The rate constant, the activation energy, and the Arrhenius parameter of a chemical reaction at `25^(@)C` are `3.0xx10^(-4)S^(-1), 104.4 KJ mol^(-1)`, and `6.0xx10^(14)S^(-1)`, respectively. The value of the rate constant as `Trarroo` 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) = ?`
`"log"(K_(2))/(K_(1)) = (E_(a))/(2.303 R)[(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)/(298) -(1)/(T)]` As `T rarr oo, (1)/(T) rarr 0`
`:. "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(3xx10^(-4))`
`=6xx10^(14)s^(-1)`
or `K = Ae^(-Ea//RT)`
As `T rarr oo`
`K = A`
`rArr 6 xx 10^(14) sec^(-1)`