Chapter 14, Problem 131AP

Interpretation Introduction

Interpretation:

The average molar mass expression and $k$ is to be determined.

Concept introduction:

The molar mass $M$

is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The basic SI unit for molar mass is $\text{kg/mol}$

c.

Average molar mass: The mass that is calculated through mole fractions $(X)$ and molar masses $(M)$ of different components is known as the average molar mass.

$\overline{M}={\displaystyle {\sum }_{i}MiXi}$

Answer to Problem 131AP

Solution: $\ln \frac{2M-\overline{M}}{\overline{M}}=-kt$

Explanation of Solution

The overall balanced equation is written as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{P}\to {\text{P}}^{*}\\ 2{\text{P}}^{*}\to {\text{P}}_{\text{2}}\end{array}}\\ \text{P}\to \frac{1}{2}{\text{P}}_{\text{2}}\end{array}$

The average molar mass is calculated as follows:

$\overline{M}=\frac{{\left[\text{P}\right]}_{\text{t}}M\text{+2}\left[{\text{P}}_{\text{2}}\right]M\frac{\text{mol}}{\text{L}}\text{×}\frac{\text{g}}{\text{mol}}}{{\left[\text{P}\right]}_{\text{t}}\text{+}\left[{\text{P}}_{\text{2}}\right]\frac{\text{mol}}{\text{L}}}$

In the above equation, $M$ is the molar mass of P. The concentration of P at a later time is represented as ${\left[\text{P}\right]}_t$. In the above equation, $\left[{\text{P}}_2\right]$ is multiplied by $2$ in the numerator as its molar mass is the twice of P.

According to the stoichiometry of the reaction, the concentration of $\left[{\text{P}}_2\right]$ can be written as follows:

$\left[{\text{P}}_2\right]=\frac{{\left[\text{P}\right]}_0-{\left[\text{P}\right]}_t}{2}$

Substitute the value of $\left[{\text{P}}_2\right]$ in the above equation as follows:

$\begin{array}{c}\overline{M}=\frac{\left({\left[\text{P}\right]}_{t}M+2\frac{{\left[\text{P}\right]}_0-{\left[\text{P}\right]}_t}{2}M\right)}{\left({\left[\text{P}\right]}_t+\frac{{\left[\text{P}\right]}_0-{\left[\text{P}\right]}_t}{2}\right)}\\ =\frac{{\left[\text{P}\right]}_t\text{M}+{\left[\text{P}\right]}_{0}M-{\left[\text{P}\right]}_t\text{M}}{\left({\left[\text{P}\right]}_t+\frac{1}{2}{\left[\text{P}\right]}_0-\frac{1}{2}{\left[\text{P}\right]}_t\right)}\\ =\frac{2\text{M}{\left[\text{P}\right]}_0}{{\left[\text{P}\right]}_0+{\left[\text{P}\right]}_t}\end{array}$

In the given mechanism, the denaturation step is the rate determining step. So, the rate is as follows:

$\text{Rate}=k\left[\text{P}\right]$

The expressions according to the first order integrated law are as follows:

$\begin{array}{c}\ln \frac{{\left[\text{P}\right]}_t}{{\left[\text{P}\right]}_0}=-kt\\ \frac{{\left[\text{P}\right]}_t}{{\left[\text{P}\right]}_0}=e^{-kt}\\ {\left[\text{P}\right]}_t={\left[\text{P}\right]}_0{e}^{-kt}\end{array}$

Substitute ${\left[\text{P}\right]}_0{e}^{-kt}$ for ${\left[\text{P}\right]}_t$ in the above equation as follows:

$\begin{array}{c}\overline{M}=\frac{2M\cancel{{\left[\text{P}\right]}_0}}{\cancel{{\left[\text{P}\right]}_0+{\left[\text{P}\right]}_0}{e}^{-kt}}\\ =\frac{2M}{1+e^{-kt}}\end{array}$

$\begin{array}{c}\frac{2M-\overline{M}}{\overline{M}}=e^{-kt}\\ \ln \frac{2M-\overline{M}}{\overline{M}}=-kt\end{array}$

By plotting $\ln \left(\frac{2M-\overline{M}}{\overline{M}}\right)$ versus $t$, the rate constant $(k)$ can be determined. A straight line and slope $\left(-k\right)$ will be given by the plot.

Conclusion

The calculated average molar mass is $\ln \frac{2M-\overline{M}}{\overline{M}}=-kt$ and by plotting $\ln \left(\frac{2M-\overline{M}}{\overline{M}}\right)$ versus $t$, the rate constant $(k)$ can be determined. A straight line and slope $\left(-k\right)$ will be given by the plot.