E. Chemical Systems and Equilibrium · Chemical EquilibriumKp, Combining Equilibrium Constants and the K Family
E3.4 E2.1 E2.4 A1.12 A1.13About 70 minutes
Learning goals- I can write Kp expressions using partial pressures of gases.
- I can convert between Kc and Kp using Kp = Kc(RT)^Δn.
- I can find K for a reversed, multiplied or combined equation.
- I can name the common equilibrium constants (Kc, Kp, Ksp, Kw, Ka, Kb) and write their expressions.
Words to know
partial pressure
The pressure one gas in a mixture would exert if it filled the container alone; the partial pressures add to the total pressure (Dalton's law).
Kp
An equilibrium constant written with the equilibrium partial pressures of gases instead of concentrations.
Δn (for Kp)
The change in the number of moles of gas: moles of gaseous products minus moles of gaseous reactants in the balanced equation.
Ksp
The solubility product: K for a slightly soluble ionic solid dissolving in water.
Ka and Kb
Equilibrium constants for the ionization of a weak acid (Ka) or weak base (Kb) in water.
Kw
The ion product of water: Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.
Pressure gauges, not test tubes
In an ammonia plant, nobody pours samples into a test tube to measure concentrations. The reactor is a steel vessel full of hot gases, and the easiest thing to measure is pressure. So for gas reactions, chemists and engineers often write the equilibrium constant with partial pressures instead of concentrations. This lesson shows how that works, how to switch between the two forms, and how to build new K values from old ones.
Remember from Grade 11
In Gases and Atmospheric Chemistry you used Dalton's law (P(total) = P₁ + P₂ + ...) and the ideal gas law, PV = nRT, with R = 8.314 kPa·L/(mol·K) and T in kelvins.
Writing Kp
For a reaction of gases, Kp uses each gas's equilibrium partial pressure in place of its concentration:
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) Kp = P(NH₃)² ÷ [P(N₂)P(H₂)³]
The rules are the same as for Kc: products over reactants, coefficients as exponents, and pure solids and liquids left out. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = P(CO₂).
In this course we use partial pressures in kPa, to match Grade 11. Many university textbooks use bar or atm, which give different numbers for the same Kp. Always check the pressure unit before comparing Kp values.
Converting between Kc and Kp
From the ideal gas law, P = (n/V)RT, and n/V is the concentration. So for every gas, P = [X]RT. Substituting into Kp and collecting the RT terms gives
Kp = Kc(RT)^Δn
where Δn = (moles of gaseous products) − (moles of gaseous reactants) from the balanced equation, R = 8.314 kPa·L/(mol·K), and T is in kelvins.
- If Δn = 0 (same number of gas moles on both sides, such as H₂ + I₂ ⇌ 2 HI), then Kp = Kc.
- Only count gases in Δn.
Worked exampleExample 1: Kc to Kp
For N₂O₄(g) ⇌ 2 NO₂(g) at 25.0 °C, Kc = 4.6 × 10⁻³. Calculate Kp (pressures in kPa).
- Δn = 2 − 1 = +1.
- T = 25.0 + 273.15 = 298.15 K, so RT = 8.314 × 298.15 = 2478.8.
- Kp = Kc(RT)^Δn = (4.6 × 10⁻³)(2478.8)¹ = 11.4.
Kp ≈ 11 (with pressures in kPa). It is bigger than Kc here because Δn is positive.
Worked exampleExample 2: Kp from partial pressures
In a sealed flask at 25 °C, the equilibrium partial pressures are P(NO₂) = 30.0 kPa and P(N₂O₄) = 79.0 kPa. Calculate Kp for N₂O₄(g) ⇌ 2 NO₂(g). Is it consistent with Example 1?
- Kp = P(NO₂)² ÷ P(N₂O₄)
- Kp = (30.0)² ÷ 79.0 = 900 ÷ 79.0 = 11.4
Kp = 11.4, which agrees with Example 1.
Building new K values from old ones
Because K comes straight from the balanced equation, changing the equation changes K in predictable ways:
- Reverse the equation: the expression flips upside down, so K(new) = 1 ÷ K.
- Multiply all coefficients by n: every exponent is multiplied by n, so K(new) = Kⁿ. (Halving the coefficients gives K^½ = √K.)
- Add two equations: the expressions multiply, so K(overall) = K₁ × K₂.
This is the equilibrium version of Hess's law, which adds ΔH values (you use it in Energy Changes and Rates of Reaction). With K, you multiply instead of add.
