Chapter 16, Problem 16.20P

(a)

Interpretation Introduction

Interpretation:

Average rate over the entire experiment 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:

$\text{Given}\text{reaction: NOBr}\text{(g)}\stackrel{}{\to }\text{NO(g)}\text{+}\frac{1}{2}{\text{Br}}_2\text{(g)}$

The average rate of the chemical reaction:

$\begin{array}{l}\text{Reaction Rate}\text{ = - }\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[reactants]}}{\text{Δt}}\text{ = +}\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[products]}}{\text{Δt}}\\ \\ \text{Reaction Rate}\text{ = -}\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}------------\left(1\right)\end{array}$

$\text{Δ}\left[\text{NOBr}\right]\text{=}\left(\text{0}\text{.0033}\right)\text{-}\left(\text{0}\text{.0100}\right)\text{=}-\text{0}\text{.0067}\text{mol/L}$;$\text{Δt}\text{=}\left(10.0\sec \right)-\left(0.00\sec \right)=10.0\sec$

Substituting in the equation (1) as follows,

$\begin{array}{l}\text{Reaction Rate}\text{ = - }\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}\\ \\ \text{=}\text{- }\left(\frac{\text{-}\left(\text{0}\text{.0067}\right)}{\text{10}\text{.0}}\right)\text{=}\text{0}\text{.00067}\text{=}6.7\times 10^{-4}mol/L.s\text{.}\end{array}$

Therefore, the average rate over the entire experiment was $6.7\times 10^{-4}mol/L.s.$

(b)

Interpretation Introduction

Interpretation:

Average rate between $\text{2}\text{.0}\text{and}\text{4}\text{.0}\text{sec}$ 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:

$\text{Given}\text{reaction: NOBr}\text{(g)}\stackrel{}{\to }\text{NO(g)}\text{+}\frac{1}{2}{\text{Br}}_2\text{(g)}$

The average rate of the chemical reaction:

$\begin{array}{l}\text{Reaction Rate}\text{ = - }\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[reactants]}}{\text{Δt}}\text{ = +}\frac{\text{1}}{\text{coefficient}}\frac{\text{Δ[products]}}{\text{Δt}}\\ \\ \text{Reaction Rate}\text{ = - }\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}------------\left(1\right)\end{array}$

$\text{Δ}\left[\text{NOBr}\right]\text{=}\left(\text{0}\text{.0055}\right)\text{-}\left(\text{0}\text{.0071}\right)\text{=}-\text{0}\text{.0016}\text{mol/L}$;$\text{Δt}\text{=}\left(4.0\sec \right)-\left(2.00\sec \right)=2.0\sec$

Substituting in the equation (1) as follows,

$\begin{array}{l}\text{Reaction Rate}\text{ = -}\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}\\ \\ \text{=}\text{- }\left(\frac{\text{-}\left(\text{0}\text{.0016}\right)}{\text{2}\text{.0}}\right)\text{=}\text{0}\text{.0008}\text{=}8.0\times 10^{-4}mol/L.s\text{.}\end{array}$

Therefore, the average rate between $\text{2}\text{.0}\text{and}\text{4}\text{.0}\text{sec}$ was $8.0\times 10^{-4}mol/L.s.$

(c)

Interpretation Introduction

Interpretation:

Initial reaction rate has to be calculated using graphical method.

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

Figure.1

Using the given values, the graph has been plotted as shown above.

The initial reaction rate is calculated as shown below,

$\begin{array}{l}\text{Reaction Rate}\text{ = }\frac{\text{Δy}}{\text{Δx}}=\text{-}\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}\\ \\ \text{=}\text{- }\left(\frac{\left(\text{0}\text{.0040 - 0}\text{.0100}\right)}{\left(4.0-0.0\right)}\right)\text{=}1.5\times 10^{-3}mol/L.s\text{.}\end{array}$

Therefore, the initial reaction rate obtained as $1.5\times 10^{-3}mol/L.s.$

(d)

Interpretation Introduction

Interpretation:

Rate of reaction at $\text{7}\text{.0}\text{sec}$ has to be calculated using the graphical method.

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 reaction rate at $\text{7}\text{.0}\text{sec}$ is calculated as shown below,

$\begin{array}{l}\text{Reaction Rate}\text{ = }\frac{\text{Δy}}{\text{Δx}}=\text{-}\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}\\ \\ \text{=}\text{- }\left(\frac{\left(\text{0}\text{.0030 - 0}\text{.0050}\right)}{\left(11.0-4.0\right)}\right)\text{=}2.857\times 10^{-4}mol/L.s\text{.}\end{array}$

Therefore, the reaction rate at $\text{7}\text{.0}\text{sec}$ obtained as $2.9\times 10^{-4}mol/L.s.$

(e)

Interpretation Introduction

Interpretation:

Particular time the instantaneous rate equal the average rate over the entire experiment has to be determined.

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 average between $\text{t}\text{=}\text{3}\text{.0}\text{sec}\text{and}\text{t = 5 sec}$ is calculated as shown below,

$\begin{array}{l}\text{Reaction Rate}\text{ = }\frac{\text{Δy}}{\text{Δx}}=\text{-}\frac{\text{Δ}\left[\text{NOBr}\right]}{\text{Δt}}\\ \\ \text{=}\text{- }\left(\frac{\left(\text{0}\text{.0050 - 0}\text{.0063}\right)}{\left(5.0-3.0\right)}\right)\text{=}6.5\times 10^{-4}mol/L.s\text{.}\end{array}$

Rate of reaction at $\text{t}\text{=}\text{4 sec}$ is approximately $\text{6}{\text{.5×10}}^{\text{-4}}\text{mol/L}\text{.s}$, hence the rates are equal at about $\text{t}\text{=}\text{4 sec}$