If a reaction has negative $\Delta G$, can we then say with certainty that the reaction will be spontaneous?

Is negative $\Delta G$ the sole determinant for spontaneity?

  • $\begingroup$ please ignore kinetics of reaction in above quo $\endgroup$ Commented May 20, 2017 at 7:21
  • $\begingroup$ What do you mean by reaction? Are we talking about whole conversion of product to reactants? Or are we talking about whether one molecule of reactant converts at all to product? $\endgroup$
    – Zhe
    Commented May 20, 2017 at 15:09
  • $\begingroup$ @Zhe. I would like to know how does that matter? Well,take my previous comment casually: so that some spontaneous reaction taking too long may not become point of argument. $\endgroup$ Commented May 22, 2017 at 13:52
  • $\begingroup$ Because the microscopic view is governed by likelihood. Likelihood compounds in samples with large numbers. If you're talking about 1 molecule, maybe it's not completely unlikely. If you're talking about a reaction, it's quite possible that given the numbers, you'll never observe an appreciable amount of the product in the sense. Though spontaneity has a very specific definition, so I guess you're right that my previous comment is moot. $\endgroup$
    – Zhe
    Commented May 22, 2017 at 19:37

1 Answer 1


For a spontaneous process at constant temperature and pressure, $\Delta G$ must (emphasis on must) be negative. 

Spontaneity of a process , explicitly, depends on the enthalpy change and the entropy change. The Gibbs free energy change includes both these factors to predict the spontaneity of a process.

So, yes it can be said that Gibbs free energy change is the main factor that decides the spontaneity of a process.

Putting into better words, $\Delta G$ being negative is a consequence of a process being spontaneous.

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    $\begingroup$ Technically, $\Delta G <0$ isn't really necessary. It's just most probable. But for any macroscopic sample the huge number of atoms/molecules stacks the odds very heavily. If this were sports, I would all $\Delta G <0$ a prohibitive favorite. $\endgroup$ Commented May 20, 2017 at 12:14
  • $\begingroup$ To add to what @OscarLanzi said, the requirement that $\Delta G$ be negative for spontaneity is just a Rule of Thumb. $\endgroup$ Commented May 20, 2017 at 12:38
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    $\begingroup$ Then again, you could just argue that "spontaneity" itself is a probabilistic label. $\endgroup$ Commented May 20, 2017 at 14:58
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    $\begingroup$ @orthocresol , you are welcome to edit my answer. $\endgroup$
    – Mitchell
    Commented May 20, 2017 at 15:17
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    $\begingroup$ You might want to mention that different state variables are used to determine spontaneity when "at constant temperature and pressure" does not hold; for instance, Helmholtz free energy for reactions taking place at constant temperature and volume but not constant pressure. $\endgroup$
    – zwol
    Commented May 20, 2017 at 15:48

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