first order reaction is:

$\begin{array}{l}\text{k =}\frac{\text{2}\text{.303}}{\text{t}}\text{ log }\frac{\text{a}}{\text{a-x}}\\ \text{k = rate constant}\\ \text{a = initial concentration}\\ \text{a-x= concentration after time t}\end{array}$

Substitute the values to calculate concentration after 1.30 hours,

$\begin{array}{l}\text{k =}\frac{\text{2}\text{.303}}{\text{t}}\text{ log }\frac{\text{a}}{\text{a-x}}\\ \text{ 6}{\text{.2× 10}}^{\text{-4}}{\text{ min}}^{\text{-1}}\text{ =}\frac{\text{2}\text{.303}}{\text{1}\text{.30×60 min}}\text{ log }\frac{\text{0}\text{.400 M}}{\text{a-x}}\\ \text{ log }\frac{\text{0}\text{.400 M}}{\text{a - x}}\text{=0}\text{.0210}\\ \frac{\text{0}\text{.400 M}}{\text{a - x}}\text{=1}\text{.622}\\ \text{a-x=}\frac{\text{0}\text{.400 M}}{\text{1}\text{.622}}\text{= 0}\text{.247M}\end{array}$

Conclusion

Thus,

$\text{k = }6.2\times {\text{ 10}}^{\text{-4}}{\text{ min}}^{\text{-1}}$

Concentration after 1.30 hours = $\text{0}\text{.247M}$