1. Tardigrade
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  3. Chemistry
  4. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol-1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln (k2/k1) is equal to: (R=8.314 J mol-1 K-1)

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Q. Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}$. If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300\, K$, then ln $(k_2/k_1)$ is equal to : $(R=8.314 \, J \, mol^{-1} K^{-1})$

JEE MainJEE Main 2017Chemical Kinetics

Solution:

$k_{1}=Ae^{-E_{a_1}/RT}$

$k_{2}=Ae^{-E_{a_2}/RT}$

$\frac{k_{2}}{k_{1}}=e^{\frac{1}{RT}\left(E_{a_1-E_{a_2}}\right)}$

In$\frac{k_{2}}{k_{1}}= \frac{E_{a_1}-E_{a_2}}{RT}$

$=\frac{10 \times 10^{3}}{8.314 \times 300} \approx 4$