Chapter 14, Problem 12PS

The reaction

2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g)

was studied at 904 °C, and the data in the table were collected.

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 for the reaction.
4. (d) Find the rate of appearance of N₂ at the instant when [NO] = 0.350 mol/L and [H] = 0.205 mol/L.

Interpretation Introduction

Interpretation: The order of the reaction for each reactant 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 12PS

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

Explanation of Solution

The order of the reaction is calculated as,

  $\begin{array}{l}\text{Reaction Rate}\text{ = k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{n}}\text{,where m, and n are orders of the reactants}\text{.}\\ \\ \text{Given}\text{reaction: 2 NO}{}_{\text{(g) }}{\text{+ 2 H}}_{\text{2}}{}_{\text{(g)}}{\text{®N}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+ 2 H}}_{\text{2}}{\text{O }}_{\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{H}}_{\text{2}}\right]}^{\text{n}}\text{, rate1 = 0}\text{.136 mol/L}\text{.s}\\ \\ \text{rate}\text{2 = k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{n}}\text{, rate2 = 0}\text{.0339 mol/L}\text{.s}\\ \\ \frac{\text{rate1}}{\text{rate}\text{2}}\text{=}\frac{\cancel{\text{k}}{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{n}}}{\cancel{\text{k}}{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{n}}}\\ \\ \frac{\text{0}\text{.136mol/L}\text{.s}}{\text{0}\text{.0339mol/L}\text{.s}}\text{=}\frac{{\left(\text{0}\text{.420}\right)}^{\text{m}}\cancel{{\left(\text{0}\text{.122}\right)}^{\text{n}}}}{{\left(\text{0}\text{.210}\right)}^{\text{m}}\cancel{{\left(\text{0}\text{.122}\right)}^{\text{n}}}}\\ \\ \text{4 = }{\left(\text{2}\right)}^{\text{m}}\\ \\ \boxed{\text{m = 2}}\\ \\ \text{Comparing}\text{last}\text{two}\text{experiments}\text{2and}\text{3,}\\ \\ \text{rate}\text{2 =}\text{k }{\left[\text{NO}\right]}^{\text{m}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{n}}\text{,rate 2 = 0}\text{.0339 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 = 0}\text{.0678 mol/L}\text{.s}\\ \\ \frac{\text{rate 2}}{\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{0}\text{.0339mol/L}\text{.s}}{\text{0}\text{.0678mol/L}\text{.s}}\text{=}\frac{\cancel{{\left(\text{0}\text{.210}\right)}^{\text{m}}}{\left(\text{0}\text{.122}\right)}^{\text{n}}}{\cancel{{\left(\text{0}\text{.210}\right)}^{\text{m}}}{\left(\text{0}\text{.244}\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 $\left[{\text{H}}_{\text{2}}\right]$ varies, and in vice-versa.  Hence, the order of reactant $\left[\text{NO}\right]$ is two and the order of reactant $\left[{\text{H}}_{\text{2}}\right]$ is one.

Interpretation Introduction

Interpretation: The rate of 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 12PS

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

Explanation of Solution

The reaction rate:

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

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

Interpretation Introduction

Interpretation: The order of the reaction for each reactant 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 12PS

The value of rate constant is $\text{6}{\text{.32 L}}^2{\text{/mol}}^2\text{.s}$

Explanation of Solution

The value of rate constant is calculated as,

  $\begin{array}{l}\text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^{\text{2}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{1}}\text{.}\\ \\ \text{Consider}\text{any one of the experiment,}\\ \\ \text{Experiment 1: }\\ \\ \text{[NO] = 0}\text{.420}\text{mol/L}\\ \\ {\text{[H}}_{\text{2}}\text{] = 0}\text{.122 mol/L}\\ \\ \text{Reaction}\text{rate = 0}\text{.136}\left(\text{mol/L}\text{.s}\right)\\ \\ \text{Therefore,}\\ \\ \text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^{\text{2}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{1}}\\ \\ \text{0}\text{.136}\left(\text{mol/L}\text{.s}\right)\text{ = k }{\left(\text{0}\text{.420}\right)}^{\text{2}}\left(\text{0}\text{.122}\right)\\ \\ \text{k = }\frac{\text{0}\text{.136 mol/L}\text{.s}}{\left(\text{0}{\text{.02152 mol}}^{\text{3}}{\text{/L}}^{\text{3}}\right)}\\ \\ \boxed{\text{k}\text{=}\text{6}{\text{.32 L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s}}\end{array}$

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 $\text{6}{\text{.32 L}}^2{\text{/mol}}^2\text{.s}$

Interpretation Introduction

Interpretation:

The rate of appearance of ${\text{N}}_{\text{2}}$ at the instant 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.

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

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 12PS

The rate when ${\text{N}}_{\text{2}}$ forming is $\text{0}\text{.136 mol/L}\text{.s}$

Explanation of Solution

Calculate the reaction rate of any one experiment:

  $\begin{array}{l}\text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^{\text{2}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{1}}\\ \\ \text{Consider}\text{any one of the experiment,}\\ \\ \text{Experiment 1: }\\ \\ \text{[NO] = 0}\text{.420}\text{mol/L}\\ \\ {\text{[H}}_{\text{2}}\text{] = 0}\text{.122 mol/L}\\ \\ \text{rate constant, k = 6}{\text{.3 L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s}\\ \\ \text{Therefore,}\\ \text{Reaction}\text{rate = k }{\left[\text{NO}\right]}^{\text{2}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{1}}\\ \\ \text{ = }\left(\text{6}{\text{.3 L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s}\right){\left(\text{0}\text{.420}\right)}^{\text{2}}\left(\text{0}\text{.122}\right)\\ \\ \text{ = 0}\text{.136 mol/L}\text{.s}\end{array}$

Calculate the rate of appearance of nitrogen:

  $\begin{array}{l}\text{Given}{\text{reaction: 2 NO}}_{\text{(g)}}{\text{ + 2 H}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{N}}_{\text{2}}{}_{\text{(g)}}\text{+ }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \\ \text{Rate}\text{of}\left[\text{NO}\right]\text{forming,}\left(\frac{\text{Δ}\left[\text{NO}\right]}{\text{Δt}}\right)\text{= 0}\text{.136 mol/L}\text{.s}\\ \\ \text{Rate}\text{of}\left[{\text{N}}_{\text{2}}\right]\text{forming }\text{= ?}\\ \\ \text{Known:}\\ \\ \text{Rate}\text{of}\text{reaction}\text{=}\text{- }\frac{\text{1}}{\text{2}}\frac{\text{Δ}\left[\text{NO}\right]}{\text{Δt}}\text{ = -}\frac{\text{1}}{\left(\text{2}\right)}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\right]}{\text{Δt}}\text{ =}\text{+}\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}\\ \\ \underset{\_}{\text{Rate}\text{of}\left[{\text{N}}_{\text{2}}\right]\text{appearance:}}\\ \\ \text{Rate}\text{of}\left[{\text{N}}_{\text{2}}\right]\text{appearance }\text{,}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}\text{=}\text{?}\\ \\ \text{0}\text{.136 mol/L}\text{.s}\text{=}\text{+}\frac{\text{1}}{\text{1}}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}\\ \\ \text{Rate}\text{of}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δt}}\text{=}\text{0}\text{.136 mol/L}\text{.s}\end{array}$

The rate ${\text{N}}_{\text{2}}$ forming is $\text{0}\text{.136 mol/L}\text{.s}$