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# Chemical Equilibria

### Skills to develop

• Identify a system, an open system, a closed system and the environment of the system.
• Define a state of equilibrium.
• Describe the mass action law.
• Apply the mass action law to write expressions for equilibrium constant.
• Write the equilibrium constant expression for any reaction equation.

# Chemical Equilibria

Heat is energy flowing from a high temperature object to a low temperature object. When the two objects are at the same temperature, there is no net flow of energy or heat. That is why a covered cup of coffee will not be colder than or warmer than the room temperature after it has been in there for a few hours. This phenomenon is known as equilibrium. In this example, we deal with the flow of energy.

Equilibria happen in phase transitions. For example, if the temperature in a system containing a mixture of ice and water is uniformly 273.15 K, the net amount of ice formed and the melt will be zero. The amount of liquid water will also remain constant, if no vapour escape from the system. In this case, three phases, ice (solid) water (liquid), and vapour (gas) are in equilibrium with one another. Similarly, equilibrium can also be established between the vapour phase and the liquid at a particular temperature. Equilibrium conditions also exist between solid phase and vapour phases. These are phase equilibria.

Chemical reactions may not be as complete as we have assumed in Stoichiometry calculations. For example, the following reaction are far short of completion.

2 NO2 = N2O4
3 H2 + N2 = 2 NH3
H2O + CO = H2 + CO2

Let us consider only the first reaction in this case. At room temperature, it is impossible to have pure NO2 or N2O4. However, in a sealed tube ( closed system), the ratio

```   [N2O4]
-------
[NO2]2
```
is a constant. This phenomenon is known as chemical equilibrium. Such a law of nature is called the law of mass action or mass action law.

Of course, when conditions, such as pressure and temperature, change, a period of time is required for the system to establish an equilibrium.

Before we introduce the mass action law, it is important for us to identify a system or a closed system in our discussion. The law provides an expression for a constant for all reversible reactions.

For systems that are not at equilibrium yet, the ratio calculated from the mass action law is called a reaction quotient Q. The Q values of a closed system have a tendency to reach a limiting value called equilibrium constant K over time. A system has a tendency to reach an equilibrium state.

## A Closed System for the Equilibrium State

In order to discuss equilibrium, we must define a system, which may be a cup of water, a balloon, a laboratory, a planet or a universe. Thus, for discussion purpose, we define an isolated portion of the universe as a system, and anything outside of the system is called environment.

When the system under consideration is isolated from its environment in such a way that there is no energy or mass transferred into or out of the system, the system is said to be a closed system.

In a closed system, changes continue, but eventually there is no NET change over time. Such a state is called an equilibrium state.

For example, a glass containing water is an open system. Evaporation let water molecules to escape into the air by absorbing energy from the environment until the glass is empty. When covered and insulated it is a closed system. Water vapour in the space above water eventually reaches a equilibrium vapour pressure.

In fact, measuring of temperature itself requires the thermometer to be at the same state as the system it measures. We read the temperature of the thermometer when heat transfer between the thermometer and the system stops (at equilibrium).

Equilibrium states are reached for physical as well as chemical reactions. Equilibrium is dynamic in the sense that changes continue, but the net change is zero.

## Reversible Chemical Reactions

Heat transfer, vapourization, melting, and other phase changes are physical changes. These changes are reversible and you have already experienced them.

Many chemical reactions are also reversible. For example

N2O4 = 2 NO2
colourless   brown
and N2 + 3 H2 = 2 NH3

are reversible chemical reactions.

## The Law of Mass Action

The law of mass action is universal, applicable under any circumstance. However, for reactions that are complete, the result may not be very useful. We introduce the mass action law by using a general chemical reaction equation in which reactants A and B react to give product C and D. a A + b B --> c C + d D where a, b, c, d are the coefficients for a balanced chemical equation.

The mass action law states that if the system is at equilibrium at a given temperature, then the following ratio is a constant.

[C]c [D]d
------------- = Keq
[A]a [B]b
The square brackets "[ ]" around the chemical species represent their concentrations. This is the ideal law of chemical equilibrium or law of mass action.

The units for K depend upon the units used for concentrations. If M is used for all concentrations, K has units

Mc+d-(a+b)

## The Reaction Quotients Q and the Equilibrium Constants K

If the system is NOT at equilibrium, the ratio is different from the equilibrium constant. In such cases, the ratio is called a reaction quotient which is designated as Q.
[C]c [D]d
------------- = Q
[A]a [B]b
A system not at equilibrium tend to become equilibrium, and the changes will cause changes in Q that its value approaches the equilibrium constant, K Q ® Keq.

## Equilibrium Expressions

The mass action law gives us a general method to write the expression for the equilibrium constant of any reaction. At this stage, you should be able to write the equilibrium expression for any reaction equation. If you are not sure from the above general theory, here are some examples. It is more important for you to understand WHY the equilibrium constants are expressed this way than what is the equilibrium expression.

