Question

The ratio of rate constant at `27^(@)C` and `37^(@)C` is `Q_(10)`. What should be the energy of activation of a reaction for which `Q_(10) = 2.5` ?
A. `71 kJ`
B. `212 kJ`
C. `35 kJ`
D. `12.1 kJ`

Answer 1

Correct Answer - A
`log Q10` or `log 2.5 = (E_(a))/(2.303R)[(1)/(T_(1)) - (1)/(T_(2))]`
Or `log 2.5 = (E_(a))/(2.303R)[(T_(2) - T_(1))/(T_(1)T_(2))]`
Subsititute the value of `R` in `kJ`. `E_(a)` comes out to be `71 kJ`