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Q.The rate of a particular reaction doubles when temperature changes from 27 degree C to 37 degree C. Calculate the energy of activation of such a reaction. (Given log 2 = 0.30)

Jharkhand JacJAC Intermediate Board 2023Subjective· 5mImportance★★★★★
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Using the two-temperature form of the Arrhenius equation with k2/k1 = 2 (rate doubles) gives the activation energy directly.

Given: rate (and hence k) doubles, so k2/k1 = 2; T1 = 27 degree C = 300 K, T2 = 37 degree C = 310 K; log 2 = 0.30; R = 8.314 J/mol/K.

Arrhenius two-point equation:

log(k2/k1) = (Ea / 2.303R) x (1/T1 - 1/T2)

1/T1 - 1/T2 = 1/300 - 1/310 = (310 - 300)/(300 x 310) = 10/93000 = 1.075 x 10^-4 K^-1

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