7.2 Equilibrium Law and
the Equilibrium Constant
Learning Goals …
… write equilibrium law equations
… calculate K given equilibrium concentrations
… calculate K given initial concentrations and one
equilibrium concentration
The Equilibrium Law Expression
• Homogeneous equilibrium: A reaction with reactants and
products in the same phase.
• Hetrogeneous equilibrium: A reaction with reactants and
products in the different phase.
• gases and aqueous solutions are included as variables in
the equilibrium law expression
• solids and liquids are not included as variables in the
equilibrium law expression, but rather into the value of
the equilibrium constant.
• The value of K will depend on temperature
Ex) Write the equilibrium law expression for the following
reactions:
a) N2 (g) + 3 H2 (g)  2 NH3 (g)
K=
[NH3(g)]2
[N2(g)][H2(g)]3
b) 2 H2O (l)  2 H2 (g) + O2 (g)
K=
[H2(g)]2[O2(g)]
c) CaCO3 (s)  CaO (s) + CO2 (g)
K=
[CO2(g)]
The Magnitude of K
The value of K helps to predict the
relative concentrations of reactants and
products in an equilibrium system.
K >> 1
o [reactants] << [products] at eqm
o Eqm lies far to the right
K=1
o [reactants] = [products] at eqm
K << 1
o [reactants] >> [products] at eqm
o Eqm lies far to the left
Krev = 1/Kfwd
Equilibrium Calculations
Ex. Calculate K for the reaction below if the equilibrium
concentration of the reactants and products
respectively are [CH3OH]= 0.00261M, [CO]= 0.105 M
and [H2] = 0.250 M
CO(g) + 2H2(g)  CH3OH(g)
K=
[CH3OH]
[CO][H2]2
K=
(0.00261)
(0.105)(0.250)2
K = 0.398
Ex) Calculate K for the reaction above if the equilibrium
moles of the reactants and products in a 2.0 L container
respectively are NH3= 0.150 moles, H2 and N2 have equal
moles of 0.300 moles.
N2(g) + 3H2(g)  2NH3(g)
[NH3]2
K=
[N2][H2]3
K=
(0.075)2
(0.150)(0.150)3
K = 11.1
Ex)
If the reaction H2(g) + F2(g)  2HF(g) begins with [H2(g)]
and [F2(g)] of 1.00 mol/L and no HF, calculate the
equilibrium constant if the eq’m concentration of [F2(g)]
is measured to be 0.24 mol/L.
H2 (g) +
F2 (g) ↔ 2 HF (g)
i
1.00
0
1.00
-x
-x
c
e
1.00 - x
(1.52)2
K=
(0.24)(0.24)
K = 40.1
1.00 - x
= 0.24 M
x = 0.76
+ 2x
2x
[H2]= 0.24 M
[HF]= 1.52 M
Ex) When ammonia is heated, it decomposed into nitrogen
gas. When 4.0 mol of NH3(g) is introduced into a 2.0L
container, its equilibrium concentration is 1.0 mol/L.
Determine the equilibrium constant.
2NH3(g)  N2(g) +
3H2(g)
4.0 mol
[initial]
0
0
2.0
2.0 L
=
2.0
M
+x
+ 3x
[change] - 2x
[eq’m]
2.0 - 2x = 1.0
- 2x = - 1.0
x = 0.5
x
= 0.5
K=
3x
=1.5
(0.5)(1.5)3
(1.0)2
K = 1.7
Self Check
How prepared am I to start my homework? Can I …
… write equilibrium law equations
… calculate K given equilibrium concentrations
… calculate K given initial concentrations and one
equilibrium concentration
HOMEWORK
p436 #1-3, 5, 6abc
p454 #3
p459 #4,5