Problem PageQuestion 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 1.3 atm of sulfur dioxide gas and 4.1 atm of oxygen gas, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.91 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.

SO₂: 0.39atm

O₂: 3.645atm

Explanation:

Based on the reaction:

2 SO₂(g) + O₂(g) → 2 SO₃(g)

2 moles of sulfur dioxide react per mole of oxygen to produce 2 moles of sulfur trioxide.

When the reaction occurs an comes to equilbrium, the partial pressure of each gas is:

SO₂: 1.3atm - 2X

O₂: 4.1atm -X

SO₃: 2X

Where X is the reaction coordinate.

As pressure at equilibrium of SO₃ is 0.91 atm:

0.91atm = 2X

Thus: X = 0.455atm.

Replacing, pressures at equilibrium of the gases are:

SO₂: 1.3atm - 2×0.455atm = 0.39 atm

O₂: 4.1atm -0.455atm = 3.645 atm

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