Question

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 of to the Nearest Integer).
(Given: $R =8.314 \,J\, mol ^{-1} K ^{-1}$ )

Answer 1

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

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

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} sec ^{-1} $