The rate of a reaction gets doubled when the temperature changes from `7^(@)C` to `17^(@)C`. By what factor will it change for the temperature change from `17^(@)C` to `27^(@)C` ?

Options

A. `1.81`
B. `1.71`
C. `1.91`
D. `1.76`

Answer

C

Explanation

`"log"(k_(2))/(k_(1)) = (E_(a))/(2.303R)[(1)/(T_(1)) - (1)/(T_(2))]`
`log 2 = (E_(a))/(2.303R)[(T_(2) - T_(1))/(T_(1)T_(2))]`
`0.3 = (E_(a))/(2.303R) xx (10)/(280 xx 290)`
`Ea = 0.3 xx 2.303 xx 280 xx 29 xx R` …(1)
Now again
`"log"(k_(2))/(k_(1)) = (E_(a))/(2.303R)[(T_(2) - T_(1))/(T_(1)T_(2))]`
`"log"(k_(2))/(k_(1)) = (0.3 xx 2.303 xx 280 xx 29 R)/(2.303 R)[(10)/(290 xx 300)] = 1.905`