Topic 7 & 17 – Equilibrium

Big ideas:

• A state of equilibrium is reached in a closed system when the rates of the forward and reverse reactions are equal.
• The equilibrium law describes how the equilibrium constant (Kc) can be determined for a particular chemical reaction.
• The magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent.
• The reaction quotient (Q) measures the relative amount of products and reactants present during a reaction at one point in time. Q is the equilibrium expression with non-equilibrium concentrations. The position of the equilibrium changes with changes in concentration, pressure and temperature.
• A catalyst has no effect on the position of equilibrium or the equilibrium constant.
• Le Châtelier’s Principle for changes in concentration can be explained by the equilibrium law.
• The position of equilibrium corresponds to a maximum value of entropy and a minimum in the value of Gibbs free energy.
• The Gibbs free energy change of a reaction and the equilibrium constant can both be used to measure the position of an equilibrium reaction and are related by the equation DG = –RTlnK.

Students will be able to…

• Describe the characteristics of chemical and physical systems in a state of equilibrium.
• Deduce the equilibrium constant expression (Kc) from an equation for a homogeneous reaction.
• Determine the relationship between different equilibrium constants (Kc) for the same reaction at the same temperature.
• Apply Le Châtelier’s Principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant.
• Solve homogeneous equilibrium problems using the expression for .
• Relate and the equilibrium constant, both qualitatively and quantitatively using

Theory of Knowledge* questions/connections:

• Scientists investigate the world at different scales; the macroscopic and microscopic. Which ways of knowing allow us to move from the macroscopic to the microscopic?

• The career of Fritz Haber coincided with the political upheavals of two world wars. He supervised the release of chlorine on the battlefield in World War I and worked on the production of explosives. How does the social context of scientific work affect the methods and findings of science? Should scientists be held morally responsible for the applications of their discoveries?

• We can use mathematics successfully to model equilibrium systems. Is this because we create mathematics to mirror reality or because the reality is intrinsically mathematical?

Introduction to Equilibrium  

So far in this course, we have concerned ourselves primarily with chemical reactions that go basically to completion. For example, hydrocarbon fuels will burn in the presence of excess oxygen to form carbon dioxide and water, leaving virtually none of the fuel behind.

However, many chemical reactions do not go to completion because a reverse reaction is simultaneously competing with the forward reaction. These reactions will eventually reach a state of chemical (dynamic) equilibrium, which we explore in this unit of study.

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Consider the following equilibrium system:

Two reactions are happening simultaneously:

• Nitrogen gas and hydrogen gas are reacting to form ammonia (the forward reaction)
• Ammonia is decomposing to form nitrogen and hydrogen gases (the reverse reaction)

Definition:  A chemical system is said to be in dynamic equilibrium when:**

• Substances continue to change from products to reactants and vice versa, but the forward and reverse rates are equal.
• The amounts (concentrations) of reactants and products are constant.

Examples of physical equilibrium systems:

a glass of ice water at 0 degrees C:
a bottle of water at 20 degrees C:
a bottle of soda water:

Examples of chemical equilibrium systems:

The ozone-oxygen equilibrium in the atmosphere:        3O2(g) ⇌** 2O3(g)
Vinegar (a weak acid) solution:  CH3COOH(aq) + H2O(l) ⇌** H3O+(aq) + CH3COO–(aq)
Oxygen transport in the human body(by a protein called hemoglobin):  

In class, we used a macro-level analogy to introduce the concept of dynamic equilibrium. The setup included two large beakers to contain the water and two smaller beakers to transfer the water between the two large beakers:  rev rxn

• Beakers A and B represent concentrations of reactants and products
• Beakers X and Y represent rates of the forward and reverse reactions

Case 1:  Start with Beaker A full, Beaker B empty. Beaker X is larger than Beaker Y. (analogous to initial conditions where only reactants are present, no products)

Case 2:  Start with Beaker A empty, but Beaker B full instead. (analogous to initial conditions where only products are present, no reactants)

Fundamental features of chemical (dynamic) equilibria:

• Dynamic equilibria can only be established in a closed system. There can be no transfer of material (mass) or energy to or from the surroundings.
• At equilibrium, the concentrations of reactants and products are constant.
  • ”NO MACROSCOPIC CHANGE”
• At equilibrium, particles continue to react in both the forward and reverse directions.
  • ”CONSTANT MICROSCOPIC CHANGE”
• At equilibrium, the rates of the forward and reverse reactions are equal.
• The same equilibrium position will be reached independent of the initial concentrations of the reactants and products (the “starting conditions”).

The Equilibrium Law

Recall from the previous section that when a chemical system is at equilibrium, reactant and product concentrations are constant. Experimentally, chemists have discovered a mathematical relationship between the concentrations of products and reactants at equilibrium, called the equilibrium law:

For the general equilibrium:  aA + bB ⇌**  cC + dD

Important

Kc is only constant for a given temperature. !
If the temperature changes, so does Kc (more on that later!)

