May 07, 2021, 09:52:29 PM
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Topic: Thermodynamics question  (Read 261 times)

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Offline Firoj

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Thermodynamics question
« on: March 17, 2021, 07:25:15 PM »
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?!

Offline Meter

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Re: Thermodynamics question
« Reply #1 on: March 18, 2021, 01:20:35 AM »
ΔG° won't be positive at T ≥ 298 K. In fact, it isn't positive for temperatures much smaller than that. Water can evaporate at temperatures lower than its boiling point.

Use values ΔH = 40.657 kj/mol and ΔS = 118.89 J/(mol·K) to verify this.

Offline mjc123

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Re: Thermodynamics question
« Reply #2 on: March 18, 2021, 07:03:15 AM »
ΔG° is not a constant, but varies with T according to ΔG° = ΔH° - TΔS°.
Note that, in this context, "standard states" does not imply a standard temperature such as 298K. It means "standard states at the temperature T, whatever that may be". You can not use ΔG°(298K) in a temperature-dependent expression like K = exp (-ΔG°/RT). That is a common error.

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