The rate of reaction triples when temperature changes form `20^(@)C` to `50^(@)C`. Calculate the energy of activation for the reaction `(R= 8.314JK^(-1)mol^(-1))`.

Options

A. `181.327J mol^(-1)`
B. `428.141 J mol^(-1)`
C. `32.4321 kJ mol^(-1)`
D. `28.8118 kJ mol^(-1)`

Answer

D

Explanation

Arrhenius equation is given by, `"log"_(10) (K_(2))/(K_(1))= (E_(a))/(2.303 xx R) [(T_(2)-T_(1))/(T_(1)T_(2))]`
Given, `(K_(2))/(K_(1))= 3, R=-8.314 JK^(-1) mol^(-1)`
`T_(1)= 20+ 273 = 293K and T_(2)= 50 + 273 = 323K`
Substituting the given values in Arrhenius equation, `log_(10)3= (E_(a))/(8.314 xx 2.303) [(323- 293)/(323 xx 293)]`
`E_(a)= (2.303 xx 8.314 xx 323 xx 293 xx 0.477)/(30)`
`=28811.8 J mol^(-1)`
`=28.8118kJ mol^(-1)`