Chapter 6, Problem 48E

a.

Interpretation Introduction

Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 2.0 moles of pure $\text{NOCl}$ in a 2.0 L flask is used.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 48E

  $\begin{array}{l}\text{[NOCl] = 1}\text{.0 - 2}\left(0.016\right)\text{ = 0}\text{.97 M}\\ \text{[NO] = 2}\left(1.6\times {\text{10}}^{-2}\right)=3.2\times {\text{10}}^{-2}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }1.6\times {\text{10}}^{-2}\text{ M}\end{array}$

Explanation of Solution

Given:

The reaction:

  $\text{2NOCl(g)}\rightleftharpoons {\text{2NO(g) + Cl}}_2(\text{g)}$ at ${\text{35 }}^{\text{o}}\text{C, K = 1}\text{.6}\times {\text{10}}^{-5}$ .

The concentration of $\text{NOCl}$ is calculated using formula:

  $\text{molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution in L}}$

Substituting the values as:

  $\begin{array}{l}\text{molarity = }\frac{\text{2}\text{.0 mol}}{\text{2}\text{.0 L}}\\ [\text{NOCl}]\text{ = 1 M}\end{array}$

The ICE table for the reaction is:

  $\begin{array}{l}\text{ 2NOCl(g) }\rightleftharpoons {\text{ 2NO(g) + Cl}}_2(\text{g)}\\ \text{Initial (M): 1}\text{.0 0 0}\\ \text{Change (M): -2x +2x +x}\\ \text{Equilibrium (M): 1}\text{.0-2x 2x x}\end{array}$

The expression for the equilibrium constant is:

  $\text{K = }\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$

Substituting the values:

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\text{x}}{{(1.0-2\text{x)}}^2}$

Let the value of x be small so, $1.0-2\text{x}\approx \text{1}\text{.0}$

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\text{x}}{{(1.0\text{)}}^2}$

Solving for x:

  $\begin{array}{l}\text{1}\text{.6}\times {\text{10}}^{-5}\text{ = }\frac{4{\text{x}}^3}{{(1.0\text{)}}^2}\\ {\text{x}}^3\text{ = }\frac{\text{1}\text{.6}\times {\text{10}}^{-5}}{4}\\ \text{x = }\sqrt[3]{\text{4}\times {\text{10}}^{-6}}\\ \text{x = }1.6\times {\text{10}}^{-2}\end{array}$

Thus, the equilibrium concentration of all the species at equilibrium is:

  $\begin{array}{l}\text{[NOCl] = 1}\text{.0 - 2}\left(0.016\right)\text{ = 0}\text{.97 M}\\ \text{[NO] = 2}\left(1.6\times {\text{10}}^{-2}\right)=3.2\times {\text{10}}^{-2}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }1.6\times {\text{10}}^{-2}\text{ M}\end{array}$

Interpretation Introduction

Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 2.0 moles of $\text{NO}$ and 1.0 mole of ${\text{Cl}}_{\text{2}}$ in a 1.0 L flask is used.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 48E

  $\begin{array}{l}\text{[NOCl] = 1}\text{.95 M}\\ \text{[NO] = }5.0\times {\text{10}}^{-2}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }2.5\times {\text{10}}^{-2}\text{ M}\end{array}$

Explanation of Solution

The concentration of $\text{NO}$ and ${\text{Cl}}_2$ is calculated using formula:

  $\text{molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution in L}}$

Substituting the values as:

  $\begin{array}{l}\text{[NO] = }\frac{\text{2}\text{.0 mol}}{\text{1}\text{.0 L}}\\ [\text{NO}]\text{ = 2}\text{.0 M}\end{array}$

  $\begin{array}{l}{\text{[Cl}}_2\text{] = }\frac{\text{1}\text{.0 mol}}{\text{1}\text{.0 L}}\\ [{\text{Cl}}_2]\text{ = 1}\text{.0 M}\end{array}$

So, the amount of NO and Cl₂ is 2.0 M and 1.0 M respectively.

