The rate law for the reaction
`C_(2) H_(4) Br_(2) + 3I^(-) to C_(2) H_(4) + 2Br^(-) + I_(3)^(-)`
is Rate = `k[C_(2)H_(4) Br_(2)][I^(-) ].` The rate of the reaction is found to be `1.1 xx 10^(-4) M//s` when the concentrations of `C_(2) H_(4) Br_(2)` and `I^(-)` are 0.12 M and 0.18 M respectively. Calculate the rate constant of the reaction.

Answer

Rate constant =`k= 5.1 xx 10^(-3) M^(-1) s^(-1)`

Explanation

Given : `C_(2) H_(4)Br_(2) + 3I^(-) to C_(2) H_(4) + 2Br^(-) + I_(3)^(-)`
By rate law ,Rate of reaction `=R = k xx [ C_(2) H_(4) Br_(2) ][I^(-)]`
`R= 1.1 xx 10^(-4) Ms^(-1)`
`[C_(2) H_(4) Br_(2) ]=0.12 M , [I^(-) ]= 0.18 M`
Rate constant = k = ?
`R= k xx [C_(2)H_(4) Br_(2) ] xx [I^(-)]`
`Ms^(-1)`
`:. k= (R)/([C_(2) H_(4) Br_(2) ] xx [I^(-)]) = (1.1 xx 10^(-4))/(underset(M)(0.12)xx underset(M)0.18) = 5.1 xx 10^(-3) M^(-1) s^(-1)`