Question

At a certain temperature, a 1.00 L flask was filled with SO2 at a pressure of 4.00 atm and O2 at a pressure of 6.00 atm. When equilibrium is reached, the total pressure of the flask is found to be 9.20 atm. What is the pressure of all species at equilibrium, and what is the value of all at this temperature?

Answer 1

The pressures are:

SO₂: 3.20 atm

O₂: 5.60 atm

SO₃: 0.40 atm

The equilibrium constant (Kc) is 1.29 x 10⁻³ atm⁻².

How can you solve the pressures?

Use the total pressure and stoichiometry to set up a system of equations:

Let x be the pressure of SO₃ at equilibrium. Then, the pressures of SO₂ and O₂ at equilibrium can be calculated based on the stoichiometry and initial pressures:

SO₂: 4.00 atm - 2x atm (2 molecules consumed for each molecule of SO₃ formed)

O₂: 6.00 atm - x atm (1 molecule consumed for each molecule of SO₃ formed)

The total pressure at equilibrium is the sum of all partial pressures:

Total pressure = [SO₂] + [O₂] + [SO₃] = 9.20 atm

4.00 atm - 2x atm + 6.00 atm - x atm + x atm = 9.20 atm

3x = 1.20 atm

x = 0.40 atm

Therefore, the equilibrium pressures are:

SO₂: 4.00 atm - 2(0.40 atm) = 3.20 atm

O₂: 6.00 atm - 0.40 atm = 5.60 atm

SO₃: 0.40 atm

Kc is the ratio of the product of the equilibrium concentrations of the products raised to their stoichiometric coefficients to the product of the reactants raised to their stoichiometric coefficients. Since pressure is directly proportional to concentration in ideal gases, we can use pressures directly:

Kc = ([SO₃]²) / ([SO₂]² * [O₂])

Kc = (0.40 atm)² / ((3.20 atm)² * (5.60 atm))

Kc ≈ 1.29 x 10⁻³ atm⁻²

Therefore:

At equilibrium, the pressures are:

SO₂: 3.20 atm

O₂: 5.60 atm

SO₃: 0.40 atm

The equilibrium constant (Kc) is 1.29 x 10⁻³ atm⁻².