Thermodynamics: Gibs free energy

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SUMMARY

The standard Gibbs free energy of formation of water vapor at 25°C is calculated using the equation ΔG = ΔH - TΔS. Given ΔH = -484 kJ/mol and ΔS = -89 J/mol·K, the calculation yields -457 kJ/mol. However, the correct value is -227 kJ/mol, which accounts for the formation of 1 mole of water vapor instead of 2 moles. Therefore, the final result requires dividing the initial calculation by 2 to obtain the accurate Gibbs free energy of formation.

PREREQUISITES
  • Understanding of thermodynamic concepts, specifically Gibbs free energy.
  • Familiarity with the equations for enthalpy (ΔH) and entropy (ΔS).
  • Knowledge of unit conversions, particularly Celsius to Kelvin.
  • Experience with stoichiometry in chemical reactions.
NEXT STEPS
  • Study the relationship between Gibbs free energy and chemical equilibrium.
  • Learn about the implications of ΔG values in predicting reaction spontaneity.
  • Explore the concept of standard state conditions in thermodynamics.
  • Investigate the calculations of Gibbs free energy for other chemical reactions.
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Chemistry students, educators, and professionals in the field of thermodynamics seeking to deepen their understanding of Gibbs free energy calculations and their applications in chemical reactions.

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Homework Statement


What is the standard Gibbs free energy of formation of water vapor at 25 C, if , for the reaction shown below under standard conditions, Δ H = -484 KJ/mol and ΔS= -89 J/mol K?
2H2(g) + O2(g) >>> 2H2O(g)

Homework Equations



ΔG = ΔH - TΔS


The Attempt at a Solution

I converted 25 C to kelvin. I then simply plugged in the values. -484000J/mol -(298K(-89J/mol k)) I get -457 kJ/mol. Answer provided is -227. I don't understand how they got this number.
 
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I think I understand. The answer that I provided is for the equation which has 2 mols of water. The correct answer is kJ/mol not per 2mols so I have to divide by 2 to get the correct answer. Can someone back me up on this?
 

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