A system can lower its own free energy while increasing total entropy. Constraints and composition determine the correct criterion.
By the end: interpret Gibbs free energy, standard reaction values, equilibrium constants, and the limits of a spontaneity claim.
Account for the surroundings through a state function
The isolated-total criterion is increasing entropy. For a system in contact with reservoirs that hold temperature and pressure fixed, it is often easier to use Gibbs free energy G. Enthalpy H is U plus pV, a state function useful when expansion work accompanies heating. G subtracts T times entropy from H. Under fixed temperature and pressure, with only pressure-volume work and the relevant composition constraints, spontaneous changes lower total system G until equilibrium.
At fixed temperature and volume, the corresponding criterion uses Helmholtz free energy, U minus TS, rather than Gibbs free energy. Neither is a universal amount of “stored heat.” Changing the constraints changes the appropriate criterion. A component’s chemical potential is its incremental Gibbs-energy cost per mole added at fixed temperature, pressure, and the other component amounts. A component that can move between phases has equal chemical potential in those phases at equilibrium, not necessarily equal concentration. A negative change in a system’s free energy accounts for the reservoir effects that are not visible if you examine the system’s entropy alone.
Derive the free-energy test from the isolated total
Draw a boundary around a closed system able to react while contacting a large thermal reservoir and a pressure reservoir. Temperature and external pressure remain fixed, and only pressure-volume work crosses the mechanical boundary. The system is not isolated, so its entropy need not increase. The isolated-total test includes the reservoirs. Gibbs energy is useful because it incorporates their contribution into a system quantity under these constraints. It does not replace the second law with a different physical law or remove the need to specify a boundary.
At fixed external pressure, expansion work done by the system is p times its volume increase. With the course convention ΔU = Q − W, heat input is ΔU + pΔV. For endpoints at the same pressure, this equals the enthalpy change ΔH because H is U + pV. Positive ΔH means heat enters in this account; negative ΔH means heat leaves. This reasoning assumes no electrical or other non-expansion work, no matter flow, and negligible changes in overall motion or height.
The Gibbs test includes heat and expansion outside the system
Schematic fixed-temperature, fixed-pressure setting. Q is heat into the system and W is work out. No matter or non-expansion work crosses the boundary.
The thermal reservoir receives the opposite heat, so its entropy change is −ΔH/T. Add the system entropy change to obtain ΔS_total = ΔS_system − ΔH/T. Multiplying by the negative number −T reverses the direction of the entropy inequality. The expression becomes ΔH − TΔS_system, which is ΔG for endpoints at the same T. Thus positive total entropy production corresponds to negative system ΔG. This sign reversal does not remove the reservoir transfers.
Consider a hypothetical change at 300 K that releases 6,000 J of enthalpy while the system entropy falls by 10 J/K. The system's entropy change is negative, yet the reservoir gains 6,000/300 = 20 J/K. The total gains 10 J/K. For the system, ΔG is −6,000 minus 300 times −10, or −3,000 J. Both descriptions permit the same direction. Looking only at the system's entropy would reject a change that increases the entropy of the isolated total.
A subsystem entropy loss can accompany a total increase
Hypothetical finite change at 300 K with ΔH = −6,000 J. Bar lengths share a signed entropy scale of 15 drawing units per J/K. The vertical line marks zero.
Now reverse those endpoint changes under the same conditions. The system would gain 6,000 J of enthalpy and 10 J/K of entropy, while the reservoir would lose 20 J/K. The combined entropy change would be −10 J/K, and system ΔG would be +3,000 J. This direction cannot occur as the sole spontaneous change. Supplying non-expansion work or coupling another process changes the complete account. The positive value alone does not permit it.
Fixed volume requires a different calculation. If a rigid system exchanges heat but does no other work, Q equals ΔU rather than ΔH. The reservoir entropy contribution is then −ΔU/T, giving the Helmholtz criterion through U − TS. Choosing Gibbs or Helmholtz energy is therefore a consequence of the boundary constraints. Neither formula says that temperature determines the system's total energy. Both combine extensive state quantities, which change with the amount of material even when temperature and pressure remain the same.
Separate standard values from the actual composition
Consider a hypothetical one-to-one conversion A to B with standard reaction enthalpy +40 kJ/mol and standard reaction entropy +100 J per mol per kelvin. Treat both as constant over 250–600 K. The standard reaction Gibbs value changes sign at 400 K. Convert entropy to kJ per mol per kelvin before combining it with a kJ enthalpy. These are invented thermodynamic parameters, not a recipe or a measured material.
