Question

For the given reaction:
`H_(2) + I_(2) rarr 2HI`
Given: `{:(T(K),1//T(K^(-1)),log k,),(769,1.3 xx 10^(-3),2.9,),(67,1.5 xx 10^(-3),1.1,):}`
The activation energy will be
A. `41.4 kcal mol^(-1)`
B. `40 kcal mol^(-1)`
C. `-41.4 kcal mol^(-1)`
D. `-40 kcal mol^(-1)`

Answer 1

Correct Answer - B
`k = Ae^(-E_(A)//RT)`
`rArr log k = log A - (E_(a))/(2.303 RT)`
`{:(log k_(1) = log A-(E_(a))/(2.303RT) ((1)/(T_(1)))),(log k_(2)=logA-(E_(a))/(2.303RT) ((1)/(T_(2)))):}} "Subtract to get" E_(a) = 40 lcal mol^(-1)`