SolveIt is under development
SolveItJEE Main
Question 118 / 166 · latest first
Chemistry · 2022

JEE Main · 27 June 2022, Shift 1 · Q86

The rate constant for a first order reaction is given by the following equation: ln k =33.24-(2.0 × 10^4 K)/T The Activation energy for the reaction…

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 \_\_\_\_ $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$. (In Nearest integer) (Given : $\displaystyle \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
ShareWhatsAppTelegram

More from Chemical Kinetics

JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.