Chapter 19, Problem 19.90QP

Interpretation Introduction

Interpretation:

From the given information, how long it take for the concentration of A to be four times that of B has to be determined.

Concept introduction:

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.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

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.

Answer to Problem 19.90QP

The time taken for the concentration of A to be four times that of B is $t=56.4\min$

Explanation of Solution

From the given information, A and B both were decomposed by first-order kinetics.

Therefore, we can represent a first-order rate equation for A and also one for B as follows.

$\ln \frac{{\left[\text{A}\right]}_t}{{\left[\text{A}\right]}_0}=-k_{\text{A}}t\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{B}\right]}_0}=-k_{\text{B}}t$

$\frac{{\left[\text{A}\right]}_t}{{\left[\text{A}\right]}_0}=e^{-k_{\text{A}}t}\frac{{\left[\text{B}\right]}_t}{{\left[\text{B}\right]}_0}=e^{-k_{\text{B}}t}$

${\left[\text{A}\right]}_t={\left[\text{A}\right]}_0{e}^{-k_{\text{A}}t}{\left[\text{B}\right]}_t={\left[\text{B}\right]}_0{e}^{-k_{\text{B}}t}$

Now, we should calculate each of the rate constant as $k_{\text{A}}\text{and}{k}_{\text{B}}$ from their corresponding half-lives as follows.

$k_{\text{A}}=\frac{0.693}{50.0\min }=0.0139{\min }^{-1}{k}_{\text{B}}=\frac{0.693}{18.0\min }=0.0385{\min }^{-1}$

Then, the initial concentrations for A and B are equal.  So, ${\text{[A]}}_{\text{0}}{\text{ = [B]}}_{\text{0}}$

Therefore, from the first order rate expressions we can write as

$\frac{{\text{[A]}}_{\text{t}}}{{\text{[B]}}_{\text{t}}}\text{ = 4}\text{= }\frac{{\text{[A]}}_0{e}^{-k_{\text{A}}t}}{{\text{[B]}}_0{e}^{-k_{\text{B}}t}}\text{= }\frac{e^{-k_{\text{A}}t}}{e^{-k_{\text{B}}t}}=e^{\left(k_{\text{B}}-k_{\text{A}}\right)t}=e^{\left(0.0385-0.0139\right)t}$

$4=e^{0.0246t}$

$\ln 4=0.0246t$

$t=56.4\min$

Conclusion

The time taken for the concentration of A to be four times that of B was $t=56.4\min$