Chapter 14, Problem 14.70QP

Interpretation Introduction

Interpretation:

From the given information, the time taken for the concentration of A to be four times that of B has to be calculated.

Concept introduction:

The first order reaction:

It is a reaction whose rate depends on the reactant concentration whose power is raised to one in the rate expression.

Consider a reaction:

$\text{A}\stackrel{}{\to }\text{Product}$

In this reaction, the reactant $\text{A}$ has been converted into the product.

The rate expression for the considered reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}$

-Ve sign indicates the decrease in concentration of the reactant A.

$\left[\text{A}\right]$- Concentration of the reactant A in molarity.

$\text{Δt}\text{-}\text{Time}\text{period}$

$\text{Δ}$- Difference between initial and final states.

The integrated form of the first order rate expression is:

${\left[\text{A}\right]}_{\text{t}}\text{=}{\left[\text{A}\right]}_{\text{o}}{\text{e}}^{{\text{-k}}_{\text{A}}\text{t}}$

${\left[\text{A}\right]}_{\text{t}}$-Concentration at any time t or final Concentration of the reactant A.

${\left[\text{A}\right]}_{\text{o}}$- Initial Concentration of the reactant A.

$\text{e}$- Exponent of natural logarithm.

${\text{k}}_{\text{A}}$- Rate constant for the reaction with the reactant $\text{A}$.

$\text{t}$-Time period.

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life. It is represented by the symbol as ${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$

The expression for the half-life period in first order kinetics:

${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$

${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$-Half-life period

$\text{k}$-Rate constant.

Answer to Problem 14.70QP

The time taken for the concentration of A to be four times that of B is $\text{t}=56.26\min$

Explanation of Solution

Given:

${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$ of A is $50.0\min$

${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$ of B is $18.0\min$

Calculating the rate constants from the given half-lives:

$\begin{array}{c}{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{k}=\frac{\text{0}\text{.693}}{{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}}\end{array}$

Let the rate constant for the reaction with the reactant A be ${\text{k}}_{\text{A}}$.

$\begin{array}{c}{\text{k}}_{\text{A}}=\frac{\text{0}\text{.693}}{{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}}\\ =\frac{\text{0}\text{.693}}{50.0\min }\\ =0.01386{\min }^{-1}\end{array}$

Therefore, ${\text{k}}_{\text{A}}=$$0.01386{\min }^{-1}$

Let the rate constant for the reaction with the reactant B be ${\text{k}}_{\text{B}}$.

$\begin{array}{c}{\text{k}}_{\text{B}}=\frac{\text{0}\text{.693}}{{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}}\\ =\frac{\text{0}\text{.693}}{18.0\min }\\ =0.0385{\min }^{-1}\end{array}$

Therefore, ${\text{k}}_{\text{B}}=0.0385{\min }^{-1}$

The initial concentrations for A and B are equal. So,

$\begin{array}{l}{\text{[A]}}_{\text{0}}{\text{ = [B]}}_{\text{0}}\\ \frac{{\text{[A]}}_{\text{0}}}{{\text{[B]}}_{\text{0}}}=1\end{array}$

The final concentrations of A is four times of B are equal. So,

$\begin{array}{l}{\text{[A]}}_{\text{t}}{\text{ = 4[B]}}_{\text{t}}\\ \frac{{\text{[A]}}_{\text{t}}}{{\text{[B]}}_{\text{t}}}=4\end{array}$

Using the integrated rate expression of first order kinetics, the initial and the final concentrations can be equated as follows:

$\begin{array}{c}\frac{{\text{[A]}}_{\text{t}}}{{\text{[B]}}_{\text{t}}}\text{= }\frac{{\text{[A]}}_{\text{0}}{\text{e}}^{{\text{-k}}_{\text{A}}\text{t}}}{{\text{[B]}}_{\text{0}}{\text{e}}^{{\text{-k}}_{\text{B}}\text{t}}}\\ \text{4}\text{= }\frac{{\text{e}}^{{\text{-k}}_{\text{A}}\text{t}}}{{\text{e}}^{{\text{-k}}_{\text{B}}\text{t}}}\because {\text{[A]}}_{\text{t}}{\text{ = 4[B]}}_{\text{t}}\text{and}{\text{[A]}}_{\text{0}}{\text{ = [B]}}_{\text{0}}\\ {\text{=e}}^{\left({\text{k}}_{\text{B}}{\text{-k}}_{\text{A}}\right)\text{t}}\\ \text{=}{\text{e}}^{\left(\text{0}\text{.0385-}0.01386\right)\text{t}}\end{array}$

$\begin{array}{c}\text{4}\text{=}{\text{e}}^{\left(\text{0}.02464\right)\text{t}}\\ \ln \text{4}=\left(\text{0}.02464\right)\text{t}\\ \text{t}=\frac{\ln \text{4}}{\left(\text{0}.02464\right)}\\ =56.26\min \end{array}$

Therefore, the time taken for the concentration of A to be four times of B is $\text{t}=56.26\min$

Conclusion

The time taken for the concentration of A to be four times has been calculated.