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 : $\displaystyle \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
Official answer
166
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JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.