Chemistry Calculating Kp of o2 and o3 equilibrium

  • Thread starter Thread starter i_love_science
  • Start date Start date
  • Tags Tags
    Equilibrium
AI Thread Summary
The discussion revolves around calculating the equilibrium constant Kp for the reaction involving O2 and O3. The initial calculations provided an incorrect Kp value of 0.550, while the correct value is 1.5. Key errors identified include the miscalculation of mole fractions, which should sum to 1.0, and the incorrect use of mass fractions instead of mole fractions. Additionally, the molar mass equation used was flawed, requiring correction to reflect the proper molecular weights. Attention to detail in scientific calculations is emphasized as crucial for accuracy.
i_love_science
Messages
80
Reaction score
2
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.
 
Physics news on Phys.org
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?
 
  • Like
Likes i_love_science
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.
 
  • Like
Likes i_love_science, jim mcnamara and Chestermiller
@i_love_science, kudos for liking my post although it criticised you. That's the sign of someone who wants to learn!
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...

Similar threads

Back
Top