The energy of achivation for the formation of hydrogen iodide is 150 kJ `mol^(-1)`. The rate constant of this reaction at 673 K is `2.3xx10^(-3)`. Calculate the rate constant 773 K.

Explanation

Formula :
`"log"(k_(2))/(k_(1))=(E_(a)(T_(2)-T_(1)))/(2.303 R T_(1)T_(2))`
Given :
Energy of activation, E = 150 kJ = 150000 J
Temperatures : `T_(1)=673 K , T_(2)=773 K`
Rate constant : `k_(1)=2.3xx10^(-3)`
Gas constant : `R = 8.314 K j^(-1) mol^(-1)`.
`"log"(k_(2))/(k_(1))=(k_(2))/(2.3xx10^(-3))=(150000(773-673))/(2.303xx8.314xx673xx773)`
`=(15000000)/(2.303xx8.314xx673xx773)`
`"log" (k_(2))/(2.3xx10^(-3))=1.505`
`(k_(2))/(2.3xx10^(-3))=` Anti log `1.5905 = 32`
`therefore k_(2)=2.3xx10^(-3)xx32 = 7.36xx10^(-2)`
`k_(2)=7.36xx10^(-2)`