Relationship between Free energy change and Equilibrium constant (K)

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Discussion Overview

The discussion revolves around the relationship between free energy change (ΔG°) and the equilibrium constant (K), particularly in the context of a specific reaction involving vapor pressures of molecular compounds. Participants explore how to derive K from ΔG° and its implications for calculating concentrations or pressures of reactants and products at equilibrium.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant questions how to relate the equilibrium constant K to the vapor pressure of a specific molecular compound in a reaction, suggesting an initial formula of 1/Bb = K.
  • Another participant points out that both products C and D, which are gases, are also part of the equilibrium constant K.
  • A participant presents a specific applied problem involving a humidity sensor and a reaction, detailing the calculation of ΔG° and K at a given temperature.
  • The participant expresses confusion over how to solve for the vapor pressure of water when multiple gases are involved, indicating a need for additional information to resolve multiple variables.
  • Another participant confirms that more information would indeed be necessary to solve for the pressure of water if multiple gases are present.

Areas of Agreement / Disagreement

Participants generally agree that the relationship between K and the vapor pressures or concentrations of reactants and products is complex, especially when multiple gases are involved. However, there is no consensus on the specific approach to derive these relationships or the implications of the calculations presented.

Contextual Notes

The discussion highlights the dependence on specific conditions such as temperature and the assumptions made regarding ΔH and ΔS being independent of temperature. There are also unresolved mathematical steps in deriving K and its application to the vapor pressure of water.

Who May Find This Useful

This discussion may be useful for students and practitioners in chemistry and related fields who are exploring thermodynamic relationships and equilibrium concepts, particularly in the context of gas reactions and vapor pressures.

Jef123
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1. I have a question regarding the equation ΔG° = -RT lnK. When solving for the equilibrium constant K, what is the relationship between the vapor pressure of each molecular compound in a reaction?

ΔG° = Free energy change
R = Universal gas constant
T = Temperature
K = equilibrium constant
*K can be solved for in concentration or pressure or solubility


2. Meaning, for the reaction aA(s) + bB(g) --> cC(g) + dD(g) where the lower case coefficients are the number of moles and the upper case coefficients are the molecular compounds. Assuming I solved for ΔG at a specific temperature and then solved for K If I wanted to find the the vapor pressure or the concentration of bB how would I relate the value of the equilibrium constant K to solve for B?

My thought is that 1/Bb = K. Is this correct?

There is a specific applied problem from where my question is derived, so if anyone would like me to post it to clarify areas where I may not be explicit enough, I can.
 
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C and D are gases as well, so they are part of K as well.

Hard to say what you are expected to do not knowing the problem, but typically you should find C and D from the stoichiometry (for example using ICE table).
 
Here's the problem:

A humidity sensor consists of a cardboard square that is colored blue in dry weather and red in humid weather. The color change is due to the reaction:

CoCl2(s) + 6 H2O(g) ⇌ [Co(H2O)6]Cl2(s)

For this reaction at 25 °C, ΔH° = -352 kJ/mol and ΔS° = -899 J/(K mol). Assuming that ΔH and ΔS are independent of temperature, what is the vapor pressure of water (in mm Hg) at equilibrium for the above reaction at 35 °C on a hot summer day?

I used the equation ΔG° = ΔH° - TΔS° at T = 35°C = 308.15 K. So, ΔG = -74973.15 J

Then, I used ΔG° = - RTlnK and solved for K at T = 308.15 Kelvin. Thus, K = 5.5 x 1012 atm.

This is where I am confused and had to look up the solution: 1/(H2O)6 = 5.5 x 1012. They then solved for H2O

I think I actually just understood what they did. wow. So if there was more than one gas in the equation then I would need more information to solve for the pressure of water because there would be more than one variable?
 
Jef123 said:
So if there was more than one gas in the equation then I would need more information to solve for the pressure of water because there would be more than one variable?

Yes.
 

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