1.

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_(a).

Answer»

Solution :Here we are GIVEN that
When `T_(1)=298K, k_(1)=k` (say)
When `T_(2)=308K, k_(2)=2k`
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
Substituting these values in the equation, we get
`"log"(2k)/(k)=(E_(a))/(2.303xx8.314J K^(-1)"mol"^(-1))xx(308K-298K)/(298K xx 308K)`
or `log 2=(E_(a))/(2.303xx8.314) xx (10)/(298xx308)`
or `0.3010=(E_(a)xx10)/(2.303xx8.314xx298xx308)`
or `E_(a)=(0.3010xx2.303xx8.314xx298xx308)/(10)`
or `E_(a)=52897.8J mol^(-1)=53.6"kJ mol"^(-1)`.


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