Question

The decomposition of the Formic acid on gold surface follows first order kinetics if the rate constant at $300\mathrm{K}$ is $1.0\times {10}^{-3}{\mathrm{s}}^{-1}$ and the activation energy${\mathrm{E}}_{\mathrm{a}}=11.488\mathrm{kJ}{\mathrm{mol}}^{-1}$, then the rate constant at $200\mathrm{K}$ is:__________×10^–5 s^–1

Answer 1

Step 1: Formula used in the first order kinetics:

$\log \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right]$

$\log \frac{1.0\times {10}^{-3}{\mathrm{s}}^{-1}}{{\mathrm{K}}_1}=\frac{11.488\times 1000}{2.303\times 8.314}\left[\frac{1}{200}-\frac{1}{300}\right]$

Hence, $\log \frac{{10}^{-3}}{{\mathrm{K}}_1}=600\times \frac{3-2}{600}$

Step 2: Finding the value of rate constant:

So, $\log \frac{{10}^{-3}}{{\mathrm{K}}_1}=1$

Therefore, $10=\frac{{10}^{-3}}{{\mathrm{K}}_1}$

= ${\mathrm{K}}_1={10}^{-4}$

So , $\mathrm{x}\times {10}^{-5}={10}^{-4}\Rightarrow \mathrm{x}=10$

Therefore, ${\mathrm{K}}_{200}=10\times {10}^{-5}{\mathrm{s}}^{-1}$.