# Are all exothermic reactions necessarily spontaneous? [closed]

I was told that the statement "all exothermic reactions have negative free energy change" is false.

But I learnt Gibbs free energy as the "energy stored inside a system capable of doing non-$$pV$$ work", so I've always thought that exothermic reactions should have a negative change in free energy, i.e. $$\Delta H < 0 \implies \Delta G < 0$$. Why isn't this true?

• Almost a duplicate (perhaps actually a duplicate) of Are all exothermic reactions spontaneous? I will leave it to the community to vote. – orthocresol Dec 22 '20 at 13:14
• If all exothermic reactions were "spontaneous" in any meaningful sense, then people would catch fire in an atmosphere containing oxygen. If byt "spontaneous" you mean something more restrictive than the normal use of the term, then you should specify what you do mean. – matt_black Dec 22 '20 at 13:31
• There is thermodynamically based spontaneity due $\Delta G \lt 0$ and there is kinetically based spontaneity assuming $\Delta G \lt 0$ AND the process is not kinetically frozen ( Otherwise all diamonds near room conditions would turn into graphite. ). By other words, for perceived spontaneity, the thermodynamic spontaneity is a mandatory but not sufficient condition. – Poutnik Dec 22 '20 at 13:37
• Separation of solid $\ce{CaCl2 . 6 H2O}$ and water from $\ce{CaCl2}$ solution would be exothermic, but is not thermodynamically spontaneous, as entropy increase of the surrounding is smaller than the entropy decrease of the system and therefore $\Delta G > 0$. Spontaneous is the opposite, endothermic process. – Poutnik Dec 22 '20 at 13:47
• Does this answer your question? Are all exothermic reactions spontaneous? – Tyberius Dec 22 '20 at 15:46

The Gibbs energy $$G$$ is defined as

$$G=H-TS$$

For a process at constant temperature, the change in Gibbs free energy $$\Delta G$$ is then defined by:

$$\Delta G^o = \Delta H^o - T\Delta S^o$$

This quantity $$\Delta G^o$$ should be negative, for a reaction to be spontaneous.

So if a reaction is exothermic then $$\Delta H^o$$ is negative, but now the conditions depend on the entropy change $$\Delta S^o$$, as well as temperature:

Case 1: If $$\Delta S^o > 0$$ then no problem.

Case 2: If $$\Delta S^o < 0$$, then up to a certain temperature the reaction is favourable by enthalpy, but above that the entropy dominates over the enthalpy making the reaction non-spontaneous.