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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 9

Lesson 2: Gibbs free energy and thermodynamic favorability

# Thermodynamic favorability and temperature

The temperature conditions under which a process is thermodynamically favored (ΔG° < 0) can be predicted from the signs of ΔH° and ΔS°. When ΔH° < 0 and ΔS° > 0, the process is favored at all temperatures. When ΔH° > 0 and ΔS° < 0, the process is favored at no temperature. When ΔH° < 0 and ΔS° < 0, the process is favored only at low temperatures. When ΔH° > 0 and ΔS° > 0, the process is favored only at high temperatures. Created by Jay.

## Want to join the conversation?

• he mentions delta entropy doesn’t change much with temperature, but wouldn’t an increase of over 1000 kelvin increase the number of micro states in a gas by a considerable amount? • I believe what you're referring to is ΔS (change in entropy), which is equal to:
k * ln(W).
In other words ΔS is directly proportional to the number of microstates, which would increase with temperature. However, as I understand it,

What's being talked about is absolute or standard entropy - ΔS⁰ (delta S nought), which is ΔS of 1 mole of a pure solid or liquid, at a pressure of 1atm and a specified temperature. In the previous video, Jay mentions that often that temperature is 25°C.

Moreover, the units of ΔS⁰ are J/(mol * K), whereas the units of ΔH are kJ/mol. (Ignore the difference in kJ and J)
So, ΔS⁰ stays constant regardless of temperature, as Kelvin is in the denominator.
That's also why we multiply it with the temperature while calculating the Gibbs Free Energy (ΔG⁰).

Hope this helps!
• What is the relationship between thermodynamic favorability and the equilibrium constant (K)?
(1 vote) • We can relate Gibbs free energy and the equilibrium constant of a reaction via this equation: ΔG° = -RTln(K). This tells us that a larger equilibrium constant value yields a more negative free energy value which means the forward reaction is favored. If we had a very small equilibrium constant value then that yields a more positive free energy value which means that the reverse reaction is favored.

Hope that helps.
• As Jay mentioned in the video and also in the description: "When ΔH° < 0 and ΔS° > 0, the process is favored at all temperatures". The example in the video plugged in a temperature of 100 degrees. However if I instead plugged in a temperature of -2000 degrees, wouldn't the ΔG°>0?  