For the reaction SO₂(g)+ NO₂(g) SO₃(g)+ NO(g), the equilibrium constant is 18.0 at 1,200ºC. If 1.0 mole of SO₂ and 2.0 moles of NO₂ are placed in a 20.L container, what concentration of SO₃ will be present at equilibrium?

Question

Quizplus · Accepted Answer

Using the equilibrium expression: K = [SO₃][NO] / [SO₂][NO₂]^2 If we let x be the concentration of SO₃ at equilibrium, then the initial concentrations will be [SO₂] = 1.0 M and [NO₂] = 2.0 M. At equilibrium, the concentrations will be: [SO₂] = 1.0 - x M [NO₂] = 2.0 - 2x M [NO] = x M Substituting these into the equilibrium expression and solving for x: 18 = (x)(2 - 2x) / (1 - x)(2 - 2x)^2 18(2 - 2x)^2 = x(1 - x)(4 - 4x) 72 - 72x + 18x^2 = 4x - 4x^2 22x^2 - 76x + 72 = 0 x = 0.48 M Therefore, the concentration of SO₃ at equilibrium is 0.48 M, and the answer is choice A.

For the Reaction SO2(g)+ NO2(g) SO3(g)+ NO(g)

For the Reaction SO₂(g)+ NO₂(g) SO₃(g)+ NO(g)