Chemical Equilibrium

Homework 4 Solutions available

Use

$$R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$$

Unless otherwise stated, pressures are given in bar.

Show the main steps of your calculations. For equilibrium-composition problems, clearly define the change in pressure before solving.

Pure solids are not included in equilibrium-constant expressions.

Problem 1 · Equilibrium Constants

Problem 1Medium

For each equilibrium below, write the expression for the equilibrium constant. If a substance does not appear in the expression, briefly explain why.

a)

$$2CO(g)+O_2(g)\rightleftharpoons2CO_2(g)$$

b)

$$C(s)+O_2(g)\rightleftharpoons CO_2(g)$$

c)

$$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$$

d)

$$2C(s)+O_2(g)\rightleftharpoons2CO(g)$$

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a)

$$K_p=\frac{P_{CO_2}^2}{P_{CO}^2P_{O_2}}$$

b)

$$K_p=\frac{P_{CO_2}}{P_{O_2}}$$

The pure solid carbon is not included in the equilibrium-constant expression.

c)

$$K_p=\frac{P_{N_2}P_{H_2}^3}{P_{NH_3}^2}$$

d)

$$K_p=\frac{P_{CO}^2}{P_{O_2}}$$

Again, the pure solid carbon is omitted.

Problem 2 · Manipulating Equilibrium Constants

Problem 2Medium

At a certain temperature,

$$N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$$

has

$$K_p=36.0$$

Without using thermodynamic data, determine the equilibrium constant for each of the following reactions.

a)

$$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$$

b)

$$\frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)\rightleftharpoons NH_3(g)$$

c)

$$NH_3(g)\rightleftharpoons\frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)$$

For each part, indicate what operation was performed on the original reaction.

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The original reaction has

$$K_p=36.0$$

a) The reaction is reversed, so

$$K_p=\frac{1}{36.0}$$ $$K_p=2.78\times10^{-2}$$

b) All stoichiometric coefficients are divided by 2, so

$$K_p=(36.0)^{1/2}$$ $$K_p=6.00$$

c) The reaction from part (b) is reversed, so

$$K_p=\frac{1}{6.00}$$ $$K_p=0.167$$

Problem 3 · Equilibrium Composition

Problem 3Hard

A rigid vessel initially contains only ammonia gas at a pressure of

$$6.00\ \mathrm{bar}$$

At constant temperature, the following equilibrium is established:

$$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$$

At equilibrium, the total pressure is

$$8.00\ \mathrm{bar}$$

Calculate:

a) the equilibrium partial pressure of each gas;

b) the equilibrium constant for the reaction.

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Let the pressure increase corresponding to the formation of nitrogen be

$$x$$

Then

$$P_{NH_3}=6.00-2x$$ $$P_{N_2}=x$$ $$P_{H_2}=3x$$

At equilibrium,

$$P_{\text{total}}=8.00\ \mathrm{bar}$$

Therefore,

$$(6.00-2x)+x+3x=8.00$$ $$6.00+2x=8.00$$ $$x=1.00$$

Thus,

$$P_{NH_3}=4.00\ \mathrm{bar}$$ $$P_{N_2}=1.00\ \mathrm{bar}$$ $$P_{H_2}=3.00\ \mathrm{bar}$$

For the reaction,

$$K_p=\frac{P_{N_2}P_{H_2}^3}{P_{NH_3}^2}$$

Substituting the equilibrium pressures,

$$K_p= \frac{(1.00)(3.00)^3}{(4.00)^2}$$

Therefore,

$$\boxed{K_p=1.69}$$

Problem 4 · Equilibrium Composition

Problem 4Hard

At a certain temperature,

$$2CO(g)+O_2(g)\rightleftharpoons2CO_2(g)$$

has

$$K_p=4.00$$

Initially,

$$P_{CO}=3.00\ \mathrm{bar}$$ $$P_{O_2}=2.00\ \mathrm{bar}$$ $$P_{CO_2}=0$$

The system is allowed to reach equilibrium at the same temperature.

Determine:

a) the equilibrium partial pressure of each gas;

b) the total pressure at equilibrium.

Set up the initial-change-equilibrium relations before solving.

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Let the reaction proceed by an amount corresponding to a pressure change

$$x$$

The equilibrium pressures are therefore

$$P_{CO}=3.00-2x$$ $$P_{O_2}=2.00-x$$ $$P_{CO_2}=2x$$

The equilibrium-constant expression is

$$K_p= \frac{P_{CO_2}^2}{P_{CO}^2P_{O_2}}$$

Substituting,

$$4.00= \frac{(2x)^2} {(3.00-2x)^2(2.00-x)}$$

The physically meaningful solution is

$$x=1.00$$

Therefore,

$$P_{CO}=1.00\ \mathrm{bar}$$ $$P_{O_2}=1.00\ \mathrm{bar}$$ $$P_{CO_2}=2.00\ \mathrm{bar}$$

The total pressure is

$$P_{\mathrm{total}} = 1.00+1.00+2.00$$

Therefore,

$$\boxed{P_{\mathrm{total}}=4.00\ \mathrm{bar}}$$

Problem 5 · Combining Equilibria

Problem 5Medium

At the same temperature, the following equilibrium constants are known:

$$2C(s)+O_2(g)\rightleftharpoons2CO(g)$$ $$K_{p,1}=5.00$$

and

$$2CO(g)+O_2(g)\rightleftharpoons2CO_2(g)$$ $$K_{p,2}=20.0$$

Using only these two reactions, determine the equilibrium constant for

$$C(s)+O_2(g)\rightleftharpoons CO_2(g)$$

Show how the given reactions must be combined.

