(a) Derive an integrated rate equation for the rate constant of a first order reaction.
(b) The specific reaction rate of a reaction quadruples when temperature changes from `30^(0)` to `50^(0)`.Calculate the energy of activation of the reaction.
[Given : R = 8.314 `JK^(-1)mol^(-1)]`.

Explanation

(a)Consider a zero order reaction
`R to P`
`-d[R]/dt= -k.dt`
where k is the velocity constant of the first order reaction
`-d[R]/[R]= -k.dt`
Integrating the equation
`intd[R]/[R] = -int k.dt`
In[R]=-kt+I …..(1)
I is constant of integration when t=0 [R]=`[R]_(0)` where `[R]_(0)` is initial concentration of the reactant .
`In[R]_(0)= -Kxx0 +1`
I = `In[R]_(0)`
Substituting I in equation..(1)
`ln[R] = -kt + ln[R](0)`
`k=2.303/t log[R](0)/[R]`
(b)`logk_(2)/k_(1)=E_(a)/2.303 xx R[T_(2)-T_(1)/T_(1)xxT_(2)]`
`log4=E_(a)/2.303 xx 8.314[323-303/(303)(323)]`
`0.6021 = E_(a)/(2.303 xx 8.314) xx 20/303xx323`
`E_(a) = (0.6021 xx 2.303 xx 8.314 xx 303 xx 323)/20`