# Calculating Kp of o2 and o3 equilibrium

i_love_science
Homework Statement:
Consider the reaction 3 O2(g) --> 2 O3(g).
At 175°C and a pressure of 128 torr, an equilibrium mixture of O2 and O3 has a density of 0.168 g/L. Calculate Kp for the above reaction at 175°C.
Relevant Equations:
PV = nRT
My solution:

PV = nRT
M = dRT / P
M is the weighted average molar mass of o2 and o3 --> M = 32x + 16(1-x), where x is the % by mass of o2 and 1-x is the % by mass of o3.

M = 32x + 16(1-x) = (0.168 g/L) (62.36 L*torr/mol*K) (448 K) / (128 torr) = 36.668 g/mol
solving the equation gives x = 0.708, 1-x = 0.282

Concentration of o2: 0.708 * 0.168 g/L * 1 mol / 32 g = 0.00372 mol/L
Concentration of o3: 0.282 * 0.168 g/L * 1 mol/48 g = 0.00102 mol/L

Kc = (0.00102 mol/L)^3 / (0.00372 mol/L)^2
Kp = Kc / (RT)^(delta n) = (0.00102 mol/L)^3 / (0.00372 mol/L)^2 / (0.08206 L*mol/atm*K)^(-1) = 0.550

My answer was wrong, the correct answer is 1.5. Could anyone explain where I went wrong? Thanks.

Mentor
Your mole fractions don't add up to 1.0. I get 0.708 for the mole fraction of oxygen, but 0.292 for the mole fraction of ozone. So, the partial pressure of oxygen is then 0.1192 atm, and the partial pressure of ozone is then 0.0492 atm. What does this give you for Kp?

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