Chapter 14, Problem 11PS

The data in the table are for the reaction of NO and O₂ at 660 K.

NO(g) + ½ O₂(g) → NO₂(g)

1. (a) Determine the order of the reaction for each reactant.
2. (b) Write the rate equation for the reaction.
3. (c) Calculate the rate constant.
4. (d) Calculate the rate (in mol/L · s) at the instant when [NO] = 0.015 mol/L and [O₂] = 0.0050 mol/L.
5. (e) At the instant when NO is reacting at the rate 1.0 × 10⁻⁴ mol/L · s, what is the rate at which O₂ is reacting and NO₂ is forming?

Interpretation Introduction

Interpretation:

The order of the reaction for each reactant 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.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

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.

Answer to Problem 11PS

The order of $\text{NO}\text{and}{\text{O}}_{\text{2}}$ is two and one respectively.

Explanation of Solution

The reaction rate of the chemical reaction is given as,

  $\begin{array}{l}\text{Reaction Rate}\text{ = k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\text{,where m, and n are orders of the reactants}\text{.}\\ \\ \text{Given}{\text{reaction: NO}}_{\text{(g)}}{\text{+ 1/2 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{NO}}_{\text{2}}{}_{\text{(g)}}\\ \\ \underset{\_}{\text{Find}\text{order}\text{of}\text{the}\text{reaction:}}\\ \\ \text{Comparing}\text{first}\text{two}\text{experiments}\text{1}\text{and}\text{2,}\\ \\ \text{rate}\text{1=}\text{k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\text{, rate1 = 2}{\text{.5×10}}^{\text{-5}}\text{mol/L}\text{.s}\\ \\ \text{rate}\text{2 = k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\text{, rate2 = 1}{\text{.0×10}}^{\text{-4}}\text{mol/L}\text{.s}\\ \\ \frac{\text{rate2}}{\text{rate}\text{1}}\text{=}\frac{\cancel{\text{k}}{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}}{\cancel{\text{k}}{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}}\\ \\ \frac{\text{1}{\text{.0×10}}^{\text{-4}}\text{mol/L}\text{.s}}{\text{2}{\text{.5×10}}^{\text{-5}}\text{mol/L}\text{.s}}\text{=}\frac{{\left(\text{0}\text{.020}\right)}^{\text{m}}\cancel{{\left(\text{0}\text{.010}\right)}^{\text{n}}}}{{\left(\text{0}\text{.010}\right)}^{\text{m}}\cancel{{\left(\text{0}\text{.010}\right)}^{\text{n}}}}\\ \\ \text{4 = }{\left(\text{2}\right)}^{\text{m}}\\ \\ \boxed{\text{m = 2}}\\ \\ \text{Comparing}\text{last}\text{two}\text{experiments}\text{1}\text{and}\text{3,}\\ \\ \text{rate}\text{1 =}\text{k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\text{,rate 1 = 2}{\text{.5×10}}^{\text{-5}}\text{mol/L}\text{.s}\\ \\ \text{rate}\text{3 = k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\text{,rate 3 = 5}{\text{.0×10}}^{\text{-5}}\text{mol/L}\text{.s}\\ \\ \frac{\text{rate 1}}{\text{rate}\text{3}}\text{=}\frac{\cancel{\text{k}}{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}}{\cancel{\text{k}}{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}}\\ \\ \frac{\text{2}{\text{.5×10}}^{\text{-5}}\text{mol/L}\text{.s}}{\text{5}{\text{.0×10}}^{\text{-5}}\text{mol/L}\text{.s}}\text{=}\frac{\cancel{{\left(\text{0}\text{.010}\right)}^{\text{m}}}{\left(\text{0}\text{.010}\right)}^{\text{n}}}{\cancel{{\left(\text{0}\text{.010}\right)}^{\text{m}}}{\left(\text{0}\text{.020}\right)}^{\text{n}}}\\ \\ \text{0}\text{.5 =}{\left(\text{0}\text{.5}\right)}^{\text{n}}\\ \\ \boxed{\text{n = 1}}\end{array}$

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 $\left[\text{NO}\right]$ is constant and $\left[{\text{O}}_{\text{2}}\right]$ varies, and in vice-versa.  Hence, the order of reactant $\left[\text{NO}\right]$ is two and the order of reactant $\left[{\text{O}}_{\text{2}}\right]$ is one.

