Kp and Equilibrium Pressures

Question

The equilibrium constant, K_p, for the following reaction is 0.160 at 298K.

2NOBr(g) 2NO(g) + Br₂(g)

If an equilibrium mixture of the three gases in a 11.8 L container at 298K contains NOBr at a pressure of 0.481 atm and NO at a pressure of 0.506 atm, the equilibrium partial pressure of Br₂ is  _____atm.

Answer 1

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2NOBr(g) < ===> 2NO(g) + Br₂(g)

0.481atm...............0.506.........x.......Equilibrium pressures

Kp = (NO)²(Br₂) / (NOBr)²

0.160 = (0.506)²(Br₂) / (0.481)²

0.160 = 0.256 (Br₂) / 0.233

(Br₂) = 0.146 atm