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 17.3 L container at 298K contains NOBr at a pressure of 0.351 atm and NO at a pressure of 0.222 atm, the equilibrium partial pressure of Br₂ is atm.

Answer 1

The exercise indicates that the equilibrium constant Kp is 0.160 at a T = 298 K

With a busy volume of 17.3 L, and we also know the partial pressures of NOBr and NO, so we proceed to work with the equation of Kp to determine partial pressure of Br2

Recall that Kp is the equilibrium constant expressed in terms of its partial pressures, for which we have, whose definition is as follows

In our case

Replaced in the equation the values of exercise and clearly PBr2

Kp=0.160

PNOBr=0.351 atm

PNO=0.222 atm