The rate constant for a first order reaction is given by the following equation: $\ln \mathrm{k}=33.24-\frac{2.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}$ The Activation energy for the reaction is given by \_\_\_\_ $\mathrm{kJ} \mathrm{mol}^{-1}$. (In Nearest integer) (Given: $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Chemical Kinetics

The rate constant for a first order reaction is given by the following equation:

ln k = 33.24 - \frac{2.0 \times 1 0 ^{4} K}{T}

The Activation energy for the reaction is given by
____

kJ mol^{- 1}

. (In Nearest integer)
(Given:

R = 8.3 J K^{- 1} mol^{- 1}

)

Answer: 166

The Arrhenius equation in logarithmic form is:

ln k = ln A - \frac{E _{A}}{R T}

Comparing with the given equation

ln k = 33.24 - \frac{2.0 \times 1 0 ^{4}}{T}

, we identify:

\frac{E _{A}}{R} = 2.0 \times 1 0^{4} K

Therefore:

E_{A} = 2.0 \times 1 0^{4} \times 8.3 = 166000 J mol^{- 1} = 166 kJ mol^{- 1}

The answer is 166.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer