Chapter 16, Problem 16.114P

(a)

Interpretation Introduction

Interpretation:

Average rate for each trial has to be determined.

Concept introduction:

Relative rates and stoichiometry: During a chemical reaction, amounts of reactant decrease with time and amounts of products increases.

$\text{Reaction Rate}\text{ = - }\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[reactants]}}{\text{Δt}}\text{ = +}\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[products]}}{\text{Δt}}$

Explanation of Solution

The reaction given is:

$\text{Given}{\text{reaction: I}}_{\text{2}}\text{(aq)}\text{+}{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\text{(aq)}\stackrel{}{\to }{\text{2I}}^{\text{-}}\text{(aq)}\text{+}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{2-}}\text{(aq)}$

The average rate of the chemical reaction:

$\begin{array}{l}\text{Reaction Rate}\text{ = - }\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[reactants]}}{\text{Δt}}\text{ = +}\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[products]}}{\text{Δt}}\\ \\ \text{Reaction Rate}\text{ = - }\frac{{\text{Δ[I}}_2\text{]}}{\text{Δt}}\text{ = -}\frac{\text{1}}{2}\frac{{\text{Δ[S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\text{]}}{\text{Δt}}\end{array}$

Molarity of ${\text{[S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\text{]}$is,

$\text{=}\frac{\left[\left(\text{10}\text{.0}\text{mL}\right)\left(\text{0}\text{.0050}\text{M}\right)\right]}{\text{50}\text{.0}\text{mL}}\text{=}\left(0.0010M\right)S_2{O}_3^{2-}.$

$\text{Reaction Rate}\text{ = - }\frac{{\text{Δ[I}}_{\text{2}}\text{]}}{\text{Δt}}\text{ = -}\frac{\text{1}}{\text{2}}\frac{{\text{Δ[S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\text{]}}{\text{Δt}}\text{=}\frac{\left[\frac{\text{1}}{\text{2}\left(\text{0}\text{.0010}\text{M}\right)}\right]}{\text{time}}\text{=}\frac{\left[\text{0}\text{.00050}\text{M}\right]}{\text{time}}\text{=}\text{rate}$

Hence, the average rates:

For experiment (1): ${\text{Rate}}_{\text{1}}\text{ = }\frac{\left(\text{0}\text{.00050}\text{M}\right)}{\text{29}\text{.0}\text{sec}}\text{=}1.7\times 10^{-5}M.s^{-1}.$

For experiment (2): ${\text{Rate}}_2\text{ = }\frac{\left(\text{0}\text{.00050}\text{M}\right)}{\text{14}\text{.5}\text{sec}}\text{=}3.45\times 10^{-5}M.s^{-1}.$

For experiment (3): ${\text{Rate}}_3\text{ = }\frac{\left(\text{0}\text{.00050}\text{M}\right)}{\text{14}\text{.5}\text{sec}}\text{=}3.45\times 10^{-5}M.s^{-1}.$

Therefore, the average rate for each trial was calculated as shown above.

(b)

Interpretation Introduction

Interpretation:

Reaction order with respect to each reactant has to be found.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

Molarity of $\left[{\text{I}}^{\text{-}}\right]\text{=}\frac{\left[\left(\text{10}\text{.0}\text{mL}\right)\left(\text{0}\text{.20}\text{M}\right)\right]}{\text{50}\text{.0}\text{mL}}\text{=}\left(\text{0}\text{.040}\text{M}\right){\text{I}}^{\text{-}}\left(\text{Experiment}\text{1}\right)$

Molarity of $\left[{\text{I}}^{\text{-}}\right]\text{=}\frac{\left[\left(\text{20}\text{.0}\text{mL}\right)\left(\text{0}\text{.20}\text{M}\right)\right]}{\text{50}\text{.0}\text{mL}}\text{=}\left(\text{0}\text{.080}\text{M}\right){\text{I}}^{\text{-}}\left(\text{Experiment}\text{2}\text{and}\text{3}\right)$

Molarity of $\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]\text{=}\frac{\left[\left(\text{20}\text{.0}\text{mL}\right)\left(\text{0}\text{.10}\text{M}\right)\right]}{\text{50}\text{.0}\text{mL}}\text{=}\left(\text{0}\text{.040}\text{M}\right){\text{I}}^{\text{-}}\left(\text{Experiment}1\text{and}2\right)$

Molarity of $\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]\text{=}\frac{\left[\left(\text{10}\text{.0}\text{mL}\right)\left(\text{0}\text{.10}\text{M}\right)\right]}{\text{50}\text{.0}\text{mL}}\text{=}\left(\text{0}\text{.020}\text{M}\right){\text{I}}^{\text{-}}\left(\text{Experiment}3\right)$

Rate law equation: $\text{Rate}\text{ = k }{\left[{\text{I}}^{\text{-}}\right]}^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}^{\text{n}}$

Let’s find the order of reactant $\left[{\text{I}}^{\text{-}}\right]$ by comparing first and second experiments as follows,