Interpretation Introduction

Interpretation:

The rate equation for the reaction has to be written.

Concept Introduction:

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

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

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.

Answer to Problem 11PS

The rate equation is $\text{k }{\left[\text{NO}\right]}^2{\left[{\text{O}}_{\text{2}}\right]}^1$

Explanation of Solution

The reaction rate is given as,

  $\begin{array}{l}\text{Given}{\text{reaction: NO}}_{\text{(g)}}{\text{+ 1/2 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{NO}}_{\text{2}}{}_{\text{(g)}}\\ \\ \text{m = 2;}\text{n = 1}\\ \\ \text{Reaction Rate}\text{ = k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\\ \\ \text{Hence,}\text{the}\text{reaction}\text{rate = k }{\left[\text{NO}\right]}^{\text{2}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{1}}\end{array}$

Hence, Rate equation is $\text{k }{\left[\text{NO}\right]}^2{\left[{\text{O}}_{\text{2}}\right]}^1$

Interpretation Introduction

Interpretation:

The rate constant 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.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

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.

Answer to Problem 11PS

The value of rate constant is ${\text{25 L}}^2{\text{/mol}}^2\text{.s}$

Explanation of Solution

The rate constant is calculated as,

  $\begin{array}{l}\text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^2{\left[{\text{O}}_{\text{2}}\right]}^1\text{.}\\ \text{consider}\text{any one of the experiment,}\\ \text{Experiment 1: }\\ \text{[NO] = 0}\text{.010}\text{mol/L}\\ {\text{[O}}_{\text{2}}\text{] = 0}\text{.010mol/L}\\ \text{Reaction}\text{rate = 2}{\text{.5×10}}^{\text{-5}}\left(\text{mol/L}\text{.s}\right)\\ \\ \text{Therefore,}\\ \text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^2{\left[{\text{O}}_{\text{2}}\right]}^1\\ \text{2}{\text{.5×10}}^{\text{-5}}\left(\text{mol/L}\text{.s}\right)\text{ = k }{\left(\text{0}\text{.010}\right)}^2\left(\text{0}\text{.010}\right)\\ \text{k = }\frac{\text{2}{\text{.5×10}}^{\text{-5}}\text{mol/L}\text{.s}}{\left({\text{1×10}}^{\text{-6}}{\text{mol}}^3{\text{/L}}^3\right)}\\ \boxed{\text{k}\text{=}{\text{25 L}}^2{\text{/mol}}^2\text{.s}}\end{array}$

The rate constant value is obtained as shown above.  By substituting the any one of the concentrations of reactants and the initial rate into the reaction equation obtained at first.

Hence, the value of rate constant is ${\text{25 L}}^2{\text{/mol}}^2\text{.s}$

Interpretation Introduction

Interpretation:

The rate in $\left(\text{mol/L}\text{.s}\right)$ 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.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

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.

