Chapter 11, Problem 19QAP

The decomposition of nitrogen dioxide is a second-order reaction. At 550 K, a 0.250 M sample decomposes at the rate of 1.17 mol/L $\cdot$

min.

(a) Write the rate expression.

(b) What is the rate constant at 550 K?

(c) What is the rate of decomposition when $\left[{\text{NO}}_2\right]=0.800M$?

Interpretation Introduction

(a)

Interpretation:

To write the expression of rate from the given statement of a chemical reaction.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 19QAP

Expression of rate is:

$\text{rate}\text{=}{\text{k[NO}}_{\text{2}}{\text{]}}^{\text{2}}$

Explanation of Solution

Nitrogen dioxide decomposes into nitrogen and oxygen gas as:

$2{\text{NO}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{(g)}\text{+}2{\text{O}}_{\text{2}}\text{(g)}$

Since the order of reaction with respect to NO₂ is second. Thus, rate law equation will look like:

$\begin{array}{l}2{\text{NO}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{(g)}\text{+}2{\text{O}}_{\text{2}}\text{(g)}\\ \text{rate}\text{=}{\text{k[NO}}_{\text{2}}{\text{]}}^{\text{2}}\end{array}$

Interpretation Introduction

(b)

Interpretation:

To determine the value of rate constant at 550 K.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 19QAP

Rate constant for given reaction, $\text{k}\text{=}18.72\text{L/mol}{\min }^{-1}$.

Explanation of Solution

Nitrogen dioxide decomposes into nitrogen and oxygen gas as:

$2{\text{NO}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{(g)}\text{+}2{\text{O}}_{\text{2}}\text{(g)}$

Since the order of reaction with respect to NO₂ is second. Thus, rate law equation will look like:

$\begin{array}{l}2{\text{NO}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{(g)}\text{+}2{\text{O}}_{\text{2}}\text{(g)}\\ \text{rate}\text{=}{\text{k[NO}}_{\text{2}}{\text{]}}^{\text{2}}\mathrm{........}(1)\end{array}$

Here we have:

Rate = 1.17 mol/L.min

[NO₂ ] = 0.250 M

Plugging value of rate of reaction in equation 1 to get the value of rate constant as:

$\begin{array}{l}\text{rate}\text{=}{\text{k[NO}}_{\text{2}}{\text{]}}^{\text{2}}\\ 1.17\text{mol/L}\text{.min}\text{=}\text{k(0}{\text{.250)}}^{\text{2}}\\ \Rightarrow \text{k}\text{=}18.72\text{L/mol}{\min }^{-1}\\ \end{array}$

Interpretation Introduction

(c)

Interpretation:

To determine the value of rate of reaction when concentration of nitrogen dioxide is 0.800 M.

Concept Introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 19QAP

Rate for given reaction when concentration of nitrogen dioxide is 0.800 M =11.981 M/min.

Explanation of Solution

Nitrogen dioxide decomposes into nitrogen and oxygen gas as:

$2{\text{NO}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{(g)}\text{+}2{\text{O}}_{\text{2}}\text{(g)}$

Since the order of reaction with respect to NO₂ is second. Thus, rate law equation will look like:

$\begin{array}{l}2{\text{NO}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{(g)}\text{+}2{\text{O}}_{\text{2}}\text{(g)}\\ \text{rate}\text{=}{\text{k[NO}}_{\text{2}}{\text{]}}^{\text{2}}\mathrm{........}(1)\end{array}$

Here we have:

Rate constant= 18.72 L/mol.min

[NO₂ ] = 0.800 M

Plugging value of rate of reaction in equation 1 to get the value of rate constant as:

$\begin{array}{l}\text{rate}\text{=}{\text{k[NO}}_{\text{2}}{\text{]}}^{\text{2}}\\ \text{rate}\text{=}\text{(18}\text{.72)×(0}{\text{.800)}}^{\text{2}}\\ \Rightarrow \text{rate}\text{=}\text{11}\text{.981}\text{mol/L}{\text{.min}}^{}\\ \end{array}$