All equilibrium concentrations known: substitute
If every equilibrium concentration (or partial pressure) has been measured, just substitute them into the K expression. Three things to watch: use equilibrium values, never initial ones; the exponents are the coefficients; and when you are given moles, divide by the volume to get concentrations first.
Only some known: use an ICE table
Often you know the initial concentrations but have measured only one equilibrium concentration. Then set up an ICE table三段式:
- I (initial): the initial concentration of each substance;
- C (change): let x be an unknown; each change = coefficient × x, negative for reactants (used up) and positive for products (formed);
- E (equilibrium): initial + change.
Use the measured substance to find x, fill in the other equilibrium concentrations, and then substitute into the K expression. The changes are in the ratio of the coefficients, but the equilibrium concentrations generally are not.
Fill in the table step by step, then compare with the K given by OpenStax:
To calculate Kp, write partial pressures (atm) in every cell of the ICE table; the method is exactly the same.
Example 1: K from equilibrium concentrations
At 2000 °C, N₂(g) + O₂(g) ⇌ 2NO(g) reaches equilibrium with [N₂] = [O₂] = 0.36 M and [NO] = 7.3 × 10⁻³ M. Find K.
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Write the K expression. The product NO has coefficient 2, which becomes an exponent:
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Substitute the equilibrium concentrations. The values given are already equilibrium values, so they go straight in:
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Work out the numerator and denominator.
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Divide.
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Check and interpret. This matches the OpenStax value at 2000 °C. K is far below 1, so reactants are favored and there is very little NO: 0.36 M of N₂ sits alongside only 0.0073 M of NO.
Example 2: Kp from partial pressures
At 400 °C, H₂ and I₂(g) are fed into a sealed container, each at an initial partial pressure of 0.200 atm. After H₂(g) + I₂(g) ⇌ 2HI(g) reaches equilibrium, the depth of the violet color shows that the partial pressure of I₂ is 0.044 atm. Find Kp.
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Set up an ICE table (partial pressures in atm). At constant temperature and volume, partial pressure is proportional to moles, so pressures change according to the coefficients, just like concentrations:
H₂ I₂ 2HI I 0.200 0.200 0 C −x −x +2x E 0.200 − x 0.200 − x 2x -
Find x from the measured pressure of I₂.
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Write the equilibrium pressures.
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Write Kp and substitute.
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Check. This matches the K of 50 at 400 °C given by OpenStax. Both sides have the same number of gas molecules (2 = 2), so Kp = Kc for this reaction.
- N₂ + 3H₂ ⇌ 2NH₃, with N₂ changing by −x. The change for H₂?
H₂’s coefficient is 3 times N₂’s.
- N₂ + 3H₂ ⇌ 2NH₃, with N₂ changing by −x. The change for NH₃?
2 mol of NH₃ form, so the sign is positive.
- 2SO₂ + O₂ ⇌ 2SO₃, with O₂ changing by −x. The change for SO₂?
SO₂’s coefficient is twice O₂’s.
- 2SO₂ + O₂ ⇌ 2SO₃, with O₂ changing by −x. The change for SO₃?
The product increases.
- N₂O₄ ⇌ 2NO₂, starting from N₂O₄, with N₂O₄ changing by −x. The change for NO₂?
Each N₂O₄ that breaks apart forms 2 NO₂.
- N₂O₄ ⇌ 2NO₂, starting from NO₂ (reaction in reverse), with N₂O₄ changing by +x. The change for NO₂?
Going in reverse, NO₂ decreases.
- H₂ + I₂ ⇌ 2HI, with H₂ changing by −x. The change for I₂?
Same coefficient.
- CO + H₂O ⇌ CO₂ + H₂, with CO changing by −x. The change for H₂?
A product with coefficient 1.
Common mistakes
Mistake 1: “Put the initial concentrations into the K expression.” K can only be calculated from equilibrium concentrations. Initial concentrations just fill the I row of the ICE table.
Mistake 2: “Substitute moles directly into K.” K uses concentrations. The volume cancels only when both sides have the same number of gas moles.
Mistake 3: “Equilibrium concentrations are in the ratio of the coefficients.” The coefficients give only the ratio of the changes. Equilibrium concentrations also depend on the starting concentrations.
