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A first order reaction is 50% completed ...

A first order reaction is `50%` completed in `30 min` at `27^(@)C` and in `10 min` at `47^(@)C`. Calculate the reaction rate constants at `27^(@)C` and the energy of activation of the reaction in `kJ mol^(-1)`.

A

`k = 0.0231 "min"^(-1), E_(a) = 43.848 kJ mol^(-1)`

B

`k = 0.017 "min"^(-1), E_(a) = 52.54 kJ mol^(-1)`

C

`k = 0.0693 "min"^(-1), E_(a) = 43.848 kJ mol^(-1)`

D

`k = 0.0231 "min"^(-1), E_(a) = 28.92 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Since `t_(1//2) = (0.693)/(k), :. K = (0.693)/(t_(1//2))`
Given `t_(1//2) = 30` min at `27^(@)C` and `t_(1//2) = 10` min at `47^(@)C`
`:. K_(26^(@)C) = (0.693)/(30) "min"^(-1) = 0.0231 "min"^(-1)`
and `k_(47^(@)C) = (0.693)/(10) = 0.693 "min"^(-1)`
We know that log `(k_(47^(@)C))/(k_(26^(@)C)) = (E_(a))/(2.303R) xx ((T_(2) - T_(1)))/(T_(2) xx T_(1))`
`:. E_(a) = (2.303 R xx (T_(1) xx T_(2))/((T_(2) - t_(1))) "log" (k_(47^(@)C))/(k_(27^(@)C))`
or `E_(a) = (2.303 xx 8.314 xx 10^(-3) xx 300 xx 320)/((320 - 300)) "log" (0.0693)/(0.0231)`
`43.848 kJ mol^(-1)`
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