# The decomposition of formic acid on gold surface follows first order kinetics.

Question:

The decomposition of 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 $\mathrm{E}_{\mathrm{a}}=11.488 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the rate constant at $200 \mathrm{~K}$ is___________ $\times 10^{-5} \mathrm{~s}^{-1}$. (Round of to the Nearest Integer).

( Given : $R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )

Solution:

(10)

$\mathrm{K}_{300}=10^{-4} \mathrm{~K}_{200}=?$

$\mathrm{E}_{\mathrm{a}}=11.488 \mathrm{KJ} /$ mole $\quad \mathrm{R}=8.314 \mathrm{~J} /$ mole $-\mathrm{K}$

so $\ln \left(\frac{\mathrm{K}_{300}}{\mathrm{~K}_{200}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{200}-\frac{1}{300}\right)$

so $\frac{\mathrm{K}_{300}}{\mathrm{~K}_{200}}=10$

$\ln \left(\frac{\mathrm{K}_{300}}{\mathrm{~K}_{200}}\right)=\frac{11.488 \times 1000 \times 100}{8.314 \times 200 \times 300}$

$=2.303$

$=\ell \mathrm{n} 10$

$\mathrm{K}_{200}=\frac{1}{10} \times \mathrm{K}_{300}=10^{-4}$

$=10 \times 10^{-5} \mathrm{sec}^{-1}$