The degree symbol in a standard Gibbs value denotes specified reference activities, not a statement that every mixture is in those conditions. Activities are dimensionless measures of effective concentration or pressure relative to a standard. For ideal A and B in this one-to-one model, the reaction quotient q is activity of B divided by activity of A. We write q in lowercase here to distinguish it from heat Q. Actual reaction free energy adds RT ln q to the standard value. This is the local driving force per mole converted at the current composition, not the energy for complete conversion of a changing mixture.
Read a reaction value as a local slope
For the hypothetical reaction A to B, one mole converted means one mole of A disappears and one mole of B appears. A reaction Gibbs value in joules per mole describes the local change in total G per small amount converted. It is the slope of total G against conversion. It is not the total height difference between the initial mixture and complete conversion. As the mixture changes composition, that slope changes too, even at fixed temperature and pressure.
Activities make the composition term dimensionless. In an ideal one-to-one mixture, the ratio q = aB/aA reduces to the amount ratio nB/nA when both components use compatible reference conventions. For example, one mole of B with nine moles of A gives q = 1/9, not a quotient with units of moles. Multiplying both amounts by two leaves q unchanged. It doubles the amount of reacting material, however, so it changes the scale of the total Gibbs-energy change available over a specified composition interval.
Conversion changes the quotient
Each equal box represents one mole in this ideal ten-mole accounting example, not molecular volume or spatial phase regions. At 400 K, K = 1 and equilibrium has equal amounts.
At 400 K the hypothetical standard reaction value is zero because 40 kJ/mol equals 400 K times 0.100 kJ per mol per kelvin. Actual reaction Gibbs energy is then RT ln q. With R in kilojoules, RT is about 3.3258 kJ/mol. If q = 1/9, ln q is about −2.1972, giving a local value about −7.31 kJ/mol. If q = 9, the logarithm has the opposite sign and the local value is about +7.31 kJ/mol.
These signs give a concrete restoring tendency. A mixture with nine A and one B favors forward conversion, increasing B and reducing A. A mixture with one A and nine B favors reverse conversion. At this temperature K equals one, so an ideal closed mixture with ten moles total reaches equilibrium at five moles of each. This equality requires K = 1 in the one-to-one ideal model. At another temperature, equal forward and reverse driving forces need not occur at equal amounts.
Imagine starting with nine moles of A and one of B at 400 K. After one mole converts forward, the mixture contains eight A and two B, so q becomes one quarter. The local reaction value is now about −4.61 kJ/mol, smaller in magnitude than before. Near five A and five B it approaches zero. Multiplying the initial −7.31 kJ/mol by all four moles needed to reach equilibrium would treat a changing slope as constant and overstate the magnitude of the total decrease.
The local driving force is the slope of total G
Ideal ten-mole A/B mixture at 400 K with ΔG° = 0. The minimum is shifted to zero for display. The curve is RT[nA ln(nA/10) + nB ln(nB/10) + 10 ln 2]. Only differences and slopes matter.
The degree symbol does not identify a standard temperature, and q = 1 does not always mean equal molar concentrations in a real mixture. Standard reaction values refer to chosen reference activities at the stated temperature. Real activities can differ from simple amount ratios because interactions matter. The lesson's algebra isolates temperature and composition using stipulated ideal behavior and constant standard enthalpy and entropy. These assumptions permit the calculations, not predictions for named materials.
Work through the mechanism
Standard reaction Gibbs energy at a fixed temperature
The reaction values are per mole converted. Keep joules and kilojoules consistent. Here ΔH° = 40,000 J/mol and ΔS° = 100 J/(mol K), both treated as constant over the model range.
Driving force at the current composition
q is the dimensionless activity quotient, not heat Q. This local driving force is zero at equilibrium; it is not the finite free energy of converting an entire changing mixture.
Equilibrium quotient
At a given temperature, q below K favors forward conversion and q above K favors reverse conversion. None of these formulas sets a reaction rate.
Compare a standard value with the actual driving force
At 500 K: q = 100, K = 11.08. Standard ΔG = -10 kJ/mol; actual ΔG = 9.14 kJ/mol. The reverse direction lowers G at this composition.
Gold dashes show the standard reaction ΔG at q = 1. Blue shows the selected composition rule; curves coincide at q = 1. Temperature presets change only T. Quotient buttons change only the composition rule; equilibrium mode tracks K as T changes. ΔH° and ΔS° are hypothetical constants. Values are local reaction driving forces in kJ/mol, not kinetic rates or complete-conversion energy yields.
Temperature
Standard ΔG
Equilibrium K
300 K
+10 kJ/mol
About 0.0182
400 K
0 kJ/mol
1
500 K
−10 kJ/mol
About 11.08
K is dimensionless. The actual driving force also depends on the current quotient q.
