A catalyst lowers the activation energy of a certain reaction from $83.314$ to $75 \mathrm{~kJ}…
A catalyst lowers the activation energy of a certain reaction from $83.314$ to $75 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $500 \mathrm{~K}$. What will be the rate of reaction as compared to uncatalysed reaction? Assume other things being equal.
Double
28 times
$7.38$ times
$7.38 \times 10^{3}$ times
Solution
$\frac{k_{2}}{k_{1}}=\frac{A e^{-E_{a_{2}} / R T}}{A e^{-E_{a_{1} / R T}}}=e^{\left(E_{a}-E_{a_{2}}ight) / R T}$
$2.303 \log \frac{k_{2}}{k_{1}}=\frac{E_{a_{1}}-E_{a}}{R T}$
$=\frac{(83.314-75) \times 10^{3}}{8.314 \times 500}=2$
$\log \mathrm{k}_{2}=\frac{2}{2.303}=0.868$
Taking Antilog $\mathrm{k}_{2}=7.38$