Confusion in relation of Gibbs free energy and equilibrium constant

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Discussion Overview

The discussion revolves around the relationship between Gibbs free energy and the equilibrium constant for the reaction SO2(g) + 1/2O2(g) ⇌ SO3(g). Participants explore how to calculate the equilibrium constant (Kp or Kc) at 298 Kelvin, considering the implications of standard states and the definitions of these constants.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates ΔG using the formula ΔG = ΔH – T x ΔS and finds ΔG = -70.01 kJ/mol, leading to a calculation of Keq.
  • Another participant questions whether the reaction is at standard state and suggests that the pressure is approximately 1 atm, which could help in determining the density of the gases.
  • Confusion arises regarding whether Keq is equal to Kp or Kc, as the standard state can refer to either partial pressure or concentration.
  • One participant states that Kp is in terms of partial pressures and mentions the relationship Kp = Kc(RT)Δn, noting that Δn = -1/2 for this reaction.
  • A later reply emphasizes that the standard state for gases is 1 bar, suggesting that in this context, Keq is Kp.
  • There is a mention of the omission of the superscript 0 in ΔG, indicating a potential oversight in notation.

Areas of Agreement / Disagreement

Participants express differing views on whether Keq should be considered as Kp or Kc, with no clear consensus reached. The discussion remains unresolved regarding the implications of standard states on the equilibrium constant.

Contextual Notes

Participants highlight the lack of information about pressure or concentration in the original problem statement, which contributes to the confusion regarding the correct value of Kp.

tbn032
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Homework Statement
for the reaction
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
find Kp at 298 Kelvin?
Relevant Equations
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
SO2(g)+1/2O2(g)SO3(g);ΔHo=-98.32KJ/mole,ΔSo=-95J/(mole-K).
find Kp at 298 Kelvin?
In given question at first Δ G will be calculated using formula ΔG = Δ H – T x ΔS, by putting the given values in formula we get ΔG = -70.01 kJ/mol.
Then Keq will be calculated using equation = Δ G = -RT ln Keq ,ln Keq = -70010J/-8.314 x 298 = 28.25 .or Keq = e28.25 = 1.86x1012.
Now my confusion is Keq=Kpor Keq=Kc(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).and
Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
No information about the pressure or concentration is given so how do I conclude that Keq=Kp or Keq=Kc and thus calculate the correct value of Kp
 
Last edited:
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Given the temperature 298 Kelvin, is this reaction at the 'standard state'? If so, the pressure is ambient or ~1 atm (100 kPa, or 1 bar).
https://en.wikipedia.org/wiki/Standard_state#Conventional_standard_states

Given the standard state, one should be able to determine the density of the gases. What does one's textbook indicate about STP, standard state and Gibbs free energy?

The problem statement one provides indicates gases.
 
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my confusion is arising from the fact that in my book it is written that
Δ G = -RT ln Keq
Keq=KporKc as the standard state is referred in terms of partial pressure or concentration
Δ G = -RT ln Kp
Δ G = -RT ln Kk
if calculate the value of Keq from the equation Δ G = -RT ln Keq using R=8.3144598J⋅K-1⋅mol-1.the value of Keq is dimensionless
so I am not able to conclude if Keq=Kp or Keq=Kc
kp≠kc because Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
20221030_174752.jpg
 
Last edited:
The standard state for gases is 1 bar. So, in this example, the Keq is Kp. It is in terms of the partial pressures of the components in bars.

Incidentally, you omitted the superscript 0 in ##\Delta G^0##
 
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tbn032 said:
Homework Statement:: for the reaction
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
find Kp at 298 Kelvin?
Relevant Equations:: SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).

SO2(g)+1/2O2(g)SO3(g);ΔHo=-98.32KJ/mole,ΔSo=-95J/(mole-K).
find Kp at 298 Kelvin?
In given question at first Δ G will be calculated using formula ΔG = Δ H – T x ΔS, by putting the given values in formula we get ΔG = -70.01 kJ/mol.
Then Keq will be calculated using equation = Δ G = -RT ln Keq ,ln Keq = -70010J/-8.314 x 298 = 28.25 .or Keq = e28.25 = 1.86x1012.
Now my confusion is Keq=Kpor Keq=Kc(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).and
Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
No information about the pressure or concentration is given so how do I conclude that Keq=Kp or Keq=Kc and thus calculate the correct value of Kp
The equation is giving it up as you have SO2(g) you are dealing with gases
 

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