stoichiometric coefficients for calculating Gibbs free energy






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Sarah Kiamanesh 1D
Posts: 30
Joined: Fri Sep 28, 2018 12:22 am

stoichiometric coefficients for calculating Gibbs free energy

Postby Sarah Kiamanesh 1D » Wed Feb 20, 2019 7:14 pm

I understand that in some equations like the formation of NH3 (N2 + 3H2 --> 2NH3), the coefficients are all divided by two to produce 1/2N2 + 3/2H2 --> NH3, while other equations keep their original coefficients. what's the procedure for this? when do we need to do this? does it make a difference?

melodyzaki2E
Posts: 61
Joined: Fri Sep 28, 2018 12:18 am

Re: stoichiometric coefficients for calculating Gibbs free energy

Postby melodyzaki2E » Wed Feb 20, 2019 8:03 pm

I think this is when you are calculating the enthalpy of formation of the standard molar entropy for 1 mole of NH3 because then you just want to produce one mole of that product. I'm not totally sure though.

Kim Tran 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

Re: stoichiometric coefficients for calculating Gibbs free energy

Postby Kim Tran 1J » Wed Feb 20, 2019 8:48 pm

Unless stated otherwise in the problem, I think you are to assume the problem is asking for the change in Gibbs Free Energy for the creation of one mole of product hence why it would be 1/2N2 + 3/2H2 --> NH3 and not N2 + 3H2 --> 2NH3.


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