Question

(a) A certain first-order reaction has an activation energy of 53 kJ mol21 . It is run twice, first at 298 K and then at 308 K (10°C higher). All other conditions are identical. Show that, in the second run, the reaction occurs at double its rate in the first run.

(b) The same reaction is run twice more at 398 K and 408 K. Show that the reaction goes 1.5 times as fast at 408 K as it does at 398 K.

Short answer

The ratio of rate constants k₁ and k₂ are close 1.5 and hence it is evident that reaction proceeds 1.5 times faster at 408K

Step-by-step solution

Arrehinus equation:

$$\begin{gathered} \mathrm{k}=\mathrm{Aexp}\left(\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\right) \\ \mathrm{Where}\mathrm{k}=\mathrm{rate}\mathrm{constant} \\ -{\mathrm{E}}_{\mathrm{a}}=\mathrm{activation}\mathrm{energy} \\ \mathrm{R}=\mathrm{universal}\mathrm{gas}\mathrm{constant} \\ \mathrm{A}=\mathrm{preexponential}\mathrm{factor} \\ \mathrm{T}=\mathrm{temaprature}\mathrm{of}\mathrm{the}\mathrm{reaction} \end{gathered}$$

Rate  of  the reaction:

As given in the quastion there are two conditions,so following eqautions are required.

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{Aexp}\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}\right).......\left(1\right) \\ {\mathrm{k}}_2=\mathrm{Aexp}\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}\right).......\left(2\right) \end{gathered}$$

Dividing 2 by 1

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{Aexp}\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}\right).......\left(1\right) \\ {\mathrm{k}}_2=\mathrm{Aexp}\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}\right).......\left(2\right) \\ \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{\exp \left(-\mathrm{Ea}/{\mathrm{RT}}_1\right)}{\exp \left(-\mathrm{Ea}/{\mathrm{RT}}_2\right)}......\left(3\right) \end{gathered}$$

Where k₁ ,k₂ are the rate constant of reaction at different conditions,E_a is the activation energy of the reaction.

Substituting the given values in the question

The given parameters are

localid="1661757882150" $$\begin{gathered} {\mathbf{\text{T}}}_{\mathbf{1}}=\mathbf{308}\mathrm{K} \\ {\mathbf{T}}_{\mathbf{2}}=\mathbf{298}\mathbf{K} \\ \mathbf{R}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{315}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}{\mathbf{K}}^{\mathbf{-}\mathbf{1}} \\ {\mathbf{E}}_{\mathbf{a}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{53}{\mathbf{kJmol}}^{\mathbf{-}\mathbf{1}} \\ {\mathbf{E}}_{\mathbf{a}\mathbf{}}\mathbf{=}\mathbf{}\mathbf{53}{\mathbf{kJmol}}^{\mathbf{-}\mathbf{1}\mathbf{}}\mathbf{x}\mathbf{}\mathbf{}\frac{\mathbf{1000}\mathbf{J}}{\mathbf{1}\mathbf{kJ}} \\ {\mathbf{E}}_{\mathbf{a}\mathbf{}}\mathbf{=}\mathbf{}\mathbf{53}\mathbf{x}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Substitue in the equation-3

$$\begin{gathered} \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{\exp \left(-53\mathrm{x}{10}^3{\mathrm{Jmol}}^{-1}/8.315{\mathrm{Jmol}}^{-1}\mathrm{x}\right)}{\exp \left(-53\mathrm{x}{10}^3{\mathrm{Jmol}}^{-1}/8.315{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}\mathrm{x}298\mathrm{K}\right)} \\ =\frac{\exp \left(-20.694\right)}{\exp \left(-21.389\right)} \\ \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{0.0000000010287}{0.000000005137} \\ \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=2.0025306599182 \end{gathered}$$

The ratio of the rate constants k1,k2 are very near to2 when all other conditions are identicle.

So it is clear that by incrasing the temparature by 10⁰ C ,reaction gets doubled to its rate then the first run.

Ratio of the constants:

b) In the above given question

$$\begin{gathered} {\mathrm{T}}_1=408\mathrm{K} \\ {\mathrm{T}}_2=398\mathrm{K} \\ \mathrm{R}=8.315\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\mathrm{and}{\mathrm{E}}_{\mathrm{a}}=53\mathrm{x}{10}^3{\mathrm{Jmol}}^{-1} \\ \mathrm{Substitute}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{eqaution}-3 \\ \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{\exp \left(-53\mathrm{x}{10}^3{\mathrm{Jmol}}^{-1}/8.315{\mathrm{Jmol}}^{-1}\mathrm{x}408\mathrm{K}\right)}{\exp \left(-53\mathrm{x}{10}^3{\mathrm{Jmol}}^{-1}/8.315{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}\mathrm{x}398\mathrm{K}\right)} \\ =\frac{\exp \left(-15.62260\right)}{\exp \left(-16.01513\right)} \\ =\frac{0.0000001641}{0.0000000110} \\ \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=1.48071 \end{gathered}$$

The ratio of rate constants k₁ and k₂ are close 1.5 and hence it is evident that reaction proceeds 1.5 times faster at 408K