Chapter 13, Problem 13.136QP

(a)

Interpretation Introduction

Interpretation:

Rate law for the given reaction has to be determined.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

The rate law expresses the relationship of the rate of a reaction to the rate constant.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Answer to Problem 13.136QP

Rate law for the given reaction is,

$Rate=k{\left[NO\right]}^2\left[O_2\right]$

Explanation of Solution

The given reaction is,

$2N{O}_{\left(g\right)}+O_{2\left(g\right)}\to 2N{O}_{2\left(g\right)}$

The rate law expresses the relationship of the rate of a reaction to the rate constant.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Similarly, rate law for the given reaction can be written as folllows,

$Rate=k{\left[NO\right]}^2\left[O_2\right]$

(b)

Interpretation Introduction

Interpretation:

Check whether it is possible to simplify the rate law for the given reaction under the given conditions if so, the simplified rate law for this reaction has to be written.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

When a reactant is made much more concentrated than the other, it is made to have no effect on the rate of that particular reaction. Such type of reaction is considered as pseudo (fake) order reaction.

Answer to Problem 13.136QP

The simplified rate law for the given reaction is,

 $Rate=k{\left[NO\right]}^2$

Explanation of Solution

The given reaction is,

$2N{O}_{\left(g\right)}+O_{2\left(g\right)}\to 2N{O}_{2\left(g\right)}$

Rate law for the given reaction can be written as folllows,

$Rate=k{\left[NO\right]}^2\left[O_2\right]$

Order of this reaction is $3$ but the concentratin of oxygen $\left[O_2\right]$ is very large compared to the $\left[NO\right]$ and so the reaction can exist in pseodo-second order reaction (the reaction which seems to be higher order reaction but actually it follows second order kinetics.)

Therefore the rate law can be simplyfied, and it is,

$Rate=k{\left[NO\right]}^2$

(c)

Interpretation Introduction

Interpretation:

Under the given condition half-life of the reaction has to be determined.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

When a reactant is made much more concentrated than the other, it is made to have no effect on the rate of that particular reaction. Such type of reaction is considered as pseudo (fake) order reaction.

Half-life is the time required for one half of a reactant to react.

For a second order reaction, $t_{\frac{1}{2}}=\frac{1}{k{\left[A\right]}_0}$

${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.

Explanation of Solution

The given reaction is,

$2N{O}_{\left(g\right)}+O_{2\left(g\right)}\to 2N{O}_{2\left(g\right)}$

Reaction follows pseudo-fist order kinetics,

Therefore the rate law can be simplyfied, and it is,

$Rate=k{\left[NO\right]}^2$

Herein a sample of air at a certain temperture is contaminated with $2.00ppm$ of $NO$ by volume. The half lifre of the reaction is found to be $6.4\times {10}^3\min$ .

If the concentration of $NO$ were $10ppm$ then the  half-life of the reaction can be calculated as follows,

Half-life for a second order reaction is,

$t_{\frac{1}{2}}=\frac{1}{k{\left[A\right]}_0}$

$\begin{array}{c}\frac{{\left({\text{t}}_{\frac{\text{1}}{\text{2}}}\right)}_{\text{1}}}{{\left({\text{t}}_{\frac{\text{1}}{\text{2}}}\right)}_{\text{2}}}\text{=}\frac{{\left(\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}\right)}_{\text{1}}}{{\left(\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}\right)}_{\text{2}}}\text{=}\frac{\left[{\left({\text{A}}_{\text{0}}\right)}_{\text{2}}\right]}{\left[{\left({\text{A}}_{\text{0}}\right)}_{\text{1}}\right]}\\ \frac{\text{6}\text{.4}\text{×}{\text{10}}^{\text{3}}\text{min}}{{\left({\text{t}}_{\frac{\text{1}}{\text{2}}}\right)}_{\text{2}}}\text{=}\frac{\text{10}\text{ppm}}{\text{2}\text{ppm}}\\ {\left(t_{\frac{1}{2}}\right)}_2=1.3\times 10^{3}min\end{array}$