**Answer:**

9.2

**Explanation:**

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm 5.1 atm 0 Initial

-2x -x +2x Reacts (stoichiometry is 2:1:2)

4.9-2x 5.1-x 2x Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2