Quick check
At a certain temperature, SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) is at equilibrium with PSO₂Cl₂ = 0.80 atm and PSO₂ = PCl₂ = 0.40 atm. What is Kp?
Kp = 0.40 × 0.40 ÷ 0.80 = 0.20. A uses only one partial pressure; B inverts the expression; C adds the two partial pressures.
The reaction 2A(g) ⇌ B(g) starts with [A] = 1.00 M. At equilibrium [B] = 0.30 M. K is about:
Forming 0.30 M of B uses 0.60 M of A, so [A] = 0.40 M at equilibrium. K = 0.30 ÷ 0.40² = 1.9. A forgets the square; C uses up A 1:1; D inverts the expression.
Pure SO₃ is placed in a container at [SO₃] = 0.200 M, where 2SO₃(g) ⇌ 2SO₂(g) + O₂(g). At equilibrium [O₂] = 0.050 M. What is K?
O₂ rises by 0.050 M, SO₂ by 0.100 M, and SO₃ falls by 0.100 M. K = 0.100² × 0.050 ÷ 0.100² = 0.050. A comes from ignoring the coefficients in the changes.
A student substitutes the equilibrium moles of N₂, H₂, and NH₃ (in a 2.0 L container) directly into K = [NH₃]² ÷ ([N₂][H₂]³). The result is:
The numerator has 2 concentrations and the denominator 4, so the volume doesn’t cancel. Each amount must be divided by 2.0 L first.
To find K from experimental data with an ICE table, which of these is NOT needed?
K needs only equilibrium concentrations. A rate constant describes how fast the reaction is and plays no part in calculating K.
The same reaction is run twice at 25 °C with different starting concentrations. The values of K calculated from the two runs:
Same temperature, same K. The equilibrium concentrations differ, but they give the same K when substituted into the expression.
AP-style free response
At a certain temperature, 0.500 mol of NOCl is placed in a sealed 1.00 L container, where 2NOCl(g) ⇌ 2NO(g) + Cl₂(g). At equilibrium, 0.0500 mol of Cl₂ is present (schematic data).
(a) Write the expression for Kc for this reaction.
(b) Set up an ICE table and find the equilibrium concentration of each substance.
(c) Calculate Kc.
(d) If 1.000 mol of NOCl were used at the same temperature, would Kc change? Would [Cl₂] at equilibrium be 0.100 M? Justify your answers.
(e) At equilibrium [NO] is twice [Cl₂], while [NOCl] is much larger than [NO]. Which of these ratios is set by the coefficients?
Model answer and scoring
(a) Kc = [NO]²[Cl₂] ÷ [NOCl]².
(b) I: [NOCl] = 0.500, [NO] = 0, [Cl₂] = 0. C: −2x, +2x, +x, with x = 0.0500. E: [NOCl] = 0.400 M, [NO] = 0.100 M, [Cl₂] = 0.0500 M.
(c) Kc = (0.100)² × 0.0500 ÷ (0.400)² = 3.1 × 10⁻³.
(d) Kc wouldn’t change, because the temperature is the same. [Cl₂] would not simply double: for Q to equal Kc again, the equilibrium concentrations don’t scale in proportion (the numerator is third order in concentration and the denominator second order).
(e) The 2 : 1 ratio of [NO] to [Cl₂] comes from the coefficients (both start at zero and form in a 2 : 1 ratio). The ratio of NOCl to NO depends on K and the starting concentration, not on the coefficients.
Scoring: (a) the expression; (b) ICE table using the coefficients; (c) the value; (d) K depends only on temperature, and equilibrium concentrations don’t scale proportionally; (e) distinguishing the ratio of changes from the ratio of equilibrium concentrations.
Summary
- K is calculated from equilibrium concentrations (or partial pressures); convert moles to concentrations first.
- ICE table: initial + change = equilibrium; the changes are in the ratio of the coefficients, and the measured substance gives x.
- K depends only on temperature, not on the starting concentrations.
AP tip: Label the units (M or atm) in your ICE table and write the C row as “coefficient × x.” When you substitute, show the expression with numbers before giving the result, so you can earn method points even if the arithmetic slips.