Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O₄]?

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Quizplus · Accepted Answer

We can use the equation: ΔG° = -RTln(K) where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant. First, we need to calculate ΔG° using the equation: ΔG° = ΔH° - TΔS° where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin. We are given ΔS° = 58.03 kJ/mol and we can assume that ΔH° is independent of temperature. Therefore, we can calculate ΔG° at 25 °C (298 K): ΔG° = (58.03 kJ/mol) - (298 K)(0.05803 kJ/mol K) = 40.67 kJ/mol Next, we can use the equation above to calculate ln(K): ln(K) = -ΔG°/RT = -(40.67 kJ/mol)/[(8.314 J/mol K)(298 K)] = -15.41 To solve for the equilibrium concentration of N₂O₄, we can use the equilibrium expression: K = [NO₂]2/[N₂O₄] And since we know that [NO₂]/[N₂O₄] = 2 (because the stoichiometric coefficient of NO₂ is 2 in the balanced equation), we can substitute to get: K = 2²/[N₂O₄] = 4/[N₂O₄] Solving for [N₂O₄], we get: [N₂O₄] = 4/K = 4/exp(-15.41) = 0.379 M

Consider the Equilibrium System: N2O4(g)⇌ 2 NO2(g)for Which Kp =

Consider the Equilibrium System: N₂O₄(g)⇌ 2 NO₂(g)for Which K_p =