Chapter 20, Problem 20.84QP

Peroxyacetyl nitrate (PAN) undergoes thermal decomposition as follows:

${\text{CH}}_3(\text{CO}){\text{OONO}}_2\to {\text{CH}}_3(\text{CO})\text{OO}+{\text{NO}}_2$

The rate constant is 3.0 × 10⁻⁴ s⁻¹ at 25°C. At the boundary between the troposphere and stratosphere, where the temperature is about −40°C, the rate constant is reduced to 2.6 × 10⁻⁷ s⁻¹. (a) Calculate the activation energy for the decomposition of PAN. (b) What is the half-life of the reaction (in minutes) at 25°C?

Interpretation Introduction

Interpretation:

The activation energy for decomposition of PAN has to be calculated.

Concept Introduction:

Arrhenius equation:

$\text{k}\text{=}{\text{Ae}}^{\frac{{\text{-E}}_{\text{a}}}{\text{RT}}}$

Where,

k is rate constant.

T is absolute temperature (kelvin)

A is pre-exponential factor.

R is gas constant

${\text{E}}_{\text{a}}$ is activation energy

The Arrhenius equation relating the rate constants ${\text{k}}_{\text{1}}{\text{ and k}}_{\text{2}}$ at temperatures ${\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}$ is used to determine activation energy is given as

$\text{ln}\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left(\frac{{\text{T}}_{\text{1}}\text{-}{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right)$

Answer to Problem 20.84QP

The activation energy for decomposition of PAN is $\text{62}\text{.6}\text{KJ/mol}$

Explanation of Solution

Given,

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{(CO)OONO}}_{\text{2}}\underset{}{\overset{}{\to }}{\text{CH}}_{\text{3}}\text{(CO)OO}\text{+}{\text{NO}}_{\text{2}}\\ \\ {\text{k}}_{\text{1}}\text{=}\text{2}{\text{.6×10}}^{\text{-7}}{\text{s}}^{\text{-1}}{\text{T}}_{\text{1}}\text{=}{\text{-40}}^{\text{o}}\text{C}\\ \\ {\text{k}}_{\text{2}}\text{=}\text{3}{\text{.0×10}}^{\text{-4}}{\text{s}}^{\text{-1}}{\text{T}}_{\text{2}}\text{=}{\text{25}}^{\text{o}}\text{C}\end{array}$

The activation energy of decomposition of PAN is calculated as

$\begin{array}{l}\text{ln}\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left(\frac{{\text{T}}_{\text{1}}\text{-}{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right)\\ \\ {\text{E}}_{\text{a}}\text{=}\frac{\text{R}{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}{{\text{(T}}_{\text{1}}{\text{-T}}_{\text{2}}\text{)}}\text{ln}\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}\\ \\ {\text{E}}_{\text{a}}\text{=}\frac{\text{8}\text{.314}\text{J/mol}K\text{×}\text{233}\text{K}\text{×}\text{298}\text{K}}{\text{233}\text{K}\text{-}\text{298}\text{K}}\text{ln}\frac{\text{2}{\text{.6×10}}^{\text{-7}}{\text{s}}^{\text{-1}}}{\text{3}{\text{.0×10}}^{\text{-4}}{\text{s}}^{\text{-1}}}\\ \\ \text{=}\text{6}\text{.26}{\text{×10}}^{\text{4}}\text{J/}\text{mol}\\ \\ \text{=}\text{62}\text{.6}\text{KJ/mol}\end{array}$

Thus, the activation energy for the decomposition reaction of PAN is $\text{62}\text{.6}\text{KJ/mol}$.

Interpretation Introduction

Interpretation:

The half-life for decomposition of PAN has to be calculated.

Concept Introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as ${\text{t}}_{\text{1/2}}$.
• Half-life is discovered by Ernst Rutherford's in 1907 from the original term half-life period.
• For the first order reaction ${\text{t}}_{\text{1/2}}\text{=}\frac{0.693}{\text{k}}$

Answer to Problem 20.84QP

The half-life for decomposition of PAN is $\text{38}\text{minutes}$

Explanation of Solution

Given,

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{(CO)OONO}}_{\text{2}}\underset{}{\overset{}{\to }}{\text{CH}}_{\text{3}}\text{(CO)OO}\text{+}{\text{NO}}_{\text{2}}\\ \\ \text{k}\text{=}\text{3}{\text{.0×10}}^{\text{-4}}{\text{s}}^{\text{-1}}\text{T}\text{=}{\text{25}}^{\text{o}}\text{C}\end{array}$

From the unit of rate constant, it is clear that the reaction is first-order and the half-life is calculated as

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{0.693}{\text{k}}\\ \\ \text{=}\frac{\text{0}\text{.693}}{\text{3}\text{.0}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}}\\ \\ \text{=}\text{2}\text{.3}{\text{×10}}^{\text{3}}\text{s}\\ \\ \text{=}\text{38}\text{minutes}\end{array}$

Thus, the half-life for the decomposition reaction of PAN is $\text{38}\text{minutes}$.