Compare fixed composition with tracked equilibrium
Select the 400 K preset and q = 1. The standard and actual curves coincide, and the selected value is zero. Before choosing q = 0.1, predict the sign of the composition term. Its logarithm is negative, so actual reaction Gibbs energy should fall below the standard curve. At 400 K it is about −7.66 kJ/mol. Select q = 100 next. The logarithm is positive and twice as large in magnitude, giving about +15.32 kJ/mol at the same temperature.
Hold q = 100 and select 500 K. The standard term becomes −10 kJ/mol, but the composition correction grows to about +19.14 kJ/mol. Their sum is about +9.14 kJ/mol, so reverse conversion remains favored. Now choose q = 0.1 without changing T. The correction becomes about −9.57 kJ/mol and the actual value about −19.57 kJ/mol. A temperature favorable under standard activities therefore does not determine the direction at every composition. Compare both curves, not the standard curve alone.
Next select equilibrium mode at 500 K. Instead of holding a quotient fixed, this mode sets q equal to the calculated K, about 11.08. Its actual reaction value is zero. Change to 300 K while retaining equilibrium mode. Now K is about 0.0182, so the quotient changes too and the actual value remains zero. This is a sequence of equilibrium compositions at different temperatures. It is not one closed sample keeping its original composition while temperature changes, nor does it trace a timed reaction.
Add standard and composition terms with matching units
T and q
Standard ΔG
RT ln q
Actual reaction ΔG
Direction
400 K, 0.1
0 kJ/mol
−7.66 kJ/mol
−7.66 kJ/mol
Forward
400 K, 100
0 kJ/mol
+15.32 kJ/mol
+15.32 kJ/mol
Reverse
500 K, 0.1
−10.00 kJ/mol
−9.57 kJ/mol
−19.57 kJ/mol
Forward
500 K, 100
−10.00 kJ/mol
+19.14 kJ/mol
+9.14 kJ/mol
Reverse
Hypothetical reaction parameters. These local values are per mole converted at the stated composition, not the energy of complete conversion.
A fixed-quotient curve has a useful algebraic check. Substituting the hypothetical standard values gives actual reaction Gibbs energy = 40,000 + T times the quantity R ln q − 100, in J/mol. Its zero occurs at T = 40,000 divided by 100 − R ln q, when that denominator is positive. For q = 0.1 the zero is about 335.7 K. For q = 100 it is about 648.2 K, outside the slider's 250–600 K range. No visible crossover is expected for that composition.
Rewrite the same driving force as RT ln of q/K using the equilibrium relation. This tests direction regardless of the standard value's sign. If q/K is below one, its logarithm is negative and forward conversion lowers G. Above one, reverse conversion lowers G. At 500 K compare q = 1 and q = 100 against K = 11.08. Their different signs follow from their positions on opposite sides of equilibrium.
Compare the current quotient with equilibrium
At 500 K, K ≈ 11.08. Position uses a base-ten logarithmic quotient scale, so equal distances mean equal ratios. Arrows indicate the favored change in q, not a rate.
The controls replace temperature or the composition rule, rather than integrating conversion over time. A fixed q button holds that quotient even while T changes. Equilibrium mode deliberately replaces q with a different number at each temperature. Read these as distinct comparisons. Neither curve is a graph of total mixture G versus conversion, and its vertical coordinate is not a stored-energy inventory. The units kJ/mol describe a local reaction slope. The horizontal axis is temperature, so the curve's own geometric slope is not the reaction Gibbs value.
Find equilibrium without claiming a reaction speed
At equilibrium the reaction driving force is zero. Solving that condition gives the equilibrium quotient K as exp of minus the standard Gibbs value divided by RT. At 500 K the model gives K about 11.08. A mixture at q = 1 is driven toward B, but one at q = 100 is driven toward A despite the same negative standard value. Equilibrium has equal forward and reverse rates, not necessarily equal amounts of A and B.
A favorable free-energy change does not imply a fast process. An activation barrier can make relaxation extremely slow; a catalyst speeds approach without changing the equilibrium constant for the same states and conditions. Coupling a conversion to another energy-releasing process can change the combined thermodynamic balance, but omitting that partner from the account gives a false conclusion. The model evaluates driving forces and never predicts reaction rates.
Locate equilibrium without confusing it with completion
At fixed temperature, K determines a ratio rather than a prescribed amount. For ideal A and B with one-to-one conversion and total amount n, let x be the amount of B. Then A has amount n − x and q = x divided by n − x. Setting this equal to K gives x = nK divided by 1 + K. Thus the equilibrium B fraction is K divided by 1 + K. This extra composition calculation translates the model's dimensionless equilibrium constant into amounts.
