The first order rate constant for the decomposition of ethyl iodide by the reaction.
`C_(2)H_(2)I (g)to C_(2)H_(4) (g)+ HI (g)at 600K is 1.60 xx10^(-5)s^(-1).` It's energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700 K.

Explanation

`k_(1)=1.60xx10^(-5)sec ^(-1)" "T_(1)=600K`
`k_(2)=?" "T_(2)=700K`
`E_(a)=209kJ mol ^(-1)=209000J mol ^(-1)`
`log ""(k_(2))/(k_(1))=(E_(a))/(2303.R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log ""(k_(2))/(1.6xx10^(-5))=(209000)/(2.303xx8.314)xx(100)/(600xx700)=(209xx10^(5))/(8041799.64)=(209xx10^(5))/(80.418xx10^(5))`
`log ""(k_(2))/(1.6xx10^(-5))=3.971xx10^(2)`
`k_(2) =1.6xx10^(5)xx3.971xx10^(2)`
`k_(2)=6.35xx10^(-3)sec ^(-1).`