Question

Which of the following expressions is correct for the rate of reaction given below?

\[\ce{5Br^- (aq) + BrO3^- (aq) + 6H^+ (aq) -> 3Br2 (aq) + 3H2O(l)}\]

Options

• `(∆["Br"^-])/(∆"t") = 5 (∆["H"^+])/(∆"t")`

• `(∆["Br"^-])/(∆"t") = 6/5 (∆["H"^+])/(∆"t")`

• `(∆["Br"^-])/(∆"t") = 5/6 (∆["H"^+])/(∆"t")`

• `(∆["Br"^-])/(∆"t") = 6 (∆["H"^+])/(∆"t")`

Answer 1

`(∆["Br"^-])/(∆"t") = 5/6 (∆["H"^+])/(∆"t")`

Explanation:

Rate law expression for the above equation can be written as

We can rewrite the rate law expression for the given equation

\[\ce{5Br- (aq) + BrO^{-}3 (aq) + 6H+ (aq) -> 3Br2 (aq) + 3H2O (l)}\]

`therefore - 1/5 (Delta ["Br"^-])/(Delta "t") = - (Delta ["BrO"_3^-])/(Delta "t") = - 1/6 (Delta ["H"^+])/(Delta "t") = 1/3 (Delta["Br"_2])/(Delta "t")`

`therefore - (Delta ["Br"^-])/(Delta "t") = - (Delta ["BrO"_3^-])/(Delta "t") = - 5/6 (Delta ["H"^+])/(Delta "t")`

`therefore (Delta ["Br"^-])/(Delta "t") = 5/6 (Delta ["H"^+])/(Delta "t")`