A first order reaction is `50%` complete in 30 minutes at `27^(@)`C and in 10 minutes at `47^(@)`C. Calculate the reaction rate constants at these temperatures and the energy of activation of the reaction in `kJ//mol` (R=8.314 J` mol^(-1)K^(-1))`

Explanation

The half period for first order reaction is given by: `t_(1//2) = 0.693/k` or `k=0.693/t_(1//2)`
Therefore, `k_(1) = 0.693/ (30 min) = 0.0231 min^(-1)`
According to Arrhenius equation,
`log k_(2)/k_(1) = E_(a)/(2.303 R) [1/T_(1) - 1/T_(2)]`
Herer, `k_(1) = 0.0231 min^(-1), k_(2) = 0.0693 min^(-1), T_(1) = 27 + 273 = 300 K, T_(2) = 47 + 273 = 320` K
`log (0.0693 min^(-1))/(0.0231 min^(-1))`
`=rArr log 3 = E_(a)/(2.303 xx 8.314 (J mol^(-1))) xx 20/(300 xx 320)`
`rArr 0.4771 = E_(a)/(2.303 xx 8.314 (J mol^(-1)) xx 4800)`
`rArr E_(a) = 43848 J mol^(-1) = 43.848 kJ mol^(-1)`