Answer to Problem 11PS

The instantaneous rate of the reaction is $\text{2}\text{.8}\times {\text{10}}^{-5}\text{mol/L}\text{.s}$

Explanation of Solution

The rate is calculated as,

  $\begin{array}{l}\text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^2{\left[{\text{O}}_{\text{2}}\right]}^1\text{.}\\ \text{consider}\text{any one of the experiment,}\\ \text{Experiment 1: }\\ \text{[NO] = 0}\text{.015}\text{mol/L}\\ {\text{[O}}_{\text{2}}\text{] = 0}\text{.0050 mol/L}\\ {\text{rate constant, k = 25 L}}^2{\text{/mol}}^2\text{.s}\\ \\ \text{Therefore,}\\ \text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^2{\left[{\text{O}}_{\text{2}}\right]}^1\\ \text{ = }\left({\text{25 L}}^2{\text{/mol}}^2\text{.s}\right){\left(\text{0}\text{.015}\right)}^2\left(\text{0}\text{.0050}\right)\\ \text{ = 2}\text{.8}\times {\text{10}}^{-5}\text{mol/L}\text{.s}\end{array}$

The instantaneous rate of the reaction is $\text{2}\text{.8}\times {\text{10}}^{-5}\text{mol/L}\text{.s}$

Interpretation Introduction

Interpretation:

The rate at which ${\text{O}}_{\text{2}}$ reacts and ${\text{NO}}_{\text{2}}$ forms has to be given.

Concept Introduction:

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

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

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.

Answer to Problem 11PS

The rate when oxygen reacting is $5.0\times {10}^{-5}\text{mol/L}\text{.s}$ and the rate when ${\text{NO}}_{\text{2}}$ forming is $1.0\times 10^{-4}mol/L.s$

Explanation of Solution

The rate is calculated as,

  $\begin{array}{l}\text{Given}{\text{reaction: NO(g)+ 1/2 O}}_{\text{2}}\text{(g)}\to {\text{NO}}_{\text{2}}\text{(g)}\\ \text{Rate}\text{of}\left[\text{NO}\right]\text{forming,}\left(\frac{\text{Δ}\left[\text{NO}\right]}{\text{Δt}}\right)\text{=}\text{1}{\text{.0×10}}^{\text{-4}}\text{mol/L}\text{.s}\\ \text{Rate}\text{of}\left[{\text{O}}_{\text{2}}\right]\text{reacting or disappearance}\text{=}\text{?}\\ \text{Rate}\text{of}\left[{\text{NO}}_{\text{2}}\right]\text{forming }\text{= ?}\\ \text{Known:}\\ \text{Rate}\text{of}\text{reaction}\text{=}\text{- }\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[\text{NO}\right]}{\text{Δt}}\text{ = -}\frac{\text{1}}{\left(\text{1/2}\right)}\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}\text{ =}\text{-}\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}\\ \\ \underset{\_}{\text{Rate}\text{of}\left[{\text{O}}_{\text{2}}\right]\text{reacting or disappearance}:}\\ \text{Rate}\text{of}\left[{\text{O}}_{\text{2}}\right]\text{reacting or disappearance}\text{=}\text{-}\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}=?\\ \text{1}{\text{.0×10}}^{\text{-4}}\text{mol/L}\text{.s}\text{= -}\frac{\text{1}}{\left(\text{1/2}\right)}\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}\\ \text{Rate}\text{of}\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}\text{=}\frac{\left(\text{1}{\text{.0×10}}^{\text{-4}}\text{mol/L}\text{.s}\right)}{2}=5.0\times 10^{-5}mol/L.s\\ \\ \underset{\_}{\text{Rate}\text{of}\left[{\text{NO}}_{\text{2}}\right]\text{forming}:}\\ \text{Rate}\text{of}\left[{\text{NO}}_{\text{2}}\right]\text{forming}\text{=}\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}=?\\ \text{1}{\text{.0×10}}^{\text{-4}}\text{mol/L}\text{.s}\text{= +}\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}\\ \text{Rate}\text{of}\frac{\text{Δ}\left[{\text{NO}}_{\text{2}}\right]}{\text{Δt}}\text{=}1.0\times 10^{-4}mol/L.s\end{array}$

The rate when oxygen reacting is $5.0\times {10}^{-5}\text{mol/L}\text{.s}$ and the rate when ${\text{NO}}_{\text{2}}$ forming is $1.0\times 10^{-4}mol/L.s$