The magnitude of Kc gives an indication of the equilibrium position:

very  very small  0.001       0.01      0.1        1        10       100       1000  large

Let’s practice writing expressions for Kc (called equilibrium expressions):

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2NO2(g) ⇌** N2O4(g)  

N2(g) + 3H2(g) ⇌** 2NH3(g) CaCO3(s) ⇌** Ca2+(aq) + CO32–(aq)  

Fe3+(aq) + SCN–(aq) ⇌ FeSCN2+(aq) CH4(g) + 2O2(g) ⇌ CO2(g) + 2H2O(g) Cu(s) + 2Ag+(aq) ⇌** Cu2+(aq) + 2Ag(s)  

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The first four examples are homogeneous equilibria (all species in the same phase); the second two are heterogeneous equilibria (two or more different phases present). In heterogeneous equilibria, solids do not appear in the equilibrium expression (neither do pure liquids) because their activities (see below) are equal to exactly one, and hence cancel out of the expression.

Although the Kc expression contains concentration values (units: mol dm-3), Kc is usually calculated as a unitless value. Strictly speaking, equilibrium constants are computed from a unitless property called activity, not concentration. However, activity and concentration are more or less equal for gases and dilute solutions, so in these cases using concentration to calculate Kc is a fair approximation.

When an equilibrium constant is calculated using activities, it is simply symbolized K (or Keq).

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Consider these two equilibria:  H2(g) + I2(g) ⇌** 2HI(g) Kc = 0.35 at 500ºC H2(g) + Cl2(g) ⇌ 2HCl(g) Kc = 4400 at 500ºC

These are similar chemical systems, but with very different values for Kc. What conclusions can we draw from these data?

When reactions are inverses or multiples of one another, the values of Kc are related mathematically. Consider these three different representation of the same equilibrium at 500ºC:

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1. H2(g) + Cl2(g) ⇌** 2HCl(g)

2. 2HCl(g) ⇌ H2(g) + Cl2(g)

3. ½H2(g) + ½Cl2(g) ⇌ HCl(g)

[ ]2

Kc,1  =    =  4400  

[ 2][ 2]

Kc,2  =

Kc,3  =

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The “reaction quotient”, Q:

Recall that the concentrations in the Kc expression are equilibrium values.  

As such, we could more accurately represent the Kc expression thusly:  **eqm          eqm **However, there are times when it is useful for a chemist to know  eqm         eqm

the numerical value for this expression under non-equilibrium conditions.

In these cases, the ratio is termed the reaction quotient:  

Q

Calculating the reaction quotient can help a chemist determine if a particular system is at equilibrium and if not, in which direction the system will react or “shift” in order to reach equilibrium.

For example, say a chemist is analyzing the equilibrium from equation (1) above and finds that at 500ºC, [H2] = [Cl2] = 0.10 M and [HCl] = 2.0 M. What conclusions can she draw?

Comparing the value of Q to the value of Kc can help us predict shift direction:

• If Q < Kc…**

• If Q > Kc…**

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Practice Problems: The Equilibrium Law

1. Write equilibrium expressions for the following systems:

a)   2SO2(g) + O2(g) ⇌** 2SO3(g) b)  4NH3(g) + 5O2(g) ⇌** 6H2O(g) + 4NO(g)

c)   3O ⇌** 2O d)  CH3COOH + H O ⇌** H O+ + CH3COO–

2(g) 3(g) (aq) 2 (l) 3 (aq) (aq)  

e)   H2O(g) + C(s) ⇌** CO(g) + H2(g) f)  H2(g) + ½O2(g) ⇌** H2O(g)

2. Which systems from #1 are homogeneous equilibria and which are heterogeneous equilibria?  

3. At 450 K, the value of Kc for the equilibrium from 1(a) is 4.0×10-4. What is the value of Kc for:

   1) SO2(g) + ½O2(g) ⇌** SO3(g) b) SO3(g) ⇌ SO2(g) + ½O2(g)

2. At 25ºC, the value of Kc for the equilibrium from 1(f) is 16. What is the value of Kc for:

   1) H2O(g) ⇌** H2(g) + ½O2(g) b) 2H2(g) + O2(g) ⇌** 2H2O(g)

2. Consider the equilibrium for the Haber Process:  N2(g) + 3H2(g) ⇌** 2NH3(g)

1. Determine the missing values in the table, given that all measurements were taken at equilibrium:

|[N2]eqm |[H2]eqm|[NH3]eqm|T |Kc |

| - | - | - | - | - |

|0.10 M |0.30 M |? |55ºC |0.20 |

|? |0.50 M |0.10 M |55ºC |? |

|

-4

7.5×10 M

|? |0.010 M |? |0.20 |

|3.0 M |0.40 M |0.015 M |125ºC |? |

|0.25 M |? |2.0 M |? |32 |

i) ii) iii) iv) v)

2. 2.0 mol N2(g), 3.0 mol H2(g), and 1.2 mol NH3(g) were put into a 1.0 dm3 flask at 55ºC. Is the system at equilibrium? If not, in which direction will it shift to reach equilibrium?

