1.

A plot of (1)/(T) Vsk for a reaction gives the slope - 1 xx 10^(4)K. The energy of activation for the reaction is (Given R = 8.314 JK^(-1) " mol"^(-1))

Answer»

`8314 "J MOL"^(-1)`
`1.202 "kJ mol"^(-1)`
`12.02 "J mol"^(-1)`
`83.14 "kJ mol"^(-1)`

Solution :Correction in the quenstion: It should be ln k vs 1/T
PLOT of ln k vs 1/T GIVES
Slope `= - (E_(a))/(R)`
`:.-1xx10^(4)= - (E_(a))/(8.314)`
`E_(a)= 8.314xx10^(4)"J mol"^(-1)= 83.14xx10^(3)"J mol"^(-1)`
or `E_(a)= 83.14 "kJ mol"^(-1)`


Discussion

No Comment Found

Related InterviewSolutions