Chapter 15, Problem 54GQ

Lanthanum oxalate decomposes when heated to lanthanum(III) oxide, CO, and CO₂.

La₂(C₂O₄)₃(s) ⇄ La₂CO₃(s) + 3 CO(g) + 3 CO₂(g)

1. (a) If, at equilibrium, the total pressure in a 10.0-L flask is 0.200 atm, what is the value of K_p?
2. (b) Suppose 0.100 mol of La₂(C₂O₄)₃ was originally placed in the 10.0-L flask. What quantity of La₂(C₂O₄)₃ remains unreacted at equilibrium at 373 K?

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{P}}$ in the decomposition reaction of lanthanum oxalate and the quantity of lanthanum oxalate remain unreacted at equilibrium if initially $\text{0}\text{.10}\text{mol}$ of lanthanum oxalate is placed in the flask has to be given.

Concept Introduction:

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

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

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}{\text{K}}_{\text{c}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{concentration}\\ {\text{K}}_{\text{p}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{pressure}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Answer to Problem 54GQ

The value of ${\text{K}}_{\text{P}}$ in the decomposition reaction of lanthanum oxalate is ${\text{1×10}}^{\text{-6}}$.

Explanation of Solution

To determine:

The value of ${\text{K}}_{\text{P}}$ in the decomposition reaction of lanthanum oxalate into lanthanum oxide, $\text{CO}\text{and}{\text{CO}}_{\text{2}}$

Given:

$\begin{array}{l}{\text{La}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)}_{\text{3}}{}_{\text{(s)}}\rightleftarrows {\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}\text{CO}{}_{\text{(g)}}\text{+}\text{3}{\text{CO}}_{\text{2}\text{(g)}}\\ \text{Total}\text{pressure}\text{=}\text{0}\text{.2}\text{atm}\\ \text{V}\text{=}\text{10}\text{L}\end{array}$

In the given reaction lanthanum oxalate and lanthanum oxide are in solid state and thus its pressure can be taken as unity. There is no role of lanthanum oxalate and lanthanum oxide in the equilibrium expression.

${\text{K}}_{\text{P}}\text{=}{\left[\text{CO}\right]}^{\text{3}}{\left[{\text{CO}}_{\text{2}}\right]}^{\text{3}}$

From the equation ${\text{La}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)}_{\text{3}}{}_{\text{(s)}}\rightleftarrows {\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}\text{CO}{}_{\text{(g)}}\text{+}\text{3}{\text{CO}}_{\text{2}\text{(g)}}$, it is clear that

Number of moles of $\text{CO}$ is equal to number of moles of ${\text{CO}}_{\text{2}}$ Therefore,

Mole fraction of $\text{CO}$ is equal to mole fraction of ${\text{CO}}_{\text{2}}$.

$\begin{array}{l}\text{Total}\text{pressure}\text{=}\text{0}\text{.2}\text{atm}\\ {\text{P}}_{\text{CO}}\text{=}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{0}\text{.2}}{\text{2}}\text{=}\text{0}\text{.1}\text{atm}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{P}}\text{=}{\left[\text{CO}\right]}^{\text{3}}{\left[{\text{CO}}_{\text{2}}\right]}^{\text{3}}\\ \text{=}\text{0}{\text{.1}}^{\text{3}}\text{×0}{\text{.1}}^{\text{3}}\\ \text{=}{\text{1×10}}^{\text{-6}}\end{array}$

The value of ${\text{K}}_{\text{P}}$ for the reaction is ${\text{1×10}}^{\text{-6}}$.

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{P}}$ in the decomposition reaction of lanthanum oxalate and the quantity of lanthanum oxalate remain unreacted at equilibrium if initially $\text{0}\text{.10}\text{mol}$ of lanthanum oxalate is placed in the flask has to be given.

Concept Introduction:

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

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

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}{\text{K}}_{\text{c}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{concentration}\\ {\text{K}}_{\text{p}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{pressure}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Answer to Problem 54GQ

The quantity of lanthanum oxalate remain unreacted if $\text{0}\text{.10}\text{mol}$ of lanthanum oxalate is initially placed in the flask is $\text{8}{\text{.91×10}}^{\text{-2}}\text{mol}$.

Explanation of Solution

To determine:

The quantity of lanthanum oxalate remains unreacted if $\text{0}\text{.10}\text{mol}$ of lanthanum oxalate is initially placed in the flask.

Using the equation ${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}$ we can find out the value of ${\text{K}}_{\text{c}}$

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{K}}_{\text{p}}}{{\left(\text{RT}\right)}^{\text{Δn}}}\text{=}\frac{{\text{1×10}}^{\text{-6}}}{{\left(\text{0}\text{.082×373}\right)}^{\text{7-1}}}\\ \text{=}\text{1}{\text{.21×10}}^{\text{-15}}\end{array}$

Given that initially $\text{0}\text{.10}\text{mol}$ of lanthanum oxalate is present in $\text{10}\text{L}$ flask.

$\begin{array}{l}\text{Initial}\text{concentration}\text{=}\frac{\text{No}\text{.}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}}\\ \text{=}\frac{\text{0}\text{.1}}{\text{10}}\\ \text{=}\text{0}\text{.01}\text{M}\end{array}$

$\begin{array}{l}{\text{La}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)}_{\text{3}}{}_{\text{(s)}}\rightleftarrows {\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}\text{CO}{}_{\text{(g)}}\text{+}\text{3}{\text{CO}}_{\text{2}\text{(g)}}\\ \text{Initial}\text{concentration}\text{0}\text{.01}\text{mol}\text{0}\text{0}\text{0}\\ \text{After}\text{reacting}\text{x}\text{moles}\text{0}\text{.01-x}\text{x}\text{3x}\text{3x}\end{array}$

For the reaction ${\text{La}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)}_{\text{3}}{}_{\text{(s)}}\rightleftarrows {\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}\text{CO}{}_{\text{(g)}}\text{+}\text{3}{\text{CO}}_{\text{2}\text{(g)}}$

$\begin{array}{l}{\text{K}}_{\text{c}}\text{=}{\left[\text{CO}\right]}^{\text{3}}{\left[{\text{CO}}_{\text{2}}\right]}^{\text{3}}\\ \text{1}{\text{.21×10}}^{\text{-15}}\text{=}{\text{3x}}^{\text{3}}{\text{×3x}}^{\text{3}}\\ \text{1}{\text{.21×10}}^{\text{-15}}\text{=}\text{729}{\text{x}}^{\text{6}}\\ \text{x}\text{=}\text{1}{\text{.08×10}}^{\text{-3}}\end{array}$

 $\begin{array}{l}\left[{\text{La}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)}_{\text{3}}\right]\text{=0}\text{.01-1}{\text{.08×10}}^{\text{-3}}\\ \text{=}\text{8}{\text{.91×10}}^{\text{-3}}\text{M}\\ \text{No}\text{.}\text{of}\text{moles}\text{=}\text{Molarity}\text{×}\text{Volume}\\ \text{=}\text{8}{\text{.91×10}}^{\text{-3}}\text{×}\text{10}\\ \text{=}\text{8}{\text{.91×10}}^{\text{-2}}\text{mol}\\ \end{array}$

Conclusion:

The value of ${\text{K}}_{\text{P}}$ in the decomposition reaction of lanthanum oxalate and the quantity of lanthanum oxalate remain unreacted at equilibrium if initially $\text{0}\text{.10}\text{mol}$ of lanthanum oxalate is placed in the flask was calculated using the data.