Chapter 19, Problem 19.33QP

(a)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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.

Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants).  It is also said to be as rate equation.

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 $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Answer to Problem 19.33QP

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

1. (b) The half life when ${\left[\text{NaOBr}\right]}_0=0.072M\text{and }{\left[\text{NaOBr}\right]}_0=0.054M$ are $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\text{17 s}$ and $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=23\text{ s}$ respectively

Explanation of Solution

To determine the concentration of $\text{NaOBr}$ after 22s

The given reaction is ${\text{2NaOBr}}_{\text{(g)}}\to {\text{2NO}}_{\text{(g)}}{\text{+Br}}_{\text{2(g)}}$

The order of the reaction is second order

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

Now, determine the concentration of $\text{NaOBr}$ after 22s as follows

We know that from 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 $\text{0}\text{.034 M}$

(b)

Interpretation Introduction

Interpretation:

The half life when ${\left[\text{NaOBr}\right]}_0=0.072M\text{and }{\left[\text{NaOBr}\right]}_0=0.054M$ has to be calculated.

Concept introduction:

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.

Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants).  It is also said to be as rate equation.

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 $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Answer to Problem 19.33QP

1. (a) The half life when ${\left[\text{NaOBr}\right]}_0=0.072M\text{and }{\left[\text{NaOBr}\right]}_0=0.054M$ are $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\text{17 s}$ and $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=23\text{ s}$ respectively

Explanation of Solution

To determine the half life when ${\left[\text{NaOBr}\right]}_0=0.072M$ is,

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}$

To determine the half-life when ${\left[\text{NaOBr}\right]}_0=0.054M$

$\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}$