Chapter 17, Problem 17.64P

Interpretation Introduction

Interpretation:

The reaction rate law for formation of bromide ion has to be calculated.

Concept Introduction:

Rate law:  The mathematical equation that relates the concentrations of the reactants to

the rate of reaction is called the rate law of a reaction.  An example of rate law of thermal decomposition of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is given by

    ${\text{rate of reaction = k[N}}_{\text{2}}{\text{O}}_{\text{5}}{\text{]}}^{\text{x}}$

The proportionality constant, k, in a rate law is called the rate constant of the reaction.

Explanation of Solution

The order of reaction with respect to ${\text{BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}\text{,}\text{and}{\text{H}}^{\text{+}}{}_{\text{(aq)}}$ are calculated as,

    $\begin{array}{l}\frac{{\text{(rate of reaction)}}_{\text{0}}\text{ for run 2}}{{\text{(rate of reaction)}}_{\text{0}}\text{ for run 3}}\text{=}\frac{\text{k[}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z\text{for run2}}{\text{k[}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z\text{for run3}}\\ \\ \frac{\text{7}\text{.56 }\text{×}{\text{10}}^{-4}}{\text{5}\text{.40}\text{×}{\text{10}}^{-4}}\text{=}\frac{{\text{(0}\text{.14)}}^{\text{x}}{\text{(0}\text{.18)}}^{\text{y}}{(\text{0}\text{.10})}^z}{{\text{(0}\text{.10)}}^{\text{x}}{\text{(0}\text{.18)}}^{\text{y}}{(\text{0}\text{.10})}^z}\\ \\ \text{(1}\text{.4)}\text{=}{\text{(1}\text{.4)}}^{\text{x}}\\ \\ \text{x}\text{=}\text{1}\text{Order}\text{of}\text{reaction}\text{with}\text{respect}\text{to}{\text{I}}^{\text{-}}{}_{\text{(aq)}}\text{is}\text{one}\\ \\ \frac{{\text{(rate of reaction)}}_{\text{0}}\text{ for run 2}}{{\text{(rate of reaction)}}_{\text{0}}\text{ for run 3}}\text{=}\frac{\text{k[}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z\text{for run}\text{2}}{\text{k[}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z\text{for run}\text{3}}\\ \\ \frac{3.00\text{×}{\text{10}}^{-4}}{\text{5}\text{.40}\text{×}{\text{10}}^{-4}}\text{=}\frac{{\text{(0}\text{.10)}}^{\text{x}}{\text{(0}\text{.10)}}^{\text{y}}{(\text{0}\text{.10})}^z}{{\text{(0}\text{.10)}}^{\text{x}}{\text{(0}\text{.18)}}^{\text{y}}{(\text{0}\text{.10})}^z}\\ \\ \text{(0}\text{.555)}\text{=}{\text{(0}\text{.555)}}^{\text{y}}\\ \\ \text{y}\text{=}\text{1}\text{Order}\text{of}\text{reaction}\text{with}\text{respect}\text{to}{\text{BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{is}\text{one}\\ \\ \frac{{\text{(rate of reaction)}}_{\text{0}}\text{ for run 3}}{{\text{(rate of reaction)}}_{\text{0}}\text{ for run 4}}\text{=}\frac{\text{k[}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z\text{for run}\text{2}}{\text{k[}\text{,}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z\text{for run}\text{3}}\\ \\ \frac{\text{5}\text{.40}\text{×}{\text{10}}^{-4}}{\text{1}\text{.67}\text{×}{\text{10}}^{-3}}\text{=}\frac{\text{(0}\text{.10)(0}\text{.18)}{(\text{0}\text{.10})}^z}{\text{(0}\text{.31)(0}\text{.18)}{(\text{0}\text{.20})}^z}\\ \\ \text{(1)}\text{=}{\text{(0}\text{.5)}}^z\\ \\ z\text{=}0\text{Order}\text{of}\text{reaction}\text{with}\text{respect}\text{to}{\text{H}}^{+}{}_{\text{(aq)}}\text{is}\text{zero}\\ \\ \text{Overall}\text{order}\text{of}\text{reaction}\text{=}\text{1}\text{+}\text{1}+0\text{=}2\end{array}$

The rate law of reaction and rate constant are given by

    $\begin{array}{l}\text{k}\text{=}\frac{{\text{(rate}\text{of}\text{reaction)}}_{\text{o}}}{{{\text{[I}}^{\text{-}}{}_{\text{(aq)}}\text{]}}_{\text{o}}^x{{\text{[BrO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}_{\text{0}}^y{{\text{[H}}^{+}{}_{\text{(aq)}}\text{]}}_o^z}\text{=}\frac{\text{3}\text{.0 ×}{\text{10}}^{\text{-4}}}{\text{(0}\text{.10)(0}\text{.10)(0}{\text{.10)}}^0}\text{=}\text{3}\text{.0 ×}{\text{10}}^{\text{-2}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \\ Ratelaw:\text{rate of reaction =}\text{3}\text{.0 ×}{\text{10}}^{\text{-2}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}{{\text{[I}}^{\text{-}}}_{{}_{\text{(aq)}}}^{\text{]}}{{\text{[BrO}}_{\text{3}}}_{{{}^{\text{-}}}_{\text{(aq)}}}^{\text{]}}{{\text{[H}}^{+}}_{{}_{\text{(aq)}}}^{\text{]}}\end{array}$