Calculating gibbs free energy

  • #1

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.
 

Answers and Replies

  • #2
Ygggdrasil
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How many moles of H2O are you forming in the original reaction?
 
  • #3
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.
 
  • #4
Ygggdrasil
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Right, so if you form 2 moles of water, ΔG = -457 kJ. That calculation is done correctly.
 
  • #5
So my book is wrong. I thought so. Thanks for checking my work, I was so confused =)
 
  • #7
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?
 
  • #8
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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.
 
  • #9
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
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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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