Q49P - Chemistry: Molecular Nature Of Matter And Change

Question

The rate constant of a reaction is $\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$ at $\mathbf{25}{\mathbf{ }}^{\mathbf{0}}\mathbf{C}$ , and the activation energy is 33.6 kJ/mol. What is k at $\mathbf{75}{\mathbf{ }}^{\mathbf{0}}\mathbf{C}$?

Step-by-Step Solution

Answer

The rate constant of a reaction is $4.7\times {10}^{-3} {\mathrm{s}}^{-1}$ at, and the activation energy is 33.6 kJ/mol. Then at 75°C, the rate constant is $0.33 {\mathrm{s}}^{-1}$ .

Step 1: The Arrhenius equation

The Arrhenius equation is written like this:

$$\begin{gathered} \mathrm{K}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}} \\ \mathrm{In} \mathrm{K}=\mathrm{In} \mathrm{A}-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT} \end{gathered}$$

 

Where k is the rate constant, A is the frequency factor,  ${\mathrm{E}}_{\mathrm{a}}$ is the reaction's activation energy at a certain temperature T, and R is the gas constant.

Step 2: What is k at 75   0 C ?

Create a new expression for the two rate constants,  ${\mathrm{K}}_1$ and  ${\mathrm{K}}_2$ , for temperatures ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$  , respectively:

$$\begin{gathered} {\mathrm{InK}}_1=\mathrm{In} \mathrm{A}-{\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1 \\ {\mathrm{InK}}_2=\mathrm{In} \mathrm{A}-{\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_2 \\ {\mathrm{InK}}_2-{\mathrm{InK}}_1=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ \mathrm{In}\left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \end{gathered}$$

 

 

Replace the values and calculate the rate constant:

$$\begin{gathered} \mathrm{In}\left(\frac{{\mathrm{K}}_2}{4.7\times {10}^{-3} {\mathrm{s}}^{-1}}\right)=\frac{33.6\mathrm{KJ}/ \mathrm{mol}}{8.314 \mathrm{J}/\mathrm{moLK}}\left(\frac{1}{(273+75)\mathrm{K}}-\frac{1}{(273+25)\mathrm{K}}\right)\times \left(\frac{1000\mathrm{J}}{1 \mathrm{KJ}}\right) \\ \mathrm{In}\left(\frac{{\mathrm{K}}_2}{4.7\times {10}^{-3} {\mathrm{s}}^{-1}}\right)=1.948515 \\ {\mathrm{K}}_2=4.7\times {10}^{-3} {\mathrm{s}}^{-1}\times \left({\mathrm{e}}^{1.948515}\right) \\ {\mathrm{K}}_2=0.33 {\mathrm{s}}^{-1} \end{gathered}$$

 

At 75°C, the rate constant is $0.33 {\mathrm{s}}^{-1}$.