Chapter 6, Problem 50E

a.

Interpretation Introduction

Interpretation: The equilibrium partial pressure of the gases needs to be calculated when flask contains only ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ at an initial pressure of 4.5 atm.

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:

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

When the equilibrium constant is expressed in terms of pressure, then it is represented as ${\text{K}}_p$ .

Answer to Problem 50E

The equilibrium partial pressure of the gases is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 4}\text{.0 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 1}\text{.0 atm}\end{array}$

Explanation of Solution

Given:

The reaction:

  ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}\rightleftharpoons {\text{2NO}}_2\text{(g)}$

  ${\text{K}}_p\text{ = }0.25$ .

The expression for the equilibrium constant for the given reaction is:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left({\text{P}}_{{\text{NO}}_{\text{2}}}\right)}^{\text{2}}}{\left({\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\right)}$ - (1)

Where ${\text{P}}_{{\text{NO}}_{\text{2}}}$ and ${\text{P}}_{{\text{N}}_2{\text{O}}_4}$ represents partial pressure of ${\text{NO}}_2\text{(g)}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$ .

The ICE table for the reaction is:

  $\begin{array}{l}{\text{ N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g) }\rightleftharpoons {\text{ 2NO}}_2\text{(g)}\\ \text{Initial (atm): 4}\text{.5 0 }\\ \text{Change (atm): -x +2x }\\ \text{Equilibrium (atm): 4}\text{.5-x 2x }\end{array}$

Substituting the values from ICE table to equation (1) as:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{4}\text{.5-x}\right)}$

Since, ${\text{K}}_p\text{ = }0.25$ so,

  $\text{0}\text{.25 = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{4}\text{.5-x}\right)}$

Solving for x:

  $\begin{array}{l}{\left(\text{2x}\right)}^{\text{2}}\text{ = 0}\text{.25}\left(\text{4}\text{.5-x}\right)\\ {\text{4x}}^{\text{2}}\text{ = 1}\text{.125 - 0}\text{.25x}\\ {\text{4x}}^{\text{2}}\text{+ 0}\text{.25x - 1}\text{.125 = 0}\end{array}$

On solving the quadratic equation, the obtained values of x are $\text{-0}\text{.56 and +0}\text{.5}$ , neglecting the negative value so, the value of $\text{x = 0}\text{.5}$ .

Hence, the equilibrium partial pressure of the gases is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 4}\text{.5 - 0}\text{.5 = 4}\text{.0 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 2}\left(0.5\right)\text{ = 1}\text{.0 atm}\end{array}$

b.

Interpretation Introduction

Interpretation: The equilibrium partial pressure of the gases needs to be calculated when flask contains only ${\text{NO}}_2$ at an initial pressure of 9.0 atm.

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:

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

When the equilibrium constant is expressed in terms of pressure, then it is represented as ${\text{K}}_p$ .

Answer to Problem 50E

The equilibrium partial pressure of the gases is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 4}\text{.0 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 1}\text{.0 atm}\end{array}$

Explanation of Solution

When flask contains only ${\text{NO}}_2$ at an initial pressure of 9.0 atm then the reaction will be:

  ${\text{2NO}}_2\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

  ${\text{K}}_p^{\text{'}}\text{ = }\frac{1}{{\text{K}}_p}\text{= }\frac{1}{0.25}$ .

The expression for the equilibrium constant for the reaction is:

  ${\text{K}}_p^{\text{'}}\text{ = }\frac{\left({\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\right)}{{\left({\text{P}}_{{\text{NO}}_{\text{2}}}\right)}^{\text{2}}}$ - (2)

Where ${\text{P}}_{{\text{NO}}_{\text{2}}}$ and ${\text{P}}_{{\text{N}}_2{\text{O}}_4}$ represents partial pressure of ${\text{NO}}_2\text{(g)}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$ .

The ICE table for the reaction is:

  $\begin{array}{l}{\text{ 2NO}}_2\text{(g) }\rightleftharpoons {\text{ N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}\\ \text{Initial (atm): 9}\text{.0 0 }\\ \text{Change (atm): -2x +x }\\ \text{Equilibrium (atm): 9}\text{.0-2x x }\end{array}$

Substituting the values from ICE table to equation (2) as:

  ${\text{K}}_p^{\text{'}}\text{ = }\frac{\left(\text{x}\right)}{{\left(\text{9}\text{.0-2x}\right)}^{\text{2}}}$

Since, ${\text{K}}_p^{\text{'}}\text{ = }\frac{1}{0.25}$ so,

  $\frac{1}{0.25}\text{ = }\frac{\left(\text{x}\right)}{{\left(\text{9}\text{.0-2x}\right)}^{\text{2}}}$

Solving for x:

  $\begin{array}{l}{\left(\text{9}\text{.0-2x}\right)}^{\text{2}}\text{ = 0}\text{.25x}\\ \text{9}\text{.0-2x = }{\left(\text{0}\text{.25x}\right)}^2\\ \text{0}{\text{.0625x}}^{\text{2}}\text{ + 2x - 9}\text{.0 = 0}\end{array}$

On solving the quadratic equation, the obtained values of x are $\text{-36 and +4}$ , neglecting the negative value so, the value of $\text{x = 4}$ .