Worked exampleExample 3: Combining equations
Suppose that at a certain temperature:
(1) N₂(g) + O₂(g) ⇌ 2 NO(g) K₁ = 2.0 × 10⁻³ (2) 2 NO(g) + O₂(g) ⇌ 2 NO₂(g) K₂ = 1.5 × 10²
Find K for (a) N₂(g) + 2 O₂(g) ⇌ 2 NO₂(g); (b) 2 NO₂(g) ⇌ N₂(g) + 2 O₂(g); (c) NO(g) ⇌ ½ N₂(g) + ½ O₂(g).
- (a) Adding (1) and (2) cancels 2 NO and gives the target. K = K₁ × K₂ = (2.0 × 10⁻³)(1.5 × 10²) = 0.30.
- (b) This is (a) reversed: K = 1 ÷ 0.30 = 3.3.
- (c) This is (1) reversed and halved. Reverse: 1 ÷ (2.0 × 10⁻³) = 5.0 × 10². Halve: K = (5.0 × 10²)^½ = 22.
(a) 0.30; (b) 3.3; (c) 22.
The K family
Every equilibrium constant follows the same equilibrium law. Chemists give the most common ones special names. You will meet Ksp, Kw, Ka and Kb in full in the unit Acid-Base and Solubility Equilibria; here is a first look at their expressions.
| Constant | Used for | Example equation | Expression |
|---|
| Kc (or K) | any equilibrium, using concentrations | H₂(g) + I₂(g) ⇌ 2 HI(g) | [HI]² ÷ ([H₂][I₂]) |
| Kp | gas equilibria, using partial pressures | N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) | P(NH₃)² ÷ (P(N₂) · P(H₂)³) |
| Ksp | a slightly soluble salt dissolving | PbI₂(s) ⇌ Pb²⁺(aq) + 2 I⁻(aq) | [Pb²⁺][I⁻]² |
| Kw | the self-ionization of water | 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) | [H₃O⁺][OH⁻] |
| Ka | a weak acid ionizing in water | CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq) | [H₃O⁺][CH₃COO⁻] ÷ [CH₃COOH] |
| Kb | a weak base reacting with water | NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) | [NH₄⁺][OH⁻] ÷ [NH₃] |
Just for funRiddle: Why did K refuse to turn upside down for the reverse reaction?
It didn't refuse. It just said, "Fine, but from now on, call me 1/K."
Hands-on activityA kitchen model of partial pressure
You need: a clear 1 L or 2 L plastic bottle, two kinds of small dried food that look different (for example at least 30 white beans and at least 15 red lentils), paper.
- Put 40 "molecules" in the bottle: 30 of kind A and 10 of kind B. Shake. Pretend the bottle is a gas mixture at a total pressure of 100 kPa.
- Each "molecule" hits the walls about equally often, so each kind exerts pressure in proportion to how many there are. Calculate each partial pressure: P(A) = (30/40)(100 kPa) = 75 kPa and P(B) = (10/40)(100 kPa) = 25 kPa.
- Now "react" them: suppose 2 A ⇌ B. Swap 10 A for 5 B (2 A make 1 B). Count again (20 A, 15 B, 35 total) and find the new partial pressures if the total pressure stays 100 kPa, then calculate P(B) ÷ P(A)².
What to look for: step 3 gives P(A) = (20/35)(100) = 57.1 kPa and P(B) = (15/35)(100) = 42.9 kPa, so P(B) ÷ P(A)² = 42.9 ÷ 3265 = 0.0131. The model shows that partial pressure is just a gas's share of the total (its mole fraction times total pressure). In a real reaction, the system would keep shifting until this ratio equalled Kp.
Your turn
1Write Kp for 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g).
Kp = P(SO₃)² ÷ [P(SO₂)²P(O₂)]
2Find Δn for (a) N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g); (b) C(s) + CO₂(g) ⇌ 2 CO(g); (c) H₂(g) + Cl₂(g) ⇌ 2 HCl(g).
(a) 2 − 4 = −2; (b) 2 − 1 = +1 (carbon is a solid, so it is not counted); (c) 2 − 2 = 0.
3Suppose Kc = 0.10 for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at 250 °C. Calculate Kp (kPa).
Δn = +1; T = 523 K. Kp = 0.10 × (8.314 × 523)¹ = 0.10 × 4348 = 4.3 × 10².
4For H₂(g) + I₂(g) ⇌ 2 HI(g), Kc = 50 at about 450 °C. What is Kp?
Δn = 0, so (RT)⁰ = 1 and Kp = Kc = 50.
5If K = 4.0 for A ⇌ 2 B, find K for 2 B ⇌ A and for 2 A ⇌ 4 B.
Reversed: 1 ÷ 4.0 = 0.25. Doubled: 4.0² = 16.
