Chemical Kinetics

Question
CBSEENCH12010582

A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK–1 mol–1)

Solution

T1 = 300 KK1 = 0.69340T2 = 320 KK2 = 0.69320logK2K1 = Ea2.303 RT2-T1T1T2log0.693200.69340 = Ea2.303 x 8.314320-300320 x 300log4020 = Ea2.303 x  8.31420320 x 300 = 0.3010 x 2.303 x 8.314 x 320 x 30020 = Ea = 27764 J= 27.76 KJ

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