Chapter 14, Problem 14.66QP

(a)

Interpretation Introduction

Interpretation: The rate law expression for the given reaction and the rate constant is to be stated.

Concept introduction: Rate law of a chemical reaction describes the relationship between concentrations of reactants and the rate of that particular reaction.

To determine: The expression that corresponds to the rate law of the given reaction.

Answer to Problem 14.66QP

Solution

The rate law is expressed by equation (6).

Explanation of Solution

Explanation

The given reaction is,

${\text{NO}}_2\left(g\right)+\text{CO}\left(g\right)\to \text{NO}\left(g\right)+{\text{CO}}_2\left(g\right)$

The rate law for the reaction will be,

$\text{Rate}=k{\left[{\text{NO}}_2\right]}^m{\left[\text{CO}\right]}^n$ (1)

Where,

• $k$ is the rate constant of the reaction.
• $m$ is the order of the reaction with respect to ${\text{NO}}_2$.
• $n$ is the order of the reaction with respect to $\text{CO}$.

The given kinetic data is,

Experiment	$\left[{\text{NO}}_2\right]\left(\text{M}\right)$	$\left[\text{CO}\right]\left(\text{M}\right)$	Initial Rate $\left(\text{M/s}\right)$
1	$0.263$	$0.826$	$1.44\times {10}^{-5}$
2	$0.263$	$0.413$	$1.44\times {10}^{-5}$
3	$0.526$	$0.413$	$5.76\times {10}^{-5}$

According to experiment $2$, the concentration of ${\text{NO}}_2$ is $0.263\text{M}$. The concentration of $\text{CO}$ is $0.413\text{M}$. The value of rate of reaction is $1.44\times {10}^{-5}\text{M/s}$.

Substitute the value of concentration of ${\text{NO}}_2$, $\text{CO}$ and rate of reaction in equation (1).

$1.44\times {10}^{-5}\text{M/s}=k{\left[0.263\text{M}\right]}^m{\left[0.413\text{M}\right]}^n$ (2)

According to experiment $3$, the concentration of ${\text{NO}}_2$ is $0.526\text{M}$. The concentration of $\text{CO}$ is $0.413\text{M}$. The value of rate of reaction is $5.76\times {10}^{-5}\text{M/s}$.

Substitute the value of concentration of ${\text{NO}}_2$, $\text{CO}$ and rate of reaction in equation (1).

$5.76\times {10}^{-5}\text{M/s}=k{\left[0.526\text{M}\right]}^m{\left[0.413\text{M}\right]}^n$ (3)

Divide equation (2) by equation (3).

$\begin{array}{c}\frac{1.44\times {10}^{-5}\text{M/s}\text{M/s}}{5.76\times {10}^{-5}\text{M/s}}=\frac{k{\left(\text{0}\text{.263}\text{M}\right)}^m{\left(\text{0}\text{.413}\text{M}\right)}^n}{k{\left(\text{0}\text{.526}\text{M}\right)}^m{\left(0.413\text{M}\right)}^n}\\ 0.25={\left(0.25\right)}^m\times {\left(1\right)}^n\\ {\left(0.5\right)}^2={\left(0.5\right)}^m\times {\left(1\right)}^0\end{array}$

Therefore, the value of $m=2$ and the value of $n=0$.

Substitute the value of $m$ and $n$ in equation (1).

$\text{Rate}=k{\left[{\text{NO}}_2\right]}^2{\left[\text{CO}\right]}^0$ (4)

The above equation represents the rate law of the equation.

(b)

Interpretation Introduction

Interpretation: The rate law expression for the given reaction and the rate constant is to be stated.

Concept introduction: Rate law of a chemical reaction describes the relationship between concentrations of reactants and the rate of that particular reaction.

To determine: The rate constant for the reaction.

Answer to Problem 14.66QP

Solution

The rate constant for the reaction is $\underset{\_}{20.87\times 10^{-5}{M}^{-1}{s}^{-1}}$.

Explanation of Solution

Explanation

According to experiment $2$, the concentration of ${\text{NO}}_2$ is $0.263\text{M}$. The concentration of $\text{CO}$ is $0.413\text{M}$. The value of rate of reaction is $1.44\times {10}^{-5}\text{M/s}$.

Substitute the value of concentration of ${\text{NO}}_2$, $\text{CO}$ and rate of reaction in equation (4).

$\begin{array}{c}1.44\times {10}^{-5}\text{M/s}=k\times {\left(0.263\text{M}\right)}^2\times {\left(0.413\text{M}\right)}^0\\ k=\frac{1.44\times {10}^{-5}\text{M/s}}{{\left(0.263\text{M}\right)}^2}\\ =\underset{\_}{20.87\times 10^{-5}{M}^{-1}{s}^{-1}}\end{array}$

(c)

Interpretation Introduction

Interpretation: The rate law expression for the given reaction and the rate constant is to be stated.

Concept introduction: Rate law of a chemical reaction describes the relationship between concentrations of reactants and the rate of that particular reaction.

To determine: The rate of formation of ${\text{CO}}_{\text{2}}$.

Answer to Problem 14.66QP

Solution

The rate of formation of ${\text{CO}}_{\text{2}}$ is $\underset{\_}{5.217\times 10^{-5}{M}^{-1}{s}^{-1}}$.

Explanation of Solution

Explanation

Given

The value of $\left[{\text{NO}}_{\text{2}}\right]=\left[\text{CO}\right]=0.500\text{M}$.

The rate of formation of ${\text{CO}}_{\text{2}}$ is equal to the rate of disappearance of ${\text{NO}}_{\text{2}}$. Substitute the value of rate constant, $\left[{\text{NO}}_{\text{2}}\right]$ and $\left[\text{CO}\right]$ in equation (1).

$\begin{array}{c}\text{Rate}=\text{20}{\text{.87×10}}^{\text{-5}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}{\left(0.500\text{M}\right)}^2\times \left(0.500\text{M}\right)\\ =\underset{\_}{5.217\times 10^{-5}{M}^{-1}{s}^{-1}}\end{array}$

Conclusion

The rate of formation of ${\text{CO}}_{\text{2}}$ is $\underset{\_}{5.217\times 10^{-5}{M}^{-1}{s}^{-1}}$.