The rate equation for the overall equation is:
`H_(3)AsO_(4) (aq)+2H^(o+) + 3I^(Θ)(aq) underset(k_(r ))overset(k_(f))hArr H_(3)AsO_(3)(aq)+I_(3)^(Θ)(eq)+H_(2)O(l)`
`(d[I_(3)^(ɵ)])/(dt)=k_(f)[H_(3)AsO_(4)][I^(ɵ)][H^(o+)]-k_(f)([H_(3)AsO_(3)][I^(ɵ)3])/{[I^(ɵ)]^(2)[H^(o+)])`
where `k_(f)= 4.7xx10^(-4)L^(2)mol^(-2)mi n^(-1)` and
`k_(r )= 3xx10^(-3)mol^(2)L^(-2)min^(-1)`
Determine the euilibrium constant for this reaction form the rate constants.

Explanation

`(d[I_(3)^(ɵ)])/(dt)=k_(f)[H_(3)AsO_(4)][I^(ɵ)][H^(o+)]-k_(f)([H_(3)AsO_(3)][I^(ɵ)3])/{[I^(ɵ)]^(2)[H^(o+)])=0`
`k=([H_(3)AsO_(3)][I_(3)^(ɵ)])/([H_(3)AsO_(3)][I^(ɵ)]^(3)[H^(o+)]^(2))`
`=(k_(f))/(k_(b))=(4.7xx10^(-4))/(3xx10^(-3))= 0.156`