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Add the two reactions:

$$2C(s)+O_2(g)\rightleftharpoons2CO(g)$$ $$2CO(g)+O_2(g)\rightleftharpoons2CO_2(g)$$

After cancelling carbon monoxide,

$$2C(s)+2O_2(g)\rightleftharpoons2CO_2(g)$$

When reactions are added, their equilibrium constants are multiplied:

$$K_p=K_{p,1}K_{p,2}$$ $$K_p=(5.00)(20.0)$$ $$K_p=100$$

Now divide the entire reaction by 2:

$$C(s)+O_2(g)\rightleftharpoons CO_2(g)$$

When all coefficients are divided by 2, the new equilibrium constant is the square root of the previous one:

$$K_p=(100)^{1/2}$$

Therefore,

$$\boxed{K_p=10.0}$$

Problem 6 · Thermodynamics and Equilibrium

Problem 6Medium

For the reaction

$$3O_2(g)\rightleftharpoons2O_3(g)$$

suppose that at

$$T=298\ \mathrm{K}$$

the thermodynamic quantities are

$$\Delta H^\circ=+285.0\ \mathrm{kJ\,mol^{-1}}$$

and

$$\Delta S^\circ=-138.0\ \mathrm{J\,mol^{-1}K^{-1}}$$

Calculate:

a) the standard Gibbs energy change at this temperature;

b) the equilibrium constant at this temperature;

c) based on the value of the equilibrium constant, state which side of the equilibrium is strongly favored under standard-state conditions.

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Use

$$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$$

First convert the enthalpy change to joules:

$$\Delta H^\circ=285000\ \mathrm{J\,mol^{-1}}$$

Then,

$$\Delta G^\circ = 285000-(298)(-138)$$ $$\Delta G^\circ = 326124\ \mathrm{J\,mol^{-1}}$$

Therefore,

$$\boxed{\Delta G^\circ\approx326.1\ \mathrm{kJ\,mol^{-1}}}$$

Now use

$$\Delta G^\circ=-RT\ln K_p$$

Therefore,

$$\ln K_p= -\frac{\Delta G^\circ}{RT}$$ $$\ln K_p= -\frac{326124}{(8.314)(298)}$$

This gives approximately

$$\ln K_p=-131.6$$

and therefore

$$\boxed{K_p\approx6.8\times10^{-58}}$$

Since

$$K_p\ll1$$

the equilibrium strongly favors the reactant side under standard-state conditions.

Problem 7 · Thermodynamics and Equilibrium

Problem 7Hard

For the reaction

$$H_2(g)+Cl_2(g)\rightleftharpoons2HCl(g)$$

the standard Gibbs energy change at

$$298\ \mathrm{K}$$

is

$$\Delta G^\circ=-5.71\ \mathrm{kJ\,mol^{-1}}$$

A rigid vessel initially contains

$$P_{H_2}=1.00\ \mathrm{bar}$$ $$P_{Cl_2}=1.00\ \mathrm{bar}$$ $$P_{HCl}=0$$

Calculate:

a) the equilibrium constant at this temperature;

b) the equilibrium partial pressure of each gas;

c) the percentage of the initial hydrogen that reacts.

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Use

$$\Delta G^\circ=-RT\ln K_p$$

Therefore,

$$\ln K_p= -\frac{\Delta G^\circ}{RT}$$

Substituting,

$$\ln K_p= -\frac{-5710}{(8.314)(298)}$$ $$\ln K_p\approx2.30$$

Therefore,

$$K_p\approx10.0$$

Let

$$x$$

bar of hydrogen and chlorine react.

At equilibrium,

$$P_{H_2}=1.00-x$$ $$P_{Cl_2}=1.00-x$$ $$P_{HCl}=2x$$

The equilibrium-constant expression is

$$K_p= \frac{P_{HCl}^2}{P_{H_2}P_{Cl_2}}$$

Therefore,

$$10.0= \frac{(2x)^2}{(1.00-x)^2}$$

Taking the positive square root,

$$\sqrt{10.0} = \frac{2x}{1.00-x}$$

Solving gives approximately

$$x=0.613$$

Thus,

$$P_{H_2}=0.387\ \mathrm{bar}$$ $$P_{Cl_2}=0.387\ \mathrm{bar}$$ $$P_{HCl}=1.23\ \mathrm{bar}$$

The fraction of the initial hydrogen that reacts is

$$\frac{0.613}{1.00}=0.613$$

Therefore,

$$\boxed{61.3\%}$$

of the initial hydrogen reacts.

Problem 8 · Combining Equilibria

Problem 8Hard

At the same temperature,

$$C(s)+O_2(g)\rightleftharpoons CO_2(g)$$

has

$$K_{p,1}=10.0$$

and

$$2CO(g)+O_2(g)\rightleftharpoons2CO_2(g)$$

has

$$K_{p,2}=20.0$$

Determine the equilibrium constant for

$$2C(s)+O_2(g)\rightleftharpoons2CO(g)$$

The final reaction must be obtained algebraically from the two given reactions. Simply writing an equilibrium-constant expression is not sufficient.

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First multiply the first reaction by 2:

$$2C(s)+2O_2(g)\rightleftharpoons2CO_2(g)$$

The equilibrium constant becomes

$$K_p=(K_{p,1})^2$$

Therefore,

$$K_p=(10.0)^2$$

Now reverse the second reaction:

$$2CO_2(g)\rightleftharpoons2CO(g)+O_2(g)$$

Its equilibrium constant becomes

$$K_p=\frac{1}{K_{p,2}}$$

Adding the two modified reactions gives

$$2C(s)+O_2(g)\rightleftharpoons2CO(g)$$

Therefore,

$$K_p= \frac{(K_{p,1})^2}{K_{p,2}}$$ $$K_p= \frac{(10.0)^2}{20.0}$$

Thus,

$$\boxed{K_p=5.00}$$