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Thermodynamics question

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HerpityDerpityDoc:
Hello, I've been brushing up on my understanding of thermodynamics, and I have come across a point I am confused about, and can't find an answer that satisfies me, so I thought I would bring my question to the fine minds here.

Specifically, I am confused by the equation that I see in many textbooks relating equilibrium constants to standard state change in free energy: K=exp(-ΔG°/RT).

Let's consider a process that has a positive ΔG°, such as boiling water ( H2O (l)  ::equil:: H2O (g) ).  If ΔG° is positive, that makes -ΔG°/RT negative at any temperature. According to the equation above, that would mean that K must be less than 1 at any temperature. But that is not the case: above 100°C, the equilibrium for that process should favor the product (water vapor).

What am I missing here?!

mjc123:
Is ΔG° positive at all temperatures?
(Note: the value for ΔG° that you use in ΔG° = -RTlnK is the value of ΔG at standard conditions at the temperature T. Not at some fixed "standard" temperature such as 298K. I've just had a long conversation with someone about this very thing.)

Enthalpy:
Is there an equilibrium in boiling water? At constant P and T, the amount of gas and water depends in the heat input. It is not imposed by the properties of water.

HerpityDerpityDoc:

--- Quote from: mjc123 on March 20, 2019, 05:42:04 AM ---(Note: the value for ΔG° that you use in ΔG° = -RTlnK is the value of ΔG at standard conditions at the temperature T. Not at some fixed "standard" temperature such as 298K. I've just had a long conversation with someone about this very thing.)

--- End quote ---

This gets to the root of the problem - I was assuming ΔG° referred to a standard fixed temperature! If that is not true, how would one actually use ΔG° = -RTlnK to calculate K values at a given temperature? Is there a way to predict ΔG° at a desired temperature, if you know it at a standard temperature like 298?

If not, is it safe to assume that ΔH° and ΔS° change negligibly with temperature, and plug literature values of  ΔH° and ΔS° at 298 into ΔG°=ΔH° - TΔS° to predict ΔG° at a desired temperature?

Corribus:

--- Quote from: HerpityDerpityDoc on March 20, 2019, 08:17:22 AM ---This gets to the root of the problem - I was assuming ΔG° referred to a standard fixed temperature! If that is not true, how would one actually use ΔG° = -RTlnK to calculate K values at a given temperature? Is there a way to predict ΔG° at a desired temperature, if you know it at a standard temperature like 298?

--- End quote ---
ΔG° specifies a point of reference based on a set of standard conditions.  It defines the direction a reaction will go when the reactant concentrations (activities) are at nonequilibrium. Because T shifts the equilibrium point, ΔG° also changes as a function of T.


--- Quote ---If not, is it safe to assume that ΔH° and ΔS° change negligibly with temperature, and plug literature values of  ΔH° and ΔS° at 298 into ΔG°=ΔH° - TΔS° to predict ΔG° at a desired temperature?
--- End quote ---
This is one standard procedure for estimating how ΔG° changes (and therefore the equilibrium constant changes) as a function of temperature. For reasonably small temperature changes, it's gives a pretty good estimate. 

This is effectively the basis of the van't Hoff equation.

https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation

There are modifications to the equation for situations where the assumption that invariant ΔH° over T fails.

Note that typically the van'T Hoff analysis is used to determine the enthalpy and entropy changes for a reaction from the equilibrium constants as a function of temperature, not the other way around. ;)

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