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Chemical Equilibrium
Topic 15
Introduction
Many reactions are reversible, meaning that they can occur in both the forward and reverse directions.
When the rate of the forward and reverse reactions are equal, the condition of chemical equilibrium exists.
A simple system at equilibrium is shown below.
CH4(g) + H2O(g) ↔CO(g) + 3H2(g)
In the forward direction, methane and water vapor combine to form carbon monoxide and
hydrogen. In the reverse direction, carbon monoxide and hydrogen combine to yield methane
and water vapor.
Chemical equilibrium is a dynamic state; both reactions are occurring simultaneously, but at the
same rate. As a result, the concentrations of reactants and products remain constant during
equilibrium. This does not mean that their concentrations are equal, only that they are constant.
At equilibrium in the system above, the concentrations of methane, water vapor, carbon
monoxide, and hydrogen would remain constant, though not equal.
Equilibrium Constant & the Law of Mass Action
At equilibrium, usually one side of the equilibrium is favored over the other. This is described
by the equilibrium constant expression. Consider a hypothetical system at equilibrium.
aA + bB ↔ cC + dD
The equilibrium constant expression for the system will always be the product of the molar
concentrations of products over the product of the molar concentration of reactant. This is
known as the law of mass action, or equilibrium law, and is shown by the expression….
Kc 
[C ]c [ D] d
[ A] a [ B]b
In the equilibrium expression, the exponents are simply the stoichiometric coefficients of each
reactant or product. The subscript “c” of Kc indicates concentration. In a gaseous reaction, the
reactants and products are expressed in partial pressures. In such cases, the equilibrium constant
is designated Kp.
For the reaction above, it is
Kp 
PCO  PH 2
3
PCH 4  PH 2O
Typically the partial pressures are expressed in atmospheres. Kp is related to Kc as follows.
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K p  K c RT 
n
R is the ideal gas constant (0.0821 L∙atm/mol∙K), T is absolute temperature, and Δn is the difference
between the beginning moles of reactants and the moles of products, i.e. the moles of reactants that are
tuned into products.
Exercise
Consider the equilibrium
N2O4 (g) ↔ 2 NO2 (g)
Write the equilibrium constant expression below.
What is the value of Δn? Explain your answer below.
Heterogeneous Equilibria
Homogeneous equilibria involve substances in a single phase. If some of the reactants or products are in
different phases, including aqueous solution, the equilibrium is heterogeneous.
Pure Solids and Liquids in Heterogeneous Equilibrium Expressions
Whenever pure liquids or pure solids appear in a heterogeneous equilibrium, they are not included in the
equilibrium expression. This is particularly applicable for solids with very low solubility (precipitates)
and for water. Some examples of such equilibria are…
Fe2S3 (s) ↔ 2 Fe3+(aq) + 3 S2- (aq)
2 H2O (l) ↔ H3O+ (aq) + OH- (aq)
Kc = [Fe3+]2[S2-]3
Kc = [H3O+][OH-]
Notice that neither the solid (Fe2S3) nor the liquid (H2O) appear in the equilibrium expression.
Exercises
Write the Kc expression for AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
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Write both the Kc and Kp expressions for CaCO3 (s) ↔ CaO (s) + CO2 (g)
Using Equilibrium Constants
For most equilibria, either the products or reactants will predominate. Because the equilibrium
constant, Kc or Kp, roughly approximates the ratio of products to reactants, its value roughly
describes the position of the equilibrium. If K >> 1, the product side is favored. If K << 1, the
reactant side is favored.
Exercise
Write the equilibrium constant expression (Kc) for each of the equations below. Predict which
side is favored and explain why.
a) 2 NO(g) + O2 (g) ↔ 2NO2 (g)
Kc = 5.0 x 1012
b) 2 HBr (g) ↔ H2 (g) + Br2 (g)
Kc = 5.8 x 10-18
If reactions a) and b) were written in reverse order, how would you determine Kc for the reverse
reactions? What would the values be? (2.0 x 10-13; 1.7 x 1017)
For some equations, coefficients have to be doubled when balancing them. What effect does
doubling the coefficients have on K? Write the Kp expression for the following equation.
