Chapter 20, Problem 20.27E

Derive an expression for the half-life of (a) a third order reaction;(b) a reaction whose order is $=-1$; (c) a reaction whose order is $\frac{1}{2}$. (In these last two cases, examples are rare but known.)

Interpretation Introduction

(a)

Interpretation:

The expression for half-life of third order reaction is to be stated.

Concept introduction:

The time required for the concentration of the reactant to reduce to half of its initial concentration gives the half-life of the reaction. The half-life of the reaction depends on the initial concentration of the reactant except for the first order reaction.

Answer to Problem 20.27E

The expression for half-life of third order reaction is $\frac{3}{k{\text{[A]}}_0{}^2}$.

Explanation of Solution

The integrated rate law for the third-order kinetics is,

$\frac{1}{{\text{[A]}}_t^2}-\frac{1}{{\text{[A]}}_0^2}=k\cdot t$ …(1)

Where,

• ${\text{[A]}}_t$ is the concentration of reactant at time $t$.

At half-life, that is, at time $t_{1/2}$ the concentration of the reactant becomes half of the initial amount. Thus, the concentration of reactant at time $t_{1/2}$ is, ${\text{[A]}}_t=\frac{{\text{[A]}}_0}{2}$.

Substitute ${\text{[A]}}_t$ in terms of ${\text{[A]}}_0$ in equation (1).

$\begin{array}{rll}\frac{1}{{\left(\frac{{\text{[A]}}_0}{2}\right)}^2}-\frac{1}{{\text{[A]}}_0^2}& =& k\cdot t_{1/2}\\ \frac{4}{{\text{[A]}}_0{}^2}-\frac{1}{{\text{[A]}}_0^2}& =& k\cdot t_{1/2}\\ \frac{3}{{\text{[A]}}_0{}^2}& =& k\cdot t_{1/2}\end{array}$

$t_{1/2}=\frac{3}{k{\text{[A]}}_0{}^2}$ …(2)

Thus, the half-life for third order kinetics is $\frac{3}{k{\text{[A]}}_0{}^2}$.

Conclusion

The expression for half-life of third order reaction is $\frac{3}{k{\text{[A]}}_0{}^2}$.

Interpretation Introduction

(b)

Interpretation:

The expression for half-life of a reaction whose order is $-1$ is to be stated.

Concept introduction:

The time required for the concentration of the reactant to reduce to half of its initial concentration gives the half-life of the reaction. The half-life of the reaction depends on the initial concentration of the reactant except for the first order reaction.

Answer to Problem 20.27E

The expression for half-life for a reaction whose order is $-1$ is $\frac{{\text{3[A]}}_0^2}{8k}$.

Explanation of Solution

The rate law for a reaction whose order is $-1$ is,

$-\frac{d\text{[A]}}{dt}=k{\text{[A]}}^{-1}$

This law is rearranged as,

$-\frac{d\text{[A]}}{{\text{[A]}}^{-1}}=k\cdot dt$

Where,

• $t$ is the time of the reaction.

• $\text{[A]}$ is the concentration of the reactant.

Integrate the given equation from ${\text{[A]}}_0$ to ${\text{[A]}}_t$ and from $t=0$ to $t$.

$\begin{array}{l}-{\displaystyle \underset{{\text{[A]}}_0}{\overset{{\text{[A]}}_t}{\int }}\frac{d\text{[A]}}{{\text{[A]}}^{-1}}}={\displaystyle \underset{t=0}{\overset{t}{\int }}k\cdot dt}\\ -{\displaystyle \underset{{\text{[A]}}_0}{\overset{{\text{[A]}}_t}{\int }}\text{[A]}d\text{[A]}}={\displaystyle \underset{t=0}{\overset{t}{\int }}k\cdot dt}\end{array}$

The constants are kept out of the integral and the equation is integrated.

$\begin{array}{rllll}-{\displaystyle \underset{{\text{[A]}}_0}{\overset{{\text{[A]}}_t}{\int }}\text{[A]}d\text{[A]}}& =& k{\displaystyle \underset{t}{\overset{}{\int }}}& =& 0tdt\\ -\frac{1}{2}{\left({\text{[A]}}^2\right)}_{{\text{[A]}}_0}^{{\text{[A]}}_t}& =& k{t|}_t^{}& =& 0t\end{array}$

Apply the limits in the above equation as shown below.

$-\frac{1}{2}\left[{\text{[A]}}_t^2-{\text{[A]}}_0^2\right]=k\left(t-0\right)$

$\frac{{\text{[A]}}_0^2}{2}-\frac{{\text{[A]}}_t^2}{2}=k\cdot t$ …(1)

Thus, equation (1) represents the integrated rate law.

