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Calculating gibbs free energy

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    ΔG = ΔH-TΔS

    3. 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.
  2. jcsd
  3. Apr 16, 2015 #2


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    How many moles of H2O are you forming in the original reaction?
  4. Apr 16, 2015 #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.
  5. Apr 16, 2015 #4


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    Right, so if you form 2 moles of water, ΔG = -457 kJ. That calculation is done correctly.
  6. Apr 16, 2015 #5
    So my book is wrong. I thought so. Thanks for checking my work, I was so confused =)
  7. Apr 16, 2015 #6
    What is 457 divided by 2?

  8. Apr 16, 2015 #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?
  9. Apr 16, 2015 #8
    No. The free energy of formation is defined on the basis of producing 1 mole of the compound from its elements.
  10. Apr 16, 2015 #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?
  11. Apr 16, 2015 #10
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