Does the ° symbol indicate that the temperature of the substance is at 298 K?
I know that ° indicates that the substance is in the standard state (1 bar, 1 mol/L for a solute), but I was confused if ° means 298 K. I thought that ΔG° could vary depending on temperature, since ΔG° = ΔH° - TΔS°, meaning that we can't assume ΔG° means the substance is 298 K.
Do ΔH° and ΔS° then also vary depending on temperature?
If that's the case, for example when we are calculating ΔG° at 900 K, are we assuming that ΔH° and ΔS° are constant as if they were at 298 K, even though they are different at 900 K?
ΔG°, ΔH°, and ΔS°
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Re: ΔG°, ΔH°, and ΔS°
the ° symbol doesn't imply that the rxn is at 298 K, but rather that this is a value calculated from the reaction at equilibrium (i believe). so yes, we can calculate these values at different temperatures. i would assume that H and S will be adjusted to different temperatures but we're normally given standard values to find these so i think it still doesn't imply 298 K
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Re: ΔG°, ΔH°, and ΔS°
sabrina ghalambor 2J wrote:the ° symbol doesn't imply that the rxn is at 298 K, but rather that this is a value calculated from the reaction at equilibrium (i believe). so yes, we can calculate these values at different temperatures. i would assume that H and S will be adjusted to different temperatures but we're normally given standard values to find these so i think it still doesn't imply 298 K
That makes sense, thank you!!
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Re: ΔG°, ΔH°, and ΔS°
The naught implies that the value for entropy, enthalpy, etc. was measured at 298K and standard pressure. In the case of the Gibbs free energy equation, we make the assumption that the change in temperature doesn't affect the delta G, S and H naught.
Re: ΔG°, ΔH°, and ΔS°
Yes, naught usually implies that this reaction happens at standard conditions at 298 K (-:
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Re: ΔG°, ΔH°, and ΔS°
Hi! Like many others have stated, usually the naught symbol means the reaction occurs under standard conditions which would be like 298K and so forth.
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