Hence, the equilibrium partial pressure of the gases is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 4}\text{.0 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 9}\text{.0 - 2}\left(4.0\right)\text{ = 1}\text{.0 atm}\end{array}$

c.

Interpretation Introduction

Interpretation: Whether the direction from which an equilibrium position is reached will matter or not should be explained from above two parts.

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:

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

When the equilibrium constant is expressed in terms of pressure, then it is represented as ${\text{K}}_p$ .

Answer to Problem 50E

No, the direction from which an equilibrium position is reached in the chemical reaction will not matter.

Explanation of Solution

The equilibrium partial pressure of the gas’s values obtained in above two parts; a and b are same that is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 4}\text{.0 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 1}\text{.0 atm}\end{array}$

Hence, the direction from which an equilibrium position is reached in the chemical reaction will not matter.

d.

Interpretation Introduction

Interpretation: The new equilibrium partial pressure needs to be calculated when the volume of the container for part a is decreased to one-half from the original volume.

Concept Introduction: The law which states the relation between pressure, $\text{P}$ and volume, $\text{V}$ under constant temperature, $\text{T}$ is said to Boyle’s law. According to this law, the pressure of an ideal gas has inverse relation with its volume:

  $\text{P}\propto \frac{1}{\text{V}}$

Answer to Problem 50E

The new equilibrium partial pressure is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 8}\text{.28 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 1}\text{.44 atm}\end{array}$

Explanation of Solution

The Boyle’s law can be expressed as for two different sets of condition for the same substance at constant temperature as:

  ${\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = P}}_2{\text{V}}_2$ - (3)

Where 1 and 2 indicates conditions at set 1 and condition at set 2.

Let ${\text{V}}_i$ be the initial volume so,

When the volume of the container for part a is decreased to one-half from the original volume, ${\text{V}}_i$ so the final volume, ${\text{V}}_{\text{f}}$ will be:

  ${\text{V}}_{\text{f}}\text{ = }\frac{{\text{V}}_{\text{i}}}{\text{2}}$

So, the new equilibrium partial pressure, ${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ of each gas can be calculated using equation (3) as:

The value of ${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}^i\text{ = 4}\text{.5 atm}$ so,

  $\begin{array}{l}\text{4}{\text{.5 atm × V}}_{\text{i}}{\text{ = P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{f}}\text{×}\frac{{\text{V}}_i}{\text{2}}\\ {\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{f}}\text{ = 4}\text{.5 atm × }2\\ {\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{f}}\text{ = 9}\text{.0 atm}\end{array}$

So, the initial partial pressure of ${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ when the volume of the container for part a is decreased to one-half from the original volume is 9.0 atm.

The reaction:

  ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}\rightleftharpoons {\text{2NO}}_2\text{(g)}$

  ${\text{K}}_p\text{ = }0.25$ .

The expression for the equilibrium constant for the given reaction is:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left({\text{P}}_{{\text{NO}}_{\text{2}}}\right)}^{\text{2}}}{\left({\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\right)}$

Where ${\text{P}}_{{\text{NO}}_{\text{2}}}$ and ${\text{P}}_{{\text{N}}_2{\text{O}}_4}$ represents partial pressure of ${\text{NO}}_2\text{(g)}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$ .

The ICE table for the reaction is:

  $\begin{array}{l}{\text{ N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g) }\rightleftharpoons {\text{ 2NO}}_2\text{(g)}\\ \text{Initial (atm): 9}\text{.0 0 }\\ \text{Change (atm): -x +2x }\\ \text{Equilibrium (atm): 9}\text{.0-x 2x }\end{array}$

Substituting the values from ICE table to equation (1) as:

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{9}\text{.0-x}\right)}$

Since, ${\text{K}}_p\text{ = }0.25$ so,

  $\text{0}\text{.25 = }\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{9}\text{.0-x}\right)}$

Solving for x:

  $\begin{array}{l}{\left(\text{2x}\right)}^{\text{2}}\text{ = 0}\text{.25}\left(\text{9}\text{.0-x}\right)\\ {\text{4x}}^{\text{2}}\text{ = 2}\text{.25 - 0}\text{.25x}\\ {\text{4x}}^{\text{2}}\text{ + 0}\text{.25x - 2}\text{.25 = 0}\end{array}$

On solving the quadratic equation, the obtained values of x are $\text{-0}\text{.78 and +0}\text{.72}$ , neglecting the negative value so, the value of $\text{x = 0}\text{.72}$ .

Hence, the equilibrium partial pressure of the gases is:

  $\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{ = 9}\text{.0 - 0}\text{.72 = 8}\text{.28 atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{ = 2}\left(0.72\right)\text{ = 1}\text{.44 atm}\end{array}$