What is the activation energy for a reaction if its rate doubles
when the temperature is raised from 20°C to 35°C? (R = 8.314
J molv K⁻¹) [2013]
(a) 269 kJ mol⁻¹ (b) 34.7 kJ mol⁻¹
(c) 15.1 kJ mol⁻¹ (d) 342 kJ mol⁻¹
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answer : option (b) 34.7 kJ/mol
Applying Arrhenius activation energy formula,
log(k2/k1) = Ea/2.303R (1/T1 - 1/T2)
where k2 = 2k1 , T1 = 20°C = 273 + 20 = 293K
T2 = 35°C = 35 + 273 = 308K and R = 8.3 J/mol/K
⇒log(2k1/k1) = Ea/(2.303 × 8.3) [1/293 - 1/308]
⇒log2 = Ea/(19.1149) [(308 - 293)/293 × 308]
⇒19.1149 log2 = Ea × 15/(293 × 308)
⇒5.7541 × 293 × 308/15 = Ea
⇒Ea = 34618.2 J/mol ≈ 34.7 kJ/mol
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