3. 12 mol N2(g), 1.5 mol H2(g), and 2.5 mol NH3(g) were put into a 5.0 dm3 flask at 55ºC.

Is the system at equilibrium?  If not, in which direction will it shift to reach equilibrium?

6. Consider the ozone-oxygen equilibrium:  3O2(g) ⇌** 2O3(g) Kc at 0ºC = 5.0×10-12

1. Discuss the significance of the magnitude of Kc in terms of the concentrations of each species present at equilibrium.  

2. Why is this equilibrium so important to human life on Earth?

3. 2.0 grams of oxygen and 0.025 mg of ozone are present in a 1.0 dm3 flask at 0ºC.

Is the system at equilibrium? If not, in which direction will it shift to reach equilibrium?

4. 1.0 kg of oxygen and 1.0 mg of ozone are present in a 5.0 dm3 flask at 0ºC. Is the system at equilibrium?  If not, in which direction will it shift to reach equilibrium?

5. Calculate the value of Kc for the following equilibria, as written:

3 (ii) 2O  3O

(i)  O2(g) ⇌** O3(g)   3(g) ⇌** 3O2(g)  (iii) O3(g) ⇌**  2(g)

2 2

7. What can we conclude about the value of Kc for a typical combustion reaction? Why?

8. State at least one experimental condition necessary before equilibrium can be achieved.

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Le Châtelier’s Principle

We previously explored chemical systems moving toward a position of equilibrium. Now we explore how chemical systems at equilibrium respond when they are disturbed by external changes.

Le Châtelier’s Principle states that if a closed system at equilibrium is subjected to a change,            

the system will respond by shifting in one direction to counteract that change.

Consider the Haber process:  N2(g) + 3H2(g) ⇌** 2NH3(g)    ΔH = –92 kJ Let’s see how this principle applies to a variety of changes (“stresses” or “disruptions”):

1. Changes in concentration (by addition or removal of one species)

• If the concentration of one species is an equilibrium is increased, the equilibrium will respond by shifting away from the addition.

• If the concentration of one species is an equilibrium is decreased, the equilibrium will respond by shifting toward the removal.

Example:  The N2/H2/NH3 system above is initially at equilibrium in a 1.0 L glass bulb when some

additional hydrogen gas is suddenly injected into the system:

[  ]  Q  rate

[H2]

[N2]

[NH3]

time  time  time

2. Changes in temperature  

• If the temperature of an equilibrium system is increased, the equilibrium will respond by shifting in the endothermic direction.

• If the temperature of an equilibrium system is decreased, the equilibrium will respond by shifting in the exothermic direction.**

Example:  The N2/H2/NH3 system above is at equilibrium when the temperature is increased. [  ]   Q  rate

[H2]

[N2]

[NH3]

time  time  time

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3. Changes in pressure:

• If the pressure on an equilibrium system is increased, the equilibrium will respond by shifting toward the side with fewer gas particles (fewer moles of gas).

• If the pressure on an equilibrium system is decreased, the equilibrium will respond by shifting toward the side with the greater number of gas particles (more moles of gas).

Example:  The N2/H2/NH3 system above is at equilibrium when the pressure is increased. [  ]  Q  rate

[H2]

[N2]

[NH3]

time  time  time

4. Adding a catalyst

Recall that catalysts increase the rate of chemical reactions by lowering the activation energy, allowing a greater fraction of particles to exceed the minimum kinetic energy required to react.

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Practice Problems: Le Châtelier’s Principle

1. Consider the equilibrium:   184 kJ + 2HCl(g) ⇌ H2(g) + Cl2(g)

In which direction would the equilibrium shift in the event of the following disruptions? a)  Increasing the temperature   b)  Injecting some HCl(g)

c)  Increasing the pressure  d)  Adding some Fe(s), which reacts with only the H2(g)

2. Consider the equilibrium:   2H2(g) + 2NO(g) ⇌ N2(g) + 2H2O(g) ΔHrxn = –84 kJ In which direction would the equilibrium shift in the event of the following disruptions?

a)  Decreasing the temperature  b)  Increasing the volume of the container

c)  Injecting nitrogen monoxide gas   d)  Removing some water vapour via condensation

e)  Compare and contrast the equilibrium response when argon (an inert gas) is added to this equilibrium when it is contained in a glass flask versus in an expandable balloon.

3. For each disruption from #1 and #2, deduce:

   1) The instantaneous change to Q (up, down or none) upon the disruption

   1) The change in Q (up, down or none) during the equilibrium shift

   1) The value of Kc after the disruption, compared to the initial Kc (greater, less, same)

3. Consider the equilibrium:   2CO(g) + O2(g) ⇌ 2CO2(g) ΔHrxn = +566 kJ Describe the net effect on Q and the concentration of carbon dioxide for each of the following

disruptions on this system, assuming it was initially at equilibrium:

a)  Increasing the temperature   b)  Injecting some oxygen gas  c)  Decreasing the pressure

d)  Injecting some carbon monoxide gas  e)  Injecting some carbon dioxide gas

5. Consider the equilibrium:   2NO2(g) ⇌ N2O4(g)

At 150ºC, the equilibrium constant is 5.0×10-5. At 25ºC, the equilibrium constant is 2.5×10-3. Deduce the sign of ΔH and rationalize it using principles of chemical bonding and structure.

