Question

A first order reaction is `50%` complete in `30` minutes at `27^(@)C` and in `10` minutes at `47^(@)C`. The rate constant at `47^(@)C` and energy of activation of the reaction in kJ`//`mole will be
A. `0.0693, 43.848 kJ mol^(-1)`
B. `0.0560,45.621 kJ mol^(-1)`
C. `0.0625, 42.926 kJ mol^(-1)`
D. `0.0660, 46.189 kJ mol^(-1)`

Answer 1

Correct Answer - A
We know that `k = (2.303)/(t) log((a)/(a-x))`
Subtituting the values at the two given conditions
`k_(27^(@)) = (0.0693)/(30) = 0.0231 , k_(47^(@)) = (0.693)/(10)= 0.0693`
We also know that
`"log" (k_(2))/(k_(1)) = (2.303)/(t)"log"((a)/(a-x))`
or `E_(a) = (2.303 R xx T_(1)T_(2))/(T_(2) - T_(1)) "log"(k_(2))/(k_(1))`
`= (2.303 xx 8.314 xx 10^(-3) xx 300 xx 320)/(320 - 300) xx "log" (0.0693)/(0.231)`
`= 43.848 kJ mol^(-1)`