Chapter 20, Problem 20.29E

Rewrite equation $20.27$ so that it has the form of a straight-line equation, and identify the expected slope(s) and intercept(s) of two possible plots.

Interpretation Introduction

Interpretation:

The equation $20.27$ is to be stated in the form of a straight-line equation. The slopes and intercepts of the two possible plots are to be to identified.

Concept introduction:

The rate of a reaction is defined as the speed by which the reaction is proceeding. The rate of reaction depends on several factors such as the concentration of reactants and temperature. The rate law equation of second order reaction is represented as,

$r=k\left[\text{A}\right]\left[\text{B}\right]$

Answer to Problem 20.29E

The equation $20.27$ in the form of a straight-line equation is represented as,

$\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}=\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)t+\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$

The slope of the above equation is $\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)$.

The intercept of the above equation is $\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$.

The equation can be further rearranged to get the equation as shown below as,

$\log \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}=\frac{\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)}{2.303}t+\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$

The slope of the above equation is $\frac{\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)}{2.303}$.

The intercept of the above equation is $\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$.

Explanation of Solution

The equation $20.27$ is represented as,

$\frac{1}{b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0}\left(\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}-\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}\right)=k\cdot t$ …(1)

Where,

• ${\left[\text{A}\right]}_0$ represents the initial concentration of reactant $\text{A}$.

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

• $t$ represents the time.

• $k$ represents the rate constant of second order reaction.

• ${\left[\text{B}\right]}_0$ represents the initial concentration of reactant $\text{B}$.

• ${\left[\text{B}\right]}_t$ represents the concentration of reactant $\text{B}$ at time $t$.

• $b$ represents the stoichiometric coefficient of reactant $\text{B}$.

• $a$ represents the stoichiometric coefficient of reactant $\text{A}$.

The general equation for a straight line is represented as,

$y=mx+c$ …(2)

Where,

• $y$ represents the y-axis coordinate.

• $x$ represents the x-axis coordinate.

• $m$ represents the slope of the line.

• $c$ represents the intercept of the graph.

Rearrange the equation (1) to get the value of $\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}$.

$\begin{array}{rll}\frac{1}{b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0}\left(\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}-\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}\right)& =& k\cdot t\\ \frac{1}{b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0}\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}-\frac{1}{b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0}\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}& =& k\cdot t\\ \left(b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0\right)\left(\frac{1}{b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0}\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}-k\cdot t\right)& =& \ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}\\ \ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}-\left(b{\left[\text{A}\right]}_0-a{\left[\text{B}\right]}_0\right)\left(k\cdot t\right)& =& \ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}\end{array}$

The above equation is rearranged to get an expression similar to equation (2).

$\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}=\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)t+\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$ …(3)

The equation (3) represents a straight line equation when a grape is plotted for $\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}$ versus $t$.

The slope of the equation (3) is $\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)$.

The intercept of the equation (3) is $\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$.

The natural logarithm of the equation (3) is converted into logarithm.

$\begin{array}{rll}2.303\log \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}& =& \left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)t+2.303\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}\\ \log \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}& =& \frac{\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)}{2.303}t+\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}\end{array}$

The slope of the above equation is $\frac{\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)}{2.303}$.

The intercept of the above equation is $\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$.

Conclusion

The equation $20.27$ in the form of a straight-line equation is represented as,

$\ln \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}=\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)t+\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$

The slope of the above equation is $\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)$.

The intercept of the above equation is $\ln \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$.

The equation can be further rearranged to get the equation as shown below as,

$\log \frac{{\left[\text{B}\right]}_t}{{\left[\text{A}\right]}_t}=\frac{\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)}{2.303}t+\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$

The slope of the above equation is $\frac{\left(a{\left[\text{B}\right]}_0-b{\left[\text{A}\right]}_0\right)\left(k\right)}{2.303}$.

The intercept of the above equation is $\log \frac{{\left[\text{B}\right]}_0}{{\left[\text{A}\right]}_0}$.