Question

For the decomposition of `HI` the following logarithmic plot is shown `: [R=1.98 cal// mol - K]`

The activation energy of the reaction is about
A. `45600 cal`
B. `13500 cal`
C. `24600 cal`
D. `32300 cal`

Answer 1

Correct Answer - 1
`logk=-(E_(a))/(2.303R)(1)/(T)+"constant"=-(E_(a))/(2.303R)xx10^(-3)xx(10^(3))/(T)+"constant"`
Thus slope of graph will be `-(E_(a)xx10^(-3))/(2.303R)= -(4)/(0.4)`
`implies" "E_(a)=2.303xx1.98xx10^(4)=45600cal`