8.2 EQUILIBRIUM CONSTANT

8.2 Equilibrium constant

Define Deduce and write qCuaanlctiutileasteofKsc,pKepcoiers
expressions for present at
equilibrium
equilibrium.
constant in terms of:

Homogeneous concentration, Kc
equilibria

Heterogeneous partial pressure, Kp
equilibria
1

LEARNING OUTCOMES

8.2 Equilibrium constant

At the end of the lesson, you should be able to:

a) Define homogeneous and heterogeneous equilibria.

b) Deduce and write expressions for equilibrium

constant in terms of concentration, Kc and partial
pressure, Kp for homogeneous and heterogenous
systems.

c) Calculate Kc, Kp or quantities of species present at
equilibrium

*calculations are limited to problems with quadratic equations only.

2

TYPES OF EQUILIBRIUM

Homogeneous Heterogeneous
Equilibria Equilibria

Reactants and products are in Reactants and products are in
the same phase the different phase / more
than one phase

Examples: Examples:

N2O4(g) 2NO2(g) CaCO3(s) CaO(s) + CO2(g)
2SO2(g) + O2(g) NH4Cl (s) NH3 (g) + HCl (g)
CH3COOH(aq) 2SO3(g) H2(g) + Br2 (l)
2HBr(g)
CH3COO–(aq)
+ H+(aq) 3

WRITING EXPRESSION OF EQUILIBRIUM
CONSTANTS FOR HOMOGENEOUS EQUILIBRIUM

aA(g) + bB(g) cC(g) + dD(g)

[C]c [D]d (PC)c (PD)d
KC = [A]a [B]b KP = (PA)a (PB)b

Subscript C in KC = P = equilibrium
concentration of the partial pressure of

reacting species the gas

4

EXAMPLES: Note:
1. Equilibrium
1. N2O4(g) 2NO2(g)
constant
[NO2]2 PNO2 2 expression
[N2O4] P (KC or KP)
KC = Kp= also called
Equilibrium law
N2O4 Expression

2. N2(g) + 3H2(g) 2NH3(g) 2. Unit of
concentration =
KC = [NH3]2 Kp= PNH 2 3
[N2][H2]3 M @ mol L-1
PN2 P
H2 3. Unit of
pressure = atm
3. CH3COOH(aq) CH3COO–(aq)
+ H+(aq)
KC = [CH3COO–][H+]
[CH3COOH] 5

WRITING EXPRESSION OF EQUILIBRIUM
CONSTANTS FOR HETEROGENEOUS EQUILIBRIUM

CaCO3(s) CaO(s) + CO2(g)

KC = [CO2] KP = PCO2

Concentration of pure solid and pure liquid
are constant and excluded in Kc and Kp

6

 A pure solid always has the same concentration at
a given temperature → same number of moles per liter

of the solid.

 Just as it has same density at any given temperature.

 Same reason applies to pure liquid. 7

CaCO3(s) CaO(s) + CO2(g)

PCO 2 = KP

PCO2does not depend on the amount of CaCO3 or CaO

8

The equilibrium constant (KC and KP)
is a dimensionless (no unit) quantity.

PCl3(g) + Cl2(g) PCl5(g)

KC = 1.67 (at 500K)
KP = 4.07 x 10–2 (at 500K)

In quoting a value for the KP or KC, you MUST specify
the balanced equation (including the phase of each

reactant/product) and the temperature.

9

EXAMPLE - 1

Write expressions for KC, and KP if applicable, for the
following reversible reactions at equilibrium:

Note: balance the equations first.

a) HF(aq) H+(aq) + F–(aq)

b) NO(g) + O2(g) NO2(g)

c) N2O5(g) NO2(g) + O2(g)

d) C3H8(g) + O2(g) CO2(g) + H2O(g)

e) NH3(g) + O2(g) NO(g) + H2O(g)

f) NO(g) N2O(g) + NO2(g)

10

EXAMPLE - 1

a) HF(aq) H+(aq) + F–(aq)

ANS: [H+] [F–]

KC = [HF]

b) 2NO(g) + O2(g) 2NO2(g)

ANS: [NO2]2 2
[NO]2 [O2]
KC = KP = (PNO2)
(PNO)2 (PO2)

c) 2N2O5(g) 4NO2(g) + O2(g)

ANS: KC = [NO2]4 [O2] =(PNO2)4 (PO2)
[N2O5]2 (PN2O5)2
KP

11

EXAMPLE - 1

d) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

ANS: = [CO2 ]3[H2O]4 34
[C3H8 ][O2 ]5 (P ) (P )CO2
H2O
KC K =P 5
P (P )C3H8
O2

e) 5 2NO(g) + 3H2O(g)

