Consider the reaction 2NOgBr2g2NOBrg Kp284 at 298 K In a rea

Consider the reaction: 2NO(g)+Br2(g)2NOBr(g) Kp=28.4 at 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 117 torr and that of Br2 is 155 torr . What is the partial pressure of NOBr in this mixture?

Solution

p(NO) = 117 torr
= 117/760 atm
= 0.154 atm

p(Br2) = 155 torr
= 155/760 atm
= 0.204 atm

use:
Kp = p(NOBr)^2 / p(NO)^2*p(Br2)
28.4 = p(NOBr)^2 / (0.154^2 * 0.204)
p(NOBr)^2 = 0.1374
p(NOBr) = 0.371 atm
= 0.371*760 torr
= 282 torr

Answer: 282 torr