Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500. mL flask with 3.3 atm of sulfur dioxide gas and 0.79 atm of oxygen gas at 31.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.47 atm Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

0.051

Explanation:

Let's consider the following reaction.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

We can compute the pressures using an ICE chart.

         2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I             3.3         0.79           0

C           -2x            -x           +2x

E         3.3-2x      0.79-x         2x

The partial pressure of sulfur trioxide gas is 0.47 atm. Then,

2x = 0.47

x = 0.24

The pressures at equilibrium are:

pSO₂ = 3.3-2x = 3.3-2(0.24) = 2.82 atm

pO₂ = 0.79-x = 0.79-0.24 = 0.55 atm

pSO₃ = 0.47 atm

The pressure equilibrium constant (Kp) is:

Kp = pSO₃² / pSO₂² × pO₂

Kp = 0.47² / 2.82² × 0.55

Kp = 0.051

answer

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