Calculate the equilibrium constant, Kp, she obtained for this reaction.

Question

A student ran the following reaction in the laboratory at 1090 K:

2SO₂(g) + O₂(g) 2SO₃(g)

When she introduced SO₂(g) and O₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of SO₂ was 4.57 atm and the initial partial pressure of O₂ was 1.50 atm, she found that the equilibrium partial pressure of SO₃ was 1.99 atm.

Calculate the equilibrium constant, K_p, she obtained for this reaction.

K_p =

Answer 1

K_p = 1.18

Step 1 — ICE table in partial pressures

2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

         SO₂      O₂     SO₃
I:  4.57   1.50     0
C: −2x   −x   +2x
E: 4.57−2x  1.50−x  2x

You can work directly in atm — K_p is defined in partial pressures, so there is no need to convert to concentrations and back.

Step 2 — find x from the one equilibrium value you were given

2x = 1.99 atm  →  x = 0.995 atm

Step 3 — fill in the rest

P(SO₂) = 4.57 − 2(0.995) = 4.57 − 1.99 = 2.58 atm
P(O₂) = 1.50 − 0.995 = 0.505 atm
P(SO₃) = 1.99 atm

A shortcut worth seeing: SO₂ and SO₃ both carry a coefficient of 2, so the SO₂ consumed equals the SO₃ formed exactly — 1.99 atm. Oxygen, with coefficient 1, changes by half that. You can often skip solving for x entirely once you notice this.

Step 4 — the K_p expression

K_p = P(SO₃)² / [ P(SO₂)² × P(O₂) ]

K_p = (1.99)² / [ (2.58)²(0.505) ] = 3.96 / [ (6.66)(0.505) ] = 3.96 / 3.36 = 1.18

Two things to get right

1. Exponents come from coefficients. Both SO₃ and SO₂ are squared; O₂ is to the first power. Forgetting to square the SO₂ term gives 3.05 — a plausible-looking number that is simply wrong.
2. Use equilibrium pressures, not initial ones. Substituting the starting 4.57 and 1.50 into the expression is the other frequent slip. That calculation gives Q at t = 0, which is 0 here since no product existed yet.

Interpreting the result: K_p ≈ 1 means neither side dominates — at 1090 K this reaction sits with substantial amounts of both reactants and products, which the numbers show directly (2.58 atm SO₂ alongside 1.99 atm SO₃). Industrially this is exactly the problem the contact process has to solve: SO₃ formation is favoured at lower temperatures, so plants run cooler with a catalyst rather than hotter.

(Formally K_p here carries units of atm⁻¹, since Δn = 2 − 3 = −1, but most courses report equilibrium constants unitless. Follow whichever convention your instructor uses.)