As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation 2.0 M of NOCl. So,

The ICE table for the reaction is:

  $\begin{array}{l}\text{ 2NOCl(g) }\rightleftharpoons {\text{ 2NO(g) + Cl}}_2(\text{g)}\\ \text{Initial (M): 2}\text{.0 0 0}\\ \text{Change (M): -2x +2x +x}\\ \text{Equilibrium (M): 2}\text{.0-2x 2x x}\end{array}$

The expression for the equilibrium constant is:

  $\text{K = }\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$

Substituting the values:

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\text{x}}{{(2.0-2\text{x)}}^2}$

Let the value of x be small so, $2.0-2\text{x}\approx 2.\text{0}$

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\text{x}}{{(2.0\text{)}}^2}$

Solving for x:

  $\begin{array}{l}\text{1}\text{.6}\times {\text{10}}^{-5}\text{ = }\frac{4{\text{x}}^3}{4}\\ {\text{x}}^3\text{ = 1}\text{.6}\times {\text{10}}^{-5}\\ \text{x = }\sqrt[3]{16.0\times {\text{10}}^{-6}}\\ \text{x = }2.5\times {\text{10}}^{-2}\end{array}$

Thus, the equilibrium concentration of all the species at equilibrium is:

  $\begin{array}{l}\text{[NOCl] = 2}\text{.0 - 2}\left(0.025\right)\text{ = 1}\text{.95 M}\\ \text{[NO] = 2}\left(2.5\times {\text{10}}^{-2}\right)=5.0\times {\text{10}}^{-2}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }2.5\times {\text{10}}^{-2}\text{ M}\end{array}$

b.

Interpretation Introduction

Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 1.0 mole of $\text{NOCl}$ and 1.0 mole of $\text{NO}$ in a 1.0 L flask is used.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 48E

  $\begin{array}{l}\text{[NOCl] = 1}\text{.0 - 2}\left(0.000016\right)\text{ = 0}\text{.99 M}\\ \text{[NO] = 1+2}\left(0.000016\right)=1.0\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }1.6\times {\text{10}}^{-5}\text{ M}\end{array}$

Explanation of Solution

The concentration of $\text{NO}$ and $\text{NOCl}$ is calculated using formula:

  $\text{molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution in L}}$

Substituting the values as:

  $\begin{array}{l}\text{[NOCl] = }\frac{\text{1}\text{.0 mol}}{\text{1}\text{.0 L}}\\ [\text{NOCl}]\text{ = 1}\text{.0 M}\end{array}$

  $\begin{array}{l}\text{[NO] = }\frac{\text{1}\text{.0 mol}}{\text{1}\text{.0 L}}\\ \text{[NO] = 1}\text{.0 M}\end{array}$

So, the amount of NOCl and NO is 1.0 M.

The ICE table for the reaction is:

  $\begin{array}{l}\text{ 2NOCl(g) }\rightleftharpoons {\text{ 2NO(g) + Cl}}_2(\text{g)}\\ \text{Initial (M): 1}\text{.0 1}\text{.0 0}\\ \text{Change (M): -2x +2x +x}\\ \text{Equilibrium (M): 1}\text{.0-2x 1}\text{.0+2x x}\end{array}$

The expression for the equilibrium constant is:

  $\text{K = }\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$

Substituting the values:

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(1.0+2\text{x)}}^2\text{x}}{{(1.0-2\text{x)}}^2}$

Let the value of x be small so, $1.0-2\text{x}\approx 1.\text{0}$ and $1.0+2\text{x}\approx 1.\text{0}$ .

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(1.0\text{)}}^2\text{x}}{{(1.0\text{)}}^2}$

Solving for x:

  $\begin{array}{l}\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(1.0\text{)}}^2\text{x}}{{(1.0\text{)}}^2}\\ \text{x = 1}\text{.6}\times {\text{10}}^{-5}\end{array}$

Thus, the equilibrium concentration of all the species at equilibrium is:

  $\begin{array}{l}\text{[NOCl] = 1}\text{.0 - 2}\left(0.000016\right)\text{ = 0}\text{.99 M}\\ \text{[NO] = 1+2}\left(0.000016\right)=1.0\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }1.6\times {\text{10}}^{-5}\text{ M}\end{array}$

c.

