50.0 g of N₂O₄ is introduced into an evacuated 2.00 L vessel and allowed to come to equilibrium with its decomposition product, N₂O₄(g) ⇌ 2NO₂(g). For this reaction K_c = 0.133. Once the system has reached equilibrium, 5.00 g of NO₂ is injected into the vessel, and the system is allowed to equilibrate once again. Calculate the mass of N₂O₄ in the final equilibrium mixture.

Question

Quizplus · Accepted Answer

The correct answer is A.The reaction is: N2O4(g) ⇌ 2NO2(g)Kc = 0.133Initial moles of N2O4 = 50.0 g / 92.01 g/mol = 0.543 molInitial moles of NO2 = 0 molInitial concentration of N2O4 = 0.543 mol / 2.00 L = 0.2715 MInitial concentration of NO2 = 0 mol / 2.00 L = 0 MChange in concentration of N2O4 = -xChange in concentration of NO2 = +2xEquilibrium concentration of N2O4 = 0.2715 M - xEquilibrium concentration of NO2 = 2xKc = [NO2]^2 / [N2O4]0.133 = (2x)^2 / (0.2715 M - x)x = 0.108 MEquilibrium concentration of N2O4 = 0.2715 M - 0.108 M = 0.1635 MMoles of N2O4 at equilibrium = 0.1635 M * 2.00 L = 0.327 molMass of N2O4 at equilibrium = 0.327 mol * 92.01 g/mol = 30.2 g

500 G of N2O4 Is Introduced into an Evacuated 2

500 G of N₂O₄ Is Introduced into an Evacuated 2