Q50P - Chemistry: Molecular Nature Of Matter And Change

Question

The rate constant of a reaction is $\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{mol}\mathbf{.}\mathbf{s}$ at  $\mathbf{195}{\mathbf{ }}^{\mathbf{0}}\mathbf{c}$and $\mathbf{3}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{mol}\mathbf{.}\mathbf{s}$  at $\mathbf{258}{\mathbf{ }}^{\mathbf{0}}\mathbf{C}$ . What is the activation energy of the reaction?

Step-by-Step Solution

Answer

The rate constant of a reaction is $4.50\times {10}^{-5} \mathrm{L}/\mathrm{mol}.\mathrm{s}$ at $195{ }^0\mathrm{c}$ and   $3.20\times {10}^{-3} \mathrm{L}/\mathrm{mol}.\mathrm{s}$ at  $258{ }^0\mathrm{C}$  . The activation energy is 139.8 kJ/mol.

Step 1: The Arrhenius equation

The Arrhenius equation is written like this:

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

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.

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:

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

Step 2: The activation energy of the reaction

At ${\mathrm{T}}_1$, $195{ }^0\mathrm{c}$  the rate constant of the reaction k₁ is $4.50\times {10}^{-5} \mathrm{L}/\mathrm{mol}.\mathrm{s}$ .

At   ${\mathrm{T}}_2$ $258{ }^0\mathrm{C}$, the rate constant of the reaction K₂ is $3.20\times {10}^{-3} \mathrm{L}/\mathrm{mol}.\mathrm{s}$ .

The reaction's gas constant is 8.314.

Substitute the variables to get the activation energy ${\mathrm{E}}_{\mathrm{a}}$.

$$\begin{gathered} \mathrm{In}\left(\frac{3.20\times {10}^{-3} \mathrm{L}/\mathrm{mol}.\mathrm{s}}{4.50\times {10}^{-5} \mathrm{L}/\mathrm{mol}.\mathrm{s}}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314 \mathrm{J}/\mathrm{moLK}}\left(\frac{1}{(273+195)\mathrm{K}}-\frac{1}{(273+258)\mathrm{K}}\right) \\ {\mathrm{E}}_{\mathrm{a}}=\frac{8.314 \mathrm{J}/\mathrm{moLK}\times \mathrm{In}\left(\frac{3.20\times {10}^{-3} \mathrm{L}/\mathrm{mol}.\mathrm{s}}{4.50\times {10}^{-5} \mathrm{L}/\mathrm{mol}.\mathrm{s}}\right)}{\left({\displaystyle \frac{1}{468\mathrm{K}}}-{\displaystyle \frac{1}{531\mathrm{K}}}\right)} \\ =139846 \mathrm{J}/\mathrm{mol} \\ =139.8 \mathrm{KJ}/\mathrm{mol} \end{gathered}$$

 

The activation energy is 139.8 kJ/mol.