Interpretation Introduction

Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 3.0 moles of $\text{NO}$ and 1.0 mole of ${\text{Cl}}_{\text{2}}$ in a 1.0 L flask is used.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 48E

  $\begin{array}{l}\text{[NOCl] = 2}\text{.0 M}\\ \text{[NO] = }1\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }6.4\times {\text{10}}^{-5}\text{ M}\end{array}$

Explanation of Solution

The concentration of $\text{NO}$ and ${\text{Cl}}_2$ is calculated using formula:

  $\text{molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution in L}}$

Substituting the values as:

  $\begin{array}{l}\text{[NO] = }\frac{\text{3}\text{.0 mol}}{\text{1}\text{.0 L}}\\ [\text{NO}]\text{ = 3}\text{.0 M}\end{array}$

  $\begin{array}{l}{\text{[Cl}}_2\text{] = }\frac{\text{1}\text{.0 mol}}{\text{1}\text{.0 L}}\\ [{\text{Cl}}_2]\text{ = 1}\text{.0 M}\end{array}$

So, the amount of NO and Cl₂ is 3.0 M and 1.0 M respectively.

As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation of NOCl takes place.

Since, the ratio of NO:Cl₂is 3:1 instead of 2:1 so, Cl₂ will limit the production of NOCl that is Cl₂ is the limiting reagent. So, the Cl₂ will react completely and amount of NO reacted will be 3.0 − 2.0 = 1.0 M and amount of NOCl formed will be 2.0 M.

The ICE table for the reaction is:

  $\begin{array}{l}\text{ 2NOCl(g) }\rightleftharpoons {\text{ 2NO(g) + Cl}}_2(\text{g)}\\ \text{Initial (M): 2}\text{.0 1}\text{.0 0}\\ \text{Change (M): -2x +2x +x}\\ \text{Equilibrium (M): 2}\text{.0-2x 1}\text{.0+2x x}\end{array}$

The expression for the equilibrium constant is:

  $\text{K = }\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$

Substituting the values:

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(1.0+2\text{x)}}^2\text{x}}{{(2.0-2\text{x)}}^2}$

Let the value of x be small so, $2.0-2\text{x}\approx 2.\text{0}$ and $1.0+2\text{x}\approx 1.\text{0}$

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(1\text{)}}^2\text{x}}{{(2.0\text{)}}^2}$

Solving for x:

  $\begin{array}{l}\text{1}\text{.6}\times {\text{10}}^{-5}\text{ = }\frac{\text{x}}{4}\\ \text{x = 1}\text{.6}\times {\text{10}}^{-5}\times 4\\ \text{x = 6}\text{.4}\times {\text{10}}^{-5}\end{array}$

Thus, the equilibrium concentration of all the species at equilibrium is:

  $\begin{array}{l}\text{[NOCl] = 2}\text{.0 - 2}\left(0.000064\right)\text{ = 1}\text{.99 M }\simeq \text{ 2}\text{.0 M}\\ \text{[NO] = 1}\text{.0+2}\left(0.000064\right)=1\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }6.4\times {\text{10}}^{-5}\text{ M}\end{array}$

d.

Interpretation Introduction

Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 2.0 moles of $\text{NOCl}$ , 2.0 moles of $\text{NO}$ and 1.0 mole of ${\text{Cl}}_{\text{2}}$ in a 1.0 L flask is used.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 48E

  $\begin{array}{l}\text{[NOCl] = 3}\text{.9 M}\\ \text{[NO] = }8{\text{×10}}^{\text{-2}}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }4{\text{×10}}^{\text{-2}}\text{ M}\end{array}$

Explanation of Solution

The concentration of $\text{NOCl}$ , $\text{NO}$ and ${\text{Cl}}_{\text{2}}$ is calculated using formula:

  $\text{molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution in L}}$

Substituting the values as:

  $\begin{array}{l}\text{[NOCl] = }\frac{\text{2}\text{.0 mol}}{\text{1}\text{.0 L}}\\ [\text{NOCl}]\text{ = 2}\text{.0 M}\end{array}$

  $\begin{array}{l}\text{[NO] = }\frac{\text{2}\text{.0 mol}}{\text{1}\text{.0 L}}\\ \text{[NO] = 2}\text{.0 M}\end{array}$

  $\begin{array}{l}{\text{[Cl}}_{\text{2}}\text{] = }\frac{\text{1}\text{.0 mol}}{\text{1}\text{.0 L}}\\ {\text{[Cl}}_{\text{2}}\text{] = 1}\text{.0 M}\end{array}$

As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation 4.0 M of NOCl. So,

The ICE table for the reaction is:

  $\begin{array}{l}\text{ 2NOCl(g) }\rightleftharpoons {\text{ 2NO(g) + Cl}}_2(\text{g)}\\ \text{Initial (M): 4}\text{.0 0 0}\\ \text{Change (M): -2x +2x +x}\\ \text{Equilibrium (M): 4}\text{.0-2x 2x x}\end{array}$

The expression for the equilibrium constant is:

  $\text{K = }\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$

Substituting the values:

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\left(\text{x}\right)}{{(4.0-2\text{x)}}^2}$

Let the value of x be small so, $4.0-2\text{x}\approx 4.\text{0}$ .

