AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-4)
Unit 7 · Topic 7.4
7.4 Calculating the Equilibrium Constant
If you know the equilibrium concentrations or partial pressures of every species, you plug them into the K expression to find K. If you know the starting amounts and only one equilibrium value, an ICE table (Initial, Change, Equilibrium) lets you work out the rest first.
Key terms
- equilibrium concentration
- ICE table
- Kc
- Kp
The direct case
K is calculated only from equilibrium values. If a problem tells you the concentration (or partial pressure) of every species once equilibrium is reached, write the expression and substitute.
Use concentrations in mol/L for Kc and partial pressures (usually atm) for Kp. If you're given moles and a volume, divide to get molarity first.
ICE tables
Often you know the starting amounts and only one equilibrium amount. An ICE table organizes the math. Write the balanced equation across the top, then three rows.
- I (Initial): the starting concentration of each species.
- C (Change): how much each concentration changes. Use the coefficients: if one species changes by x, a species with coefficient 2 changes by 2x. Reactants and products change in opposite directions.
- E (Equilibrium): Initial + Change.
- Use the one known equilibrium value to solve for x, fill in the E row, then plug into K.
Finding x from other kinds of data
Sometimes the problem gives a percentage instead of an equilibrium concentration. If 20.0% of a 0.500 M reactant with coefficient 1 reacts, then x = 0.200 × 0.500 = 0.100 M.
For gases in a rigid container, a change in total pressure can give you x. For N₂O₄(g) ⇌ 2NO₂(g) starting with 1.00 atm of N₂O₄ and no NO₂, the equilibrium pressures are 1.00 − x and 2x, so the total is 1.00 + x. If the total pressure settles at 1.30 atm, then x = 0.30 atm, P(N₂O₄) = 0.70 atm and P(NO₂) = 0.60 atm.
K doesn't depend on where you start
At a given temperature, K is the same no matter what amounts you start with. Different experiments end up with different equilibrium concentrations, but the ratio in the K expression comes out the same. The table shows three experiments for H₂(g) + I₂(g) ⇌ 2HI(g) at the same temperature.
| Experiment | Starting mixture | [H₂] (M) | [I₂] (M) | [HI] (M) | Kc |
|---|---|---|---|---|---|
| 1 | Equal H₂ and I₂ | 0.0220 | 0.0220 | 0.156 | 50.3 |
| 2 | Extra I₂ | 0.0100 | 0.0400 | 0.141 | 49.7 |
| 3 | Only HI | 0.0350 | 0.0350 | 0.248 | 50.2 |
Checking your answer
Every equilibrium concentration must be positive. If a value in your E row is negative, the change row is set up wrong (often a missed coefficient or the wrong direction).
K values are often very large or very small, so keep scientific notation and the right number of significant figures.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Kp from equilibrium pressures
For N₂O₄(g) ⇌ 2NO₂(g) at a certain temperature, the equilibrium partial pressures are P(NO₂) = 0.40 atm and P(N₂O₄) = 0.20 atm. Calculate Kp.
Show the solutionHide the solution
- Step 1: Kp = (P_NO₂)² / P_N₂O₄.
- Step 2: Kp = (0.40)² ÷ 0.20 = 0.16 ÷ 0.20 = 0.80.
Answer: Kp = 0.80
- Example 2Calculator allowed
ICE table with one known equilibrium value
A flask starts with 0.200 M H₂ and 0.200 M I₂ at a certain temperature. At equilibrium, [HI] = 0.300 M. Find Kc for H₂(g) + I₂(g) ⇌ 2HI(g).
Show the solutionHide the solution
- Step 1: I row: [H₂] = 0.200, [I₂] = 0.200, [HI] = 0.
- Step 2: C row: −x, −x, +2x (HI has coefficient 2).
- Step 3: E row: 0.200 − x, 0.200 − x, 2x.
- Step 4: Use the known value: 2x = 0.300, so x = 0.150. Then [H₂] = [I₂] = 0.200 − 0.150 = 0.050 M.
- Step 5: Kc = [HI]² / ([H₂][I₂]) = (0.300)² ÷ (0.050 × 0.050) = 0.0900 ÷ 0.0025 = 36.
Answer: Kc = 36
- Example 3Calculator allowed
Trap: moles are not concentrations
0.800 mol SO₂ and 0.400 mol O₂ are placed in a 2.00 L flask. At equilibrium there is 0.600 mol SO₃. Find Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Show the solutionHide the solution
- Step 1: Convert to molarity first: [SO₂]₀ = 0.800 ÷ 2.00 = 0.400 M, [O₂]₀ = 0.200 M, and equilibrium [SO₃] = 0.600 ÷ 2.00 = 0.300 M.
- Step 2: ICE: SO₂ changes by −2x, O₂ by −x, SO₃ by +2x. Since 2x = 0.300, x = 0.150.
- Step 3: Equilibrium: [SO₂] = 0.400 − 0.300 = 0.100 M; [O₂] = 0.200 − 0.150 = 0.050 M; [SO₃] = 0.300 M.
- Step 4: Kc = [SO₃]² / ([SO₂]²[O₂]) = (0.300)² ÷ ((0.100)² × 0.050) = 0.0900 ÷ 0.00050 = 180.
- Step 5: If you plug in moles instead of molarities, you get 90, which is wrong. Because the exponents on top and bottom don't add to the same total, the volume doesn't cancel.
Answer: Kc = 180 (not 90)
Common mistakes
- Plugging initial concentrations, instead of equilibrium concentrations, into K.
- Using moles instead of molarity when the volume isn't 1 L.
- Ignoring coefficients in the Change row: a coefficient of 2 means a change of 2x.
- Forgetting to square (or cube) terms in the K expression after finding the equilibrium values.
On the exam
- Free-response questions often walk you through an ICE table: write the expression, find x from the given data, then calculate K. Show each row so partial credit is possible.
- Questions may give data from several trials at one temperature and ask why K is the same: K depends only on temperature, not on starting amounts.
Connected topics
Videos
Check yourself
4 questions on 7.4 Calculating the Equilibrium Constant. Pick an answer to see if you got it, and why.
For H₂(g) + I₂(g) ⇌ 2HI(g) at a certain temperature, an equilibrium mixture contains [H₂] = 0.10 M, [I₂] = 0.20 M and [HI] = 1.0 M. What is Kc at this temperature?
PCl₅ is placed in an empty, rigid flask at an initial concentration of 1.00 M and decomposes: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium, [Cl₂] = 0.40 M. What is Kc?
Pure HI(g) is placed in a rigid container at an initial pressure of 1.00 atm and decomposes: 2HI(g) ⇌ H₂(g) + I₂(g). At equilibrium, P(H₂) = 0.10 atm. What is Kp?
An equilibrium mixture of N₂O₄(g) ⇌ 2NO₂(g) in a 2.0 L flask contains 0.60 mol of N₂O₄ and 0.20 mol of NO₂. What is Kc?
0 of 4 answered