6. At a particular temperature, Kc for the Haber Process equilibrium we worked with on pp. 7-8 is known to be 125. The temperature is then changed and the new Kc is calculated to be 32. Is this new temperature higher or lower than the initial temperature? How do you know?

Answer questions#7-9 on the [  ]/Q/rate graph templates provided. Choose arbitrary initial values for the concentrations of each species, Kc and rates, then plot the effects of each disruption. Assume that each system is initially at equilibrium and takes three minutes to return to equilibrium after each disruption.

7. Consider the equilibrium:   3O2(g) ⇌ 2O3(g) ΔHrxn = +286 kJ

a)  At t = 5 min, some oxygen gas is injected   b)  At t = 10 min, temperature is decreased

c)  At t = 15 min, a copper plate is introduced to the system, which reacts with the ozone only

8. Consider the equilibrium:   2SO2(g) + O2(g) ⇌ 2SO3(g) ΔHrxn = –197 kJ

a)  At t = 5 min, temperature is increased  b)  At t = 10 min, some SO2 is injected

c)  At t = 15 min, the pressure is increased by decreasing the volume of the container

9. Consider the equilibrium:  N2(g) + 3H2(g) ⇌** 2NH3(g)    ΔH = –92 kJ

a)  At t = 5 min, pressure is decreased  b)  At t = 10 min, temperature is increased

c)  At t = 15 min, a condenser is introduced to the system, removing only ammonia

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Industrial application: The Haber Process

The Haber Process is one equilibrium of particular significance that we will study in detail:

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = –92 kJ/mol  

Between 1886 and 1911, Fritz Haber developed a process for the direct combination   of nitrogen and hydrogen gases over an iron catalyst to produce ammonia.  

Large quantities of ammonia were needed at the time for two main applications:  

• Bombs:  Ammonia was used to produce nitric acid (HNO3), a key chemical in  the manufacture of munitions (bombs, ammunition, other weapons)  

• Bread:  Ammonia is a key fertilizer; applying ammonia to soil improves crop yields.      

During WWI, Germany’s main supply of ammonia was from mining efforts in Chile, but the Allies blockaded German ships and cut off the supply, crippling production of German munitions and food.

It is estimated that billions are alive today because Haber found a way to transform relatively inert atmospheric nitrogen into ammonia (“bread from the air”), but that tens of millions have died as a direct result of explosives created from Haber-manufactured ingredients.

Talk about a complicated legacy! Let’s now explore the chemistry of the Haber process:

According to Le Chatelier’s Principle, which temperature and pressure conditions will result in the highest yield of ammonia?

In practice, these temperature and pressure conditions are not used.  Why?

• The reaction is typically conducted at a temperature near 500ºC (~800 K).

• The pressure in the reaction vessel is typically close to 200 atm.

200      400       600      800     1000

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Environmental application: The ozone-oxygen equilibrium

Most of the ozone in Earth’s atmosphere exists in the layer known as the stratosphere (15-50 km above Earth’s surface). Stratospheric ozone absorbs most of the ultraviolet radiation from the Sun. Without ozone, the Sun’s intense UV radiation would sterilize the Earth’s surface.

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Ozone is formed due to the photoreaction of molecular oxygen. When an O2 molecule absorbs a photon of light (hν), it dissociates into highly reactive, monoatomic oxygen. These O atoms then immediately react with another O2 molecule to form ozone:

Ozone is also destroyed by a photoreaction. The absorption of ultraviolet light breaks an O–O bond in the ozone molecule, generating O2 and a highly reactive oxygen atom, which reacts with another ozone molecule to produce two more O2 molecules:

O2 + hν → O + O O2 + O → O3

O3 + hν → O + O2 O3 + O → O2 + O2

Equilibrium-10

Because of these competing chemical processes, an equilibrium is  3O2(g) ⇌ **2O3(g) **established between oxygen and ozone in our atmosphere:

In the kinetics unit, we explored how NO catalyzes ozone depletion via a three-step mechanism (p. 14).

Ozone depletion can also be catalyzed by chlorine radicals released

from chlorofluorocarbons (CFCs) after they undergo homolytic fission:  e.g.  CCl2F2 → Cl•  +   •CClF2  

In the mid-20th century, CFCs were in common use as refrigerants and

fire-fighting materials. In the 1970s, scientists discovered that atmospheric  Cl• + O3 → ClO• + O2 ozone concentrations were in rapid decline and showed that CFCs were  ClO• + O → Cl• + O2

largely responsible. By 1987, many nations had signed a treaty called the

Montreal Protocol, which severly limited CFC use and production.    O3 + O → O2

UV light from the Sun is classified into three groups, according to their relative energies:

UV-A: 400-315 nm   UV-B: 315-280 nm   UV-C: 100-280 nm

Answer the following questions to explore how the ozone-oxygen equilibrium in the Earth’s atmosphere plays a key role in the absorption of UV light from the Sun:

1. Draw Lewis structures for the oxygen and ozone molecules. Which of the two molecules exhibits resonance? State the O–O bond order in each molecule.

