Question

15 Question (3 points) e See page 648 In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (.790 atm N2, 0.2 10 atm O2). What are the equilibrium partial pressures of N2, O2, and NO if Kp = 0.0520 for the following reaction: N2(8)+02(g) 2NO (g) 1st attempt See Periodic Table Part 1 (1 point) See Hint P(N2) = atm Part 2 (1 point) P(O2) atm See Hint Part 3 (1 point) P(NO) = atm

Answer 1

ICE Table:

p(N2) p(02) p(NO) initial 0.79 0.21 change - 1x -1x +2x equilibrium 0.79-1x 0.21–1x +2x

Equilibrium constant expression is

Kp = p(NO)^2/p(N2)*p(O2)

0.052 = (4*x^2)/((0.79-1*x)(0.21-1*x))

0.052 = (4*x^2)/(0.1659-1*x + 1*x^2)

8.627*10^-3-5.2*10^-2*x + 5.2*10^-2*x^2 = 4*x^2

8.627*10^-3-5.2*10^-2*x-3.948*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -3.948

b = -5.2*10^-2

c = 8.627*10^-3

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 0.1389

roots are :

x = -5.379*10^-2 and x = 4.062*10^-2

since x can't be negative, the possible value of x is

x = 4.062*10^-2

At equilibrium:

p(N2) = 0.79-1x = 0.79-1*0.04062 = 0.74938 atm

p(O2) = 0.21-1x = 0.21-1*0.04062 = 0.16938 atm

p(NO) = +2x = +2*0.04062 = 0.08124 atm

Answer:

1)

0.749 atm

2)

0.169 atm

3)

0.0812 atm