The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is $1.0 \times 10^{-3} s^{-1} $ and the activation energy $E_a = 11.488 kJ.mol^{-1} $, the rate constant at 200 K is _ _ _ _ $\times 10^{-5} s^{-1} $. (Round off to the nearest integer)

[Given $R = 8.314 J mol^{-1} K^{-1} $]

Solution

We have, $\log \frac{{{K_2}}}{{{K_1}}} = \frac{{{E_a}}}{{2.303R}}\left( {\frac{{{T_2} – {T_1}}}{{{T_1}{T_2}}}} \right)$

$ \Rightarrow \log \frac{{{K_2}}}{{1.0 \times {{10}^{ – 3}}}} = \frac{{11.488 \times 1000}}{{2.303 \times 8.314}}\left( {\frac{{200 – 300}}{{200 \times 300}}} \right)$

$ \Rightarrow \log \frac{{{K_2}}}{{1.0 \times {{10}^{ – 3}}}} =  – \frac{{11.488 \times 5}}{{2.303 \times 8.314 \times 3}} =  – 1$

$ \Rightarrow {K_2} = {10^{ – 4}} = 10 \times {10^{ – 5}}{s^{ – 1}}$

Answer: 10