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\left(\text{x}\right)}{{(4.0\text{)}}^2}$

Solving for x:

  $\begin{array}{l}\text{1}{\text{.6×10}}^{\text{-5}}\text{=}\frac{\text{(4}{\text{.0x}}^{\text{2}}\text{)}\left(\text{x}\right)}{{\text{(4}\text{.0)}}^{\text{2}}}\\ {\text{x}}^{\text{3}}=\frac{\text{1}{\text{.6×10}}^{\text{-5}}\times 16}{4.0}\\ \text{x}=\sqrt[3]{64{\text{×10}}^{\text{-6}}}\\ \text{x}=4{\text{×10}}^{\text{-2}}\end{array}$

Thus, the equilibrium concentration of all the species at equilibrium is:

  $\begin{array}{l}\text{[NOCl] = 4}\text{.0 - 2}\left(4{\text{×10}}^{\text{-2}}\right)\text{ = 3}\text{.9 M}\\ \text{[NO] = 2}\left(4{\text{×10}}^{\text{-2}}\right)=8{\text{×10}}^{\text{-2}}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = }4{\text{×10}}^{\text{-2}}\text{ M}\end{array}$

e.

Interpretation Introduction

Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when the concentration of all the gases is 1.00 mol/L.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 48E

  $\begin{array}{l}\text{[NOCl] = 2}\text{.0 M}\\ \text{[NO] = }11.4\times {\text{10}}^{-3}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = 0}\text{.51 M}\end{array}$

Explanation of Solution

As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation of NOCl takes place.

Since, the ratio of NO:Cl₂is 1:1 instead of 2:1 so, NO will limit the production of NOCl that is NO is the limiting reagent. So, the NO will react completely and amount of Cl₂ reacted will be 1.0 − 0.5 = 0.5 M and amount of NOCl formed will be 1.0 + 1.0 = 2.0 M.

The ICE table for the reaction is:

  $\begin{array}{l}\text{ 2NOCl(g) }\rightleftharpoons {\text{ 2NO(g) + Cl}}_2(\text{g)}\\ \text{Initial (M): 2}\text{.0 0 0}\text{.5}\\ \text{Change (M): -2x +2x +x}\\ \text{Equilibrium (M): 2}\text{.0-2x +2x 0}\text{.5+x}\end{array}$

The expression for the equilibrium constant is:

  $\text{K = }\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$

Substituting the values:

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\left(\text{0}\text{.5+x}\right)}{{(2.0-2\text{x)}}^2}$

Let the value of x be small so, $2.0-2\text{x}\approx 2.\text{0}$ and $0.5+\text{x}\approx 0.5$

  $\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\left(\text{0}\text{.5}\right)}{{(2.0\text{)}}^2}$

Solving for x:

  $\begin{array}{l}\text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{{(2\text{x)}}^2\left(\text{0}\text{.5}\right)}{{(2.0\text{)}}^2}\\ \text{1}\text{.6}\times {\text{10}}^{-5}\text{=}\frac{2{\text{x}}^2}{4}\\ {\text{x}}^2\text{ = 1}\text{.6}\times {\text{10}}^{-5}\times 2\\ \text{x = }\sqrt{\text{32}\times {\text{10}}^{-6}}\\ \text{x = 5}\text{.7}\times {\text{10}}^{-3}\end{array}$

Thus, the equilibrium concentration of all the species at equilibrium is:

  $\begin{array}{l}\text{[NOCl] = 2}\text{.0 - 2}\left(0.0057\right)\text{ = 1}\text{.99 M }\simeq \text{ 2}\text{.0 M}\\ \text{[NO] = 2}\left(\text{5}\text{.7}\times {\text{10}}^{-3}\right)=11.4\times {\text{10}}^{-3}\text{ M}\\ {\text{[Cl}}_{\text{2}}\text{] = 0}\text{.5+}0.0057\text{ = 0}\text{.51 M}\end{array}$