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=298K`,
`E_(a)=104.4"kJ mol"^(-1)=104.4xx10^(3)"J mol"^(-1)` ,
`k_(1)=3xx10^(-4) .k_(2)=?`
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`"log"(k_(2))/(3xx10^(-4))=(104.4xx10^(3)"J mol"^(-1))/(2.303xx(8.314"JK"^(-1)"mol"^(-1)))`
`[(1)/(298)-(1)/(T)]"As" T rarr infty,1//Trarr0`
`:. "log"(k_(2))/(3xx10^(-4))=(104.4xx10^(3))/(2.303xx8.314xx298)`
`"log"(k_(2))/(3xx10^(-4))=18.297`
`(k_(2))/(3xx10^(-4))=1.98xx10^(18)`
or `k_(2)=(1.98xx10^(18))xx(3xx10^(-4))`
`=6 xx 10^(14)`.