How Is Standard Gibbs Free Energy Calculated for Water Vapor Formation?

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Discussion Overview

The discussion revolves around the calculation of the standard Gibbs free energy of formation for water vapor at 25°C, specifically addressing the reaction involving hydrogen and oxygen to form water. Participants explore the relationship between enthalpy, entropy, and Gibbs free energy, while clarifying the implications of standard states and mole ratios in thermodynamic calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant presents a calculation for the standard Gibbs free energy using given values for enthalpy and entropy, but arrives at a different result than expected.
  • Another participant questions the number of moles of water formed in the reaction, clarifying that 2 moles of water are produced.
  • There is a discussion about whether the Gibbs free energy calculation should be divided by 2, with some participants asserting that the standard free energy of formation is defined for 1 mole of product.
  • One participant expresses confusion about the calculations and seeks confirmation on whether their understanding of the relationship between ΔH and the mole ratio is correct.
  • Another participant confirms that the standard Gibbs free energy of formation for 1 mole of water would indeed be half of the value given for 2 moles.

Areas of Agreement / Disagreement

Participants generally agree on the need to adjust the Gibbs free energy calculation to reflect the formation of 1 mole of water. However, there is some uncertainty regarding the initial calculations and whether the book's values are correct.

Contextual Notes

Participants discuss the implications of standard states and the definition of free energy of formation, which may not be fully resolved in terms of how it applies to the specific reaction presented.

brake4country
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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, ΔH = -484 kJ/mol and ΔS=-89 J/mol K?
2H2 + Os→2H2O

Homework Equations


ΔG = ΔH-TΔS

The Attempt at a Solution


Usually I can do these problems with no issue but I cannot seem to get the correct answer (which is -229 kJ/mol).
First, I converted -89 J/mol K to kJ = 0.089 J/mol K. Plugging this into the equation gives me:

ΔG = (-484 kJ/mol) - 298 K(-.089 kJ/mol K) = -457 kJ/mol.

I did this problem twice, making sure my units were correct. Is my book wrong or did I do something wrong in the calculations? Thanks in advance.
 
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How many moles of H2O are you forming in the original reaction?
 
Sorry looks like a typo above. It should be oxygen gas as a reactant. But 2 moles of water are formed. The reactants are both gas and the product is liquid.
 
Right, so if you form 2 moles of water, ΔG = -457 kJ. That calculation is done correctly.
 
So my book is wrong. I thought so. Thanks for checking my work, I was so confused =)
 
What is 457 divided by 2?

Chet
 
Why do I have to divide by 2 if the ΔH for the reaction is given for that equation? Does it have to do with the standard states?
 
brake4country said:
Why do I have to divide by 2 if the ΔH for the reaction is given for that equation? Does it have to do with the standard states?
No. The free energy of formation is defined on the basis of producing 1 mole of the compound from its elements.
 
Ok I see. That's where my knowledge gap was. So standard free energy requires 1 mol of product formation. Thus in this problem if ΔH = -484 kJ/mol is for 2 mol of H2O, the 1 mol (standard formation) would be half that = -229 kJ/mol. Am I understanding this correctly?
 
  • #10
brake4country said:
Ok I see. That's where my knowledge gap was. So standard free energy requires 1 mol of product formation. Thus in this problem if ΔH = -484 kJ/mol is for 2 mol of H2O, the 1 mol (standard formation) would be half that = -229 kJ/mol. Am I understanding this correctly?
Yes.
 

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