Gibbs Energy Calculator: Spontaneity, Crossover Temperature, Equilibrium Constant, Enthalpy and Internal Energy
Decide whether a change runs by itself, and find the temperature at which it turns over.
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How this calculator works
Choose the Gibbs case and enter the change in enthalpy and in entropy and the temperature. The calculator finds delta G = delta H - T delta S and says whether the change is spontaneous.
It gives the crossover temperature delta H over delta S, where delta G is zero, and the equilibrium constant K = exp(-delta G over RT).
The heating case gives the internal energy and enthalpy change of an ideal gas at constant volume and constant pressure. The isothermal case gives the greatest work, which is minus the change in Helmholtz energy.
Formulas used
- \Delta G = \Delta H - T\Delta S
- T_c = \frac{\Delta H}{\Delta S}
- K = e^{-\Delta G/RT}
- H = U + PV,\quad F = U - TS
- W_{max} = nRT\ln\frac{V_2}{V_1}
Worked example
With the default numbers in the calculator above:
- Gibbs energy change: 92.00 J/mol (0.0920 kJ/mol)
- Does the change run by itself?: No: delta G is positive, so it runs by itself in the other direction
- Crossover temperature, where delta G is zero: 273.18 K (0.0318 C)
- Equilibrium constant at this temperature: 0.96 (the products are favoured when K is above 1)
- The three parts of the free energy: Heat term (delta H) 6010.00 J/mol; entropy term (T delta S) 5918.00 J/mol
Change the numbers in the calculator to see the full working, every step, for your own case.
A second worked example
A different case, with these numbers: What do you want? Heat an ideal gas: internal energy, enthalpy and the work against the piston.
- Heat at constant volume (a rigid box): all of it goes into the internal energy: 1039.31 J (1.04 kJ)
- Heat at constant pressure (a free piston): the internal energy rises and the piston is lifted: 1455.03 J (1.46 kJ)
- Work done on the piston at constant pressure: 415.72 J
- Check: enthalpy change is the internal energy change plus the work: 1039.31 + 415.72 = 1455.03 J
- Ratio of the heats: 1.40 (1.40 times more heat is needed at constant pressure)
Common questions
When is a change spontaneous?
When delta G is negative at constant temperature and pressure.
Why does ice melt above 0 C?
Melting takes in heat (delta H positive) but it raises the disorder (delta S positive). Above 273 K the T delta S term wins and delta G becomes negative.
What is the difference between U, H, F and G?
U is the energy stored in the system, H adds the work to make room at constant pressure, F subtracts the heat term at constant temperature and volume, and G does both for constant temperature and pressure.
What does the equilibrium constant tell me?
Above 1, the products are favoured at equilibrium. It is 1 when delta G is zero.
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Got a question about this? Ask an engineering question and get a plain-language answer with references.
Further reading: Atkins, Physical Chemistry (chemical thermodynamics); Cengel and Boles, Thermodynamics: An Engineering Approach.
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