The specific reaction rate of a reaction quadruples when temperature changes from 30°C to 50°C. Calculate the energy of activation of the reaction [Given : `R = 8.314 "JK"^(-1) "mol"^(-1)` ]

Explanation

(a) In first order reaction the rate of the reaction is proportional to first power of the concentration of the reactants.
Let us consider the following first order reaction :
`R to P`
The rate is given by :
`(-d[R])/(dt) prop [R]`
`(-d[R])/(dt) = k[R]`
where k is rate constant
`(-d[R])/([R]) = kdt`
multiply is given by -ve sign
`(d[R])/([R]) = - kdt`
on integration `int(d[R])/([R]) = -k int dt`
`In [R] = -kt + I`
Whent `t = 0, R = [R]_(0)`, where `[R]_0` = Initial concentration of the reactant
`:. [R]_(0) = -k(0) + I " " .......(1)`
`I = ln [R]_(0)`
Substituting the value of I in the above eqn,
`ln [R] = -kt + In [R]`
`kt = In[R]_0 - In [R]`
`kt-ln [R]_(0) - ln [R]`
`kt = ln ([R]_0)/([R])`
`k = 1/t ln ([R]_0)/([R])`
Changing into common log
`k = (2.303)/(t) "log" ([R]_0)/([R])" ".........(2)`
(b) `"log" (k_2)/(k_1) = (E_a)/(2.303 R) [(T_2 - T_1)/(T_1T_2)]`
Given : `(k_2)/(k_1) = 4, T_1 = 30^@C = 30 + 273 = 303 K, T_2 = 50^@C = 50 + 273 = 273 = 323 K , R = 8.314 K^(-1) mol^(-1)`
`"log" 4 = (E_a)/(2.303 xx 8.314) [(323 - 303)/(303 xx 323)] `
`E_a = ("log " 4 xx 2.303 xx 8.314 xx 303 xx 323)/(20 ) = 56,414 J//mol = 56.414 KJ//mol` .