Question

At 4124 oC the equilibrium constant for the reaction: 2 NO(g) N2(g) + O2(g) is KP = 7.16. If the initial pressure of NO is 0.00663 atm, what are the equilibrium partial pressures of NO, N2, and O2?

p(NO) = .

p(N2) = .

p(O2) = .

Answer 1

ICE Table:

   

Equilibrium constant expression is
Kp = p(N2)*p(O2)/p(NO)^2
7.16 = (1*x)^2/(6.63*10^-3-2*x)^2
sqrt(7.16) = (1*x)/(6.63*10^-3-2*x)
2.67582 = (1*x)/(6.63*10^-3-2*x)
1.774*10^-2-5.352*x = 1*x
1.774*10^-2-6.352*x = 0
x = 0.00279

At equilibrium:
p(NO) = 0.00663-2x = 0.00663-2*0.00279 = 0.00104 atm
p(N2) = +1x = +1*0.00279 = 0.00279 atm
p(O2) = +1x = +1*0.00279 = 0.00279 atm

Answer:
p(NO) = 0.00104 atm
p(N2) = 0.00279 atm
p(O2) = 0.00279 atm