2NH3(g) + 2 O2(g)

ANS: KC = [NO]2 [H 2 O]3 KP = (PNO )2 (PH2O )3

5 5

[NH3 ]2[O2 ]2 (PNH3 )2 (PO2 )2

f) 3NO(g) N2O(g) + NO2(g)

ANS: KC = [N2O][NO2 ] KP = (PN2O )(PNO2 )
[NO]3 (PNO )3
12

EXAMPLE - 2

Write equilibrium constant expression for KC and
KP for the formation of nickel tetracarbonyl, which
is used to separate nickel from other impurities:

Ni(s) + 4CO(g) Ni(CO)4(g)

ANS: [Ni(CO)4] P Ni(CO)4
[CO]4 PCO 4
KC = KP =

13

EXAMPLE - 3

Balance each of the following equations and write
its equilibrium constant expression, KC and KP:

a) Na2O2(s) + CO2(g) Na2CO3(s) + O2(g)

b) H2O(l) H2O(g)

c) NH4Cl(s) NH3(g) + HCl(g)

d) H2O(l) + SO3(g) H2SO4(aq)

e) KNO3(s) KNO2(s) + O2(g)

f) S8(s) + F2(g) SF6(g) 14

EXAMPLE - 3

a) 2Na2O2(s) + 2CO2(g) 2Na2CO3(s) + O2(g)

ANS: KP = (PO2)
(PCO2) 2
KC = [O2]
[CO2]2

b) H2O(l) H2O(g)

ANS: KP = P
KC = [H2O] H2O

c) NH4Cl(s) NH3(g) + HCl(g)
ANS:

KC = [NH3][HCl] KP = (PNH3 )(PHCl) 15

EXAMPLE - 3

d) H2O(l) + SO3(g) H2SO4(aq)

ANS: 1
PSO3
KC = [H2SO4 ] KP =
[SO3 ]

e) 2KNO3(s) 2KNO2(s) + O2(g)
ANS:

KC = [O2 ] K P = PO2

SF6(g)

f) S8(s) + 3F2(g)

ANS: KC = [SF6 ] KP = PSF6
[F2 ]3 (PF2 )3
16

EXERCISE - 1

Balance each of the following equations and write
its equilibrium constant expression, KC and KP:

a) NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)

b) SnO2(s) + H2(g) Sn(s) + H2O(g)

c) H2SO4(l) + SO3(g) H2S2O7(l)

d) AgCl(s) Ag+(aq) + Cl–(aq)

e) CO2(s) CO2(g)
f) N2O5(s) NO2(g) + O2(g)

17

CALCULATION OF EQUILIBRIUM PROBLEMS

If given data are If given data are
the equilibrium quantities the value of KC or KP
(concentrations / partial pressure)

Solve for KC or KP Solve for equilibrium quantities
(concentrations / partial pressure)

18

EXAMPLE - 4

The following equilibrium concentrations are

measured at 483K: [CO] = 1.03 M, [H2] = 0.322 M
and [CH3OH] = 1.56 M. Determine the equilibrium
constant, Kc.

CO(g) + 2H2(g) CH3OH(g)

Solution: 1.56
(1.03)(0.322)2
Kc = [CH3OH] =
[CO][H2]2

= 14.5 19

EXAMPLE - 5

The following equilibrium process has been studied

at 230oC:

2NO(g) + O2(g) 2NO2(g)

In one experiment the concentration of the reacting

species at equilibrium are found to be [NO] = 0.0542

M, [O2] = 0.127 M, and [NO2] = 15.5 M. Calculate the
equilibrium constant (KC) of the reaction at this
temperature.

Solution: [NO2]2 = (15.5)2

Kc = [NO]2[O2] (0.0542)2 x 0.127 = 6.44 x 105

20

EXAMPLE - 6

The partial pressure of PCl3, O2 and POCl3 at
equilibrium are 5.05 kPa, 2.50 kPa and 101 kPa

respectively. Calculate the equilibrium constant,

Kp for the reaction.

2PCl3(g) + O2(g) 2POCl3(g)

Solution: (101)2
(5.05)2(2.5)
Kp = (PPOCl3)2 =
(PPCl3)2(PO2)

= 160 21

EXAMPLE - 7

Consider the following equilibrium at 295 K:

NH4HS(s) NH3(g) + H2S(g)

The partial pressure of each gas is 0.265 atm.
Calculate KP for the reaction?

