Why is My Free Energy Calculation from Ksp Incorrect?

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SUMMARY

The discussion centers on the discrepancy between the calculated free energy change (\Delta G) for calcium hydroxide (Ca(OH)2) and the value derived from its solubility product constant (Ksp). The Ksp for Ca(OH)2 is stated as 6.5E-6, while the calculated \Delta G using the equation \Delta G=-RTln(Ksp) yields 29.59 kJ/mol, which contradicts the expected \Delta G of -898.5 kJ/mol. Participants emphasize the importance of distinguishing between the solubility constant and the \Delta G of formation, highlighting the need to consider the specific reactions involved in these calculations.

PREREQUISITES
  • Understanding of thermodynamics, specifically Gibbs free energy.
  • Familiarity with solubility product constants (Ksp).
  • Knowledge of the ideal gas constant (R) and its application in calculations.
  • Basic principles of chemical equilibrium and reaction stoichiometry.
NEXT STEPS
  • Review the derivation and implications of the Gibbs free energy equation.
  • Study the relationship between Ksp and \Delta G for various compounds.
  • Learn about the standard Gibbs free energy of formation for common substances.
  • Explore the concept of chemical equilibrium and its effect on solubility.
USEFUL FOR

Chemistry students, researchers in physical chemistry, and professionals involved in thermodynamic calculations will benefit from this discussion.

elitespart
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Okay so I'm confused. In my book it says that ksp for Ca(OH)2 = 6.5E-6 and \Delta G=-898.5 yet when I use the equation \Delta G=-RTln(Ksp) the value I get is 29.59KJ/mol. I used R=8.314 and T=298K. What's up with that? Thanks.
 
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Hint: Consider the implications of the solubility constant versus the
\Delta G\
of formation (which is the value you have given), specifically the reactions involved.
 

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