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reaction:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}\text{1}\text{and}\text{2,}\\ \\ \text{rate}\text{1}\text{=}\text{k }{\left[{\text{I}}^{\text{-}}\right]}_{\text{1}}^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_{\text{1}}^{\text{n}}\text{, rate1 = 1}{\text{.7 ×10}}^{\text{-5}}\text{mol/L}\text{.s}\\ \text{rate}\text{2 = k}{\left[{\text{I}}^{\text{-}}\right]}_2^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_2^{\text{n}}\text{, rate2 = 3}\text{.4}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}\\ \\ \frac{\text{rate}2}{\text{rate}1}\text{=}\frac{\text{k}{\left[{\text{I}}^{\text{-}}\right]}_2^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_2^{\text{n}}}{\text{k }{\left[{\text{I}}^{\text{-}}\right]}_{\text{1}}^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_{\text{1}}^{\text{n}}}\\ \\ \frac{\text{3}\text{.4}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}}{\text{1}{\text{.7 ×10}}^{\text{-5}}\text{mol/L}\text{.s}}\text{=}\frac{{\left(0.0800\right)}^m\bcancel{{\left(0.040\right)}^n}}{{\left(0.0400\right)}^m\bcancel{{\left(0.040\right)}^n}}\\ \\ \text{2 = }{\left(2\right)}^m\\ \\ \boxed{\text{m = 1 }}\end{array}$

The order with respect to $\left[{\text{I}}^{\text{-}}\right]$ is FIRST order.

Let’s find the order of reactant $\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]$ by comparing second and third experiments as follows,

$\begin{array}{l}\underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reaction:}}\\ \\ \ast \text{ Comparing}\text{first}\text{two}\text{experiments}\text{2 and}\text{3,}\\ \\ \text{rate}\text{3}\text{=}\text{k }{\left[{\text{I}}^{\text{-}}\right]}_3^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_3^{\text{n}}\text{, rate3 = 3}{\text{.4 ×10}}^{\text{-5}}\text{mol/L}\text{.s}\\ \text{rate}\text{2 = k}{\left[{\text{I}}^{\text{-}}\right]}_2^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_2^{\text{n}}\text{, rate2 = 3}\text{.4}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}\\ \\ \frac{\text{rate}3}{\text{rate 2}}\text{=}\frac{\text{k}{\left[{\text{I}}^{\text{-}}\right]}_3^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_3^{\text{n}}}{\text{k}{\left[{\text{I}}^{\text{-}}\right]}_2^{\text{m}}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}_2^{\text{n}}}\\ \\ \frac{\text{3}\text{.4}\times {\text{10}}^{-5}\text{ mol/L}\text{.s}}{\text{3}{\text{.4 ×10}}^{\text{-5}}\text{mol/L}\text{.s}}\text{=}\frac{\bcancel{{\left(0.040\right)}^m}{\left(0.0200\right)}^n}{\bcancel{{\left(0.040\right)}^m}{\left(0.0400\right)}^n}\\ \\ \text{1 = }{\left(0.5\right)}^m\Rightarrow \ln \left(1\right)=n\ln \left(0.5\right)\\ \\ \boxed{\text{n = 0 }}\end{array}$

The order with respect to $\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]$ is ZERO order.

In order to figure out the reaction equation the order of the reactants needed, which is calculated by comparing any two experiments where the concentration of two reactants are constant and another varies, and in vice-versa. Hence, Rate equation is

$\text{Reaction rate = k }{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}^0$.

The respective order of reactant $\left[{\text{I}}^{-}\right]$ is FIRST ORDER.

The respective order of reactant $\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]$ is ZERO ORDER.

Hence, the reaction rate becomes,

$\text{Reaction rate =}k{\left[I^{-}\right]}^1{\left[S_2{O}_8^{2-}\right]}^0.$

(c)

Interpretation Introduction

Interpretation:

Rate constant of the reaction has to be calculated.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^{\text{p}}\text{,where 'm, n and p' are orders of the reactants}\text{.}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

Considering the experiment (1),

$\text{rate1 = 1}{\text{.7 ×10}}^{\text{-5}}\text{mol/L}\text{.s ;}\left[{\text{I}}^{\text{-}}\right]\text{=0}\text{.0400}\text{mol/L}\text{;}\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]\text{=}\text{0}\text{.040mol/L}\text{.}$

Given rate law: $\text{Reaction rate = k }{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}^0$

Calculating the rate constant of reaction as follows,

$\begin{array}{l}\text{Reaction rate = k }{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2-}}\right]}^0\\ \\ \left(\text{1}{\text{.7 ×10}}^{\text{-5}}\text{mol/L}\text{.s}\right)\text{=}\text{k}{\left(\text{ 0}\text{.0400}\text{mol/L}\right)}^1\\ \\ \text{k}\text{=}\frac{\text{1}{\text{.7 ×10}}^{\text{-5}}\text{mol/L}\text{.s}}{{\left(\text{ 0}\text{.0400}\text{mol/L}\right)}^1}\text{=}42.5\times 10^{-5}L/mol.s.\end{array}$

Therefore, the rate constant of reaction at given concentration is $42.5\times 10^{-5}L/mol.s.$

(d)

Interpretation Introduction

Interpretation:

Rate law of the overall reaction has to be calculated.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^{\text{p}}\text{,where 'm, n and p' are orders of the reactants}\text{.}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

The overall reaction law is obtained by substituting the rate constant value as follows,

$\text{Reaction rate =}k{\left[I^{-}\right]}^1{\left[S_2{O}_8^{2-}\right]}^0.$

The rate constant of reaction at given concentration is $42.5\times 10^{-5}L/mol.s.$

$\text{Reaction rate =}\left(42.5\times 10^{-5}L/mol.s.\right){\left[I^{-}\right]}^1.$