For the reaction 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g) Kp = 2.62 × 10⁻⁶ at 400 °C. A 1.50 L vessel at 400 °C is filled with SO₃(g) at an initial pressure of 3.00 atm and allowed to come to equilibrium...

Question

For the reaction 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g) Kp = 2.62 × 10⁻⁶ at 400 °C. A 1.50 L vessel at 400 °C is filled with SO₃(g) at an initial pressure of 3.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of O₂(g) at equilibrium?

Answer 1

Set up an ICE table:

2SO3(g) <===> 2SO2(g) + O2(g)

3.00........................0................0.............Initial

-2x........................+2x.............+x............Change

3.00-2x..................2x................x.............Equilibrium

Kp = (SO₂)²(O₂) / (SO₃)²

2.62x10^-6 = (2x)²(x) / (3-2x)²

2.62x10^-6 = 4x² / 4x² -12x + 9 and simplifying this by considering 2.62x10^-6 to be insignificant, we have..

4x² = 4x² - 12x + 9

x = 0.75 atm = pressure of O₂ at equilibrium