Thermodynamics: Gibs free energy

AI Thread Summary
The discussion focuses on calculating the standard Gibbs free energy of formation for water vapor at 25°C using the equation ΔG = ΔH - TΔS. The user initially calculated ΔG as -457 kJ/mol but later realized this value corresponds to the formation of 2 moles of water. To find the standard Gibbs free energy per mole, the user acknowledges the need to divide the initial result by 2, leading to the correct answer of -227 kJ/mol. This highlights the importance of considering stoichiometry in thermodynamic calculations. The conversation emphasizes clarity in unit conversions and the significance of understanding the reaction's mole ratio.
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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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