[Needaask]

ΔG° would be for the temperature specified that is not necessarily 298K and also depends on the stoichiometric amount of products and reactant like if A->2B then the ΔG°=ΔG°f(2B)-ΔG°f(A).

While ΔG would just be the general term for it and might not have the specific conditions imposed by the °.

So I was wondering about a phase change problem. In this case, ΔG=0 as there is a equilibrium here. So, for example H₂O(l) H₂O(g) the ΔG° would not be equal to the actual ΔG (which is equal to 0). So I don't really understand why we can use the formula ΔG=ΔH-TΔS and substitute the ΔH° and ΔS° values and equate it to 0. If we substitute them into the formula, won't we get ΔG° which in this case would not be equal to 0?

Also, usually we would be asked on whether melting is favorable at 0°C or 10°C. Then we would have to use the formula ΔG=ΔH-TΔS to show if its more than or less than 0. But again isn't this the ΔG° of the reaction and not the actual ΔG of the reaction which should be 0 after a while since equilibrium would be achieved?

Thanks in advance

[ricky99]

The degree symbol indicates standard conditions. For example, this means that all solutes are present at 1 molar concentration and all gases are at 1 atm pressure (or 1 bar, depending on your convention). Below, felmarah mentioned the classic equation accounting for non-standard concentrations. Standard conditions also implies 298 K (25 C). The common approximation for correcting for temperature is DeltaG = DeltaHo - T*DeltaSo . The (relatively good) approximation lies in assuming that, for example, DeltaH at the nonstandard temperature is that same as it is under standard conditions.

So, for example, DeltaGo tells you whether a reaction is spontaneous under standard conditions. DeltaG (without the degree symbol) tells you whether the reaction is spontaneous, under the conditions that you actually have right now.

If DeltaGo = 0, then DeltaG = 0 only under the special case of actually being at standard conditions. A simple example of this is a concentration gradient. For example, suppose you evacuate all the gas from a flask, then open it to air again. The "reaction" Air(outside) -> Air (inside) is clearly spontaneous; DeltaG < 0 for this process. However, if the air inside and outside the flask had been at 1 atm in both places, no net movement of air would have taken place; DeltaGo = 0.

thanks.

[Needaask]

The degree symbol indicates standard conditions. For example, this means that all solutes are present at 1 molar concentration and all gases are at 1 atm pressure (or 1 bar, depending on your convention). Below, felmarah mentioned the classic equation accounting for non-standard concentrations. Standard conditions also implies 298 K (25 C). The common approximation for correcting for temperature is DeltaG = DeltaHo - T*DeltaSo . The (relatively good) approximation lies in assuming that, for example, DeltaH at the nonstandard temperature is that same as it is under standard conditions.

So, for example, DeltaGo tells you whether a reaction is spontaneous under standard conditions. DeltaG (without the degree symbol) tells you whether the reaction is spontaneous, under the conditions that you actually have right now.

If DeltaGo = 0, then DeltaG = 0 only under the special case of actually being at standard conditions. A simple example of this is a concentration gradient. For example, suppose you evacuate all the gas from a flask, then open it to air again. The "reaction" Air(outside) -> Air (inside) is clearly spontaneous; DeltaG < 0 for this process. However, if the air inside and outside the flask had been at 1 atm in both places, no net movement of air would have taken place; DeltaGo = 0.

thanks.

Ohh so we are not getting temperature where dG is 0 but actually dGo=0 but they are close enough that the temperature is about the same as the phase change?