Chemistry equilibrium concentration

Question

A mixture of 3.51 mol N2 and 35.21 g NO is heated in a closed vessel to 2000C. After heating, the total pressure of the mixture at equilibrium is 3.14 atm. Calculate the equilibrium partial pressures of N2, O2 and NO at 2000C.

kp = 0.101

N2 + O2 = (equilibrium sign) 2NO

(I’m not sure how I got 1.00 for O2 but will try again to get the correct moles. I used the ICE diagram to find the equilibrium concentrations for the products and reaction.)

I assume O2 has zero moles of initial concentration since it’s not given.

Answer 1

Ahhh...now we have the value of Kp, so we can use the ICE table. And yes, your assumption of no moles of O₂ is correct as it is not present initially.

N₂ + O₂ <==> 2NO ... Kp = 0.101 Since Kp is < 1, it tells us the equilibrium favors the reactants

moles N2 = 3.51

moles O2 = 0

moles NO = 35.21 g NO x 1 mol / 30.0 g = 1.174

N₂ + O₂ <==> 2NO

3.51......0............1.17.......Initial mols

+x........+x............-2x.........Change in mols

3.51+x....x...........1.17-2x...Equilibrium mols

Given Kp, we can use Kp = Kc(RT)^∆n to find Kc

Since the number of moles of gas are the same on the reactant and product side, this reduces to Kp = Kc.

Kc = 0.101 = [NO]² / [N₂][O₂]

0.101 = (1.17-2x)² / (3.51-x)(x)

0 = 4x² - 5.04x + 1.37 (be sure to check this to ensure it is correct)

x = 0.863 (discard as 2x this is more than the initial moles of NO)

x = 0.397 moles

Moles @ equilibrium:

N₂ = 3.51 + 0.397 = 3.91 mols

O₂ = 0 + 0.397 = 0.397 mols

NO = 1.17 - 0.794 = 0.376 mols

Total mols of gas = 4.68 mols

Mole fraction of gases @ equilibrium:

N₂ = 3.91 / 4.68 = 0.848

O₂ = 0.397 / 4.68 = 0.0848

NO = 0.376 / 4.68 = 0.0803

Partial pressures will be mole fraction x total pressure

Partial pressure N₂ = 0.848 x 3.14 atm = 2.66 atm

Partial pressure O₂ = 0.0848 x 3.14 atm = 0.266 atm

Partial pressure NO = 0.0803 x 3.14 atm = 0.252 atm