1.

In the Arrhenius plot of ln k vs (1)/(T) , a linear plot is obtained with a slope of -2 xx 10^(4) K . The energy of activationof the reaction (in kJ "mole"^(-1)) is (R value is 8.3 JK^(-1) mol^(-1))

Answer»

83
166
249
332

Solution :`k =AE^(-E_(a) //RT)`
ln k= ln A - `E_(a)//RT`For ln k vs 1/T
ln A = INTERCEPT
`-E_(a)//R` = slope = `-2 xx 10^(4)` K
`E_(a) = R xx 2 xx 10^(4) K = 8.3 xx 2 xx 10^(4) J "MOL"^(-1)`
`16.6 xx 10^(4) J mol^(-1)` or 166 kJ `mol^(-1)`


Discussion

No Comment Found

Related InterviewSolutions