A first order reaction is 50% complete in 40 minutes at `30^(@)` C and 20 minutes at `50^(@)` C calcualte the reaction rate constant at these temperatures and the energy of acitivation in kj/ mol

Explanation

The half-life period for first order reaction is given by,
`t_(1//2) = (0.693)/(k)`, or `k = (0.693)/(t_(1//2))` or
`k_(1) = (0.693)/("40 min") = 0.01733 "min"^(-1)`
and `k_(2) = (0.693)/("20 min") = 0.03465 "min"^(-1)`
According to Arrhenius equation,
`log""(k_(2))/(k_(1)) = (E_(a))/(2.303 R)[(1)/(T_(1)) - (1)/(T_(2))]`
Here, `k_(1) = 0.01733 "min"^(-1)` at `T_(1) = 30 + 273 = 303 K` and `k_(2) = 0.03465 "min"^(-1)` at `T_(2) = 50 + 273 = 323 K`
`log""(0.03465 "min"^(-1))/(0.01733 "min"^(-1)) = (E_(a))/(2.303 xx 8.314 J mol^(-1)K^(-1))[(1)/(303 K) - (1)/(323 K)]`
`log 1.9994 = (E_(a))/(2.303 xx 8.314 J mol^(-1)) xx (20)/(303 xx 323)`
`0.3009 = (E_(a) xx 20)/(2.303xx 8.314 J mol^(-1) xx 303 xx 323)`
`E_(a) = (0.3009 xx 2.303 xx 8.314 J mol^(-1) xx 303 xx 323)/(20)`
`E_(a) = 28193 J mol^(-1) = 28.193 kJ mol^(-1)`