Chapter 13, Problem 13.27QP

The rate constant for the second-order reaction

$2\text{NOBr}(g)\stackrel{}{\to }2\text{NO}(g)+{\text{Br}}_2(g)$

is 0.80/M · s at 10°C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]₀ = 0.072 M and [NOBr]₀ = 0.054 M.

Interpretation Introduction

Interpretation:

The concentration of $\text{NaOBr}$ after 22s from the given information has to be calculated.

Concept introduction:

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

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

The rate of the reaction is proportinal to the concentration of A to the power of x, is ${\left[A\right]}^x$

The rate of the reaction is proportional to the concentration of B to the power of y is ${\left[B\right]}^y$

Then the rate equation becomes,

$Rate=k{\left[A\right]}^x{\left[B\right]}^y$

Order of this reaction is the sum of the powers to which all reactant concentrations appearing in the rate law are  raised.

$\text{Order}\text{= }\text{x+y}$

Answer to Problem 13.27QP

The concentration of $\text{NaOBr}$ after 22s from the given information is $\text{0}\text{.034 M}$

Explanation of Solution

The given reaction is

${\text{2NaOBr}}_{\text{(g)}}\to {\text{2NO}}_{\text{(g)}}{\text{+Br}}_{\text{2(g)}}$

The reaction follows second order kinetics.

Rate constant of the given reaction is $k=0.80M^{-1}\cdot s^{-1}$

The concentration of $\text{NaOBr}$ after $22s$ can be determined as follows,

For a second order reaction the relationship between concentrations of reactant and time is,

$\begin{array}{l}\frac{1}{{\left[\text{NaOBr}\right]}_t}=kt+\frac{1}{{\left[\text{NaOBr}\right]}_0}\\ \\ \frac{1}{{\left[\text{NaOBr}\right]}_t}=\left(0.80/\text{M}\cdot \text{s}\right)\left(22s\right)+\frac{1}{0.086\text{M}}\\ \\ \frac{1}{{\left[\text{NaOBr}\right]}_t}=29{\text{M}}^{-1}\\ \\ \text{[NOBr] = 0}\text{.034 M}\end{array}$

Therefore, the concentration of $\text{NaOBr}$ after 22s is $\text{0}\text{.034 M}$

Interpretation Introduction

Interpretation:

The half-life when ${\left[\text{NaOBr}\right]}_{\text{0}}\text{=0}\text{.072}\text{M}\text{and }{\left[\text{NaOBr}\right]}_{\text{0}}\text{=0}\text{.054}\text{M}$ has to be calculated.

Concept introduction:

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

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

The rate of the reaction is proportinal to the concentration of A to the power of x, is ${\left[A\right]}^x$

The rate of the reaction is proportional to the concentration of B to the power of y is ${\left[B\right]}^y$

Then the rate equation becomes,

$Rate=k{\left[A\right]}^x{\left[B\right]}^y$

Order of this reaction is the sum of the powers to which all reactant concentrations appearing in the rate law are  raised.

$\text{Order}\text{= }\text{x+y}$

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

Half-life for a second order reaction is

${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}\text{where}{\left[\text{A}\right]}_{\text{0}}\text{is}\text{the}\text{initial}\text{concentration}\text{of}\text{reactant}\text{A}$

Answer to Problem 13.27QP

The half-life when ${\left[\text{NaOBr}\right]}_{\text{0}}\text{=0}\text{.072}\text{M}\text{and }{\left[\text{NaOBr}\right]}_{\text{0}}\text{=0}\text{.054}\text{M}$ is $23s$

Explanation of Solution

The given reaction is

${\text{2NaOBr}}_{\text{(g)}}\to {\text{2NO}}_{\text{(g)}}{\text{+Br}}_{\text{2(g)}}$

The reaction follows second order kinetics.

Rate constant of the given reaction is $k=0.80M^{-1}\cdot s^{-1}$

The concentration of $\text{NaOBr}$ after $22s$ can be determined as follows,

We know that for a second order half-life reaction and half-life of a second order reaction is dependent on initial concentration

$\begin{array}{l}{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{k{\left[\text{A}\right]}_0}\\ \\ t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\left(0.80/M\cdot s\right)\left(0.072M\right)}\\ \\ t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\text{17 s}\end{array}$

The half-life when ${\left[\text{NaOBr}\right]}_0=0.054M$ can be determined as follows,

$\begin{array}{l}{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{k{\left[\text{A}\right]}_0}\\ \\ t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\left(0.80/M\cdot s\right)\left(0.054M\right)}\\ \\ t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=23\text{ s}\end{array}$