Enough NO gas was placed into a 1.00 liter container such that the pressure was 0.750 atm, and the following reaction was allowed to reach equilibrium:
2 NO (g) N₂(g) + O₂(g)
If the equilibrium pressure of O₂was 5.79 10 - 2 atm, what is the value of K p for this reaction?

Question

Enough NO gas was placed into a 1.00 liter container such that the pressure was 0.750 atm, and the following reaction was allowed to reach equilibrium:
2 NO (g)  N₂(g) + O₂(g)
If the equilibrium pressure of O₂was 5.79 10 - 2 atm, what is the value of K p for this reaction?

Quizplus · Accepted Answer

First, we need to calculate the equilibrium concentration of NO. We can use the ideal gas law to do this:

PV = nRT

n(NO) = (P(NO) × V)/(R × T) = (0.750 atm × 1.00 L)/(0.0821 L·atm/mol·K × 298 K) = 0.0298 mol

The initial concentration of NO is 0.0298 mol/L.

At equilibrium, the molar ratio of N₂:O₂ is 1:1 (from the balanced equation), which means that the equilibrium concentration of O₂ is also 0.0298 mol/L. Using the ideal gas law again, we can calculate the equilibrium partial pressures of N₂ and O₂:

P(N₂) = (n(N₂) × R × T)/V = (0.0298 mol × 0.0821 L·atm/mol·K × 298 K)/1.00 L = 0.709 atm

P(O₂) = (n(O₂) × R × T)/V = (0.0298 mol × 0.0821 L·atm/mol·K × 298 K)/1.00 L = 0.709 atm

Now we can use the equilibrium partial pressures to calculate Kp:

Kp = (P(N₂) × P(O₂))/²(P(NO))
    = (0.709 atm × 0.709 atm)/(0.750 atm)²
    = 8.33×10^-3

Therefore, the answer is D.

Enough NO Gas Was Placed into a 1