Lesson 13.2 The Equilibrium Constant II
Suggested Reading

Zumdahl Chapter 13 Section 13.1 - 13.4
Essential Question

How is the equilibrium constant calculated?
Learning Objectives:


Deduce the equilibrium constant expression (Kc or Kp) for the equation for a
homogenous reaction.
Solve problems for Kc or Kp given appropriate data.
Introduction
In the previous lesson you learned about some of the characteristics of chemical equilibrium
and how to write the expression for the equilibrium constant. Recall that the equilibriumconstant expression for a reaction is an expression obtained by multiplying the concentrations
of products, dividing by the reactants, and raising each term to the power equal to the coefficient
in the balanced chemical equation. The equilibrium constant, Kc (often denoted by just K), is
the value obtained from the equilibrium constant expression when equilibrium concentrations
are substituted. In this lesson, you will use the equilibrium constant expression to obtain the
equilibrium constant
Calculating Kc
The value of the equilibrium constant at a given temperature can be calculated if we know the
equilibrium concentrations of the reactants and products for a given reaction. Equilibrium
constants are customarily given without units, so we will not use units for K. When a prompt
states the equilibrium concentrations we can simply plug these values into the equilibrium
constant expression and solve for K. However, some problems do not give all of the equilibrium
concentrations, which complicates things a bit. The next two examples will illustrate these two
problem types.
Example: Calculating the Equilibrium Constant from Equilibrium Concentrations
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 equilibrium.
CO2(g) + H2(g) ⇌ CO(g) + H2O(g)
Solution: First, write the equilibrium constant expression.
Second, express the given equilibrium amounts in moles per liter.
[CO]
[H2O]
[CO2]
[H2]
= 0.0092 moles / 2.00 L = 0.0046
= 0.0092 moles / 2.00 L = 0.0046
= 0.1908 moles / 2.00 L = 0.0954
= 0.0908 moles / 2.00 L = 0.0454
Lastly, substitute each concentration into the equilibrium expression and calculate the value of
the equilibrium constant.
Example: Calculating the Equilibrium Constant from Reaction Composition
Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation,
2HI(g) ⇌ H2(g) + I2(g). When 4.00 mol HI was placed in a 5.00 L vessel at 458∘C, the equilibrium
mixture was found to contain 0.442 mole I2. What is the value of Kc for the decomposition of HI
at this temperature?
Solution: The problem gives the starting amount of HI and the equilibrium amount of I2, but we
must determine the molar concentration at equilibrium for all reactants and products in order to
solve the problem. Setting up a table of initial, change and equilibrium values is a convenient
way to make sense of the givens and unknowns. Once you know the equilibrium values you can
calculate Kc.
Step 1: Obtain the starting concentration of HI and the equilibrium concentration of I2 using
information given in the prompt.
Starting concentration of HI = 4.00 mol / 5.00 L = 0.800 M
Equilibrium concentration of I2 = 0.442 mol / 5.00 L = 0.0884 M
Step 2: Find the equilibrium concentrations for the other substances by setting up the following
table (ICE it!).
You fill in the knowns using the values given in the prompt (at the beginning the concentration of
products is 0, so this is also a known). You then must describe the changes in the starting
concentrations. You are not given explicit values for the changes that occur, so you let x be the
molar change. The reactant decreases by x moles multiplied by the coefficient from the
balanced chemical equation while each product increases by x moles multiplied by the
coefficient from the balanced chemical equation. The decrease is indicated by the negative sign.
Equilibrium values are equal to the starting values plus the changes.
Concentration (M)
Initial
Change
Equilibrium
2HI(g) ⇌
0.800
-2x
0.800 - 2x
H2(g)
0
+x
x
I2(g)
0
+x
x = 0.0884
The equilibrium concentrations can be determined from the expressions given in the last line of
this table. You know that the equilibrium concentration of I2 is 0.0884 M and that this equals x.
Therefore, at equilibrium
[H2] = [I2] = x = 0.0884
[HI] = (0.800 - 2x) M, substituting gives (0.800 - 2(0.0884)) M = 0.623 M
Step 3: Write the equilibrium constant expression and substitute.
The problem above may look complicated, but it is really pretty formulaic. Once you get the
hang of it, you will see these problems are not too bad.
Calculating Equilibrium Constant Kp from Kc
Recall from the previous lesson that when you express an equilibrium constant for a gaseous
reaction in terms of partial pressures, you call it Kp. For catalytic methanation, CO(g) + 3H2(g) ⇌
CH4(g) + H2O(g), the equilibrium expression in terms of partial pressures becomes
The value of Kp is different from Kc. From the relationship n/V = P/RT we can show that
Kp = Kc(RT)∆n
where ∆n is the sum of the coefficients of the gaseous products minus the sum of the
coefficients of the gaseous reactants.
∆n = ∑nproducts - ∑nreactants
For the reaction, CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g), Kc = 3.92 and T = 1200 K
∆n = (1+1) - (1 + 3) = -2
Substituting into Kp = Kc(RT)∆n gives Kp = (3.92)(0.0821 x 1200)-2 = 4.04 x 10-4. The usual unit for
partial pressures is atm, so the value of R is 0.0821 L ∙ atm / (K ∙ mol). However, like Kc, Kp is
customarily reported without units.
Equilibrium Constant for the Sum of Reactions
It is possible to determine the equilibrium constants for various chemical reactions and then use
them to obtain the equilibrium constants for other reactions. A useful rule is as follows: If a given
chemical equation can be obtained by taking the sum of other equations, the equilibrium
constant for the given equation equals the products of the equilibrium constants of the other
equations.
For example, consider the following reactions at 1200 K,
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g); K1 = 3.92
CH4(g) + 2H2S(g) ⇌ CS2(g) + 4H2(g); K2 = 3.3 x 104
When you sum these reactions you get
CO(g) + 2H2S(g) ⇌ CS2(g) + H2O(g) + H2(g)
According to the rule, the equilibrium constant K3 = K1K2 = (3.92)(3.3 x 104) = 1.3 x 105
Homework: Practice exercises 15.7-15.9