Thermodynamic Favorability (College Board AP® Chemistry)

Study Guide

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Gibbs free energy provides an effective way of focusing on a reaction system at constant temperature and pressure to determine whether the reaction is thermodynamically favorable
A thermodynamically favorable reaction is one that occurs without the need for additional energy input after the reactants are mixed
- A thermodynamically favored reaction will have a negative ΔG^o value
For a reaction to be spontaneous, Gibbs free energy must be a negative value (ΔG^o ≤ 0)

Historically, a thermodynamically favorable reaction has been referred to as a spontaneous reaction.

Thermodynamically favored is the preferred term as it avoids the common misunderstanding that the reaction happens suddenly or without cause.

The temperature conditions for a process to be thermodynamically favored (whereby ΔG^ois negative and less than 0) can be predicted from the signs of ΔH^oand ΔS^o

If ΔH^ois negative (exothermic reaction) and ΔS^o is positive:
- This will result in a negative ΔG^o
- The reaction is always thermodynamically favored
- No calculation of ΔG^o would be necessary to determine this
If ΔH^ois negative (exothermic reaction) and ΔS^o is negative:
- At high temperatures -TΔS^owill be very large and positive and will overcome ΔH^o
- At high temperatures ΔG^ois positive
- The reaction is thermodynamically favored at low temperatures

If ΔH^ois negative and If ΔS^o is negative the reaction will only be thermodynamically favored at low temperatures

If ΔH^ois positive (endothermic reaction) and ΔS^o is negative:
- Both ΔH^oand -TΔS^owill be positive
- This results in a positive ΔG^o
- Therefore, regardless of the temperature, endothermic with a negative ΔS^o is not thermodynamically favored
- No calculation of ΔG^o is necessary to determine that the process is thermodynamically unfavored
If ΔH^ois positive (endothermic reaction) and ΔS^o is positive
- ΔH^ois positive and -TΔS^owill be negative
- At low temperatures -TΔS^o cannot overcome the larger ΔH^o
- At low temperatures, ΔG^o is positive and the reaction is thermodynamically favored
- At higher temperatures, the second term will become negative enough to overcome the ΔH resulting in a negative ΔG^o

If ΔH^ois positive and If ΔS^o is positive the reaction will only be thermodynamically favored at a high temperatures

ΔH^o	ΔS^o	ΔG^o < 0 favoured at:	Reaction is:
< 0	> 0	all temperatures	thermodynamically favored
> 0	< 0	no temperatures	thermodynamically unfavored
> 0	> 0	high temperatures	thermodynamically favoured as temperature increases
< 0	< 0	low temperatures	thermodynamically unfavored as temperature increases

For a reaction with the following enthalpy and entropy changes:

a) Calculate ΔG^oat 298 K.

b) Calculate the temperature the reaction will become thermodynamically favored.

Answer:

a) At 298 K:

Convert ΔS^ofrom J K^-1mol^-1 into kJ K^-1mol^-1
- = $\frac{- 75.8}{1000}$
- = – 0.0758 kJ K^-1mol^-1
ΔG^o= ΔH^o – TΔS^o
- ΔG^o = -13.65 – 298 x (–0.0758)
- ΔG^o= + 8.94 kJ mol^-1

b) The temperature at which the reaction will become thermodynamically favored:

Rearrange ΔG^o= ΔH^o – TΔS^o to find T
- When ΔG^o= 0, then $T = \frac{Δ H^{o}}{Δ S^{o}}$
- T= $\frac{- 13.65}{- 0.0758} = 180 K$
When ΔH^oand ΔS^oare both negative, decreasing the temperature causes reactions to be thermodynamically favored
Therefore below 180 K, the reaction is thermodynamically favored, and above 180 K the reaction is not thermodynamically favored

Last updated: 24 August 2024

the (exam) results speak for themselves:

Did this page help you?