Question

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at $300K$ is $1.0\times 10^{-3}{s}^{-1}$ and the activation energy $E_a=11.488kJmo{l}^{-1}$, the rate constant at $200K$ is $\times 10^{-5}{s}^{-1}$. (Round of to the Nearest Integer).
(Given: $R=8.314Jmo{l}^{-1}{K}^{-1}$ )

Answer 1

Answer: 10

$K_{300}=10^{-4}{K}_{200}=?$

$E_a=11.488KJ/moleR=8.314J/mole-K$

so $\ln \left(\frac{K_{300}}{K_{200}}\right)=\frac{E_a}{R}\left(\frac{1}{200}-\frac{1}{300}\right)$

$\ln \left(\frac{K_{300}}{K_{200}}\right)=\frac{11.488\times 1000\times 100}{8.314\times 200\times 300}$

$=2.303$

$=\ln 10$ so $\frac{K_{300}}{K_{200}}=10$

$K_{200}=\frac{1}{10}\times K_{300}=10^{-4}$

$=10\times 10^{-5}se{c}^{-1}$