Chapter 13, Problem 13.98QP

Interpretation Introduction

Interpretation:

The average rate constant of the decomposition of Nitrogen dioxide has to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time. The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$ .

The equation that relates the reaction rate to the reactants concentrations that is raised to various powers is called as rate law.

Rate law can be determined by the slow step or otherwise called as rate-determining step.

Rate constant is defined as proportionality constant in the connection between rate and concentrations.

To calculate the average rate constant of the decomposition of Nitrogen dioxide

Answer to Problem 13.98QP

The average rate constant of the decomposition of Nitrogen dioxide is $\text{3}{\text{.8×10}}^{\text{-3}}\text{/}\left(\text{s}\right)$ .

Explanation of Solution

Given,

$\begin{array}{l}\begin{array}{cc}\text{Time}& \left[{\text{NO}}_{\text{2}}\right]\\ \text{0}\text{.0min}& \text{0}\text{.1103}\text{M}\end{array}\\ \begin{array}{cc}\text{1}\text{.0min}& \text{0}\text{.1076}\text{M}\\ \text{2}\text{.0min}& \text{0}\text{.1050}\text{M}\end{array}\\ \begin{array}{cc}\text{3}\text{.0}\text{min}& \text{0}\text{.1026}\text{M}\end{array}\end{array}$

$\begin{array}{l}\begin{array}{cc}\text{Time}& \text{Average}\text{rate}\text{of}\text{Decomposition}\\ \text{1}\text{.0}& \text{4}{\text{.5×10}}^{\text{-5}}\text{M/s}\end{array}\\ \begin{array}{cc}\text{2}\text{.0}& \text{4}{\text{.3×10}}^{\text{-5}}\text{M/s}\\ \text{3}\text{.0}& \text{4}{\text{.0×10}}^{\text{-5}}\text{M/s}\end{array}\end{array}$

The average concentration and division of the rate by the average concentration are equal for all three time intervals.

The calculation of the rate constant for each interval is done below,

For 10 minutes,

$\begin{array}{l}{\text{k}}_{\text{1}\text{.0}\text{min}}\text{=}\frac{\text{rate}}{\text{average}\text{concentration}}\\ \\ {\text{k}}_{\text{1}\text{.0}\text{min}}\text{=}\frac{\text{4}\text{.50}{\text{×10}}^{\text{-5}}\text{M/s}}{{\left(\frac{\text{0}\text{.1103}\text{M+0}\text{.1076}\text{M}}{\text{2}}\right)}^{\text{2}}}\\ \\ {\text{k}}_{\text{1}\text{.0}\text{min}}\text{=3}{\text{.79×10}}^{\text{-3}}\text{/}\left(\text{s}\right)\end{array}$

For 20 minutes,

$\begin{array}{l}{\text{k}}_{\text{2}\text{.0}\text{min}}\text{=}\frac{\text{rate}}{\text{average}\text{concentration}}\\ \\ {\text{k}}_{\text{2}\text{.0}\text{min}}\text{=}\frac{\text{4}\text{.30}{\text{×10}}^{\text{-5}}\text{M/s}}{{\left(\frac{\text{0}\text{.1076}\text{M+0}\text{.1050}\text{M}}{\text{2}}\right)}^{\text{2}}}\\ \\ {\text{k}}_{\text{2}\text{.0}\text{min}}\text{=3}{\text{.83×10}}^{\text{-3}}\text{/}\left(\text{s}\right)\end{array}$

For 30 minutes,

$\begin{array}{l}{\text{k}}_{\text{3}\text{.0}\text{min}}\text{=}\frac{\text{rate}}{\text{average}\text{concentration}}\\ \\ {\text{k}}_{\text{3}\text{.0}\text{min}}\text{=}\frac{\text{4}\text{.00}{\text{×10}}^{\text{-5}}\text{M/s}}{{\left(\frac{\text{0}\text{.1050}\text{M+0}\text{.1026}\text{M}}{\text{2}}\right)}^{\text{2}}}\\ \\ {\text{k}}_{\text{3}\text{.0}\text{min}}\text{=3}{\text{.71×10}}^{\text{-3}}\text{/}\left(\text{s}\right)\end{array}$

Hence, the average rate constant can be given as,

$\begin{array}{l}\underset{\_}{\begin{array}{l}\begin{array}{ccc}\text{Time}& \text{Rate}& \text{k}\\ \text{1}\text{.0}\text{min}& \text{4}{\text{.50×10}}^{\text{-5}}\text{M/s}& \text{3}{\text{.79×10}}^{\text{-3}}\text{/}\left(\text{s}\right)\\ \text{2}\text{.0}\text{min}& \text{4}{\text{.33×10}}^{\text{-5}}\text{M/s}& \text{3}{\text{.83×10}}^{\text{-3}}\text{/}\left(\text{s}\right)\end{array}\\ \begin{array}{ccc}\text{3}\text{.0}\text{min}& \text{4}{\text{.00×10}}^{\text{-5}}\text{M/s}& \text{3}\text{.71}\end{array}{\text{×10}}^{\text{-3}}\text{/}\left(\text{s}\right)\end{array}}\\ \text{Average}\text{k}\text{=}\text{3}{\text{.8××10}}^{\text{-3}}\text{/}\left(\text{s}\right)\end{array}$

The average rate constant of the decomposition of Nitrogen dioxide = $\text{3}{\text{.8×10}}^{\text{-3}}\text{/}\left(\text{s}\right)$

Conclusion

The average rate constant of the decomposition of Nitrogen dioxide was calculated and found to be $\text{3}{\text{.8×10}}^{\text{-3}}\text{/}\left(\text{s}\right)$.