Chapter 15, Problem 63GQ

A sample of N₂O₄ gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, 20.0% of the N₂O₄ has been convened to NO₂ gas.

1. (a) Calculate K_p.

If the original pressure of N₂O₄ is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier’s principle?

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{P}}$ for the dissociation reaction of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ to ${\text{NO}}_{\text{2}}$ with a pressure of $\text{1}\text{atm}$ has to be calculated.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

  Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

• Ideal gas equation:$\text{PV}\text{=}\text{nRT}$

Equilibrium constant in terms of pressure${\text{[K}}_{\text{p}}\text{]}$: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

• $\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}_{\text{C}}{}^{\text{c}}\text{×}{\text{P}}_{\text{D}}{}^{\text{d}}}{{\text{P}}_{\text{A}}{}^{\text{a}}\text{×}{\text{P}}_{\text{B}}{}^{\text{b}}}\end{array}$

Le Chatelier’s principle: If an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Answer to Problem 63GQ

The ${\text{K}}_{\text{P}}$ for the given reaction is $\text{0}\text{.2}$.

Explanation of Solution

Given:

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\rightleftarrows \text{2}{\text{NO}}_{\text{2}}\\ \text{Pressure}\text{=}\text{1}\text{atm}\\ \text{At}\text{equilibrium,}\text{20\%}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{converts}\text{to}{\text{NO}}_{\text{2}}\end{array}$

Given that initial pressure is $\text{1}\text{atm}$

Given the reaction:

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\rightleftarrows \text{2}{\text{NO}}_{\text{2}}\\ \text{Initial}\text{pressure}\text{1}\text{atm}\text{0}\\ \text{Change}\text{-x}\text{+2x}\\ \text{At}\text{equilibrium}\text{1}\text{-x}\text{2x}\end{array}$

At equilibrium 20% ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ decomposed and 80% remains. Thus we can write

$\text{1-x}\text{=}\text{0}\text{.8}\text{atm}\text{[80\%]}$

$\text{x}\text{=}\text{0}\text{.2}\text{atm}$

$\begin{array}{l}{\text{K}}_{\text{P}}\text{=}\frac{\text{P}{\left[{\text{NO}}_{\text{2}}\right]}^{\text{2}}}{\text{P}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right]}\\ \text{=}\frac{{\left[\text{2×0}\text{.2}\right]}^{\text{2}}}{\text{1-0}\text{.2}}\\ \text{=}\frac{\text{0}\text{.16}}{\text{0}\text{.8}}\\ \text{=}\text{0}\text{.2}\end{array}$

Interpretation Introduction

Interpretation:

The percent of dissociation of gas if the original pressure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\text{0}\text{.10}\text{atm}$ has to be given.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

  Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

• Ideal gas equation:$\text{PV}\text{=}\text{nRT}$

Equilibrium constant in terms of pressure${\text{[K}}_{\text{p}}\text{]}$: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

• $\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}_{\text{C}}{}^{\text{c}}\text{×}{\text{P}}_{\text{D}}{}^{\text{d}}}{{\text{P}}_{\text{A}}{}^{\text{a}}\text{×}{\text{P}}_{\text{B}}{}^{\text{b}}}\end{array}$

Le Chatelier’s principle: If an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Answer to Problem 63GQ

The percent dissociation of the gas is $\text{50}\text{\%}$.

Explanation of Solution

Given:

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\rightleftarrows \text{2}{\text{NO}}_{\text{2}}\\ \text{Pressure}\text{=}\text{0}\text{.10}\text{atm}\end{array}$

Given the reaction:

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\rightleftarrows \text{2}{\text{NO}}_{\text{2}}\\ \text{Initial}\text{pressure}\text{0}\text{.1}\text{atm}\text{0}\\ \text{Change}\text{-x}\text{+2x}\\ \text{At}\text{equilibrium}\text{0}\text{.1}\text{-x}\text{2x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{P}}\text{=}\frac{\text{P}{\left[{\text{NO}}_{\text{2}}\right]}^{\text{2}}}{\text{P}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right]}\\ \text{0}\text{.2}\text{=}\frac{{\left[\text{2x}\right]}^{\text{2}}}{\text{0}\text{.1-x}}\\ {\text{4x}}^{\text{2}}\text{+0}\text{.2x}\text{-0}\text{.02}\text{=}\text{0}\end{array}$

Solving the quadratic equation we get $\text{x}\text{=}\text{0}\text{.05}$

Therefore the percent of dissociation at equilibrium $\begin{array}{l}\text{=}\left[\frac{\text{0}\text{.05}\text{atm}}{\text{0}\text{.1}\text{atm}}\right]\text{×100}\\ \text{=}\text{50}\text{\%}\end{array}$

Here we can see that when the pressure decreases from $\text{1}\text{atm}\text{to}\text{0}\text{.1}\text{atm}$, the percent of dissociation increases and more amount of product is formed.

According to Le Chatelier’s principle, if an equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

If pressure of the system decreases, system will shift to the direction which having more number of molecules. Here product side has more number of molecules and thus the dissociation percent increases.

The result is in agreement with Le Chatelier’s principle.