The activation energy for a hypothetical reaction `A rarr X` is `12.49 kcal mol^(-1)`. If temperature is raised to `305` form `295 K`, the reaction rate increased by `0.002 kcal L^(-1) mol^(-1)` is almost equal to

Options

A. `approx60%`
B. `approx 50%`
C. `approx 100%`
D. unpredictable

Answer

C

Explanation

`E_(a)=12.49"kcal mol"^(-1)`
`T_(1)=295K`
`T_(2)=305K`
`R = 0.002"kcal K"^(-1)"mol"^(-1)`
`log.(k_(2))/(k_(1))=(E)/(2.303K)[(T_(1)-T_(1))/(T_(1)T_(2))]`
`log.(k_(2))/(k_(1))=(12.49)/(2.303xx0.002)[(305-295)/(305xx295)]`
`(k_(2))/(k_(1))="antilog"0.306=2.02`
`k_(2)=2.02k_(1)`
`therefore` Increase in rate `=(2.02-1)=1.01k_(1)`
`%` Increase in rate `=(1.01k_(1)xx100)/(k_(1))=101%`
`~~100%`