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.

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)

Kp = p SO₃² / (p SO₂² x p O₂)

The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm SO₂ O₂ SO₃

I 3.3 0.79 0

C - 2x - x 2x

E 3.3 - 2x 0.79 - x 2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:

p SO₂ = 3.3 - 0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm =.56 atm

Now we can calculate Kp:

Kp = 0.47² / [ (2.83) ² x 0.56 ] = 0.049 (rounded to 2 significant figures)

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.