Two reactions $R_{1}$ and $R_{2}$ have identical pre-exponential factors. Activation energy of $R_{1}$ exceeds that of $R_{2}$ by $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ are rate constants for reactions $R_{1}$ and $R_{2}$ respectively at 300 K, then $\ln \left(\mathrm{k}_{2} / \mathrm{k}_{1}\right)$ is equal to : ($\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$)

JEE Main · 2017 · 02 Apr 2017 · Offline · Chemical Kinetics

Two reactions

R_{1}

and

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

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}

)

Answer: (D)

With equal pre-exponential factors the Arrhenius ratio is

ln \frac{k _{2}}{k _{1}} = \frac{E _{a 1} - E _{a 2}}{R T}

. The difference in activation energies is

10000

J mol

^{- 1}

, so

ln \frac{k _{2}}{k _{1}} = \frac{10000}{8.314 \times 300} = \frac{10000}{2494} \approx 4

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option