Question

0.75 moles of O₂ gas is mixed with 0.45 moles of N₂ gas in a 2.5 L tank at 1970 K. The two gases react to form NO gas and the system is allowed to react equillibrium. The equillibrium constant for this reaction is equal to 4.14 x 10^-4. What are the quillibrium concentrations of all three components?

I would really apreciate help with this question. thank you!

Answer 1

Given

Intial moles of O2 = 0.75 moles

Intial moles of N2 = 0.45 moles

Volume = 2.5 L

Inital [O2] = 0.75 moles / 2.5 L = 0.3 mol/L (M)

Inital [N2] = 0.45 moles / 2.5 L = 0.18 mol/L (M)

p = C* R*T

intial pO2 = 0.3 mol/L * 0.08206 L.atm/mol.K * 1970 K = 48.5 atm

intial pN2 = 0.18 mol/L * 0.08206 L.atm/mol.K * 1970 K = 29 atm

reaction is

O2 (g) + N2 (g) <-------> 2NO (g)

i am considering the given constant as Kc = 4.14 * 10^-4

O2 N2 NO

Intial 0.3 0.18 -

Converted -x -x 2x

Equilibrium 0.3-x 0.18-x 2x

Kc = (2x)² / (0.3-x) (0.18-x) = 4.14 * 10^-4

x = 2.34 * 10^-3 mol/L

equilibrium concentrations

[O2] = 0.3 - x = 0.29766 mol/L

[N2] = 0.18 -x = 0.17766 mol/L

[NO] = 2x = 0.00468 mol/L