Question

At 200 degrees Celsius, the equiLiBrium constant (Kp) for the reaction 2NO(g) = N₂(g) + O₂(g) is 2.40*10³. A closed vessel is charged with 36.1 atm of NO. At equiLiBrium, the partial pressure of O₂ is what?

Answer 1

To find the equilibrium partial pressure of O2, we calculate it using the equilibrium constant Kp and the reaction stoichiometry for 2NO(g) = N2(g) + O2(g). The equilibrium constant expression is set up based on the change in moles of gases and solved for the equilibrium partial pressure of O2.

Step-by-step explanation:

The student's question pertains to finding the equilibrium partial pressure of O2 when 36.1 atm of NO is placed in a closed vessel and allowed to reach equilibrium at 200 degrees Celsius for the reaction 2NO(g) = N2(g) + O2(g), given that the equilibrium constant Kp is 2.40×10³. To solve for the partial pressure of O2, we use the expression for the Kp which includes the partial pressures of the products and reactants raised to the power of their stoichiometric coefficients.

Let's assume the equilibrium partial pressure of N2 is x atm and the equilibrium partial pressure of O2 is y atm. Since the reaction consumes 2 moles of NO for every mole of O2 and N2 produced, the equilibrium partial pressure of NO will be 36.1 - 2y atm. Kp is given by the expression: Kp = (x · y)/(36.1 - 2y)². Plugging the value of Kp into the equation and solving for y will give the equilibrium partial pressure of O2.