Why Does the Position of Psat(P) Change in Kp,eq Formula?

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Discussion Overview

The discussion revolves around the positioning of Psat(P) in the Kp,eq formula, particularly in the context of two different reactions presented in homework questions. Participants seek to understand the reasoning behind why Psat(P) appears in different positions in the formula for each reaction, as well as the implications of changes in the number of moles in the balanced equations.

Discussion Character

  • Homework-related
  • Debate/contested
  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about why Psat(P) is placed differently in the Kp,eq formula for two separate reactions.
  • There are inquiries about the change in the number of moles in the balanced equations for both reactions and its relevance to Kp,eq.
  • Some participants mention the relationship between partial pressures and mole fractions, suggesting that P1 can be expressed as a function of total pressure P.
  • There are discussions about the definition and calculation of Δv, with some participants attempting to clarify its role in the conversion between Kc and Kp.
  • Some participants challenge the understanding of the relationship between moles and volume, referencing Avogadro's law.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the reasons for the different placements of Psat(P) in the Kp,eq formula. Multiple competing views and uncertainties remain regarding the implications of mole changes and the definitions of Δv.

Contextual Notes

Limitations include unclear assumptions about the reactions and the definitions of terms like Δv. The discussion also reflects varying levels of understanding among participants regarding the mathematical relationships involved.

williamcarter
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Homework Statement


question1-sentence
https://www.physicsforums.com/attachments/105257
question1-answer
https://www.physicsforums.com/attachments/105258

question2-sentence
https://www.physicsforums.com/attachments/105259
question2-answer
https://www.physicsforums.com/attachments/105260

Homework Equations


Kp,eq=Pproduct^stoichiometric/Preactant^stoichiometric=yproduct^stoichiometric/yreactant^stoichiometric

The Attempt at a Solution


I did not understand why in question 1 why Kp,eq is done like that and why Psat(P) is underneath.
And why in question 2 Psat(P) is at the top in Kp,eq formula.

Why one time P is below and 1 time P is at the top in Kp,eq formula?
 
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Please post HW questions in the proper HW forum.
 
williamcarter said:

Homework Statement


question1-sentence
https://www.physicsforums.com/attachments/105257
question1-answer
https://www.physicsforums.com/attachments/105258

question2-sentence
https://www.physicsforums.com/attachments/105259
question2-answer
https://www.physicsforums.com/attachments/105260

Homework Equations


Kp,eq=Pproduct^stoichiometric/Preactant^stoichiometric=yproduct^stoichiometric/yreactant^stoichiometric

The Attempt at a Solution


I did not understand why in question 1 why Kp,eq is done like that and why Psat(P) is underneath.
And why in question 2 Psat(P) is at the top in Kp,eq formula.

Why one time P is below and 1 time P is at the top in Kp,eq formula?
What is the change in the number of moles in the balanced equation for the first reaction?
What is the change in the number of moles in the balanced equation for the second reaction?
 
Chestermiller said:
What is the change in the number of moles in the balanced equation for the first reaction?
What is the change in the number of moles in the balanced equation for the second reaction?

Thank you for your quick response.
I would really appreciate it if you could answer why they changed the P position in the formula.

I am in kind of a rush to understand this because I will have an exam soon on those things, and there is not much time to focus on.

I need to be fast, and understand when to put P on top and when to put it on the bottom.

I don't really get what you want to mean with "the change in number of moles in the balanced equation"
I don't get what has the change of number of moles to do with the kp,eq.
I want to understand why they changed P , why 1 time is underneath in formula and why 1 time is on the top.

For the 1st reaction extent of reaction ξ=0.873 mol
For the 2nd reaction extent of reaction ξ=0.877 mol

Q1
Capture.JPG

Q1-Answer
Capture2.JPG
Q2-
Capture3.JPG


Q2-Answer
Capture4.JPG
 
What is Δv in these formulas? Do you understand where it comes from?

(Actually it is exactly the same question Chestermiller asked).
 
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In problem 1, the give you the mole fractions of the 3 species in the final state. What are the partial pressures of these three species algebraically in terms of the total pressure P?
 
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Borek said:
What is Δv in these formulas? Do you understand where it comes from?

(Actually it is exactly the same question Chestermiller asked).
Thank you for your answer, I do not know what it is and where it comes from.
I really need help to understand this as soon as possible.
Thank you
 
Chestermiller said:
In problem 1, the give you the mole fractions of the 3 species in the final state. What are the partial pressures of these three species algebraically in terms of the total pressure P?
We can interpret the partial pressure P1 let's say as
P1=x1*P1,sat=y1*P
 
williamcarter said:
I do not know what it is and where it comes from

Sorry then, go back to your book and notes.

Look for a ways to convert between Kc and Kp.
 
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  • #10
Borek said:
Sorry then, go back to your book and notes.

Look for a ways to convert between Kc and Kp.
Thank you, it does not say anywhere what that is, I looked everywhere,and tomorrow I have exam on this, please help me with this.
It is much appreciated
 
  • #11
I know partial pressure P1=y1*P
 
  • #12
Have you tried googling for "converting Kc to Kp"?
 
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  • #13
Borek said:
Have you tried googling for "converting Kc to Kp"?
Thanks for the answer ,I know that Kc=Product (P/RT)^delta v /Reactant (P/RT)^delta v
 
  • #14
I need to understand when to know when delta v= -1 and when delta v=1
 
  • #15
Come on, look at any page where the conversion is described and check what is the definition of the Δv and how it is calculated. It is really trivial.
 
  • #16
williamcarter said:
I know partial pressure P1=y1*P
Okay. Now substitute this into the equation for ##K_p## (i.e., with ##K_p## is expressed in terms of partial pressures) for your A+B=C problem. What do you get? (And please use LaTex).
 
  • #17
williamcarter said:
We can interpret the partial pressure P1 let's say as
P1=x1*P1,sat=y1*P
This applies to vapor-liquid equilibrium, not to chemical reactions. But, the y part is correct. What do you get when you substitute this for A, B, and C into the equilibrium relationship.
 
  • #18
williamcarter said:
I need to understand when to know when delta v= -1 and when delta v=1
delta v = total number of moles of gases on product side - total number of moles of gases on the reactant side
like 2A(g)+B(g)---> 3C (g)
in this case products moles = 3
reactants = 3
delta v= 0
similarly it can be negative or positive
Remember mole is equal to volume if gases are at the same temp and pressure.
 
  • #19
Dr Uma Sharma said:
Remember mole is equal to volume if gases are at the same temp and pressure.
Did you really want to say this?
 
  • #20
Sorry they are not equal but proportional
Moles ratio of gases and volume ratio of the gases are same if temp and pressure are kept constant.
Avogadro's law
...that equal volumes of gases at the same temperature and pressure contain equal numbers of moles.
 
  • #21
Dr Uma Sharma said:
Sorry they are not equal but proportional
Moles ratio of gases and volume ratio of the gases are same if temp and pressure are kept constant.
Avogadro's law
...that equal volumes of gases at the same temperature and pressure contain equal numbers of moles.
I think what you are trying to say is that the partial pressure of each gas is equal to the total pressure times the mole fraction of that gas?