For the reaction $$\mathrm{BrO}_{3}^{-}+5 \mathrm{Br}^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_{2}+3 \mathrm{H}_{2} \mathrm{O}$$ the experimentally determined rate law is: $$\text { Rate }=k\left[\mathrm{BrO}_{3}^{-}\right]\left[\mathrm{Br}^{-}\right]\left[\mathrm{H}^{+}\right]^{2}$$ (a) What is the order of this reaction with respect to \(\mathrm{Br}^{-}\) ? (b) What is the order of this reaction with respect to \(\mathrm{H}^{+} ?\) (c) What is the overall order of the reaction? (d) What happens to the rate of this reaction when you double the \(\mathrm{H}^{+}\) concentration?

The rate law of the reaction is given by: $$\text{Rate} = k[\mathrm{BrO}_{3}^{-}][\mathrm{Br}^{-}][\mathrm{H}^{+}]^{2}$$ The order of the reaction with respect to \(\mathrm{Br}^{-}\) is represented by the exponent of \(\mathrm{Br}^{-}\) in the equation, which is 1.

In the rate law, $$\text{Rate} = k[\mathrm{BrO}_{3}^{-}][\mathrm{Br}^{-}][\mathrm{H}^{+}]^{2}$$ The order of the reaction with respect to \(\mathrm{H}^{+}\) is represented by the exponent of \(\mathrm{H}^{+}\) in the equation, which is 2.

The overall order of the reaction is the sum of the exponents of all reactants in the rate law: Overall order = Order with respect to \(\mathrm{BrO}_{3}^{-}\) + Order with respect to \(\mathrm{Br}^{-}\) + Order with respect to \(\mathrm{H}^{+}\) In this case, the overall order is: Overall order = 1 (Order with respect to \(\mathrm{BrO}_{3}^{-}\), as it is not mentioned) + 1 (Order with respect to \(\mathrm{Br}^{-}\)) + 2 (Order with respect to \(\mathrm{H}^{+}\)) Overall order = \(1+1+2 = 4\)

When the \(\mathrm{H}^{+}\) concentration is doubled, let the new rate be represented as Rate': Original rate: $$\text{Rate} = k[\mathrm{BrO}_{3}^{-}][\mathrm{Br}^{-}][\mathrm{H}^{+}]^{2}$$ Increased rate: $$\text{Rate}' = k[\mathrm{BrO}_{3}^{-}][\mathrm{Br}^{-}][2\mathrm{H}^{+}]^{2}$$ Divide the increased rate by the original rate: $$\frac{\text{Rate}'}{\text{Rate}} = \frac{k[\mathrm{BrO}_{3}^{-}][\mathrm{Br}^{-}][2\mathrm{H}^{+}]^{2}}{k[\mathrm{BrO}_{3}^{-}][\mathrm{Br}^{-}][\mathrm{H}^{+}]^{2}}$$ After canceling out the constants and other reactants, we are left with: $$\frac{\text{Rate}'}{\text{Rate}} = \left(\frac{2\mathrm{H}^{+}}{\mathrm{H}^{+}}\right)^{2} = \left(2\right)^{2} = 4$$ So when the \(\mathrm{H}^{+}\) concentration is doubled, the rate of the reaction quadruples.