Rate constant of reaction at 300 K and 400 K are `0.0345 S^(-1)` and `0.1365 S^(-1)` respectively. Calculate the activation energy for the reaction.
[Given : `R = 8.314 J K^(-1) mol^(-1)`]

Explanation

(A) `logK_(2)/K_(1) = E_(a)/2.303R[T_(2)-T_(1)/T_(1)T_(2)]`
`log0.136/0.034 = E_(a)/2.303 xx 8.314[400-300/300xx400]`
`E_(a) = 13.8 kJ//mol`
(b) Consider a zero order reaction
`R rarr P`
Zero order reaction is one in which rate is proportional to zeroth power of reactant concentration.
According to rate law for zero order reaction,
`Rate = k[R]^(0)` Rate = `k xx 1` where k is rate constant or velocity constant.
But rate is defined by,
Rate `= -d[R]/dt :. -d[R]/dt = k`
Rearranging the equation, we get `d[R] = -kdt`
On integration
` int d[R] = -kint dt [R] = -kt + I`
where I is called integration constant
To get, I, when t = 0,`[R] = [R]_(0)` the initial concentration of the reactant
`[R]_(0) = -k xx 0 + I I = [R]_(0)`
Substitute I value in equation(1), we get
`[R] = -kt + [R]_(0) [R]_(0)-[R] = kt`
`k = [R]_(0) - [R]/t`
On rearranging, `[R] = -kt + [R]_(0)`
This equation is in y = mx + c form and suggests that
A plot of [R] versus .t. gives a straight line with a slope = -k and y-intercept =` [R]_(0)`