#### 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

t_{1/2} = 40 min at temperature (T_{1}) = 300 K

t_{1/2} = 20 min at temperature (T_{2}) = 320 K

For first-order reaction the half lifetime t_{1/2}

`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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#### APPEARS IN

Solution 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) Concept: Temperature Dependence of the Rate of a Reaction.