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# 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) - Chemistry

#### Question

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

Given

t1/2 = 40 min at temperature (T1) = 300 K

t1/2 = 20 min at temperature (T2) = 320 K

For first-order reaction the half lifetime t1/2

t_"1/2" = 0.693/k_1 at  300 k and t_"1/2" = 0.693/k_2 at temperature 320 K

k_1 = 0.693/40 and k_2 = 0.693/20

k_2/k_1 = (0.693/20)/(0.693/40) = 40/20 = 2

log 2 = E_a/(2.303xx8.314) [(320-300)/(300xx320)]

(2.303xx8.314xx300xx320 log 2)/20  =  E_a

E_a = (2.303xx8.314xx300xx320xx0.3010)/20

= 27663.79 J mol^(-1)

= 27.66379  kJ mol^(-1)

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