Question

Consider the reaction:

2 NO(g) + Br2(g) ↔ 2 NOBr(g)  Kp = 28.4 atm-1 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 173 Torr and that of Br2 is 142 Torr. What is the partial pressure (in Torr) of NOBr in this mixture?

P_NOBr = ___ Torr

Answer 1

        2 NO(g) + Br2(g) ----------------- 2 NOBr(g)

Kp = 28.4 atm-1

1 torr = 0.001316 atm

partail pressure of NO = 173 torr = 0.2277 atm

partial pressure of Br2= 142 torr = 0.1869 atm

Kp = P^2NOBr/P^2NO xPBr2

28.4 = P^2 NOBr / ( 0.2277)^2 x(0.1869)

PNOBr = 0.525 atm

Partial pressur of NOBr = 0.525 atm = 398.94 torr

Partial pressure of NOBr = 398.94 torr.