JEE Main 2026 Chemistry question
For reaction A → P, rate constant k = 1.5 × 10³ s⁻¹ at 27°C. If activation energy for the above reaction is 60 kJ mol⁻¹, then the temperature (in °C) at which rate constant, k = 4.5 × 10³ s⁻¹ is . (Nearest integer) [Given: log 2 = 0.30, log 3 = 0.48, R = 8.3 J K⁻¹ mol⁻¹, ln 10 = 2.3]
Answer: 41
Solution
ln 3 = 2.3 × 0.48 = 1.104. From ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂): 1/T₂ = 1/300 −
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