SO2 (g) + ½ O2 (g) ↔ SO3 (g)
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Kp = 1.85 @ 1000K
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Rewrite the equation, doubling the coefficients, then write the new Kp expression. What is the
effect upon Kp? What is the value of Kp? (3.42)
Calculating K when Equilibrium Amounts are Known
When equilibrium quantities are known, write the equilibrium expression, calculate the molar
concentrations if necessary, substitute and solve.
Exercise:
Calculate the value of the equilibrium constant, Kc , for the system shown, if 0.1908 moles of
CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapor were present in a
2.00 L reaction vessel were present at equilibrium1. (4.9 x 10-3)
CO2 (g) + H2 (g) ↔CO (g) + H2O (g)
Calculating K when Some Equilibrium Quantities are Unknown
When some equilibrium quantities are known and some are not, then K is determined using the
available information and the stoichiometry of the equation. This information is organized in an
“ICE” chart. Equilibrium amounts are determined algebraically from initial concentrations and
changes in concentrations of reactants and products.
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This problem, and many other helpful explanations and examples can be found at
http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_Equilibrium_Constants.htm#Kone
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As before, begin 1) by writing the equilibrium expression and 2) determining molar
concentrations or partial pressures of all species. Then, 3) construct and ICE chart to organize
your information and to continue your problem solving.
Constructing an ICE Chart
“ICE” is an acronym that expresses the Initial concentration, the Change in concentration, and
the Equilibrium concentration for each species in the reaction. From the chart you can determine
the changes in the concentrations of each species and the equilibrium concentrations. Your
analysis must be consistent with the balanced chemical equation.
Example:
When 0.40 mol N2 and 0.96 mol H2 are placed in a 2.00 L container at a constant temperature,
the equilibrium concentration of NH3 is 0.14 M. Find the equilibrium constant, Kc.
The equation is: N2 + 3 H2 ↔ 2 NH3
The equilibrium expression is: Kc = [NH3]2/[N2][H2]3
Known molar conc’s are:
[N2]i = 0.40 mol/2.00 L
[H2]i = 0.96 mol/2.00 L
[NH3] eq
= 0.20 M
= 0.48 M
= 0.14M
Construct an ICE table. Use the known concentrations and the stoichiometry of the equation to
determine the equilibrium concentrations of N2 and H2.
Molar conc
N2
H2
NH3
Initial (i)
0.20
0.48
0
Change
-x
-3x
+2x
Equilibrium (eq) 0.20 - x 0.48 – 3x 2x
Since the equilibrium concentration of NH3 is known to be 0.14 M,
2x = 0.14 M
x = 0.07 M
so
Knowing x, the equilibrium concentrations of N2and H2 can be determined.
[N2] eq = 0.20-0.07
[H2] eq = 0.48 - 3(0.07)
[NH3] eq (given)
= 0.13M
= 0.27M
= 0.14M
Kc can now be calculated.
Kc = (0.14)2/[(0.13)(0.27)3] = 7.66
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Visualizing the Equilibrium
A plot of the concentrations of all species in the problem above would appear as shown below.
Note two things in particular. 1) Once equilibrium is achieved, the concentration of all species
remains constant, though not equal. 2) Equilibrium is achieved from either direction. The
problem above began with N2 and H2 and no NH3. This is represented hy the graph on the left.
Equilibrium could be achieved by beginning with NH3 alone and allowing some of it to
decompose into N2 and H2. The result would be the same.
Challenge
What initial concentration of NH3 would be required to achieve the same equilibrium
concentrations in the problem above?
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The Law of Mass Action: Applications of Equilibrium Constants
As stated previously, the equilibrium constant expression for a system is the multiplicative product of
the concentrations of products divided the multiplicative product of the concentration of reactants. This
is known as the law of mass action, or equilibrium law, and is shown by the expression….
Kc 
[C ]c [ D] d
[ A] a [ B]b
The Reaction Quotient
In accordance with the law of mass action, if a system is not at equilibrium, it will move toward
equilibrium. The reaction quotient, Q, is a quantity that is used to predict the direction of a reaction if
the system is not at equilibrium. Q is calculated exactly the same way as the equilibrium constant.
Q
[C ]c [ D] d
[ A] a [ B]b
When the values of Q and K are compared, the direction of the reaction can be predicted. In other
words, Q is useful in prediction the direction (forward or reverse) are reaction will take to achieve
equilibrium.
When….
Q=K
Q<K
Q>K
Direction of Reaction
No net change;
System is at equilibrium
Toward product (right)
Toward reactant (left)
Exercise
Determine if the systems below are in equilibrium. If not, predict the direction in which the reaction will
proceed.
1. H2 (g) + I2 (g) ↔ 2HI (g)
Kc = 49
[H2] = 0.10 M; [I2] = 0.10 M; [HI] = 0.70 M
2. Br2 (g) + Cl2 (g) ↔ 2 BrCl (g)
Kc = 6.9
[Br2] = 0.10 M; [Cl2] = 0.20 M; [BrCl] = 0.45 M
3. HF (aq) + H2O (l) ↔ F- (aq) + H3O+ (aq)
Kc = 6.8 x 10-4
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[HF] = 0.20 M; [F-] = 2.0 x 10-4 M; [H3O+] = 2.0 x 10-4 M
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LeChâtelier’s Principle
Chemical equilibria have a degree of stability and are sensitive to change. When a chemical equilibrium
is disturbed (stressed), it responds so as to minimize the change.
When an equilibrium is disturbed, the forward and reverse rates are no longer equal. The system
responds in a manner to re-establish the equality of the forward and reverse rates. A shift toward the
product side is a shift to the right, a shift toward the reactant side is a shift to the left.
In order to predict the response of an equilibrium to a stress, think of K as a desired number for the
system. The system will respond in a manner so as to try to return to that number.
Changes that disturb equilibria
1. Change in concentrations of reactants or products
2. Changes in volume or pressure for gaseous equilibria
3. Changes in temperature
Changes in concentrations
[ NH 3 ]2
Kc 
[ N 2 ][ H 2 ] 3
N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
If the concentration of NH3 is increased, then Kc becomes too large. The equilibrium will shift to the left
to return toward Kc. By shifting to the left, the concentration of NH3 is decreased, and the
concentrations of N2 and H2 are increased. This lowers Kc back toward the optimum value
Changes in volume/pressure
Kp 
PNH 3
2
PN 2  PH 2 3
When volume is decreased, pressure increases. An equilibrium will respond to an increased pressure in
a manner which reduces pressure. Since total pressure is a function of total moles of a gas, the
equilibrium will shift toward the side with fewer total moles. For example, if the volume of the
equilibrium
N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
is reduced (i.e. pressure increased), it will shift to the right, toward the lesser number of moles/particles.
Changes in temperature
Consider heat as a reactant for endothermic reactions and as a product of exothermic reactions. The
effect of heating or cooling upon a reaction is analogous to changing the concentration of a reactant or
product.
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Consider an example of an exothermic reaction:
A + B ↔ C + D + heat
Cooling such a reaction would be like removing a product (heat). This would drive the reaction
to the right, in attempt to replace the removed product (heat). Increasing the temperature of
the equilibrium would drive it to the left, in an attempt to consume the excess product (heat)
Effect of a Catalyst upon an equilibrium
A catalyst has no effect upon the position of an equilibrium. The value of the equilibrium constant is
unaffected. However, the rate at which the equilibrium is achieved.
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