Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Firoj 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?!
-
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.
-
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.