which answer best describes the reaction when 1.50 mol of P4(g) and 1.15 mol of P2(g) are mixed in a 25.0 L, closed container at 400 ºC?

Question

Given that Kp = 1.39 at 400 ºC for the reaction, P4(g) <=> 2 P2(g), which answer best describes the reaction when 1.50 mol of P4(g) and 1.15 mol of P2(g) are mixed in a 25.0 L, closed container at 400 ºC? [REMEMBER: it's Kp]

The correct answer is Q_p = 1.95, the reaction proceeds to the left

Answer 1

PV=nRT use ideal gas law

P=nRT/V solve for P for reactant and product

P_P4 = 1.5mol x .08206 atm L/mol K x 673K/25L = 3.31atm remember to convert from celsius to kelvin

P_P2 = 1.15mol x .08206 atm L/mol K x 673K/25L = 2.54atm

Q_p = (P_P2)² / (P_P4) = 2.54² / 3.31 = 1.95 > 1.39 Solve for Q_p and compare to K_p

Because Q_p > K_p reaction proceeds to the left in favor of reactants.