Chapter 19, Problem 19.132QP

Interpretation Introduction

Interpretation:

From the given information how to determine rate constant k from molar mass measurements has to be described.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Answer to Problem 19.132QP

From the given information how to determine rate constant k from molar mass measurements is described.

Explanation of Solution

We can write the balanced reaction from the given information as follows

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

The average molar mass can be given by

$\stackrel{-}{\mu }=\frac{\left({\left[\text{P}\right]}_t\mu +2\left[{\text{P}}_{\text{2}}\right]\mu \right)\frac{\text{mol}}{\text{L}}\times \frac{\text{g}}{\text{mol}}}{\left[P\right]t+\left[{\text{P}}_{\text{2}}\right]\frac{\text{mol}}{\text{L}}}(1)$

Based on the stoichiometry of the reaction, the concentration of $\left[{\text{P}}_{\text{2}}\right]$ is

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

Substitute back into equation (1) it comes as

$\stackrel{-}{\mu }=\frac{\left({\left[\text{P}\right]}_t\mu +\frac{{\left[{\text{P}}_{\text{2}}\right]}_0-{\left[{\text{P}}_{\text{2}}\right]}_t}{2}\right)}{\left(\left[P\right]t+\frac{{\left[{\text{P}}_{\text{2}}\right]}_0-{\left[{\text{P}}_{\text{2}}\right]}_t}{2}\right)}=\frac{\left({\left[\text{P}\right]}_t\mu +{\left[\text{P}\right]}_0\mu -{\left[\text{P}\right]}_t\mu \right)}{\left({\left[\text{P}\right]}_t+\frac{1}{2}{\left[\text{P}\right]}_0-\frac{1}{2}{\left[\text{P}\right]}_t\right)}=\frac{2\mu {\left[\text{P}\right]}_0}{{\left[\text{P}\right]}_0+{\left[\text{P}\right]}_t}$ (2)

From the proposed mechanism, the denaturation step is rate-determining step, thus

$\text{rate = k[P]}$

Now, look back at the change in the concentration over time, for that we use first-order integrated rate law.

$\frac{{\left[\text{P}\right]}_{\text{t}}}{{\left[\text{P}\right]}_{\text{0}}}\text{ = -kt}$

$\frac{{\left[\text{P}\right]}_{\text{t}}}{{\left[\text{P}\right]}_{\text{0}}}{\text{ =e}}^{\text{-kt}}$

${\left[\text{P}\right]}_{\text{t}}\text{=}{\left[\text{P}\right]}_{\text{0}}{\text{e}}^{\text{-kt}}$

Substitute this in equation (2),

$\begin{array}{l}\stackrel{-}{\mu }=\frac{2\mu {\left[\text{P}\right]}_0}{{\left[\text{P}\right]}_0+{\left[\text{P}\right]}_{\text{0}}{\text{e}}^{\text{-kt}}}=\frac{2\mu }{1+{\text{e}}^{\text{-kt}}}\\ or\\ \frac{2\mu -\stackrel{-}{\mu }}{\stackrel{-}{\mu }}={\text{e}}^{\text{-kt}}\\ \ln \left(\frac{2\mu -\stackrel{-}{\mu }}{\stackrel{-}{\mu }}\right)=-kt\end{array}$

Therefore, the rate constant, k, is determined by plotting a graph $\left(\frac{2\mu -\stackrel{-}{\mu }}{\stackrel{-}{\mu }}\right)$ versus time.  The plot will obtain as straight line with a slope of -k

Conclusion

From the given information rate constant k from molar mass measurements was described.