The activation energy of a reaction can be determined from the slope of which of the following graphs ?

Options

A. `(T)/(In k)vs. (1)/(T)`
B. `In k vs. T`
C. `(In k)/(T)vs. T`
D. `In k vs. (1)/(T)`

Answer

D

Explanation

According to the Arrhenius equation: ltbr. `K=Ae^(-E_(a)//RT)`
where `A` is the Arrhenius factor ot or the frequency factor. It is also called pre-exponetial factor. It is a constant specific to ia particular temperature. `R` is gas content and `E_(a)` is activation energy measured in joules/mole `(jmol^(-1))`
Taking natural logarithm of both sides of Arrhenius equation gives
`{:(lnk=,lnA,-,(E_(a))/R,.1/T),(darr,darr,+,darr,darr),(y=,c,+,m,x):}`
It corresponds to an equation of straight line `(y=c+mx)` .
Therefore, the plot of `Inkvs.1//T` gives a straight line as shown below: