Ch 16.5 The Equilibrium Constant
1. In 1864 two Norwegian chemists, Guldberg and Waage, proposed the law of chemical
equilibrium (also called the law of mass action), as a general description of the
equilibrium condition.
For a reaction of the type
aA + bB <==> cC + dD
where A, B, C and D represent chemical species (i.e., compounds, elements) and a, b, c
and d are their coefficients in a balanced equation, the law of chemical equilibrium is
represented by the following equilibrium expression:
K
[C ]c [ D]d
[ A] a [ B]b
K = equilibrium constant
-- The brackets [ ] represent the concentrations of the chemical species at equilibrium
in units of moles/liters.
-- Notice that equilibrium expression is a special ratio of the concentrations of the
products to the concentrations of the reactants.
Ex: Write the equilibrium expression for the following reaction:
N2(g) + 3H2(g)  2NH3(g)
K
[ NH 3 ] 2
[ N 2 ][ H 2 ] 3
-- Remember to place the concentrations of the products in the numerator and the
concentration of the products in the denominator. The coefficients of the balanced
equation become the exponents (powers of 10).
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Ex: Write the equilibrium expression for the following reaction
4NH2(g) + 7O2(g) <==> 4NO2(g) + 6H2O(g)
2. The meaning of the equilibrium expression.
-- For a given reaction at a given temperature, the ratio of the concentrations of
products to reactants as defined by the equilibrium expression will always be equal to the
same number – i.e., the equilibrium constant (K).
-- See Table 16.1 showing experimental results for the following reaction occurring at
500 oC
N2(g) + 3H2(g)  2NH3(g)
● The results for Trial 1 show an initial concentration of 1.000 M for N2 and 1.000
M for H2. When the reaction reached equilibrium the final concentrations of all
reactants are 0.921 M for N2, 0.736 M for H2, and 0.157 M for NH3.
● Placing these values into the equilibrium expression for the reaction allows you
to calculate the equilibrium constant (K):
K
[ NH 3 ] 2
[ N 2 ][ H 2 ] 3

[0.157] 2
[0.921][0.763]3
 0.0602  6.02 102
● Note that all trials of this experiment conducted at the same temperature
produce the same calculated K. Therefore, K for a particular reaction is constant as long
as the reaction occurs at the same temperature. A new K must be calculated if the
reaction occurs at a different temperature.
● Furthermore, the system always comes to the same equilibrium position, such
that K = 0.0602 for this particular reaction, regardless of the amounts of the reactants
and products that are mixed together initially.
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● It is important to note that the equilibrium concentrations are not always the
same (see Table 16.1). However, the equilibrium constant (K), which depends on the
ratio of concentrations, remains the same for a given temperature.
● See example 16.2 in the text. Note that there is some rounding off due to
experimental error.
3. How to interpret K for a reaction.
-- K represents a ratio of concentrations of products to concentrations of reactants.
-- If K > 1 the mixture contains greater concentrations of products than reactants.
-- If K < 1 the mixture contains greater concentrations of reactants than products.
-- If K = 1 the mixture contains approximately equal concentrations of reactants and
products.
4. Solve problems 17 and 21 in your text (page 511) and any other practice problems.
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Ch 16.6 Heterogeneous Equilibria
1. The examples and problems in Ch 16.5 show examples of homogeneous equilibria.
This means that all reactants and products are in a gaseous state.
2. However, many reactions involve reactants and/or products in other than the gaseous
state. For example:
CaCO3(s) <==> CaO(s) + CO2(g)
-- The concentrations of pure solids or pure liquids are not included in an equilibrium
expression for the reaction. Why? The amount of a solid or liquid present does not
affect the equilibrium position of the reaction.
-- A simple example shows this:
H2O(l) <==> H2O(g)
At a given temperature, the concentration of water vapor is a constant no matter how
much liquid water there is. Thus, the position of the equilibrium does not change when
you add or subtract water, and so the equilibrium expression will have no term for liquid
water.
K = [H2O]
-- Similarly, for the following reaction: CO2(g) + H2(g) <==> CO(g) + H2O(l)
K
[CO]
[CO2 ][ H 2 ]
3. Write the equilibrium expressions for the reactions in Self-Check Exercise 16.2 on page
496.
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