Verified Solution • 2025–26 Session moderate

QJEEM20Chc029: JEE Main 2020 Chemistry - Class 12 JEEM20Chc029 Solution: The rate of a reaction decreased by 3.555 times when the…

The rate of a reaction decreased by 3.555 times when the temperature was changed from 40oC to 30oC. The activation energy (in kJ mol–1) of the reaction is _______.

Take; R = 8.314 J mol–1 K–1 ln 3.555 = 1.268

Step-by-Step Verified Solution

k = A e−EaRT{e^{ - {{{E_a}} \over {RT}}}}

∴\therefore ln⁡k2k1=EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)

⇒\Rightarrow ln (3.555) = Ea8.314(1303−1313){{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)

⇒\Rightarrow Ea = 1.268×8.314×3.3×31310{{1.268 \times 8.314 \times 3.3 \times 313} \over {10}}

= 99980.7 = 99.98 kJ/mol ≃\simeq 100 kJ/mol

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