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Topic: Standard free enery vs. standard Gibbs free energy  (Read 5193 times)

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Offline Ai

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Standard free enery vs. standard Gibbs free energy
« on: August 19, 2008, 03:18:45 PM »
Can any one tell me the the difference between the two?

I was doing this problem:

2H2(g) + O2(g) -> 2H2O

What is the standard Gibbs free energy of formation of water vapor at 25oC, if for the reaction shown below, deltaH = -484 kJ/mol and deltaS = -89 J/mol K?

Now I did the calculation as: deltaG = (-484000)-([273+25]*-89] = -457 kJ/mol

But the answer came to be: -457/2 = -229kJ/mol

My Questions are:

A) Why do you divide by 2? Isn't Gibbs free energy an extensive property?

B) If the question had said "What is the standard free energy.." then the answer would be -457kJ/mol?


Offline Hunt

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Re: Standard free enery vs. standard Gibbs free energy
« Reply #1 on: August 19, 2008, 04:15:36 PM »
Free energy is an extensive property but the molar free energy is intensive. Notice that here you calculate the molar gibb's free energy. Why divide by 2 ? because you wish to find dela G for the reaction per 1 mole not 2 moles of H2O.

-457kJ would be the answer for 2 moles i.e. -457kJ/ 2 mol

Offline Hunt

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Re: Standard free enery vs. standard Gibbs free energy
« Reply #2 on: August 19, 2008, 04:17:54 PM »
I do not understand what you mean by free energy versus Gibb's free energy. In the example you give both seem to identify delta G . You can call it either of the two names , but it makes no difference. The only difference here is the molar quanity.

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