[Firoj]

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)  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?!

[Meter]

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.

[mjc123]

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.