Verified Solution • 2025–26 Session moderate

QJEEM21Ch418e: JEE Main 2021 Chemistry - Class 12 JEEM21Ch418e Solution: The decomposition of formic acid on gold surface follows first…

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 ×\times 10 −- 3 s −- 1 and the activation energy Ea = 11.488 kJ mol −- 1, the rate constant at 200 K is ____________ ×\times 10 −- 5 s −- 1. (Round off to the Nearest Integer). (Given : R = 8.314 J mol −- 1 K −- 1)

Step-by-Step Verified Solution

log⁡k2k1=Ea2.303R[1T1−1T2]\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]

k1 (at 200 K) = ?

k2 (at 300 K) = 1×10−3s−11 \times {10^{ - 3}}{s^{ - 1}}

log⁡1×10−3k1=11.488×1032.303×8.314[1600]=1\log {{1 \times {{10}^{ - 3}}} \over {{k_1}}} = {{11.488 \times {{10}^3}} \over {2.303 \times 8.314}}\left[ {{1 \over {600}}} \right] = 1

⇒\Rightarrow 1×10−3k1=10{{1 \times {{10}^{ - 3}}} \over {{k_1}}} = 10

⇒\Rightarrow k1=10×10−5s−1{k_1} = 10 \times {10^{ - 5}}{s^{ - 1}}

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