Calculating Kp of o2 and o3 equilibrium

  • #1
i_love_science
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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.
 

Answers and Replies

  • #2
22,319
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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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  • #3
mjc123
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Do note that x is mole fraction, not mass fraction (and certainly not % mass). And your equation should be
M = 32x + 48(1-x)
(I think you just made a typo, you seem to have calculated x correctly.)
And you've been doing this long enough to know that dioxygen is O2, not o2. If you think o2 is acceptable shorthand - it isn't! It makes you look like someone who isn't taking it seriously. It's of a piece with your little errors in calculation. Science is about taking the trouble to get things correct, not rushing and cutting corners and making silly mistakes.
 
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  • #4
mjc123
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@i_love_science, kudos for liking my post although it criticised you. That's the sign of someone who wants to learn!
 

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