Chapter 17, Problem 17.65P

Interpretation Introduction

Interpretation:

The rate law for the reaction described by following equation at $\text{300}\text{k}$ is $\text{rate}\text{of}\text{reaction}\text{=}\text{(5}\text{.0}\text{×}{\text{10}}^{\text{-3}}{\text{M}}^{\text{-1}}\cdot {\text{s}}^{\text{-1}}{\text{)[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}{\text{][I}}^{\text{-}}\text{]}$.  The missing entries in the given table has to be filled.

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

Given, $\text{rate}\text{of}\text{reaction}\text{=}\text{(5}\text{.0}\text{×}{\text{10}}^{\text{-3}}{\text{M}}^{\text{-1}}\cdot {\text{s}}^{\text{-1}}{\text{)[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}{\text{][I}}^{\text{-}}\text{]}$.

For run-1:

  $\begin{array}{l}{\text{[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}\text{=}\text{0}\text{.20}\text{M}{\text{[I}}^{\text{-}}\text{]}\text{=}\text{0}\text{.20}\text{M}\\ \text{Rate}\text{=}{\text{k[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}{\text{[I}}^{\text{-}}\text{]}\\ \text{=}\text{5}\text{.0}\text{×}{\text{10}}^{\text{-3}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{×}\text{0}\text{.20}\text{M}\text{×}\text{0}\text{.20}\text{M}\\ \text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{M}{\text{s}}^{\text{-1}}\end{array}$

For run-2:

  $\begin{array}{l}{\text{[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}\text{=}\text{0}\text{.20}\text{M}{\text{[I}}^{\text{-}}\text{]}\text{=}\text{?}\\ \text{Rate}\text{=}\text{4}\text{.0}\text{×}{\text{10}}^{\text{-4}}{\text{Ms}}^{\text{-1}}\\ \text{Rate}\text{=}{\text{k[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}{\text{[I}}^{\text{-}}\text{]}\\ {\text{[I}}^{\text{-}}\text{]}\text{=}\frac{\text{Rate}}{{\text{k[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}}\\ \text{=}\frac{\text{4}\text{.0}\text{×}{\text{10}}^{\text{-4}}{\text{Ms}}^{\text{-1}}}{\text{5}\text{.0}\text{×}{\text{10}}^{\text{-3}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{×}\text{0}\text{.20}\text{M}}\\ \text{=}\text{0}\text{.4}\text{M}\end{array}$

For run-3:

  $\begin{array}{l}{\text{[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}\text{=}\text{?}{\text{[I}}^{\text{-}}\text{]}\text{=}\text{0}\text{.20}\text{M}\\ \text{Rate}\text{=}\text{8}\text{.0}\text{×}{\text{10}}^{\text{-4}}{\text{Ms}}^{\text{-1}}\\ \text{Rate}\text{=}{\text{k[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}{\text{[I}}^{\text{-}}\text{]}\\ {\text{[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}\text{=}\frac{\text{Rate}}{{\text{k[I}}^{\text{-}}\text{]}}\\ \text{=}\frac{\text{8}\text{.0}\text{×}{\text{10}}^{\text{-4}}{\text{Ms}}^{\text{-1}}}{\text{5}\text{.0}\text{×}{\text{10}}^{\text{-3}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{×}\text{0}\text{.20}\text{M}}\\ \text{=}\text{0}\text{.8}\text{M}\end{array}$

The given table is filled as follows:

Run	${{\text{[C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Br}}_{\text{2}}\text{]}}_{\text{0}}\text{/M}$	${{\text{[I}}^{\text{-}}\text{]}}_{\text{0}}\text{/M}$	$\begin{array}{l}\text{Initial}\text{rate}\text{of}\text{formation}\\ \text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}\text{(g)}}\text{/M}{\text{s}}^{\text{-1}}\end{array}$
1	$\text{0}\text{.20}$	$\text{0}\text{.20}$	$\text{2}\text{.0}\text{×}{\text{10}}^{\text{-4}}$
2	$\text{0}\text{.20}$	$\text{0}\text{.4}$	$\text{4}\text{.0}\text{×}{\text{10}}^{\text{-4}}$
3	$\text{0}\text{.8}$	$\text{0}\text{.20}$	$\text{8}\text{.0}\text{×}{\text{10}}^{\text{-4}}$