At half-life, that is, at time $t_{1/2}$ the concentration of the reactant becomes half of the initial amount. Thus, the concentration of reactant at time $t_{1/2}$ is, ${\text{[A]}}_t=\frac{{\text{[A]}}_0}{2}$.

Substitute ${\text{[A]}}_t$ in terms of ${\text{[A]}}_0$ in equation (1).

$\begin{array}{rll}\frac{1}{2}\left\{{\text{[A]}}_0^2-{\left(\frac{{\text{[A]}}_0}{2}\right)}^2\right\}& =& k\cdot t_{1/2}\\ \frac{{\text{3[A]}}_0^2}{8}& =& k\cdot t_{1/2}\end{array}$

$t_{1/2}=\frac{{\text{3[A]}}_0^2}{8k}$ …(2)

Thus, the half-life for a reaction whose order is $-1$ is $\frac{{\text{3[A]}}_0^2}{8k}$.

Conclusion

The expression for half-life for a reaction whose order is $-1$ is $\frac{{\text{3[A]}}_0^2}{8k}$.

Interpretation Introduction

(c)

Interpretation:

The expression for half-life of a reaction whose order is $\frac{1}{2}$ is to be stated.

Concept introduction:

The time required for the concentration of the reactant to reduce to half of its initial concentration gives the half-life of the reaction. The half-life of the reaction depends on the initial concentration of the reactant except for the first order reaction.

Answer to Problem 20.27E

The expression for half-life for a reaction whose order is $\frac{1}{2}$ is $\frac{2^{1/2}{\text{[A]}}_0^{1/2}\left\{2^{1/2}-1\right\}}{k}$.

Explanation of Solution

The rate law for a reaction whose order is $\frac{1}{2}$ is,

$-\frac{d\text{[A]}}{dt}=k{\text{[A]}}^{1/2}$

This law is rearranged as,

$-\frac{d\text{[A]}}{{\text{[A]}}^{1/2}}=k\cdot dt$

Where,

• $t$ is the time of the reaction.

• $\text{[A]}$ is the concentration of the reactant.

Integrate the given equation from ${\text{[A]}}_0$ to ${\text{[A]}}_t$ and from $t=0$ to $t$.

$-{\displaystyle \underset{{\text{[A]}}_0}{\overset{{\text{[A]}}_t}{\int }}\frac{d\text{[A]}}{{\text{[A]}}^{1/2}}}={\displaystyle \underset{t=0}{\overset{t}{\int }}k\cdot dt}$

The constants are kept out of the integral and the equation is integrated.

$\begin{array}{rllll}-{\displaystyle \underset{{\text{[A]}}_0}{\overset{{\text{[A]}}_t}{\int }}\frac{d\text{[A]}}{{\text{[A]}}^{1/2}}}& =& k{\displaystyle \underset{t}{\overset{}{\int }}}& =& 0tdt\\ -2{\left({\text{[A]}}^{1/2}\right)}_{{\text{[A]}}_0}^{{\text{[A]}}_t}& =& k{t|}_t^{}& =& 0t\end{array}$

Apply the limits in the above equation as shown below.

$-2\left[{\text{[A]}}_t^{1/2}-{\text{[A]}}_0^{1/2}\right]=k\left(t-0\right)$

${\text{2[A]}}_0^{1/2}-2{\text{[A]}}_t^{1/2}=k\cdot t$ …(1)

Thus, equation (1) represents the integrated rate law.

At half-life, that is, at time $t_{1/2}$ the concentration of the reactant becomes half of the initial amount. Thus, the concentration of reactant at time $t_{1/2}$ is, ${\text{[A]}}_t=\frac{{\text{[A]}}_0}{2}$.

Substitute ${\text{[A]}}_t$ in terms of ${\text{[A]}}_0$ in equation (1).

$\begin{array}{l}{\text{2[A]}}_0^{1/2}-2{\left(\frac{{\text{[A]}}_0}{2}\right)}^{1/2}=k\cdot t_{1/2}\\ {\text{2[A]}}_0^{1/2}-2^{1/2}{\left({\text{[A]}}_0\right)}^{1/2}=k\cdot t_{1/2}\\ 2^{1/2}{\text{[A]}}_0^{1/2}\left\{2^{1/2}-1\right\}=k\cdot t_{1/2}\end{array}$

$t_{1/2}=\frac{2^{1/2}{\text{[A]}}_0^{1/2}\left\{2^{1/2}-1\right\}}{k}$ …(2)

Thus, the half-life for a reaction whose order is $\frac{1}{2}$ is $\frac{2^{1/2}{\text{[A]}}_0^{1/2}\left\{2^{1/2}-1\right\}}{k}$.

Conclusion

The expression for half-life for a reaction whose order is $\frac{1}{2}$ is $\frac{2^{1/2}{\text{[A]}}_0^{1/2}\left\{2^{1/2}-1\right\}}{k}$.