2. Compare and contrast the bond strengths and bond lengths in oxygen versus ozone.

3. The average bond enthalpy for the oxygen-oxygen bond in O2 is given in the Data Booklet in units of kJ mol-1. Calculate this bond enthalpy in units of Joules per bond.

4. Using your answer from #4, calculate the maximum wavelength of the photon that has enough energy to cleave (break) the oxygen-oxygen bond in O2.

5. The bond enthalpy for the oxygen-oxygen bond in ozone is 380 kJ mol-1. Repeat #3 and #4 for the oxygen-oxygen bond in ozone to deduce the maximum wavelength of light that they absorb.

6. Which class(es) of UV light (A, B, C) are most effectively absorbed by oxygen? By ozone?

7. Explain which class of UV light is the most energetic and hence, most dangerous to human health.

8. Peak concentrations of about 8 ppm (8 ozone molecules per million total atmospheric molecules) occur about 25km above the Earth’s surface. What can we conclude about the value of Kc for the ozone-oxygen equilibrium (as written in the box, above-right)?

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Equilibrium (Kc) Calculations

Previously, we explored how Le Châtelier’s Principle can be applied to predict shift direction in the event of changes in concentration, temperature or pressure. Now, we will learn how the equilibrium law can be applied to quantify the effects of such changes.

Recall the Haber Process for producing ammonia:     N2(g) + 3H2(g) ⇌ 2NH3(g) + 92 kJ Example 1:  At 100ºC, a 5.0 L flask contains 1.0 mol of N2(g), 2.5 mol of H2(g), and 12 mol of NH3(g)

in a state of chemical equilibrium. Calculate Kc.

Example 2:  5.0 mol of ammonia were added to an empty 10.0 L bulb and the system was allowed to reach equilibrium at 250ºC. The equilibrium concentration of H2(g) was 0.36 M. Calculate Kc.

N2 (g)      +   3H2 (g)   ⇌ 2NH3 (g)

|I (initial) ||||

| - | :- | :- | :- |

|S (shift) ||||

|E (equil.) ||||

Example 3:  Some NH3(g) was introduced to an empty 1.0 L flask. After the system reached equilibrium at 600ºC, 0.40 mol of N2(g) were present. If Kc = 0.0072, how many moles of NH3(g) were added initially?

N2 (g)      +   3H2 (g)   ⇌ 2NH3 (g)

|I (initial) ||||

| - | :- | :- | :- |

|S (shift) ||||

|E (equil.) ||||

How can Kc be different for Examples 1-3 if we are examining the same equilibrium system?

Example 4:  For the equilibrium below, Kc = 42 at 275ºC. If 1.0 mol of CO2 and 1.0 mol of H2 are injected into an initially empty 1.0 L flask and the system is allowed to reach equilibrium, calculate the concentrations of all species once equilibrium is reached.

 CO(g)    +   H2O(g)   ⇌ CO2(g)   +    H2 (g)

|I (initial) |||||

| - | :- | :- | :- | :- |

|S (shift) |||||

|E (equil.) |||||

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Example 5:  At a temperature of 225ºC, Kc = 133 for the equilibrium below. If 5.0 mol of CO2, 5.0 mol of H2, 1.0 mol of H2O and 1.0 mol of CO are injected into a 1.00 L flask at 75ºC, in which direction will the equilibrium shift?  Determine the equilibrium concentrations of each species.  

 CO(g)    +   H2O(g)   ⇌ CO2(g)   +    H2 (g)

|I (initial) |||||

| - | :- | :- | :- | :- |

|S (shift) |||||

|E (equil.) |||||

Considering data from Examples 4 and 5, deduce whether the equilibrium is exothermic or endothermic as written.

Calculate DG at both temperatures (225ºC and 275ºC) and comment on the differences between the two values. Flip ahead a page or two for help!

Example 6:  At a different temperature, the same system reached equilibrium in a 1.0 L flask and was found to contain 2.0 mol of CO, 2.0 mol of H2O, 1.0 mol of CO2 and 1.0 mol of H2. If 1.0 mol of CO was injected into this system, determine the amount of H2 present after equilibrium re-establishes.

 CO(g)    +   H2O(g)   ⇌ CO2(g)   +    H2 (g)

|I (initial) |||||

| - | :- | :- | :- | :- |

|S (shift) |||||

|E (equil.) |||||

Equilibrium-13

The Equilibrium Constant (Kc) and Spontaneity (DGº)

Recall that the Gibbs free energy change (DGº) determines whether or not a reaction is spontaneous. The Gibbs free energy change is related to the equilibrium constant by van’t Hoff’s equation:

DGº = –RTlnK

Let’s examine this mathematical relationship in broad terms:

• When K is greater than 1…

• When K is less than 1…

Consider the Haber Process (p. 9):     N2(g) + 3H2(g) ⇌ 2NH3(g)     ΔGº = –33 kJ/mol

• The overall Gibbs free energy (G) of a system depends on how much of each substance is present.

