What a learner can do afterwards
- Calculates a free energy change from enthalpy and entropy values with the units made consistent first
- Reads the sign to say whether a reaction is feasible at that temperature
- Finds the temperature at which a reaction becomes feasible by setting the free energy change to zero
- Explains why a feasible reaction can still be far too slow to see
1 · Read
Gibbs free energy combines enthalpy and entropy into one test. Delta G equals delta H minus T times delta S, with T in kelvin. Negative means the reaction is feasible at that temperature, positive means it is not. For pure ice and water at normal pressure, that is why melting runs on its own just above 0 degrees Celsius while freezing runs on its own just below.
Watch the units before you touch the maths. Enthalpy usually arrives in kilojoules per mole while entropy comes in joules per mole per kelvin. Convert one of them first or the answer lands wrong by a factor of a thousand.
Set delta G to zero and solve: T equals delta H divided by delta S. Endothermic changes that gain entropy turn feasible when hot. Exothermic changes that lose entropy stay feasible only when cold.
Feasible never means fast. Diamond turning to graphite is feasible at room temperature yet far too slow to see, because few particles clear the activation barrier.
Balance enthalpy against temperature weighted entropy, fix the units first, and never read speed from the sign.
The rewritten readings were written for this dataset.
More to read
Using Gibbs free energy
At constant temperature and pressure, Gibbs free energy, G, combines enthalpy and entropy to test thermodynamic favorability: delta G = delta H - T times delta S. Negative delta G favors the forward reaction under those conditions; positive delta G does not.
Use T in kelvin and compatible energy units. If delta H uses kJ/mol and delta S uses J/(mol K), convert before calculating. Otherwise, the mismatched terms differ in scale by a factor of 1,000.
Favorability and reaction speed
Even a reaction with strongly negative delta G may proceed too slowly to notice. Its particles must overcome an activation-energy barrier. At room temperature, conversion of diamond to graphite is thermodynamically favorable but unobservably slow.
Industry needs both thermodynamics, which tests favorability, and kinetics, which describes reaction rates. Catalysts lower activation barriers and increase rates without changing thermodynamic favorability. Heating usually increases reaction speed, but it also changes the free-energy balance and can reverse which direction is favorable.
When temperature changes favorability
With nonzero delta S, solving delta G = 0 gives T = delta H / delta S, in kelvin. This assumes nearly constant delta H and delta S. A positive threshold separates favorable and unfavorable temperatures.
Endothermic reactions that increase entropy favor higher temperatures. Exothermic reactions that decrease entropy favor lower temperatures. If both contributions favor reaction, it remains favorable; if both oppose reaction, it remains unfavorable within that approximation.
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.