Solution:

Kp = (PNH3)(PH2S)

= 0.265 x 0.265

= 0.0702

22

EXERCISE - 2

Consider the following equilibrium process at 700oC:

2H2(g) + S2(g) 2H2S(g)

Analysis shows that at equilibrium, there are 2.50
mol of H2, 1.35 x 10–5 mol of S2, and 8.70 mol of H2S
present in a 12.0 L flask. Calculate the equilibrium

constant KC for the reaction.

ANS:
1.08 x 107

23

EXERCISE - 3

At a certain temperature, KC = 1.8 x 104 for the
reaction

N2(g) + 3H2(g) 2NH3(g)

If the equilibrium concentrations of N2 and NH3 are
0.015 M and 2.00 M, respectively, what is the

equilibrium concentrations of H2?

ANS:
0.25 M

24

EXERCISE - 4

The equilibrium constant Kp for the reaction

2NO2 (g) 2NO (g) + O2 (g)

is 158 at 1000K. What is the equilibrium pressure
of O2 if the PNO2 = 0.27 atm and PNO = 0.40 atm?

ANS:
71.99 atm

25

EXERCISE - 5

The equilibrium constant KP for the decomposition of
phosphorus pentachloride to phosphorus trichloride

and molecule chlorine is found to be 1.05 at 250oC.

PCl5(g) PCl3(g) + Cl2(g)

If the equilibrium partial pressure of PCl5 and PCl3
are 0.875 atm and 0.463 atm, respectively, what is
the equilibrium partial pressure of Cl2 at 250oC.

ANS:
1.98 atm

26

EXERCISE - 6

At equilibrium in the following reaction at 60oC,

the partial pressure of the gases are found to be
PHI = 3.65 x 10–3 atm and PH2S = 0.996 atm.

What is the value of KP for the reaction?

H2S(g) + I2(s) 2HI(g) + S(s)

ANS:
1.34 x 10–5

27

The value of KC and KP depend on how the
equilibrium equation is written and balanced.

N2O4(g) 2NO2(g) (at 25oC)

KC = [NO2]2 = 4.63 x 10–3
[N2O4]

2NO2(g) N2O4(g) (at 25oC)

KC’ = [N2O4] = 216
[NO2]2
28

The value of KC and KP depend on how the
equilibrium equation is written and balanced.

N2O4(g) 2NO2(g) (at 25oC)

KC = [NO2]2 = 4.63 x 10–3
[N2O4]

1 N2O4(g) NO2(g) (at 25oC)
2

KC” =[NO2] = 0.0680 29
[N2O4]1/2

EXERCISE - 6

For the ammonia–formation reaction,

N2(g) + 3H2(g) 2NH3(g)

KC = 2.4 x 10–3 at 1000 K. If we change the coefficients
of this equation, what are the values of KC for the
following balanced equations?

a) 1 N2(g) + H2(g) 2 NH3(g)
3 3
ANS:
b) NH3(g) 1 N2(g) + 3 H2(g) a) 0.13
2 2 b) 20.4

30

CALCULATION OF EQUILIBRIUM PROBLEMS

If given data are If given data are
initial quantities initial quantities
(concentrations / partial pressure) (concentrations / partial pressure)

along with one of the along with value KC or KP
concentration or partial pressure

at equilibrium

Solve for KC or KP Solve for equilibrium quantities
(concentrations, partial pressure)

Note: Solving problems involve ICE (Initial- 31
Change-Equilibrium) Table

Guidelines of solving problems
( involving ICE Table)

STEP 1: Balance the chemical equation

STEP 2: Construct ICE (Initial-Change-Equilibrium) Table

STEP 3: Determine x (change in concentration)
from ICE Table

STEP 4: Calculate KC or KP or quantities at equilibrium
(depends on the given problem)

32

USING A REACTION (ICE) TABLE

Tips of use ICE table:
1. The system lack of information:
A + B → C only [A] & [B] given, [C] is not given

2. Keywords that show the system is still not achieve 33

equilibrium:

• Initially placed
• Is/are allowed to reach equilibrium
• Are introduced into flask
• Is placed in flask
• After a period of time the equilibrium is achieve
• Calculate degree of dissociation

EXAMPLE - 6

At a certain temperature, a mixture of H2 and I2 was

prepared by placing 0.200 mol of H2 and 0.200 mol

of I2 into a 2.00 L flask. initial
After a period of time the equilibrium was

established. mol

H2(g) + I2(g) 2HI(g)

At equilibrium, the concentration of I2 concentration

had dropped to 0.020 M. Equi. [ ]

What is the value of KC for this reaction at this 34
temperature?

EXAMPLE - 6

STEP 1& 2: Construct ICE (Initial-Change-Equilibrium) Table using a
balanced chemical equation.

