# Calculating gibbs free energy

1. Apr 16, 2015

### brake4country

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. Apr 16, 2015

### Ygggdrasil

How many moles of H2O are you forming in the original reaction?

3. Apr 16, 2015

### brake4country

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. Apr 16, 2015

### Ygggdrasil

Right, so if you form 2 moles of water, ΔG = -457 kJ. That calculation is done correctly.

5. Apr 16, 2015

### brake4country

So my book is wrong. I thought so. Thanks for checking my work, I was so confused =)

6. Apr 16, 2015

### Staff: Mentor

What is 457 divided by 2?

Chet

7. Apr 16, 2015

### brake4country

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. Apr 16, 2015

### Staff: Mentor

No. The free energy of formation is defined on the basis of producing 1 mole of the compound from its elements.

9. Apr 16, 2015

### brake4country

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. Apr 16, 2015

Yes.