Chapter 13, Problem 13.146QP

(a)

Interpretation Introduction

Interpretation:

The rate constant, half life and the concentration of $\left[\text{A}\right]$ have to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

Integrated rate law for second order reactions:

Taking in the example of following reaction,

$\text{aA}\to \text{products}$

And the reaction follows second order rate law,

Then the relationship between the concentration of $\text{A}$ and time can be mathematically expressed as,

$\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

The above expression is called as integrated rate for second order reactions.

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To calculate the rate constant of the reaction

Answer to Problem 13.146QP

Answer

The rate constant of the reaction is $\text{2}{\text{.8×10}}^{\text{-2}}\text{L/}\left(\text{mol}\text{.s}\right)$ .

Explanation of Solution

The plot of $\frac{\text{1}}{\left[\text{A}\right]}$ (vs) time gives a straight line and the reaction is said to be second order. In this case, the value of the slope is $\text{2}{\text{.8×10}}^{\text{-2}}\text{L/}\left(\text{mol}\text{.s}\right)$ that is said to be same as the rate constant of the reaction. Hence, the rate constant of the reaction is $\text{2}{\text{.8×10}}^{\text{-2}}\text{L/}\left(\text{mol}\text{.s}\right)$.

(b)

Interpretation Introduction

Interpretation:

The rate constant, half life and the concentration of $\left[\text{A}\right]$ have to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

Integrated rate law for second order reactions:

Taking in the example of following reaction,

$\text{aA}\to \text{products}$

And the reaction follows second order rate law,

Then the relationship between the concentration of $\text{A}$ and time can be mathematically expressed as,

$\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

The above expression is called as integrated rate for second order reactions.

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To calculate the half life of the reaction

Answer to Problem 13.146QP

Answer

The half period of the reaction is $\text{1}{\text{.9×10}}^{\text{5}}\text{s}$ .

Explanation of Solution

Initial concentration = $\text{1}{\text{.9×10}}^{\text{-4}}\text{M}$

Rate constant= $\text{2}{\text{.8×10}}^{\text{-2}}\text{L/}\left(\text{mol}\text{.s}\right)$

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\left(\text{2}{\text{.8×10}}^{\text{-2}}\text{L}\left(\text{mol}\text{.s}\right)\text{×1}{\text{.9×10}}^{\text{-4}}\text{M}\right)}\\ \\ {\text{t}}_{\text{1/2}}\text{=1}{\text{.88×10}}^{\text{5}}\text{=1}{\text{.9×10}}^{\text{5}}\text{s}\end{array}$

The half period of the reaction = $\text{1}{\text{.9×10}}^{\text{5}}\text{s}$ .

(c)

Interpretation Introduction

Interpretation:

The rate constant, half life and the concentration of $\left[\text{A}\right]$ have to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

Integrated rate law for second order reactions:

Taking in the example of following reaction,

$\text{aA}\to \text{products}$

And the reaction follows second order rate law,

Then the relationship between the concentration of $\text{A}$ and time can be mathematically expressed as,

$\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

The above expression is called as integrated rate for second order reactions.

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To calculate the concentration of $\left[\text{A}\right]$ after 275 seconds

Answer to Problem 13.146QP

Answer

The concentration of $\left[\text{A}\right]$ is $\text{2}{\text{.1×10}}^{\text{-3}}\text{M}$.

Explanation of Solution

The equation for second order reaction is given as,

$\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

Initial concentration = $\text{2}{\text{.1×10}}^{\text{-2}}\text{M}$

Rate constant= $\text{2}{\text{.8×10}}^{\text{-2}}\text{L/}\left(\text{mol}\text{.s}\right)$

The concentration of $\left[\text{A}\right]$ after 275 seconds =

$\begin{array}{l}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=2}{\text{.8×10}}^{\text{-2}}\text{L/}\left(\text{mol}\text{.s}\right)\text{×275}\text{s+}\frac{\text{1}}{\text{2}{\text{.1×10}}^{\text{-3}}\text{mol/L}}\\ \\ \frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=7}\text{.70}\text{L/mol+476}\text{.2L/mol=483}\text{.9}\text{L/mol}\\ \\ \frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=}\frac{\text{1}}{\text{483}\text{.9}\text{L/mol}}\\ \\ \frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=2}{\text{.06×10}}^{\text{-3}}\approx \text{2}{\text{.1×10}}^{\text{-3}}\text{M}\end{array}$

The concentration of $\left[\text{A}\right]$= $\text{2}{\text{.1×10}}^{\text{-3}}\text{M}$.