Concentration H2(g) + I2(g) 2HI(g)
Initial (M) 0.000
0.200 mol 0.200 mol
2.00 L
2.00 L
= 0.100 = 0.100

Change (M) –x –x + 2x

Equilibrium (M) 0.100 – x 0.100 – x 0.000 + 2x
= 0.020 = 2x

STEP 3: From the data given, 0.100 – x = 0.020
Determine x
from ICE Table x = 0.080 35

EXAMPLE - 6

STEP 4: Calculate KC [HI] = 2 x 0.080 M
= 0.160 M
Therefore, at equilibrium:

[I2] = 0.020 M
[H2] = (0.100 – 0.080) M

= 0.020 M

H2(g) + I2(g) 2HI(g)

KC = [HI]2 (0.160)2
[H2] [I2] =

0.020 x 0.020

= 64.0 36

EXAMPLE - 7

The atmospheric oxidation of nitric oxide,

2NO(g) + O2(g) 2NO2(g)

was studied at 184oC with pressure of 1.000 atm of

NO and 1.000 atm of O2. Initial P
At equilibrium, PO2= 0.506 atm.
Calculate KP.

Equi. P

37

EXAMPLE - 7

STEP 1& 2: Construct ICE (Initial-Change-Equilibrium) Table using a
balanced chemical equation.

Pressure 2NO(g) + O2(g) 2NO2(g)
Initial (atm) 0.000
1.000 1.000
Change (atm) + 2x
– 2x – x 0.000 + 2x
Equilibrium = 2x
(atm) 1.000 – 2x 1.000 – x
= 0.506

STEP 3:
Determine x
from ICE Table From the data given, 1.000 – x = 0.506

x = 0.494

38

EXAMPLE - 7

STEP 4: Calculate Kp

Therefore, at equilibrium: PNO2= 2x0.494 atm
= 0.988 atm
PNO = (1.000 – 2x0.494) atm
= 0.012 atm

PO2 = 0.506 atm

2NO(g) + O2(g) 2NO2(g)

Kp = (PNO2)2 ) (0.988)2
(PNO)2 (PO2 =

(0.012)2 x 0.506

= 1.34 x 104 39

EXAMPLE - 8

At 1280oC the equilibrium constant (Kc) for the
reaction

Br2(g) 2Br(g)

is 1.1 x 10–3. If the initial concentrations are [Br2]
= 0.063 M and [Br] = 0.012 M, calculate the

concentrations of these species at equilibrium.

40

EXAMPLE - 8

STEP 1& 2: Construct ICE (Initial-Change-Equilibrium) Table using a
balanced chemical equation.

Concentration Br2(g) 2Br(g)
Initial (M) 0.063 0.012
Change (M) + 2x
Equilibrium (M) –x 0.012 + 2x
0.063 – x

41

EXAMPLE - 8

STEP 3: [Br]2
Determine x Kc = [Br2]
from ICE Table

1.1 x 10-3 = (0.012 + 2x)2
0.063 – x

Rearrange the equation to its quadratic form: ax2 + bx + c =0

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

Use the quadratic formula to find the x = -b ±  b2 – 4ac
value of x: 2a

x = –0.0105 x = –0.00178

42

EXAMPLE - 8

If x = – 0.0105 :
[Br2] = 0.0705 , [Br] = – 0.009 Not acceptable

if x = – 0.00178 :

[Br2] = 0.0648 , [Br] = 0.00844 correct

Note: Only one x value makes sense chemically!

At equilibrium, [Br] = 0.00844 M

[Br2] = 0.0648 M 43

Tip – The value of x = - 0.0105 is not acceptable because the
concentration of all reacting species should be always positive.

EXERCISE - 8

The reaction

CO(g) + H2O(g) CO2(g) + H2(g)

has KC = 4.06 at 500oC. If 0.100 mol of CO and
0.100 mol of H2O(g) are placed in a 1.00 L reaction
vessel at this temperature, what are the
concentration of the reactants and products when
the system reaches equilibrium?

ANS:
[CO] = 0.033 M; [H2O] = 0.033 M;
[CO2] = 0.0668 M; [H2] = 0.0668 M 44

EXERCISE - 9

Phosgene is a potent chemical warfare agent that is
now outlawed by international agreement. It
decomposes by the reaction

COCl2(g) CO(g) + Cl2(g)
KC = 8.3 x 10–4 (at 360oC)

Calculate [CO], [Cl2], and [COCl2] when the following
5.000 mol of COCl2 of phosgene decompose and
reach equilibrium in a 10 L flask:

ANS: [COCl2] = 0.4804 M, [CO] = [Cl2] = 0.0196 M

45