The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at `25^(@)C` are `3.0 xx 10^(-4) s^(-1), 104 "kJ mol"^(-1)` and `6.0 xx 10^(14)s^(-1)` respectively. The value of the rate constant as `T rarr oo` is

Options

A. `2.0 xx 10^(18) s^(-1)`
B. `6.0 xx 10^(14) s^(-1)`
C. Infinity
D. `3.6 xx 10^(30) s^(-1)`

Answer

b

Explanation

`T_(2) = T`(say) , T = `25^(@)C = 298 K`
`E_(a) = 104.4 kJ mol^(-1) = 104.4 xx 10^(3) J mol^(-1)`
`k_(1) = 3 xx 10^(-4), k_(2) = ? implies log (k_(2))/(k_(1)) = (E_(a))/(2.303R) [(1)/(T_(1)) - (1)/(T_(2))]`
log `(k_(2))/(3 xx 10^(-4)) = (104.4 xx 10^(3) J mol^(-1))/(2.303 xx (8.314 J k^(-1) mol^(-1)))`
`[(1)/(298K) - (1)/(T)]` As `T implies oo (1)/(T) to 0`
`therefore` log `(k_(2))/(3 xx 10^(-4)) = (104.4 xx 10^(3) J mol^(-1))/(2.303 xx 8.314 xx 298)`
log `(k_(2))/(3 xx 10^-4) = 18.297 , (k_(2))/(3 xx 10^(-4)) = 1.98 xx 10^(18)` or
`k_(2) = (1.98 xx 10^(18)) xx (3 xx 10^(-4)) = 6 xx 10^(14) s^(-1)`.