Question

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 1.5 L flask with 0.59 atm of sulfur dioxide gas and 2.9 atm of oxygen gas at 35.0 °C . He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.53 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=

Answer 1

The equilibrium reaction is given below

2SO₂(g) + O₂(g) <-------> 2SO₃(g)

From the balanced reaction it can be observed that 2 atm of SO₂(g) react with 1 atm of O₂(g) to form 2 atm of SO₃(g)

In other words 1 atm of SO₂(g) react with 1/2 atm of O₂(g) to form 1 atm of SO₃(g)

Thus 1 atm of SO₃(g) is formed when 1 atm of SO₂(g) is consumed

At equilibrium amount of SO₃(g) present = 0.53 atm

Thus amount of SO₂(g) consumed = 0.53 atm

Hence the equilibrium concentration of SO₂(g) = 0.59 atm - 0.53 atm = 0.06 atm

Similarly amount of O₂(g) consumed = 0.53 / 2 = 0.26 atm

Amount of O₂(g) left at equilibrium = 2.9 atm - 0.26 atm = 2.64 atm

Kp = [SO₃(g)]² / [SO₂(g)]² [O₂(g)] = (0.53 atm)² / (0.06)² atm X (2.64 atm) = 29. 55 atm