For ten total moles at 500 K, K about 11.08 gives a B fraction about 0.9172. The equilibrium mixture therefore contains about 9.172 moles of B and 0.828 mole of A. A large preference for B still leaves A present. At 300 K, K about 0.0182 gives only about 0.178 mole of B and 9.822 moles of A. These amounts require the specified ideal mixture and stoichiometry. The standard value alone does not fix them.
An equilibrium ratio need not mean complete conversion
Each strip represents ten total moles of an ideal one-to-one A/B mixture. Blue is B and gold is A. Widths encode mole fractions, not material volume fractions.
The graph of total mixture G has its minimum at that equilibrium composition. On the A-rich side its local reaction slope is negative; on the B-rich side it is positive. At the minimum the local slope is zero. Zero reaction driving force does not mean that total G is zero, just as a horizontal tangent need not occur at zero graph height. Nor does equilibrium require every possible reaction or molecular transition to stop. It requires no net allowed conversion.
A barrier can delay reaching that minimum. Two systems with the same endpoint thermodynamics can have different activation barriers and therefore different relaxation times. A catalyst supplies an alternative pathway with a lower kinetic barrier for the same overall reaction, accelerating the approach in both directions without moving the equilibrium minimum. It does not make the positive reaction Gibbs value of a product-rich mixture negative. The local slope and the pathway barrier answer different questions, so an equilibrium calculation cannot supply a rate constant or a completion time.
A different path can lower a barrier without changing endpoints
Schematic reaction-coordinate profiles, not total mixture G versus composition and not a measured reaction. Both routes have the same endpoint difference. Dashed gold indicates an alternative catalyzed route; vertical and horizontal scales are arbitrary.
Coupling also requires a complete boundary. If one permitted conversion has a local reaction value of +5 kJ/mol and a coupled partner has −8 kJ/mol, one mole of each in the specified stoichiometric combination has combined local value −3 kJ/mol. This permits the combined direction without proving a coupling mechanism. The partner's value can change as it is consumed. Count the partner's changing composition and any supplied work before claiming the original unfavorable conversion occurs without another change.
The equilibrium calculation finally connects back to entropy. A system can move toward a Gibbs minimum while its own entropy falls, because the reservoirs belong in the isolated total. At equilibrium the allowed local conversion cannot further increase that total entropy under the specified constraints. Changing temperature, pressure, conserved amounts, or permitted reactions changes the equilibrium problem. This is why a standard reaction number is insufficient by itself. The actual state determines driving force; kinetic information describes the available route toward equilibrium.
Exercises
Check the units and the sign
For the stated hypothetical reaction at 300 K and q = 1, calculate standard reaction ΔG. Is the forward direction favored at that composition?
Hint
Convert 100 J per mol per kelvin to 0.1 kJ per mol per kelvin.
Worked answer
Step
Value
Entropy in matching units
0.1 kJ/(mol K)
TΔS at 300 K
30 kJ/mol
ΔG at q = 1
+10 kJ/mol
A joule/kilojoule mismatch would produce a thousandfold error.
The entropy contribution TΔS is 30 kJ/mol. Subtracting it from 40 kJ/mol gives +10 kJ/mol. At q = 1, the logarithmic composition term is zero, so forward conversion is not favored. The reverse direction lowers G at that composition. This conclusion is not a rate prediction.
Account for accumulated product
At 500 K the standard reaction ΔG is −10 kJ/mol. Evaluate the direction at q = 100 using R ≈ 8.314 J/(mol K).
Hint
The composition correction is RT ln 100, about +19.14 kJ/mol.
Worked answer
Term
Approximate value
Standard ΔG
−10 kJ/mol
RT ln q
+19.14 kJ/mol
Actual ΔG
+9.14 kJ/mol
Direction at this composition
Reverse, toward smaller q
Standard favorability is not a promise of complete forward conversion.
The actual reaction driving force is about +9.14 kJ/mol, so reverse conversion is favored. The large product-to-reactant activity ratio outweighs the negative standard value. Since K is only about 11.08, q = 100 lies on the product-rich side of equilibrium.
Find the composition correction
R is given in J/(mol K). Dividing by 1,000 converts the resulting J/mol to kJ/mol; q is dimensionless.
Add the standard and composition terms
The positive local driving force favors reverse conversion at this composition, despite the negative standard value.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.
Selected Gibbs energy, reaction quotient, equilibrium constant, and temperature/composition distinctions. Our reaction has invented parameters. Interpret work with an explicit sign convention: this course counts output as positive. Reaction driving force is local to the current composition, not a finite complete-conversion energy.
Secondary reference: selected canonical weights and partition-function normalization. The two-level spectrum and degeneracies are constructed examples, not a material fit.