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The rate constant, the activation energy and the arrhenius paremeter of a chemical reaction at 25^(@) are 3.0times10^(-4)s^(-), 104.4kJ*mol^(-1) and 6.0times10^(14)s^(-1) respectively. The value of the rate constant as T to infty is

Answer»

`2.0times10^(18)s^(-1)`
`6.0times10^(14)s^(-1)`
Infinity
`3.6times10^(30)s^(-1)`

Solution :`T_(2)=T(SAY), T_(1)=25^(@)=298K, E_(a)=104.4kJ*mol^(-1)=104.4times10^(3)J*mol^(-1)`
`k_(1)=3times10^(-4), k_(2)=?`
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`"log"(K_(2))/(3times10^(-4))=(104.4times10^(3)J*mol^(-1))/(2.303times(8.314J*K^(-1)*mol^(-1))[(1)/(298)-(1)/(T)]`
As `T to INFTY, (1)/(T)to 0`
`therefore "log"(K_(2))/(3times10^(-4))=(104.4times10^(3)J*mol^(-1))/(2.303times8.314times298)`
`"log"(k_(2))/(3times10^(-4))=18.297, (k_(2))/(3times10^(-4)=1.98times10^(18)`
`k_(2)=6times10^(14)s^(-1)`


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