• The equilibrium mixture represents the composition of reactants and products that gives a minimum value of the Gibbs free energy.

Practice:

1. For the equilibrium PCl3(g) + Cl2(g) ⇌ PCl5(g) at 275ºC, DG = –99 kJ mol-1. Calculate Kc at this temperature.

2. Using data from Table 12 of the Data Booklet and Sº(H ) = 131 J mol-1 K-1, calculate ΔHº, ΔSº and ΔGº for 2

the equilibrium system CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g). Calculate Kc and comment on its magnitude.

3. Predict the signs of ΔHº and ΔSº for the equilibrium system N2O4(g) ⇌ 2NO2(g), explaining your reasoning for each. Then, predict and explain the effect of increasing temperature on ΔGº and Kc.

4. At 450ºC, Kc = 2.4×104 for the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g). Given ΔS = –189 J mol-1 K-1, calculate ΔG and ΔH at this temperature.

1. 2.7×109; 2. ΔHº = 205 kJ mol-1, ΔSº = 216 J mol-1 K-1, ΔGº = 141 kJ mol-1, K = 2.0×10-25;  

c

3. ΔHº>0, ΔSº> 0. Increasing T causes a decrease in ΔG and an increase in K ; 4. ΔG = –61 kJ mol-1, ΔH = +76 kJ mol-1. c

Equilibrium-14

Equilibrium Calculations – Practice

1. Consider the following equilibrium:   2NO(g) + O2(g) ⇌ 2NO2(g)

In an experiment, 0.30 mol of NO and 0.80 mol of NO2 are placed in a 5.0 L container at 10ºC.  When equilibrium is reached, it is found that [O2] = 0.020 M. Calculate Kc and ΔG.

2. At 25ºC, Kc = 3.00 for the following equilibrium:  N2(g) + 3H2(g) ⇌ 2NH3(g)

Some NH3 was added to a 1.0 L container and after equilibrium was established, the container was found to contain 2.5 mol of N2. How many moles of NH3 were originally introduced? What mass?

3. At 25ºC, Kc = 49.5 for the following equilibrium:  H2(g) + I2(g) ⇌ 2HI(g)

1. If 0.250 mol of H2 and 0.250 mol of I2 are placed into a 10.0 L vessel at 25ºC and permitted to react, what will be the concentration of each substance at equilibrium?

2. If 0.150 mol of HI, 0.0340 mol of H2, and 0.0960 mol of I2 are introduced to a 8.00 L vessel, in which direction will the reaction proceed in order to reach equilibrium?**  

4. At 0ºC, Kc = 720 for the following equilibrium:   2NO2(g) ⇌ N2O4(g)

Predict the shift direction when the following systems are introduced into a 10.0 L bulb: a) no NO2, 1.5 mol N2O4  b) 0.050 mol NO2, 0.500 mol N2O4  c) 10 g NO2, 50 g N2O4

Calculate DG for this equilibrium at 0ºC and predict the effect of increasing temperature on the value of ΔG, explaining your reasoning.

5. At 55ºC, Kc = 35.0 for the following equilibrium:  PCl5(g) ⇌ PCl3(g) + Cl2(g)

If [PCl5] = 1.34 × 10-3 M and [PCl3] = 0.205 M at equilibrium in a 6.00 L vessel, what is the equilibrium concentration of Cl2(g)? Calculate ΔG at this temperature.

6. Consider this equilibrium system:   2NO(g)  + O2(g) ⇌2NO2(g)

In an experiment, 1.2 mol of NO and 1.8 mol of O2 were placed in a 5.0 L container at 10ºC. When equilibrium was reached, it was found that [NO2] = 0.060 M. Calculate Kc for this reaction at 10ºC.

7. Kc = 5.0 at 298 K for the equilibrium:   2SO2(g) + O2(g)  ⇌ 2SO3(g)

A certain amount of SO3(g) was placed in a 2.0 L reaction vessel. At equilibrium the vessel contained 0.30 mol of O2(g). How many moles of SO3(g) were originally placed in the vessel?

8. Consider this equilibrium system:   H2 (g)  + I2 (g) ⇌ 2 HI(g)

At equilibrium, there are found to be 2.00 mol of H2 and 2.00 mol of I2 in equilibrium with 4.00 mol of HI in a 10.0 L container. If 1.00 mol of HI is then added and the system is allowed to re-establish equilibrium, find the new [HI].  

9. A student obtained the following data at 25ºC when studying the equilibrium below:   2+

|Container volume |Moles Tl+ |

2+

Moles Cd

|

| - | - | - |

|1.00 L |0.316 |0.414 |

|5.00 L |? |0.339 |

2Tl+(g) + Cd(s) ⇌ 2Tl(s) + Cd

(g)

1. Calculate the number of moles of Tl+ present in the second data set.  

2. Calculate DG for the equilibrium at 25ºC and deduce whether the position of the equilibrium lies closer to the reactants or products side of the equation.