6Write the expression for Ksp of CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq).
Ksp = [Ca²⁺][F⁻]² (the solid is left out).
More practice: warm-up
1What does Kp use instead of concentrations?
Equilibrium partial pressures of the gases.
When Δn = 0 (the same number of moles of gas on both sides).
3If an equation is reversed, what happens to K?
It becomes its reciprocal, 1/K.
4If two equations are added, how are their K values combined?
They are multiplied: K = K₁ × K₂.
More practice: on your own
1For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = 1.2 kPa at a certain temperature. What is the partial pressure of CO₂ at equilibrium? Does adding more CaCO₃ change it?
Kp = P(CO₂), so P(CO₂) = 1.2 kPa. Adding more solid does not change it, because solids are not in the expression.
2For 2 NO(g) + O₂(g) ⇌ 2 NO₂(g), the equilibrium partial pressures are P(NO) = 2.0 kPa, P(O₂) = 50.0 kPa and P(NO₂) = 40.0 kPa. Calculate Kp.
Kp = (40.0)² ÷ [(2.0)²(50.0)] = 1600 ÷ 200 = 8.0
3For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), suppose Kc = 0.0579 at 500 °C. Calculate Kp (kPa).
Δn = −2; T = 773 K; RT = 8.314 × 773 = 6427. Kp = 0.0579 × (6427)⁻² = 0.0579 ÷ (4.131 × 10⁷) = 1.40 × 10⁻⁹.
4For ½ N₂(g) + 3⁄2 H₂(g) ⇌ NH₃(g), what is Kc if Kc = 0.0579 for N₂ + 3 H₂ ⇌ 2 NH₃?
The coefficients are halved, so K = (0.0579)^½ = 0.241.
5Write the Ka expression for HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F⁻(aq), and say why water is left out.
Ka = [H₃O⁺][F⁻] ÷ [HF]. Water is the solvent (a pure liquid), so its concentration is essentially constant and is built into Ka.
Challenge
1Given at some temperature: (1) H₂(g) + ½ O₂(g) ⇌ H₂O(g), K₁; (2) C(s) + ½ O₂(g) ⇌ CO(g), K₂. Find an expression for K of C(s) + H₂O(g) ⇌ CO(g) + H₂(g) in terms of K₁ and K₂.
Target = (2) + reverse of (1): C + ½ O₂ + H₂O ⇌ CO + H₂ + ½ O₂, and the ½ O₂ cancels. K = K₂ × (1/K₁) = K₂ ÷ K₁.
2Derive Kp = Kc(RT)^Δn for N₂O₄(g) ⇌ 2 NO₂(g) by substituting P = [X]RT into Kp.
Kp = P(NO₂)² ÷ P(N₂O₄) = ([NO₂]RT)² ÷ ([N₂O₄]RT) = ([NO₂]² ÷ [N₂O₄]) × (RT)²⁻¹ = Kc(RT)¹. Here Δn = 2 − 1 = 1.
3Suppose Kp for 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) is 3.5 × 10⁻³ (kPa) at 1000 K. Find Kc, and Kc for SO₃(g) ⇌ SO₂(g) + ½ O₂(g).
Δn = 2 − 3 = −1, so Kc = Kp ÷ (RT)^Δn = Kp × RT = (3.5 × 10⁻³)(8.314 × 1000) = 29. The second equation is reversed and halved: Kc = (1/29)^½ = 0.19.
Quiz
1. What is Kp for 2 NO₂(g) ⇌ N₂O₄(g)?
2. For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), Δn is:
3. For which reaction does Kp = Kc?
4. If K = 1.0 × 10⁴ for A + B ⇌ C, what is K for C ⇌ A + B?
5. If K = 3.0 for X ⇌ Y, what is K for 3 X ⇌ 3 Y?
6. Which expression is Ksp for Ag₂CrO₄(s) ⇌ 2 Ag⁺(aq) + CrO₄²⁻(aq)?
Big idea (E3.4, E2.4): gas equilibria can be described with Kp (partial pressures). Kp = Kc(RT)^Δn. Reversing an equation inverts K, multiplying by n raises K to the nth power, and adding equations multiplies the K values. Several special K's (Ksp, Kw, Ka, Kb) follow the same law.
How to help: ask your teen to explain where (RT)^Δn comes from (P = [X]RT). When combining equations, have them write the target equation first and tick off each step (reverse? multiply? add?).
Common difficulties:
- Counting solids or liquids in Δn.
- Forgetting to convert °C to kelvins.
- Adding K values instead of multiplying them when equations are added.
- Comparing Kp values that use different pressure units (kPa, bar, atm).