What is Kp for this reaction?

Question

An evacuated reaction vessel is filled with 0.80 atm of N₂ and 1.68 atm of Br₂. When equilibrium is established according to the reaction below, there are 1.42 atm of Br₂ remaining in the vessel. What is Kp for this reaction?



2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g)

Answer 1

2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g)

0.................0.80.........1.68

+2x.............(-x).........(-3x)

2x...........(0.80-x)......(1.68-3x)

[Br₂]_eq=1.42, so

1.68-3x=1.42

Solving for x:

x≈0.0866667

Use the x value to find the equilibrium concentrations of the other things:

NBr₃ equilibrium=2x=0.1733334

N₂ equilibrium=0.80-x=0.713333

Now solve for Kp

Kp=(P_N2)(P_Br2³) / (P_NBr3 ²)

Kp=(0.713333 (1.42³) / (0.1733334²)

Kp=67.981856363