Equilibrium-15

10. Kc = 5.0 at 298 K for the equilibrium:    2SO2(g) + O2(g)  ⇌ 2SO3(g) + 196 kJ

At another temperature, Kc is determined to be 0.50. Is this other temperature above or below 298 K?  Explain your reasoning.  

11. At a certain temperature, the Kc is 6.80 x 10-2 for the following reaction: H2(g) + S(s) ⇌ H2S(g)

At this temperature, if 0.200 mol of hydrogen and 2.00 mol of sulphur are allowed to come to equilibrium in a 1.00 L flask, what will the final concentration of H2(g) be in the flask?

12. At 300ºC, Kc = 10.0 for the following equilibrium:

CO(g) + H2O(g) ⇌ H2 (g) + CO2 (g)

1. If 2.00 mol of CO(g) and 2.00 mol of H2O(g) are initially present in a 5.0 L flask, determine the number of moles of each species present after the system reaches equilibrium.

2. Given that at 20ºC, Kc = 0.22 for the same equilibrium, state and explain the sign of ΔH.

13. Kc = 1.60 x 10-4 at 298 K for the equilibrium:  2HCl(g) ⇌ H2(g) + Cl2(g)

Some HCl(g) was placed inside an empty 4.00 L flask and the contents reached equilibrium at 298 K.   If [H2] = 0.0150 M at equilibrium, how many grams of HCl were originally added to the flask?

14. Consider the following equilibrium system:  CO(g) + H2O(g) ⇌ H2 (g) + CO2 (g)

In a 1.0 L flask, the system was at equilibrium with:  

[CO] = 0.20 M, [H2O] = 0.30 M, [H2] = 0.50 M, [CO2] = 1.0 M.  

How many moles of H2 (g) needs to be added to the system in order to increase the H2O(g) concentration to 0.40 M? (Hint: Calculate Kc first!)

15. Consider the equilibrium at the right:     CO(g) + H2O(g) ⇌ H2 (g) + CO2 (g) The equilibrium concentrations of a particular mixture in a 2.0 L reaction vessel are:   [CO] = 0.10 M, [H2O] = 0.10 M, [H2] = 0.20 M, [CO2] = 0.20 M

How many moles of CO2 must now be injected into the vessel in order to create a new equilibrium where the [CO] = 0.20 M?

Answers:

1.  72, –10 kJ mol-1 2.  61 mol, 1.0 kg   3. a) [H2] = [I2] = 0.0055 M, [HI] = 0.039 M, b) Q = 6.9, shift right

4.  left, left, right, DG = –15 kJ/mol   5.  0.229 M, –9.7 kJ mol-1         6.  0.34           7.  1.1 mol  8.  0.45 M

9. (a) 0.639 mol   (b) DG = –3.5 kJ/mol; products side (but only slightly…why?)  

10. Above 298 K; according to Le Châtelier’s Principle, excess heat should result in a shift to the left.

11. 0.187 M

12a) 0.48 mol CO, 0.48 mol H2O, 1.52 mol H2, 1.52 mol CO2  

12b) The sign of ΔH has to be positive because the equilibrium is endothermic as written (the “heat” term is on the left).  We know this because as temperature is decreased, the equilibrium shifts to the left resulting in a lower Kc.

13.  177 g  14.  0.71 mol  15.  3.0 mol

Equilibrium-16

Equilibrium Review Problems

1. Consider the following equilibrium:  CO(g) + H2O(g)  ⇌  CO2(g) + H2(g)        ΔH = –41 kJ

In which direction will the equilibrium shift if:

a)  Some CO2(g) is injected?  b)  Some H2O(g) is removed using a drying agent?

c)  The temperature is increased?  d)  The volume of the vessel is increased?

2. Draw a graph showing the behaviour of [CO2] as a function of time as each of the changes in #1 occur.  (You can just draw one graph; assume change (a) takes place at t = 5 minutes, (b) at 10 minutes, (c) at 15 minutes, and (d) at 20 minutes.  Assume the equilibrium takes 2 minutes to re-establish itself).

3. For each of the equilibria below, state three changes that will cause a shift toward the products side.

   1) N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = –92 kJ

   1) 3O2(g) ⇌ 2O3(g) ΔH = 286 kJ

   1) H2(g) + S(s) ⇌ H2S(g) ΔH = –73 kJ

   1) H2O(l) ⇌ H2O(g)

2. For each of the changes in #3, state the effect on the value of the equilibrium constant, Kc.

3. Analyze the graph below and determine the changes that took place at time = 10, 20 and 30 minutes.   PCl5(g)  ⇌  PCl3(g) + Cl2(g);  DH > 0

6. Explain the significance of the reaction quotient, Q. Why would we ever need to calculate it? Draw a graph of reaction quotient (Q) versus time that corresponds to the graph from #5.

7. Consider the following equilibrium:  2NO(g) + O2(g) ⇌ 2NO2(g)

1.20 mol of NO2 are placed in a 5.00 L container at 15ºC. When equilibrium is reached, it is found that      [NO] = 0.180 M.  Calculate Kc and DG and deduce whether the position of equilibrium favours the reactants or the products side of the equation.

