Question

If energy of activation of the rection is 53.6kJ`mol^(-1)` and the temperature changes from `27%^(@)` to `37^(@)` C, then the value of
`k_(37^(@)C)/(k_(27^(@)C)` is
A. 2.5
B. 1
C. 2
D. 1.5

Answer 1

log `(k_(2)/k_(1)) = E_(a)/(2.303 R) (T_(2)-T_(1))/(T_(1)T_(2))`
`E_(a) = 53.6 xx 10^(3) J mol^(-1), T_(1)=300K, T_(2)=310K`
`log (k_(2)/k_(1)) = (53.6 xx 10J mol^(-1))/(2.303 xx (8.314 JK^(-1)mol^(-1))) xx (310-300)/(310 xx 300)`
`=0.3010`
`k_(2)/k_(1) = "Antilog"(0.3010)=2`