Chapter 3.8, Problem 22E

(a)

To determine

To find: The rate of reaction at time $t$.

Answer to Problem 22E

The rate of reaction with respect to time $\frac{d\left[C\right]}{dt}$ is $\frac{a^{2}k}{{\left(akt+1\right)}^2}$.

Explanation of Solution

Given:

The expression is given as below.

$\left[C\right]=\frac{a^{2}kt}{akt+1}$ (1)

Here, the variable $\left[C\right]$ is the product, and $a$ is the initial concentration of $\left[A\right]=\left[B\right]$ in $\text{moles}/\text{L}$.

Calculation:

Differentiate equation (1) with respect $t$.

$\begin{array}{c}\frac{d\left[C\right]}{dt}=\frac{\left[\left(akt+1\right)\left(a^{2}k\right)\right]-\left(a^{2}kt\right)\left(ak\right)}{{\left(akt+1\right)}^2}\\ =\frac{\left(a^3{k}^{2}t+a^{2}k\right)-\left(a^3{k}^{2}t\right)}{{\left(akt+1\right)}^2}\end{array}$

$\frac{d\left[C\right]}{dt}=\frac{a^{2}k}{{\left(akt+1\right)}^2}$ (2)

Hence, the rate of reaction with respect to time $\frac{d\left[C\right]}{dt}$ is $\frac{a^{2}k}{{\left(akt+1\right)}^2}$.

(b)

To determine

To show: The equation $\frac{dx}{dt}=k{\left(a-x\right)}^2$, where $\left[C\right]=x$.

Explanation of Solution

Proof:

The equation is given as below.

$\frac{dx}{dt}=k{\left(a-x\right)}^2$ (3)

Here the condition $x=\left[C\right]$ applies.

Substitute the value of $\frac{a^{2}kt}{akt+1}$ for $x$ from equation (1) in (3).

$\begin{array}{c}\frac{dx}{dt}=k{\left(a-\frac{a^{2}kt}{akt+1}\right)}^2\\ =k{\left(\frac{a\left(akt+1\right)-\left(a^{2}kt\right)}{akt+1}\right)}^2\\ =k{\left(\frac{\left(a^{2}kt+a\right)-\left(a^{2}kt\right)}{akt+1}\right)}^2\end{array}$

$\frac{dx}{dt}=k{\left(\frac{a}{akt+1}\right)}^2$

$\frac{dx}{dt}=\frac{k{a}^2}{{\left(akt+1\right)}^2}$ (4)

Compare equation (2) and (4).

Thus, the condition $\frac{dx}{dt}=k{\left(a-x\right)}^2$ is true.

(c)

To determine

To show: The concentration if time $t$ tends to $\infty$.

Answer to Problem 22E

The concentration as $t$ tends to $\infty$ is $a$.

Explanation of Solution

Determine the concentration as $t$ tends to $\infty$.

Substitute $t$ tends to $\infty$ $\left(t\to \infty \right)$ in equation (1).

$\begin{array}{c}\underset{t\to \infty }{\lim }\left[C\right]=\underset{t\to \infty }{\lim }\frac{a^{2}kt}{akt+1}\\ =\frac{a^{2}k\left(\infty \right)}{ak\left(\infty \right)+\infty }\\ =\frac{a^{2}k}{ak}\times \frac{1}{1}\\ =a\end{array}$

Therefore, the concentration value as $t$ tends to $\infty$ is $a$.

(d)

To determine

To find: The rate of reaction if time $t$ tends to $\infty$.

Answer to Problem 22E

The rate of reaction as $t$ tends to $\infty$ is 0.

Explanation of Solution

Determine the rate of reaction as $t$ tends to $\infty$.

Substitute $t$ tends to $\infty$ $\left(t\to \infty \right)$ in equation (2).

$\begin{array}{c}\underset{t\to \infty }{\lim }\frac{d\left[C\right]}{dt}=\underset{t\to \infty }{\lim }\frac{a^{2}k}{{\left(akt+1\right)}^2}\\ =\frac{a^{2}k}{{\left(ak\left(\infty \right)+\infty \right)}^2}\\ =\frac{a^{2}k}{\infty }\\ =0\end{array}$

Therefore, the rate of reaction as $t$ tends to $\infty$ is 0.

(e)

To determine

To compare: The practical means of part (c) and (d).

Answer to Problem 22E

Compared the practical means of part (c) and (d).

Explanation of Solution

The result from part (c) suggests that the concentration value as $t$ tends to $\infty$ reduces to zero.

The result from part (d) suggests that the rate of reaction value as $t$ tends to $\infty$ attains zero and no further reaction takes place.