The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is `4 xx 10^(10)" s"^(-1)`. Calculate `k` at 318 `K and E_(a)`.

Explanation

(i) Calculation of activation energy `(E_a)`
For first order reaction, `k= (2.303)/( t) log"" ([R]_0)/( [R])`
At `298K, k_1 = (2.303)/( t) log"" (100)/( 90) " "...(i)`
At `308K, k_2 = (2.303)/( t) log"" (100)/(75)" "...(ii)`
On dividing Eq. (ii) by (i),
`(k_2)/( k_1) = (log (100// 75) )/( log (100//90) ) = (0.1249)/( 0.0458) = 2.73`
According to Arrhenius theory
`log (k_2)/( k_1) = (E_a)/( 2.303 R) xx (T_2 - T_1)/( T_1 T_2)`
`log 2.73 = (E_a)/( 2.303 R) [ (308 - 298)/( 298 xx 308) ]`
`E_a = (0.4361 xx 2.303 xx (8.314" JK"^(-1)" mol"^(-1) ) xx 298xx 308)/(10)`
`E_a = 76640"J mol"^(-1) = 76.640" kJ mol"^(-1)`
(ii) Calculation of rate constant `(k)`
According to Arrhenius equation,
`log k = log A - (E_a)/( 2.303 RT)`
`log k = log (4xx10^(10) )`
`- (76640"J mol"^(-1) )/( 2.303 xx (8.314" J mol"^(-1)" K"^(-1) ) xx (318" K" ) )`
`log k = 10.6021 - 12.5870 = - 1.9849`
`k = " antilog" (-1.9849) = " antilog" (2.0151) `
`= 1.035 xx 10^(-2)" s"^(-1)`
So, `E_a = 76.640" kJ mol"^(-1)`
`k = 1.035 xx 10^(-2)" s"^(-1)`