For the reaction:
`[Cr(H_(2)O)_(6)]^(3) + [SCN^(ɵ)] rarr [Cr(H_(2)O)_(5)NCS]^(2+)H_(2)O`
The rate law is `r = k[Cr(H_(2)O)_(6)]^(3+)[SCN^(ɵ)]`.
The value of `k` is `2.0 xx 10^(-6) L mol^(-1) s^(-1)` at `14^(@)C` and `2.2 xx 10^(-5) L mol^(-1)s^(-1)` at `30^(@)C`. What is the value of `E_(a)` ?

Options

A. `26 "kcal mol"^(-1)`
B. `2.6 "kcal mol"^(-1)`
C. `2600 "kcal mol"^(-1)`
D. `260 "kcal mol"^(-1)`

Answer

A

Explanation

Use Arrhenius equation
`T_(1) = 273 + 14 = 287 K, T_(2) = 273 + 30 = 303 K`
`log.(k_(2))/(k_(1)) = (E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1) xx T_(2))]`
`log.(2.2 xx 10^(-5))/(2.0 xx 10^(-6)) = (E_(a))/(2.303 xx 2 xx 10^(-3)) ((16)/(287 xx 303))`
`E_(a) = (log(11) xx 2.303 xx 2 xx 10^(-3) xx 287 xx 303)/(16)`
`= (1.0414 xx 2.303 xx 2 xx 287 xx 303)/(16000)`
`~~ 26 kcal mol^(-1)`