Chapter 18: Chemical Equilibria

Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been completely used up.“Only” the forward reaction occurs.

Reversible reactions:Both the forward and the reverse reaction occur to a significant degree.

Au2O3 (s) + 2 Fe (s) 2 Au (s) + Fe2O3 (s)Ex:

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

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

Forward:

Reverse:

Equilibrium: A reaction is at equilibrium when

Rate forward rxn = Rate reverse rxn

3 H2 (g) + N2 (g) ⇋ 2 NH3 (g)

anim

General form of equilibrium constant, Keq:

aA + bB ⇋ cC + dD

ba

dc

eq[B] [A]

[D] [C]][Reactants

[Products]K

If K > 1, then more products than reactants present at equilibrium

If K < 1, then more reactants than products present at equilibrium

Time

Con

cent

ratio

nN2 (g) + 3 H2 (g) 2 NH3 (g)⇋

H2

NH3

N2

Keq =[NH3]

[N2] [H2]

2

3

Heterogeneous equilibria• Occurs when the reactants and products are in more than one state

• Because the molar concentration of solids and liquids do not change, SOLIDS AND LIQUIDS NEVER appear in equilibrium constant calculations (they have a concentration value of 1).

Ex: BaCl2 (s) ⇋ Ba2+ (aq) + 2 Cl- (aq)

22eq ]][Cl[BaK

• Since this particular type of equilibrium involves the solubility of the product, it is given a special designation:

Ksp = solubility product constant

Ksp

Heterogeneous equilibria (cont.)

Ex: H2O (l) ⇋ H2O (g)

O][HK 2eq

What are the equilibrium constants for the following:

1. C10H8 (s) ⇋ C10H8 (g)

2. CaCO3 (s) ⇋ CaO (s) + CO2 (g)

3. C (s) + H2O (g) ⇋ CO (g) + H2 (g)

4. FeO (s) + CO (g) ⇋ Fe (s) + CO2 (g)

Keq = [C10H8]

Keq = [CO2]Keq =

[CO][H2]

[H2O]

Keq =[CO2]

[CO]

The equilibrium constant for the reaction below at 700K is 0.44. What is the concentration of the carbon dioxide if [H2O] = 0.16 M, [CO] = 0.15 M and [H2] = 0.14 M?

H2O(g) + CO(g) H⇋ 2(g) + CO2(g)

1. Set up the equilibrium expression

][Reactants

[Products]K

O][CO][H

]][CO[H

2

22

2. Substitute in the known values and solve for the unknown.

M] M][0.15 [0.16

]M][CO [0.1444.0 2 [CO2] = 0.075 M

Molar solubility: the number of moles of the solute per liter in a saturated solution

Ex 1: What is the molar solubility of lead(II) iodide if its Ksp is 8.7 x 10–9 ? PbI2(s) Pb2+ + 2 I–

x 2xx = the molar solubility

Ksp = [Pb2+][I–]2

8.7 x 10–9 = [x][2x]2

8.7 x 10–9 = 4x3

x = 1.3 x 10–3 M

Ex 2: What is the molar solubility of PbI2 if 0.2 M KI is added?

[I–] = 0.2 M 8.7 x 10–9 = [x][0.2]2 x = 2.2 x 10–7 M

Ex 3: What is the Ksp of Pb3(PO4)2 if it has a molar solubility of 1.5 x 10-9 M?

1) Write out reaction equation: Pb3(PO4)2 3 Pb2+ + 2 PO43-

3x 2x

2) Write out Ksp expression: Ksp = [Pb2+]3 [PO43-]2

Ksp = (3x)3 (2x)2

Ksp = (27x3)(4x2)

Ksp = 108x5 x = 1.5 x 10-9

Ksp = (108)(1.5 x 10-9 )5

Ksp = 8.2 x 10-43

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 ⇋ Dark red

Initial color

Left shift = more reactants (color is lighter)

Right shift = more products (color is darker)

Other ways to cause a Le Châtelier Shift:

3 H2 (g) + N2 (g) + heat ⇋ 2 NH3 (g)

What kind of shift would you see if:

*Pressure increased?

[NH3] increased?

Heating temperature increased?

Right shift

Left shift

Right shift

*A gas–phase equilibrium will shift to the side of the reaction that takes up less space (smaller coefficient sum) when pressure is increased.

CH4 (g) + 2 Cl2 (g) ⇋ CCl4 (g) + 2 H2 (g) + heat

What kind of shift would you see if:

Pressure increased?

Heating temperature increased?

No Change

Left shift

1 32 4

2 drops 0.05 M NaSCN2 drops 0.01 M Fe(NO3)3

3 drops H2O

+ 2 drops H2O

+ 2 drops 0.1 M Fe(NO3)3

+ 2 drops 0.05 M NaSCN

+ 1 drop 1 M NaNO3

0.1 MFe(NO3)3

1 MNaNO3

0.05 MNaSCN

0.01 MFe(NO3)3

H2O