Question

You place 3.00 mol of pure SO₃ in an 8.00-L flask at 1150 K. At equilibrium, 0.580 mol of O₂ has been formed. Calculate Kc for the following reaction at 1150 K:

2SO₃ (g)⇌2SO₂ (g)+O₂ (g)

A) 2.3×10⁻³

B) 4.6×10⁻³

C) 9.2×10⁻³

D) 1.8×10⁻²

Answer 1

To find Kc for the equilibrium reaction, we calculate the equilibrium concentrations of SO₃, SO₂, and O₂, and then plug these values into the equilibrium constant expression. After calculation, Kc is approximately 7.2''10³, so the closest given option is C) 9.2''10³.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), we first determine the concentration of each species at equilibrium. Originally, we have 3.00 mol of SO₃ in an 8.00-L flask, which gives an initial concentration of SO₃ as (3.00 mol/8.00 L = 0.375 M). At equilibrium, 0.580 mol of O₂ has formed. Since the reaction shows that 2 moles of SO₃ produce 1 mole of O₂, then 2((0.580 mol O₂) = 1.160 mol of SO₃ have reacted.

The concentration of SO₃ at equilibrium is (initial concentration of SO₃ - concentration of SO₃ reacted) = 0.375 M - (1.160 mol/8.00 L) = 0.375 M - 0.145 M = 0.230 M. The concentration of SO₂ will be the same as that of O₂ because they are produced in a 1:1 ratio from SO₃. Therefore, the concentration of SO₂ will also be (0.580 mol/8.00 L = 0.0725 M). The concentration of O₂ at equilibrium is (0.580 mol/8.00 L = 0.0725 M).

Using the balanced chemical equation, we write the equilibrium constant expression as:
Kc = ([SO₂]² * [O₂])/[SO₃]². Substituting the equilibrium concentrations, we get: Kc = ((0.0725)² * (0.0725)/((0.230)² = ((0.00526)*(0.0725)/(0.0529) = ((0.000381)/(0.0529) = 7.208''10³ which is approximately equal to 7.2''10³, so the closest option is C) 9.2''10³.