Library / Foundations: Chemistry / FCH-18
Foundations: Chemistry
Gibbs free energy
Tells you whether a reaction can happen spontaneously at a given temperature.
- Ref
- FCH-18
- Rev
- v1
- Field
- Foundations: Chemistry
- Result
- ΔG [kJ/mol]
T in kelvin. Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ (divide by 1000) so all energy terms are in kJ/mol. A negative ΔG means the reaction is thermodynamically feasible.
A reaction is favoured if it releases heat or increases disorder; temperature sets the balance between these two.
Calculator
- Crossover temperature ΔH/ΔS464 K
- T ΔS term−59.21 kJ/mol
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- For ammonia synthesis ΔH and ΔS are both negative, so the reaction is feasible only below about 464 K, which is why a catalyst and high pressure are used.
- Feasible does not mean fast: a negative ΔG says nothing about rate.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kJ/mol) | Local step | Change in result |
|---|---|---|---|---|---|---|
| ΔH Enthalpy change | −2,000 to 2,000 kJ/mol | −1,941 | 2,059 | 4,000 | +10% | +28 % |
| ΔS Entropy change | −1,000 to 1,000 J/(K mol) | −390.2 | 205.8 | 596 | +10% | −18 % |
| T Temperature | 100 to 2,000 K | −72.33 | 305.2 | 377.5 | +10% | −18 % |
Sweep of Temperature
| Temperature (K) | Free energy change (kJ/mol) |
|---|---|
| 100 | −72.33 |
| 337.5 | −25.14 |
| 575 | 22.05 |
| 812.5 | 69.24 |
| 1,050 | 116.4 |
| 1,288 | 163.6 |
| 1,525 | 210.8 |
| 1,763 | 258 |
| 2,000 | 305.2 |
Sweep one variable
Variables and how each one contributes
Free energy change
What the equation gives you. Everything else feeds it.
Enthalpy change
Reaction enthalpy (N₂ + 3H₂ → 2NH₃ = −92.2). Raises ΔG one-for-one.
Temperature
Absolute temperature. Scales the entropy term.
Entropy change
Reaction entropy change. A positive ΔS lowers ΔG as T rises.
Worked example with the values above
- T ΔS = 298 × −0.1987 = −59.21 kJ/mol.
- ΔG = ΔH − T ΔS = −92.2 − (−59.21).
- ΔG = −32.99 kJ/mol (negative, so feasible).
Assumptions and limits
- ΔH and ΔS are constant with temperature.
- Standard conditions.
Common mistakes
- Using ΔS in J without converting to kJ.
- Using °C for T.
Statistics and uncertainty
ΔH and ΔS are tabulated at 298 K and change slightly with temperature, so extrapolating a long way from 298 K is an approximation.
Where it is used
- Judging whether a process can run at a given temperature.
- Explaining why propellant reactions proceed.
Maths you need
- Enthalpy: ΔH
- Entropy: disorder