### Examples

1. Write the the equilibrium constant expression for the reaction equation: NH3 + HOAc = NH4+ + OAc-. Hint
```      [NH4+] [OAc-]
---------------------  =  K    (unitless constant)
[NH3] [HOAc]
```

2. For the ionization of an acid, H2SO4 = 2 H+ + SO42- what is the equilibrium constant expression?

Hint

The equilibrium constant is

```   [H+]2  [SO42-]
--------------  = K M2
[H2SO4]
```
where M = mol/L. Note the unit for K.

3. For the reaction equation: Cu2+ + 6 NH3 = Cu(NH3)62+ what is the equilibrium constant expression?

Hint

The expression is

```   [Cu(NH3)62+]
-----------------  =  K  M-6
[Cu2+] [NH3]6
```
The equilibrium constant depends on the written equation.

4. Discuss the ionization of oxalic acid, H2C2O4, in two stages.

Hint:

Experimentally, it has been shown that

H2C2O4 = H+ + HC2O4- - - - - - (1)
```    [HC2O4-] [H+]
-------------- = K1 = 0.059 M
[H2C2O4]
```
The second ionization constant is much smaller: HC2O4- = H+ + C2O42- - - - - - (2)
```    [H+]  [C2O42-]
--------------- = K2 = 0.000064 M
[HC2O4-]
```
The overall ionization can be obtained by adding (1) and (2) to give (3). H2C2O4 = 2 H+ + C2O42- - - - - - (overall)

and the equilibrium constant is

```    [H+]2 [C2O42-]
--------------  = Koverall M2
[H2C2O4]
```
It is obvious that Koverall = K1 * K2

= 3.8x10-6 M2

Please confirm the above obvious relationship to satisfy yourself.

5. At some temperature, the equilibrium constant is 4.0 for the reaction CO(g) + 3 H2(g) = CH4(g) + H2O(g). What is the equilibrium constant expression and value for the reaction, CH4(g) + H2O(g) = CO(g) + 3 H2(g)?

Hint:

 [CO] [H2]3 ------------------ = 1/4.0 = 0.25 [CH4] [H2O]

6. At some temperature, the equilibrium constant is 4.0 for the reaction CO(g) + 3 H2(g) = CH4(g) + H2O(g). What is the equilibrium constant expression and value for the reaction, 1/3 CO(g) + H2(g) = 1/3 CH4(g) + 1/3 H2O(g)?

Hint:

 [CH4]1/3 [H2O]1/3 ------------------------- = 4.01/3 = 1.59 [CO]1/3 [H2]
The application of the mass action law leads to the method to write the expression for the equilibrium constant. The law is given in a general form, and these examples should help you grasp the method.

### Confidence Building Questions

• A closed container with N2O4 and NO2 gases in it, and it has been placed in the lab for many days. What would you consider the container and the gases to be?
a. an open system
b. a closed system
c. not a system

Discussion... Since it has been in the lab, the temperature of the system is the same as its environment.

• A closed container with N2O4 and NO2 gases in it, and it has been placed in the lab for many days. Is the system at an equilibrium state for the chemical reaction?
N2O4 = 2 NO2

Discussion... Only a few minute is required for the reaction to reach an equilibrium state. NO2 is a brown gas

• A closed container with N2O4 and NO2 gases in it, and it has just been placed in an ice/water bath. Is the system at an equilibrium state for the chemical reaction?
N2O4 = 2 NO2

Discussion...
Heat will be extracted from the container, and that cause the equilibrium to shift. At 0 deg C, the equilibrium is shifted to have lots of N2O4.

• If the equilibrium constant for the reaction equation HCOOH + CN- = HCN + HCOO- is 5E5, what is the equilibrium constant for the reaction equation HCN + HCOO- = HCOOH + CN-?

Discussion...
Since the reaction equation is reversed, use the relationship

Kreverse = 1/Kforward. If you reverse the reaction equation, take the reciprocal of the previous equilibrium constant.

• For reactions taking place in the gas phase, the equilibrium constant is normally expressed in terms of partial pressures of the reactants and products. If C represents the concentration, and other symbols of the ideal gas equation are used, which of the following is correct?
(a) C = RT/(PV)
(b) C = RT/P
(c) C = RT/V
(d) C = RT/(PM)
(e) C = P/(RT)
(f) C = PV/(RT).

Discussion...
By definition, C = n / V. Derive it from n R T = P V.

• If Kc represents the equilibrium constant in terms of concentration, and Kp represents that in terms of partial pressure, which of the following is correct?
(a) Kc = Kp
(b) Kc is proportional to Kp
(c) Kc = 1/Kp
(d) Kc is inversely proportional to Kp
(e) Kc * Kp = RT

Discussion...
Statements (c), (d), and (e) may be true for special cases, but they are not generally true. In this list, Kc is proportional to Kp is true, but a more quantative relationship will be derived.

• For which of the following gas-phase reaction equations are the equilibrium constants unitless quantities?
(a) 2 H2 + O2 = 2 H2O
(b) 2 NO = N2 + O2
(c) COCl2 = CO + Cl2
(d) CO + H2O = CO2 + H2.