At a certain temperature, a 29.5-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 22.6 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients) ?

Question

Alexzander Irwin

Physics

5 October, 03:52

At a certain temperature, a 29.5-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 22.6 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients) ?

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Makhi Kerr · Answer 1

Kc = conc. products / conc. reactants Check their mole ratios: Equation: SO2 (g) + N2O (g) ⇌ SO3 (g) + N2 (g) All 1:1 ratios, so Kc = [SO3][N2]/[SO2][N2O] Moles: 4.6g SO2 x 1 mol SO2 / 64.066g = 0.0718 moles SO2 0.98g N2O x 1 mol N2O / 44.0g = 0.0223 moles N2O 3.5g SO3 x 1 mol SO3 / 80.1g = 0.0437 moles SO3 16.8g N2 x 1 mol N2 / 28.0g = 0.600 moles N2 Concentrations: 0.0718 mol / 22L = 3.26 x 10^-3 M SO2 0.0223 mol / 22L = 1.01 x 10^-3 M N2O 0.0437 mol / 22L = 1.99 x 10^-3 M SO3 0.600 mol / 22L = 2.73 x 10^-2 M N2 Kc = (1.99x10^-3) (2.73x10^-2) / (1.01x10^-3) (3 ... = 16.5