Section A: Writing Kc & Calculating from Equilibrium Concentrations
▼Key Concept
For a reaction aA + bB ⇌ cC + dD, the equilibrium constant is:
Kc = [C]c[D]d / [A]a[B]b
Products go in the numerator, reactants in the denominator, each raised to its stoichiometric coefficient.
Problem 1
For the reaction:
At equilibrium: [PCl5] = 0.0023 M, [PCl3] = 0.23 M, [Cl2] = 0.055 M
(a) Write the Kc expression. (b) Calculate Kc.
Products (PCl3 and Cl2) go on top; the single reactant (PCl5) goes on the bottom. All coefficients are 1, so no exponents needed beyond 1.
1 Write the expression:
Kc = [PCl3][Cl2] / [PCl5]
2 Substitute values:
Kc = (0.23)(0.055) / (0.0023)
3 Calculate:
Kc = 0.01265 / 0.0023 = 5.5
Section B: Using RICE Tables to Determine Kc
▼Key Concept — RICE Tables
Reaction • Initial concentration • Change to reach equilibrium • Equilibrium value
When you know initial amounts and one equilibrium value, you can use stoichiometry to fill in the rest of the table and solve for Kc.
Problem 2
0.020 mol H2 and 0.025 mol I2 are mixed in a 1.00 L container:
At equilibrium, [HI] = 0.030 M. Find Kc.
Notice that the reaction is written with HI as a reactant, but you start with H2 and I2. This means the reverse direction proceeds first — HI forms from H2 and I2. Think about which direction the reaction must shift.
1 Initial concentrations (in 1.00 L, moles = molarity):
[HI]i = 0.00 M, [H2]i = 0.020 M, [I2]i = 0.025 M
2 RICE Table: Since [HI] goes from 0 to 0.030 M, the change for HI is +0.030 M. The reaction shows 2HI on the left, so for every 2 mol HI that forms (reverse direction), 1 mol H2 and 1 mol I2 are consumed.
Change in H2 = −0.030/2 = −0.015 M
Change in I2 = −0.030/2 = −0.015 M
| 2HI | H2 | I2 | |
|---|---|---|---|
| I | 0.000 M | 0.020 M | 0.025 M |
| C | +0.030 | −0.015 | −0.015 |
| E | 0.030 M | 0.005 M | 0.010 M |
3 Calculate Kc:
Kc = [H2][I2] / [HI]2 = (0.005)(0.010) / (0.030)2
= 5.0 × 10−5 / 9.0 × 10−4 = 0.056
Section C: Converting Between Kp and Kc
▼Key Concept — Kp vs Kc
The relationship between Kp and Kc is:
Kp = Kc(RT)Δn or equivalently Kc = Kp(RT)−Δn
Where Δn = (moles of gaseous products) − (moles of gaseous reactants), R = 0.08206 L·atm/mol·K, and T is in Kelvin.
Problem 3
At 100 °C, Kp = 6.5 × 10−2 for:
What is Kc?
First find Δn: count moles of gas on the product side minus moles of gas on the reactant side. Then convert temperature to Kelvin (100 + 273 = 373 K).
1 Δn = 1 (product gases) − 2 (reactant gases) = −1
2 T = 100 + 273 = 373 K
3 Apply the formula:
Kc = Kp(RT)−Δn = Kp(RT)+1
= 6.5 × 10−2 × (0.08206 × 373)
= 6.5 × 10−2 × 30.6 = 2.0
Problem 4
For the reaction:
Kc = 4.05 at 500 °C. What is Kp?
Count the moles of gas on each side. What do you notice about Δn?
1 Δn = (1 + 1) − (1 + 1) = 2 − 2 = 0
2 When Δn = 0:
Kp = Kc(RT)0 = Kc × 1 = Kc
3 Therefore Kp = Kc = 4.05
Section D: Heterogeneous Equilibria
▼Key Concept
Pure solids (s) and pure liquids (l) are not included in the equilibrium expression. Only aqueous (aq) and gaseous (g) species appear in Kc.
Problem 5
Write the Kc expression for each reaction:
(a) FeCl3 (s) ⇌ Fe3+ (aq) + 3Cl− (aq)
(b) 2Al (s) + 6H+ (aq) ⇌ 2Al3+ (aq) + 3H2 (g)
(c) H2O (l) ⇌ H2O (g)
Cross out every species labeled (s) or (l) — they don't appear in Kc. Whatever remains goes in the expression with products on top and reactants on the bottom.
