Question

Calculate the partial pressure (in atm) of NO at equilibrium when 1.05 atm of NO2 dissociates at 250 K according to the following chemical equilbrium:

2NO2(g) ⇌ 2NO(g) + O2(g)

Kp = 2.96×10-12

If the 5% approximation is valid, use the assumption to compute the partial pressure. Report your answer to three significant figures in scientific notation.

Answer 1

Apply

Kp = PNO^2 * PO2 / (PNO2^2)

Initially

PNO2 = 1.05

PNO = 0

PO2 = 0

in equilbirium due to stoichiomerty

PNO2 = 1.05 - 2x

PNO = 0 + 2x

PO2 = 0 + x

Substitute in Kp

Kp = (2x)^2(x)/(1.05 - 2x)^2

kP = 2.96*10^-12

then

2.96*10^-12 = (2x)^2(x)/(1.05 - 2x)^2

2.96*10^-12 = (4x^3)/(1.05 - 2x)^2

then

(2.96*10^-12)/4 = x^3 / (1.05-2x)^2

apply the 5% rule, since Kp is too low (i.e. 10^-12) then assume the concnetration in equilbilium will not be shifted toward reactants, rather to product so

1.05 - 2x may be assumed to be 1.05 - 2*0 = 1.05

then

(2.96*10^-12)/4 = x^3 / (1.05-2x)^2

turns to

(2.96*10^-12)/4 = x^3 / (1.05)^2

solve for x

x^3 = (1.05^2)(2.96*10^-12)/4

x = ((1.05^2)(2.96*10^-12)/4)^(1/3)

x = 0.00009344084

Substitute x in all concnetrations in equilibrium

PNO = 0 + 2x = 2*0.00009344084 = 0.00018688168 or 1.8*10^-4

NOTE that this makes sense since Kp <<1 meaning it favours reactants by a high factor