Gibbs Free Energy and Thermodynamic Favorability
Introduced| English |
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| Gibbs free energy/ɡɪbz friː ˈenədʒi/ |
Will it happen on its own?
- Some reactions run by themselves; others need a constant push.
- Two factors decide it: heat released and disorder created.
- A single number combines them into a verdict.
- Its sign tells you whether nature will do it for free.
The free-energy equation
- Gibbs free energy 吉布斯自由能 combines enthalpy and entropy:
- Here $T$ is the temperature in kelvin.
With $\Delta H = -100\ \text{kJ}$, $T = 300\ \text{K}$, $\Delta S = 0.2\ \text{kJ/K}$, find $\Delta G$ (in kJ).
$\Delta G = \Delta H - T\Delta S = -100 - (300)(0.2) = -160\ \text{kJ}$.
In $\Delta G = \Delta H - T\Delta S$, the temperature must be in...
Thermodynamic equations need an absolute temperature in kelvin.
Negative means favourable
- $\Delta G < 0$ means the reaction is thermodynamically favourable.
- $\Delta G > 0$ means it is not favourable on its own.
- $\Delta G = 0$ means the system is at equilibrium.
A reaction is thermodynamically favourable when $\Delta G$ is...
$\Delta G < 0$ means the reaction is favourable.
When $\Delta G = 0$, the system is at ____.
$\Delta G = 0$ marks equilibrium.
Temperature can flip it
- Exothermic plus more disorder is favourable at all temperatures.
- Endothermic plus less disorder is never favourable.
- The other two cases flip with temperature.
A reaction with $\Delta H < 0$ and $\Delta S > 0$ is favourable...
Both terms make $\Delta G$ negative regardless of $T$.
A reaction has $\Delta H < 0$ and $\Delta S > 0$. Is it favourable?
- Both terms push $\Delta G$ negative.
- So it is favourable at every temperature.
Free energy versus temperature
ΔG = ΔH - TΔS is a straight line in temperature. Where it crosses zero, the reaction turns favourable.
A thermodynamically favourable reaction is always fast.
Favourability is about direction, not speed; kinetics sets the rate.
"Favourable" ($\Delta G < 0$) means thermodynamically allowed, not fast -- kinetics decides the speed. The temperature is in kelvin in $\Delta G = \Delta H - T\Delta S$. And the $T\Delta S$ term grows with temperature, so heating can flip the sign of $\Delta G$.
Gibbs free energy $\Delta G = \Delta H - T\Delta S$ combines heat and disorder into one verdict: $\Delta G < 0$ is thermodynamically favourable, $> 0$ is not, and $= 0$ is equilibrium. Because of the $T\Delta S$ term, temperature can flip the sign for two of the four cases.