Gibbs free energy change
Introduced| English | Meaning |
|---|---|
| Gibbs free energy/ɡɪbz friː ˈenədʒi/ | A thermodynamic quantity equal to enthalpy minus temperature times entropy, used to assess spontaneity at constant temperature and pressure. |
| enthalpy/enˈθælpi/ | A state function equal to internal energy plus the product of pressure and volume. |
| entropy/ˈentrəpi/ | A thermodynamic measure related to energy dispersal and the number of microscopic arrangements consistent with a system's macroscopic state. |
| feasible/ˈfiːzɪbl/ | Thermodynamically possible under the stated conditions; this does not establish that a reaction is fast. |
| kelvin/ˈkelvɪn/ | The SI unit of thermodynamic temperature, with zero at absolute zero. |
Will it actually happen?
- Gibbs free energy 吉布斯自由能 decides whether a reaction can happen on its own.
- It combines enthalpy 焓 and entropy 熵 with temperature.
- A negative $\Delta G$ means the reaction is feasible 可行.
Gibbs free energy lab
deltaG = deltaH - T deltaS
Move temperature and see when deltaG becomes negative.
A reaction is feasible when the Gibbs free energy change ΔG is ______.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
ΔG = ΔH − TΔS.
The equation
- $T$ is the temperature in kelvin 开尔文.
- A reaction is feasible (can happen) when $\Delta G \leq 0$.
When a reaction is feasible from the signs of enthalpy and entropy change
The Gibbs free energy change is calculated as:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
ΔG° = ΔH° − TΔS°, with T in kelvin.
A reaction is feasible when:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A reaction can happen on its own when ΔG ≤ 0.
Match each Gibbs idea.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Each item links the term to its correct meaning.
Feasibility and temperature
| $\Delta H$ | $\Delta S$ | Feasible |
|---|---|---|
| negative | positive | at all temperatures |
| positive | positive | only at high temperature |
| negative | negative | only at low temperature |
| positive | negative | never |
- The changeover temperature is found by setting $\Delta G = 0$, giving $T = \dfrac{\Delta H}{\Delta S}$.
A gaseous atom gains an electron, usually releasing energy
A reaction with negative ΔH and positive ΔS is feasible:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
With −ΔH and +ΔS, ΔG = ΔH − TΔS is always negative, so it is feasible at all temperatures.
The changeover temperature (where feasibility begins) is found by setting:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
At ΔG = 0 the reaction is on the verge of feasibility; rearranging gives T = ΔH/ΔS.
A feasibility calculation
Worked example. ΔG = ΔH − TΔS. If ΔH = +20 kJ/mol and TΔS = +30 kJ/mol, then ΔG = 20 − 30 = −10 kJ/mol, so the reaction is feasible.
You've got it
- $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ (T in kelvin)
- a reaction is feasible when $\Delta G \leq 0$
- $-\Delta H$ & $+\Delta S$ → feasible at all T; $+\Delta H$ & $-\Delta S$ → never
- changeover temperature: set $\Delta G = 0$, so $T = \dfrac{\Delta H}{\Delta S}$