Chapter 15: Chemical Equilibrium
N2O4(g)
⇋ 2 NO2(g)
A reaction is in equilibrium when the rate of the forward
reaction: ratefor = kfor[N2O4]
equals the rate of the reverse reaction: raterev = krev[NO2]2
ratefor = raterev
kfor[N2O4] = krev[NO2]2
[NO2 ]2
Kc 

krev [N2O 4 ]
k for
Kc = the equilibrium constant for the reaction when Molar
concentrations are used to calculate K.
Note that it is only possible to establish an equilibrium for a
reaction that proceeds via a single step mechanism. Hence the
concentration of each reactant and product is raised to its
corresponding coefficient.
N 2 O4
⇋
2 NO2
It is important to remember that even though the [reactants]
and [products] are constant at equilibrium the reaction hasn’t
stopped  Chemical equilbrium is a dynamic process.
What is Kc for the following reactions?
N2(g) + 3 H2(g)
H2O(l)
⇋ 2 NH3(g)
⇋ H2O(g)
Kc 
[ NH 3 ]2
[ N 2 ][H 2 ]3
Kc  [ H 2O]


[
H
][
HCO
3]
CO2(g) + H2O(l) ⇋ H+(aq) + HCO3(aq) K c 
[CO2 ]
2
Ag+(aq)
+ CO3
2-(aq)
⇋ Ag2CO3(s)
Kc 
1
[ Ag  ]2[CO32 ]
Note that reactants and products in the solid or liquid state do
not appear in the equilibrium constant expression because
their molarity does not change as a result of the position of
the equilibrium.
Sometimes, it is more convenient to determine the equilibrium
constant for gas phase reactions in terms of the partial pressures
of the gasses.
N2O4(g) ⇋ 2 NO2(g)
Kp

PNO 2

2
PN 2O4
Kp and Kc are related through the ideal gas law as follows:
PNO2V = nNO2RT, so PNO2 = (nNO2/V)RT = [NO2]RT
Similarly:
Kp
PN2O4 = [N2O4]RT

[ NO2 ]RT 2

K
[ N 2O4 ]RT
In general:
c RT
Kp = KcRTn
(A) 2 NOBr(g)
⇋ 2 NO(g) + Br2(g)
(B) What is Kc for 2 NO(g) + Br2(g)
Kc 
[ NOBr]2
[ NO ]2 [ Br2 ]
Kc 
[ NO ]2 [ Br2 ]
2
 0.014
[ NOBr ]
⇋ 2 NOBr(g)

1
 71
0.014
Which reaction, A or B, favors product formation?
Rxn B because Kc > 1
Q: What is the relative value of Kc if reactants are favored?
Kc < 1 for reactant favored rxns
Q: If neither reactants nor products are favored? Kc = 1
Combining equilibria a la Hess’s Law:
If you know that:
2 NOBr(g)
And that
⇋ 2 NO(g) + Br2(g)
K c1 
[ NO ]2 [ Br2 ]
[ NOBr ]
2
 0.014
[ BrCl ]2
 7.2
Br2(g) + Cl2(g) ⇋ 2 BrCl(g) K c 2 
[ Br2 ][ Cl2 ]
What is Kc for the combined reaction?
2 NOBr(g) + Cl2(g)
⇋ 2 NO(g) + 2BrCl(g)
 [ NO]2 [ Br2 ]  [ BrCl ]2 


Kc 

[ NOBr]2 [ Br2 ][Cl2 ]  [ NOBr]2  [ Br2 ][Cl2 ] 
[ NO]2 [ Br2 ][BrCl ]2
Kc = (Kc1)(Kc2) = 0.10
Calculations with Equilibrium Constants: ICE Tables
15.24 A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a
2.00 L vessel at 700K. These substances react as follows:
H2(g) + Br2(g) ⇋ 2 HBr(g)
At equilibrium, the vessel is found to contain 0.566 g of H2. What
are the equilibrium partial pressures of each species?
Initial mol H2 = 1.374 g/(2.01588 g/mol) = 0.68159 mol
Initial mol Br2 = 70.31 g/(159.808 g/mol) = 0.43997 mol
Eq mol H2 = 0.566 g/(2.01588 g/mol) = 0.2808 mol
H2(g)
0.68159 mol
Initial
-0.4008 mol
Change
Equilibrium 0.2808 mol
+
Br2(g)
⇋
2 HBr(g)
0.43997 mol
0 mol
-0.4008 mol
+0.8016 mol
0.0392 mol
0.8016 mol
Use the number of moles of each gas present at equilibrium to
determine their partial pressures.
PH2 = nRT/V = (0.2808 mol)(0.0821Latm)(700K)/(2.00 L)
PH2 = 8.07 atm
PBr2 = nRT/V = (0.0392 mol)(0.0821Latm)(700K)/(2.00 L)
PBr2 = 1.1 atm
PHBr = nRT/V = (0.8016 mol)(0.0821Latm)(700K)/(2.00 L)
PHBr = 23.0 atm
d) What is Kp?
e) What is Kc?

PHBr 2

23.0atm2
= 58
Kp 

PH 2 PBr2  8.07atm1.1atm
Kp = KcRTn
n = 0, so Kp = Kc
Stoichiometry and Equilibrium ICE Tables
Ex: A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of
I2 at 448°C. The value of the equilibrium constant Kc for the
reaction shown below at 448°C is 50.5. What are the equilibrium
concentrations of H2, I2, and HI in moles per liter?
H2(g)
I
C
E
E
+
I2(g) ⇋
2 HI(g)
1.000 M
2.000 M
0
-x
-x
+2x
1.000 - x
2.000 - x
2x
0.444 M
1.444 M
1.11 M
2
[2 x]2
50.5 
[1.000 x][2.000 x]
[ HI ]
Kc 
[ H 2 ][ I 2 ]
Solve quadratic and get that x = 0.556 M
Using the reaction quotient to predict equilibrium shifts
H2(g) + I2(g) ⇋ 2 HI(g)
Kc = 50.5
Ex: Predict in which direction the reaction will proceed to reach
equilibrium if we start with 2.0 × 10−2 mol of HI, 1.0 × 10−2 of H2,
and 3.0 × 10−2 of I2 in a 2.00-L container.
Initial concentrations:
[HI] = (0.020 mol HI)/2.00 L = 0.010 M
[H2] = (0.010 mol H2)/2.00 L = 5.0 x 10−3 M
[I2] = (0.030 mol H2)/2.00 L = 0.015 M
[0.010 M] 2
[ HI ]2

 1.3
Qc 

3
[ H 2 ][ I 2 ] [5.0 x 10 M][0.015 M]
Qc < Kc so not enough products
Right shift
Le Chatelier’s Principle: When a stress is applied to an
equilibrium, the equilibrium will shift to alleviate the stress.
Fe+3 (aq) + SCN-1 (aq) ⇋ FeSCN+2 (aq)
Colorless
Left shift =
lighter color
⇋
Dark red
Right shift =
darker color
Initial color
Other ways to cause a Le Châtelier Shift:
N2(g) + 2 O2(g) ⇋ 2 NO2(g)
H = 67.7 kJ
What kind of shift would you see if:
Pressure increased?
Right shift 
Volume increased?
 Left shift
Heating temperature increased?
Right shift 
CH4 (g) + 2 Cl2 (g) ⇋ CCl4 (g) + 2 H2 (g)
H = -32 kJ
What kind of shift would you see if:
Pressure increased?
Heating temperature increased?
H2 removed?
CH4 removed?
Catalyst added?
No Change
 Left shift
Right shift 
 Left shift
No Change