Relationship between Free energy change and Equilibrium constant (K)

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SUMMARY

The relationship between free energy change (ΔG°) and the equilibrium constant (K) is defined by the equation ΔG° = -RT lnK, where R is the universal gas constant and T is the temperature in Kelvin. In the context of the reaction CoCl2(s) + 6 H2O(g) ⇌ [Co(H2O)6]Cl2(s), the equilibrium constant K can be calculated at different temperatures, such as 35 °C, resulting in K = 5.5 x 1012 atm. To find the vapor pressure of water at equilibrium, one must consider the stoichiometry of the reaction and the number of gaseous components involved, as multiple variables may complicate the calculation.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically ΔG°, ΔH°, and ΔS°.
  • Familiarity with the ideal gas law and equilibrium concepts.
  • Knowledge of stoichiometry and ICE (Initial, Change, Equilibrium) tables.
  • Ability to perform calculations involving natural logarithms and gas pressures.
NEXT STEPS
  • Learn how to derive the equilibrium constant K from ΔG° for various reactions.
  • Study the impact of temperature changes on equilibrium constants and reaction dynamics.
  • Explore the use of ICE tables for complex reactions involving multiple gaseous components.
  • Investigate the relationship between vapor pressure and equilibrium constants in different chemical systems.
USEFUL FOR

Chemistry students, chemical engineers, and researchers involved in thermodynamics and reaction equilibrium analysis will benefit from this discussion.

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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