Calculating Kp of o2 and o3 equilibrium

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In summary, the conversation discusses calculating the concentration and partial pressures of oxygen and ozone using the ideal gas law and the weighted average molar mass equation. However, there were some errors in the calculations, leading to an incorrect answer for Kp. It is important to pay attention to details and use correct units and notation in scientific calculations.
  • #1
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.
 
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  • #2
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
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
@i_love_science, kudos for liking my post although it criticised you. That's the sign of someone who wants to learn!
 

1. How do you calculate the equilibrium constant (Kp) for the reaction between O2 and O3?

To calculate Kp for the reaction between O2 and O3, you need to first write out the balanced chemical equation for the reaction. Then, use the partial pressure of each gas at equilibrium and plug them into the equation Kp = (P(O3)^2)/(P(O2)^3). The units for Kp will depend on the units used for the partial pressures.

2. What is the significance of the equilibrium constant (Kp) in this reaction?

The equilibrium constant (Kp) represents the ratio of products to reactants at equilibrium for the reaction between O2 and O3. It is a measure of the extent to which the reaction proceeds towards products or reactants. A larger Kp value indicates a higher concentration of products at equilibrium, while a smaller Kp value indicates a higher concentration of reactants at equilibrium.

3. How does temperature affect the equilibrium constant (Kp) for this reaction?

According to Le Chatelier's principle, an increase in temperature will shift the equilibrium towards the endothermic reaction, which in this case is the formation of O3. This will result in an increase in the partial pressure of O3 and a decrease in the partial pressure of O2, leading to an increase in the value of Kp. Similarly, a decrease in temperature will shift the equilibrium towards the exothermic reaction, resulting in a decrease in the value of Kp.

4. Can the equilibrium constant (Kp) be used to predict the direction of the reaction?

Yes, the equilibrium constant (Kp) can be used to predict the direction of the reaction. If the calculated value of Kp is greater than 1, it indicates that the equilibrium lies towards the products, and the reaction will proceed in the forward direction. If the calculated value of Kp is less than 1, it indicates that the equilibrium lies towards the reactants, and the reaction will proceed in the reverse direction. A value of Kp equal to 1 indicates that the reaction is at equilibrium.

5. How can the equilibrium constant (Kp) be affected by changes in pressure?

Changes in pressure will only affect the value of Kp if the number of moles of gas on each side of the reaction equation is not equal. If the number of moles of gas on each side is equal, then changes in pressure will not affect the value of Kp. However, if the number of moles of gas is not equal, then an increase in pressure will shift the equilibrium towards the side with fewer moles of gas, while a decrease in pressure will shift the equilibrium towards the side with more moles of gas. This will result in a change in the value of Kp.

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