+

Unit 9: Lecture 2
Heterogeneous Equilibria
ICE Charts

+
Changing Kc
Like

H, when we manipulate a chemical reaction, the value of the equilibrium constant changes.

Three

possible changes:

1.Running

Reaction in Reverse

2.Multiplying
3.Adding

Reaction by Common Factor

Equilibriums

+Option 1: Running Reactions in Reverse
 If

we have the equilibrium
N2O ⇌ 2NO2

 Then

the value of the equilibrium constant is:
2

[NO2 ]
Kc 
1  0.212
[N2O4 ]
 If

we reverse the equilibrium, then it is now:
2NO ⇌ N2O

 

 Therefore,

the new equilibrium constant is
1

[N2O4 ]
1
Kc 
2 
[NO2 ]
0.212

When a reaction is run in reverse, the equilibrium constant is

+
Option 2: Multiplying by a Constant
 Like

Hess’s Law, we can multiply equilibriums in order to add them.

 If

we multiply the previous equilibrium by two, then we get:

⇌

2N2O

4NO2

2
2
4 the we  need to raise
K
to



[NO2 ]
[NO2 ]
2
value weKmultiplied by:  
1  
2

 Therefore,

c

 

 [N2O4 ]  

[N2O4 ]

+
Option 3: Adding Equilibriums
 Like

Hess’s Law, we can add equilibriums to obtain the desired reaction

 When

we add equilibriums, the equilibrium constants are multiplied.

+
Changing Kc Summary
1.

Kc of a rxn in the reverse direction is the inverse of the Kc of the forward rxn.

2.

The Kc of a rxn that has been multiplied by a number is the Kc raised to a power equal to that number.

3.

The Kc for a net rxn made up of two or more steps is the product of the equilibrium constants for the individual steps.

+Class Example

 Given

the reaction:

HF (aq) ⇌

H+ (aq) + F- (aq)

Kc = 6.8 x 104

H2C2O4 (aq) ⇌ 2 H+ (aq) + C2O42- (aq)
Kc =
3.8 x 10-6

 Determine

the value of Kc for the reaction:

22 HF
(aq)
+
C
O
(aq)
⇌
2
F
(aq)
+
H
C
O
2
4
2
2
4
Kc= 0.12
(aq)

+Table Talk
 Given

the reaction:

H2 (g) + I2 (g) ⇌ 2HI (g)
N2 (g) + 3 H2 (g) ⇌ 2NH3 (g)
10-4
 Determine

Kp = 1.04 x

the value of Kp for the reaction:

2 NH3 (g) + 3 I2 (g) ⇌

Kp= 15.1

Kp = 54.0

6 HI (g) + N2 (g)

Stop and Jot
+
 The

following equilibrium was obtained at
823 K:
CoO (s) + H2 (g)

CoO (s) + CO (g)

⇌

Co (s) + H2O (g)
⇌ 67

Kc =

Co (s) + CO2 (g)
490

Kc =

⇌

 Based

on these equilibrium, calculate the equilibrium constant for:
H2 (g) + CO2 (g)

Kc= 0.14
 Assume

CO (g) + H2O

it is also at 823 K.

+
Homogeneous vs.
Heterogeneous Equilibria
 Homogeneous
 When

Equilibria

all substances are in the same phase

 Heterogeneous
 When

Equilibria

substances are in different phases
 Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium constant expression for the reaction +
Example 1
 Write

the equilibrium-constant expression Kc for the following rxns:

a)

CO2 (g) + H2O (g) ⇌ CO(g) + H2O (l)

b)

SnO2 (s) + 2CO (g) ⇌ Sn(s) + 2CO2
(g)

+Calculating Equilibrium Concentrations
Oftentimes,

we know the amount of substance that we begin with, but the equilibrium amounts are unknown.

We

use stoichiometry to help solve for the equilibrium concentrations using the initial conditions and the value of Kc.

+
ICE, ICE, Baby
 Use

an ICE Table to solve for equilibrium concentrations. Initial
Change
Equilibrium

N2 (g) +
(g)

Initial concentrations go here 3H2 (g) 

Coefficient times x goes here 2NH3

Result of adding and subtracting
I/C

+Class Example
A

1.000 L flask if filled with 1.000 mol of H2 and
2.00 mole of I2 at 448 oC. The value of the equilibrium constant Kc, for the reaction at 448 oC is 50.5.
What are the equilibrium concentrations of H2, I2, and HI in moles per liter?
H2 (g) + I2 (g) ⇌

2HI (g)

+Class Example
A

1.000 L flask if filled with 1.000 mol of H2 and
2.00 mole of I2 at 448 oC. The value of the equilibrium constant Kc, for the reaction at 448 oC is
50.5.
What are the equilibrium concentrations of H2, I2, and HI in moles per liter?
H2 (g) + I2 (g) ⇌ 2HI (g)

1.

Make an ICE table…

Initial
2HI (g)
Change
Equilibrium

H2 (g) +

I2 (g)