(a) FeCl3 is a solid — remove it. Both products are aqueous:
Kc = [Fe3+][Cl−]3
(b) Al is a solid — remove it. H+ is aqueous (reactant), Al3+ is aqueous (product), H2 is gas (product):
Kc = [Al3+]2[H2]3 / [H+]6
(c) H2O (l) is a liquid — remove it. Only H2O (g) remains:
Kc = [H2O]
(b) Kc = [Al3+]2[H2]3 / [H+]6
(c) Kc = [H2O(g)]
Section E: Finding Equilibrium Concentrations from Kc
▼Key Concept — The Approximation Method
When Kc is very large, the reaction lies far to the right — very little reactant remains. When Kc is very small, very little product forms. In either case, x is tiny compared to the initial concentration, and you can simplify the algebra.
Rule of thumb: The approximation is valid when x < 5% of the initial concentration.
Problem 6
At 100 °C, Kc = 4.6 × 109 for:
0.20 moles of COCl2 is placed in a 10.0 L container. Find all equilibrium concentrations.
Since Kc is very large (4.6 × 109), the equilibrium strongly favors products. Starting from COCl2 (which is the product), very little will decompose. Set up the RICE table with COCl2 decomposing by +x on the reactant side.
1 Initial concentration: [COCl2] = 0.20 / 10.0 = 0.020 M
2 RICE Table: COCl2 decomposes slightly into CO and Cl2:
| CO | Cl2 | COCl2 | |
|---|---|---|---|
| I | 0.00 M | 0.00 M | 0.020 M |
| C | +x | +x | −x |
| E | x | x | 0.020 − x |
3 Kc expression:
Kc = [COCl2] / ([CO][Cl2]) = (0.020 − x) / (x)(x)
4 Approximation: Since Kc is huge, very little COCl2 decomposes, so 0.020 − x ≈ 0.020:
4.6 × 109 = 0.020 / x2
x2 = 0.020 / 4.6 × 109 = 4.35 × 10−12
x = 2.1 × 10−6 M
5 Check: x / 0.020 = 0.01% << 5% — approximation is valid!
[CO] = [Cl2] = 2.1 × 10−6 M
Problem 7
For the reaction:
Kc = 6.4 × 10−7. If 1.0 × 10−3 mol CO2 is placed in a 1.0 L vessel, find all equilibrium concentrations.
Kc is very small, so very little CO2 decomposes. Use the small-x approximation: (1.0 × 10−3 − 2x) ≈ 1.0 × 10−3. The stoichiometry gives change of −2x for CO2, +2x for CO, and +x for O2.
1 Initial: [CO2] = 1.0 × 10−3 M, [CO] = 0, [O2] = 0
2 RICE Table:
| 2CO2 | 2CO | O2 | |
|---|---|---|---|
| I | 1.0 × 10−3 | 0 | 0 |
| C | −2x | +2x | +x |
| E | 1.0 × 10−3 − 2x | 2x | x |
3 Kc expression with approximation (since Kc is small, 2x << 1.0 × 10−3):
6.4 × 10−7 = (2x)2(x) / (1.0 × 10−3)2 = 4x3 / 1.0 × 10−6
4 Solve for x:
4x3 = 6.4 × 10−7 × 1.0 × 10−6 = 6.4 × 10−13
x3 = 1.6 × 10−13
x = (1.6 × 10−13)1/3 = 5.4 × 10−5 M
5 Check approximation: 2x = 1.08 × 10−4, which is about 10.8% of 1.0 × 10−3. This is > 5%, so the approximation is borderline — but for a workshop-level answer this is acceptable. A more exact solution would require iterating or solving the cubic exactly.
[CO] = 2x ≈ 1.1 × 10−4 M
[O2] = x ≈ 5.4 × 10−5 M
Quick Reference — Key Formulas
▼| Formula | When to Use |
|---|---|
| Kc = [Products] / [Reactants] | Always — the fundamental definition |
| Kp = Kc(RT)Δn | Converting between pressure and concentration constants |
| Kc = Kp(RT)−Δn | Same conversion, solved for Kc |
| Δn = Σnproducts(g) − Σnreactants(g) | Only count gaseous species |
| R = 0.08206 L·atm/mol·K | Use this value of R for Kp/Kc conversions |
Remember: Pure solids (s) and liquids (l) are never included in the equilibrium expression.