It is generalized that a `10^(@)C` increases in temperature casues the rate of reaction to double. Applied to a reaction at `295 K`, what is the value of `E_(a)`?

Options

A. `120 "kcal mol"^(-1)`
B. `1200 "kcal mol"^(-1)`
C. `1.2 "kcal mol"^(-1)`
D. `12 "kcal mol"^(-1)`

Answer

D

Explanation

Only reaction whose `E_(a)` falls in the range of `50-55 kJ mol^(-1)` or `12-13 kcal mol^(-1)` are found to double their rate for `10^(@)C` rise in temperature i.e., form `298` to `308 K`.
Alternatively
Use Arrheinius equation
`T_(1) = 295 K, T_(2) = 305 K`.
`log.(k_(2))/(k_(1)) = (E_(a))/(2.3 xx 2 xx 10^(-3))((10)/(295 xx 305))`
`E_(a) = (log(2) xx 2.3 xx 2 xx 10^(-3) xx 295 xx 305)/(10)`
`E_(a) = (0.3 xx 4.6 xx 295 xx 305)/(10^(4))`
`= 12.4 kcal mol^(-1)`
`~~ 12 kcal mol^(-1)`