1. Equilibrium Applications
How to use the equilibrium expression to determine quantities of substances AT equilibrium.

2. Note on Keq’s Keq refers to all classes of equilibrium constants:
Ksp = equilibrium of solubility of salts
IE: NaCl (s)  Na+ (aq) + Cl- (aq)
Kc refers to concentrations of aqueous solutions.
Kp refers to equilibrium systems dealing with gases in terms of their partial pressures.
This value is NOT the same as a Keq for the same reaction in terms of concentration [Kc].
Ka refers to the solubility constant of acids
Kb refers to the solubility constant of bases

3. Remember: meaning of Keq values
For a general equilibrium expression:
Keq = [product]
[reactant]
Therefore:
for Keq >>> 1 the chemical reaction is heavily favoured to the right - product side.
Keq ~ 1 the reaction is favoured in neither direction.
Keq <<< 1 the reaction is heavily favoured to the left – reactant side

4. Equilibrium expression uses:
Given concentrations of react/prod: calculate the Keq for a reaction.
[ ] MUST be at equilibrium.
Given Ksp: determine the [ ] of aqueous ions for a salt at equilibrium.
Ksp = solubility product
Given Keq and initial concentrations: determine the equilibrium concentrations of react/products
Works as well for equilibrium systems in terms of partial pressures.

5. 1. Determine the Keq Kc = [NH3]2 = [0.157]2 = 0.06025
For the reaction:
N2 (g) + 3H2 (g)  2NH3 (g)
Provided: remember [ equilibrium
[N2] = M; [H2] = 0.763; [NH3] = M
Determine the Kc!
Kc = [NH3]2 = [0.157]2 =
[N2][H2]3 [0.921][0.763]3

6. Kp application: N2 (g) + 3H2 (g)  2NH3 (g)
It can be shown that the previous situation could be converted to pressures assuming a 1.0 Litre container. Therefore: using PV=nRT
0.921 moles of N2  22.5 atm
0.763 moles of H2  atm
0.157 moles of NH3  3.84 atm
Kp = P2NH3 = [3.84]2 =
PN2P3H2 [22.5][18.65]3

7. Ksp : Solubility of a salt:
Because all salts are solids at room temp:
SaYLtZ (s)  YSa (aq) ZLt (aq)
Ksp = [Sa]*Y[Lt]*Z
Therefore if you are provided the solubility product - Ksp – then you can easily calculate the [] of the aqueous ions.

8. Example of Ksp ZnCl2 (s)  Zn+2 (aq) + 2Cl- (aq)
Ksp = 1.5 x = [Zn][Cl-]2
Therefore: 1.5 x = [x][2x]2 = 4x3
x = ∛1.5 x / 4
x = *10-4 mol/L
The solubility of ZnCl2 (s) is therefore *10-4 mol/L

9. 3. Determine Eq [] given initial [reactant] and [product]
Hypothesize the following:
CO (g) + H2O (g)  CO2 (g) + H2 (g) Keq = [CO2][H2]
[CO][H2O]
Vcontainer = 0.125
nCO and H2O = 0.250
Therefore the [ ] of each reactant:
Molarity = moles / volume (L)
= / 0.125
= 2.00 M
Cond’t

10. How to set up the previous Q!
CO (g) + H2O (g)  CO2 (g) + H2 (g)
I M M
C x x +x +x
E – x – x x x
Substitute: Keq = (x)(x) = 1.56
(2.00-x)(2.00-x)
= x2 = 1.56
(2.00 – x)2

11. Finishing the question:
Taking the square root of both sides provides:
(x) = √1.56 = ±1.25
(2.00 – x)
cross multiply, gather like terms, solve for x
2.25x = 2.50
x = 1.11 mol/L
Finally see the next slide:

12. The finale: CO (g) + H2O (g)  CO2 (g) + H2 (g) If x = 1.11 mol/L
I M M
C x x +x x
E – x – x x x
If x = 1.11 mol/L
Then the equilibrium concentrations are:
E M 0.89 M M M