1.

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K.Calculate E_(a). (R=8.314 J K^(1) mol^(-1))

Answer»

Solution :`log(k_(2))/(k_(1))=(E_(a))/(2.303 R)((1)/(T_(2))-(1)/(T_(2)))`
`(E_(a))/(2.303R)(T_(2)-T_(1))/(T_(1)T_(2))`
Where ,`T_(1)=298 K,T_(2)=(298+10)=308 K`
R=8.314 J`K^(-1) MOL^(-1)`
`(k_(2))/(k_(1))=2.0` (`because` Rate becomes double)
`therefore log 2.0=(E_(a)XX(308-298)K)/(2.303xx8.314 JK^(-1)mol^(-1)(308xx298))`
`therefore E_(a)=(0.3010xx8.314xx308xx298xx2.303)/(10)`
52898 J`mol^(-1)`


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