8. Consider the reaction:  2NO(g) + Cl2(g) ⇌ 2NOCl(g)

1. A mixture of NO and Cl , each originally at a concentration of 0.60 mol dm-3 in a 5.0 dm3 container at

2

325ºC is allowed to reach equilibrium.  When the system reaches equilibrium, 10.0% of the original NO has reacted. Calculate the equilibrium constant (Kc) and Gibbs free energy change (DG) for the reaction. Discuss

whether the position of equilibrium lies closer to the reactants or products side of the equation.

2. 1.00 mol of NOCl was introduced into a 2.00 dm3 flask and allowed to come to equilibrium.                      If 0.200 mol of Cl2 was found at equilibrium, what is Kc for the reaction?

9. In a reaction:  A2(g) + B2(g) ⇌ 2AB(g)

There are found to be 1.00 mol of A and 1.00 mol of B in equilibrium with 5.00 mol of AB in a 1.00 dm-3

2 2

container.  If 1.00 mol of AB is then added and the system is allowed to re-establish equilibrium, find the new concentration of AB.  (Hint: determine Kc first!)

10. At 25ºC, Kc = 7.5 for the following equilibrium:  2H2(g) + S2(s) ⇌ 2H2S(g)

A certain amount of H2S was added to a 1.00 L flask and allowed to come to equilibrium.  At equilibrium, 3.206 g of S2 were found.  What mass of H2S was originally added to the flask?

11. At 25ºC, Kc = 30.0 for the following equilibrium:  N2(g) + 3H2(g) ⇌ 2NH3(g)

Enough NH3 was added to a 5.00 L container such that, after equilibrium was established, the container was found to contain 0.250 mol of N2.  How many moles of NH3 were originally introduced into the container?

12. At 25ºC, Kc = 48.0 for the following equilibrium:  H2(g) + I2(g) ⇌ 2HI(g)

1. If 0.250 mol of H2 and 0.250 mol of I2 are placed into a 2.00 L vessel at 25ºC and permitted to react, what will be the concentration of each substance at equilibrium?

2. If 0.350 mol of HI, 0.120 mol of H2, and 0.885 mol of I2 are introduced to a 5.00 L vessel, in which direction will the reaction proceed in order to reach equilibrium?

13. Consider the following equilibrium:  H2(g) + Cl2(g) ⇌ 2HCl(g)

0.150 mol of H2, 0.150 mol of Cl2, and 0.870 mol of HCl are at equilibrium in a 10.0 L flask.  If 0.400 mol of HCl are now added to this system and the system is allowed to come to equilibrium again, what will the new concentrations of H2, Cl2, and HCl be?

14. Consider the following equilibrium:    2NO2(g) ⇌ N2O4(g)

1.00 mol of N2O4(g) was placed in a 1.00 L flask at 298 K and allowed to come to equilibrium.  Once equilibrium was reached, 1.36 mol of NO2(g) were in the flask.

1. Determine Kc for the above equilibrium at 298 K.

2. Plot concentration vs. time as the system comes to equilibrium, clearly illustrating [NO2] and [N2O4].

3. Plot rate vs. time as the system comes to equilibrium, clearly illustrating both the fwd and rev rates.

4. Is the above equilibrium exothermic or endothermic as written?  Explain.

5. Assume that at some time after equilibrium was reached, the temperature was increased.  Continue your graphs from (b) and (c) and plot the effects of this change.

15. Consider a N2/H2/NH3 system at equilibrium at 55ºC and t = 0 min.  At t = 5 min, some NH3(g) was injected into the system.  At 10 min, the temperature was decreased to 20ºC.  At 15 min, the temperature was increased to 100ºC.  Plot the reaction quotient, Q, as a function of time, assuming that the system takes 2 minutes to return to equilibrium after each disturbance.

Answers

1.  a) left, b) left, c) left, d) none

3. a) add N2 or H2, remove NH3, decrease T, increase P  b) add O2, remove O3, increase T, increase P

   100) add H2, remove H2S, decrease T  d) remove H2O(g), increase T, decrease P

3. Did you say they all cause an increase in Kc? Remember, only temperature changes affect the value of Kc!  

4. At 10 mins, some PCl5 was added; at 20 mins, volume was increased (pressure decreased); at 30 mins, temp was increased

5. Q is the value of the equilibrium expression for non-equilibrium concentrations (i.e. when we are not at equilibrium).

Q can be compared to Kc to determine the direction that the system will shift.

7. 1.23, –496 J/mol (0.496 kJ/mol), very slightly favours the products side  

8. a) K = 0.022, DG = 19 kJ/mol, reactants side,  b) 22.5  9.  5.71 mol dm-3 10.  12.7 g (0.374 mol)

11.  0.856 mol  12.  a) [H2] = [I2] = 0.028 M, [HI] = 0.194 M, b) right

13.  [H2], [Cl2] = 0.0201 M, [HCl] = 0.117 M  14. a) 0.173, d) exo (bonds formed)