Chemistry - Chemical Kinetics Question with Solution | TestHub

If $\mathrm{K}$ is known at two different temperatures the activation energy can be calculated as:

At temperature 1: $\ln {\mathrm{K}}_1 = -\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1} + \ln \mathrm{A}$

At temperature 2: $\ln {\mathrm{K}}_2 = -\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2} + \ln \mathrm{A}$

We can subtract one of these equations from the other:

$\ln {\mathrm{K}}_1 - \ln {\mathrm{K}}_2 =\left( -\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1} + \ln \mathrm{A}\right) - \left(-\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2} + \ln \mathrm{A}\right)$

This equation can further simplified to:

$\ln \frac{{\mathrm{K}}_1}{{\mathrm{K}}_2} = \frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2} - \frac{1}{{\mathrm{T}}_1}\right)$ 

$\ln \frac{0.05}{0.03}=\frac{\mathrm{Ea}}{2.305\times 8.314}\times \left[\frac{1}{200}-\frac{1}{300}\right]$

$\Rightarrow 0.22 =\frac{\mathrm{Ea}}{2.305\times 8.314}\times \left[\frac{1}{600}\right]$

$\Rightarrow {\mathrm{E}}_{\mathrm{a}} = 2.3 \times 8.3 \times 600 \times 0.22$

$\Rightarrow {\mathrm{E}}_{\mathrm{a}}=2519.88 \mathrm{J}\Rightarrow {\mathrm{E}}_{\mathrm{a}}=2520 \mathrm{J}$