1. and Equilibrium Problems
Keq
and Equilibrium Problems

2. C12-4-04 EQUILIBRIUM AND Keq
OUTCOME QUESTION(S):
C EQUILIBRIUM AND Keq
Relate the concept of equilibrium to physical and chemical systems.
Include: conditions necessary to achieve equilibrium.
Write equilibrium law expressions from balanced chemical equations and solve problems involving equilibrium constants.
Include: ICE tables
Vocabulary & Concepts

3. *Always make sure the equation is balanced, FIRST*
1. Plug
and solve
*Always make sure the equation is balanced, FIRST*

4. Always check the volume
At 225°C, a 2.0 L container holds moles of N2, 0.15 moles of H2 and 0.50 moles of NH3. If the system is at equilibrium, calculate KC.
N2(g) + 3 H2(g) NH3(g)
1. Change all into concentrations - mol/L
0.040 mol N2
= M N2
2.0 L
Always check the volume
= M H2
= M NH3
2. Write the equilibrium expression
Kc =
[N2][H2]3
[NH3]2

5. Don’t worry about the units for the constant
3. Substitute the concentrations and calculate K.
0.020 M N2
Kc =
[N2][H2]3
[NH3]2
0.075 M H2
M NH3
Kc =
[0.020][0.075]3
[0.25]2
Don’t worry about the units for the constant
Kc = 7.4 x 103

6. The equilibrium concentrations of N2 and NO are 1. 40 mol/L and 5
The equilibrium concentrations of N2 and NO are 1.40 mol/L and 5.20 mol/L. Calculate the KC.
2 NO(g) N2(g) + O2(g)
5.20 M
1.40 M
1.40 M
According to the equation stoichiometry – the same amount of O2 will be produced as N2 (1:1 ratio)
Kc =
[N2][O2]
[NO]2
Kc =
[1.40][1.40]
[5.20]2 =
0.0725

7. 2. Rearrange
and solve

8. Concentrations are already in mol/L (M)
At 210°C, the Kc is At equilibrium,
[N2] = mol/L and [O2] = 0.60 mol/L.
Calculate the [NO]eq.
Concentrations are already in mol/L (M)
N2(g) + O2(g) NO(g)
Kc =
[N2][O2]
[NO]2
Temperature isn’t needed here – but remember the constant changes if temperature changes
Write Eqbm expression =
Kc[N2][O2]
[NO]2
2. Rearrange for missing value =
Kc[N2][O2]
[NO]

9. = Kc[N2][O2] [NO] = (64.0)(0.40)(0.60) [NO] = 3.9 mol/L [NO]
3. Substitute concentrations then solve =
Kc[N2][O2]
[NO] =
(64.0)(0.40)(0.60)
[NO] =
3.9 mol/L
[NO]
These question types are not difficult, so to increase difficulty look for:
Unbalanced equations
Volume values
Unmatched units
Stoichiometry assumptions

10. Calculate the [S] and [O2] at equilibrium.
At 10.0°C, the Kc is The equilibrium concentrations of SO2 is 9.40 mol/L.
Calculate the [S] and [O2] at equilibrium.
SO2(g) S(g) + O2(g)
9.40 M
x M
x M
Assume values should be the same for sulfur and oxygen – but we will need to use a variable here…
Kc =
[S][O2]
[SO2]
Write Eqbm expression
2. Rearrange for missing value =
[S][O2]
[SO2]Kc

11. [S]eq and [O2] eq = 45.0 M = [S][O2] [SO2]Kc = [x][x] [9.40](215) =
3. Substitute concentrations then solve
1. Write Eqbm
expression =
[S][O2]
[SO2]Kc
2. Rearrange for
missing value =
[x][x]
[9.40](215) =
[x]2
(2021) =
[x]
(2021) =
[x]
45.0 M
[S]eq and [O2] eq = 45.0 M

12. The 4th type – “hard” ICE tables – will appear in another unit
3. Simple I.C.E. Tables
The 4th type – “hard” ICE tables – will appear in another unit

13. 1.00 M of hydrogen and 1.00 M of fluorine are allowed to react at 150.0°C. At equilibrium, the [HF] is 1.32 M. Calculate KC
H2(g) F2(g) HF(g)
[Initial]
ICE table questions involve reaction progress – to identify this type, look for expressions showing the passage of time
Tables are used to calculate the equilibrium concentrations from known progress values to use in the equilibrium expression
[Change]
[Eqlbm]

14. 1.00 M of hydrogen and 1.00 M of fluorine are allowed to react at 150.0°C. At equilibrium, the [HF] is 1.32 M. Calculate KC
H2(g) F2(g) HF(g) 1 1
[Initial]
1.00
[Change]
- 0.66
- 0.66
+ 1.32
[Eqlbm]
0.34
0.34
1.32
Enter available data into the ICE table.
– if no initial value is given assume a value of zero
Find the change for one species and make the appropriate changes in the others.
Notice: the changes in concentration reflect the stoichiometry of the reaction.

15. (What is favoured? How much?)
H2(g) F2(g) HF(g)
[Eqlbm]
0.34
0.34
1.32
Use the calculated equilibrium values to solve the equilibrium expression for the constant.
Kc =
[H2][F2]
[HF]2
Kc =
0.1156
1.742 =
15.1 Kc
Kc =
[0.34][0.34]
[1.32]2
Look at the constant value – how does it reflect the concentrations at equilibrium
(What is favoured? How much?)

16. Be mindful of the “tricks”
Initially 2.0 mol of SO2, 1.0 mol of O2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O2 are found to remain. Calculate the Kc.
2 SO3 (g) SO2 (g) + O2 (g)
[I]
0.67
0.33
[C]
[E]
0.067
2.0 mol SO2
= M SO2
3.0 L
= M O2
Be mindful of the “tricks”
= M O2

17. Initially 2. 0 mol of SO2, 1. 0 mol of O2 are mixed in a 3
Initially 2.0 mol of SO2, 1.0 mol of O2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O2 are found to remain. Calculate the Kc.
2 SO3 (g) SO2 (g) + O2 (g) 1
[I]
0.67
0.33
[C]
0.52 +
0.52 – –
0.26
[E]
0.52
0.15
0.067
This reaction is being run “backwards” – this only affects the ICE table and how information is recorded:
make sure the appropriate side is being made or used up (+/-) to reflect the progress of the reaction

18. Kc = 5.6 x 10-3 2 SO3 (g) 2 SO2 (g) + O2 (g) [E] 0.52 0.15 0.067 Kc =
A “backwards” reaction only affects the ICE table and how information is recorded:
Write the equilibrium expression for reaction as written
Kc =
[SO2]2[O2]
[SO3]2
Kc =
Kc = 5.6 x 10-3
Kc =
[0.15]2[0.067]
[0.52]2
Again…look at the constant value – how does it reflect the concentrations at equilibrium

19. C12-4-04 EQUILIBRIUM AND Keq
CAN YOU / HAVE YOU?
C EQUILIBRIUM AND Keq
Relate the concept of equilibrium to physical and chemical systems.
Include: conditions necessary to achieve equilibrium.
Write equilibrium law expressions from balanced chemical equations and solve problems involving equilibrium constants.
Include: ICE tables
Vocabulary & Concepts