Does ΔG° for reactions vary by temperature?

Nikitin

I thought it was only ΔG which varied by temperature,with the equation ΔG = ΔH - TΔS ?

It makes no sense to me why ΔG° (= change of gibbs free energy at standard conditions during a reaction) should vary with temperature, but I've gotten an assignment here to calculate ε° for a reaction at 90°C (ε is a function of ΔG), which suggests that ΔG° does indeed vary!

How can this be? I thought Gibbs free energy at standard conditions was always constant, with the temperature being 25°C?

Borek

ΔG° at 90°C, when ΔG° is defined for 25°C - something doesn't add in what you wrote.

Nikitin

Is ΔG° for a reaction defined at 25 degrees, or is it temperature independent? This is what I am unsure about..

Borek

Defined at 25°C if you ask me.

Not that I am always right when it comes to thermodynamics.

Darwin123

Conjectures:
1) Maybe the superscript "°" doesn't represent all "standard conditions." Maybe "°" in this case it only represents "standard pressure".

Maybe ε° is the internal energy at standard pressure. Maybe the temperature is allowed to vary for internal energy.

2) Maybe in your system the internal energy, ε, doesn't vary with temperature. Maybe it only varies with pressure.

Just because ΔG varies with both temperature and pressure doesn't mean that the internal energy has to vary with both temperature and pressure. The variation with temperature may be subtracted out.

I admit number 2 is a bit of more speculative then the other one. It would take a strange equation of state to make the internal energy NOT vary with temperature. Number 1 is just a matter of convention. Not everyone is careful with the use of subscripts.

By the way: I am used to seeing internal energy designated by "ΔU", not "ΔG". I had to go through Wikipedia to find out that ε sometimes represents internal energy.

DrDu

Maybe the exact text of the assignment would be helpful.

Nikitin

It was asking for the emf° (=ε°) at 90°C for a certain reaction. ε is the international sign for emf in chemistry, right?

Anyway, I think delta G does indeed vary by temperature. According to this equation, (ΔG = ΔG°+R*T*ln(K), at equilibrium ΔG° = -R*T*ln(K). This implies ΔG°=0 when K=1 at equilibrium, thus ΔG° must vary with temperature (since it can be both 0 and other values, depending on K). Can anyone see the flaw in my reasoning?

Indeed, when K=1 then all the substances in the equilibrium have an activity of 1. This means that ΔG = ΔG° only when that is the case (or if T=0).

Darwin123: Internal energy (=U) and G are two entirely different things. Internal energy represents the energy of the molecules (=1.5*R*T for an ideal gas), G represents energy free to do work during reversible processes. Enthalpy is a function of U. H = U + P*V. Gibbs free energy is a function of H, T and S. G = H - T*S