For the reaction SO₂(g) + NO₂(g) $\rightleftharpoons$ 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

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Quizplus · Accepted Answer

Using the equilibrium constant expression: Kc = [SO₃][NO]/[SO₂][NO₂]^2 Substituting the values given: 18.0 = [SO₃][NO]/[SO₂][NO₂]^2 We are interested in the equilibrium concentration of SO₃, which we can represent as x. Therefore, the equilibrium concentrations of SO₂ and NO₂ will be (1 - x) and (2 - x), respectively. Substituting these expressions into the equilibrium constant expression: 18.0 = x*[(2 - x)]/[(1 - x)]^2 Simplifying: 18.0 = (2 - x)/(1 - x)^2 18.0(1 - x)^2 = 2 - x Expanding and rearranging: 18.0x^2 - 35.0x + 16.0 = 0 Solving for x using the quadratic formula: x = 0.476 mol/L (rounded to three significant figures) Therefore, the concentration of SO₃ at equilibrium is 0.48 mol/L.

For the Reaction SO2(g) + NO2(g) ⇌\rightleftharpoons⇌ SO3(g) + NO(g), the Equilibrium Constant Is 18

For the Reaction SO₂(g) + NO₂(g) $\rightleftharpoons$ SO₃(g) + NO(g), the Equilibrium Constant Is 18