Entropy and Gibbs Free Energy: Predicting Which Way a Process Goes

Spontaneous means thermodynamically allowed, not fast

A spontaneous process is one that, once started, continues without a continuing input of energy from outside. Iron rusting and heat flowing from a hot object to a cold one are spontaneous. Their reverses are not: rust does not turn back into iron by itself.

Spontaneity says nothing about speed. Diamond turning into graphite at room temperature is spontaneous, but it is so slow that diamonds last indefinitely. Thermodynamics tells you which direction is downhill. Kinetics tells you how fast the system gets there.

Many spontaneous processes release heat, but not all. Ice melting above 0 °C, water evaporating and ammonium nitrate dissolving are all endothermic and still spontaneous. Something besides enthalpy is involved, and that is entropy.

Entropy counts the ways energy can be spread

Entropy, S, measures how many microscopic arrangements (microstates) are consistent with a system's observed state. Boltzmann's relation is S = k ln W, where W is the number of microstates and k is the Boltzmann constant. More freedom of motion and more places to put energy mean more microstates and higher entropy.

The third law of thermodynamics states that a perfect crystal at 0 K has zero entropy. That gives entropy an absolute zero point, so tables list absolute standard molar entropies, S°, not entropies of formation. Unlike ΔH°f, the S° of an element is not zero: OpenStax Appendix G gives 130.7 J/(mol·K) for H₂(g) and 205.2 J/(mol·K) for O₂(g).

You can usually predict the sign of ΔS before calculating it:

• Gases have much more entropy than liquids, and liquids more than solids. Appendix G lists 70.0 J/(mol·K) for H₂O(l) and 188.8 J/(mol·K) for H₂O(g).
• A reaction that increases the number of moles of gas usually has a positive ΔS. One that decreases it usually has a negative ΔS.
• Raising the temperature of a substance raises its entropy.
• Dissolving a solid in a liquid usually increases entropy, although ions that strongly order the surrounding water can reverse this.

Calculating ΔS° and applying the second law

The standard entropy change of a reaction follows the same pattern as Hess's law: ΔS° = Σ n S°(products) − Σ m S°(reactants), using the coefficients of the balanced equation.

Worked example 1. For N₂(g) + 3 H₂(g) → 2 NH₃(g), Appendix G gives S° = 191.6 for N₂, 130.7 for H₂ and 192.8 for NH₃, all in J/(mol·K). ΔS° = 2(192.8) − [191.6 + 3(130.7)] = 385.6 − 583.7 = −198.1 J/K. The negative sign fits the prediction, since four moles of gas become two.

A negative ΔS for the system does not rule out a spontaneous reaction. The second law says that the entropy of the universe increases in every spontaneous process: ΔS_univ = ΔS_sys + ΔS_surr > 0. Heat released into the surroundings raises their entropy by ΔS_surr = −ΔH_sys/T. Appendix G gives ΔH° = 2(−45.9) = −91.8 kJ for this reaction. At 298.15 K, ΔS_surr = +91 800 J ÷ 298.15 K = +307.9 J/K. ΔS_univ = −198.1 + 307.9 = +109.8 J/K, which is positive, so the reaction is spontaneous under standard conditions at 25 °C.

Gibbs free energy and the four sign combinations

Tracking the surroundings separately is awkward. Multiplying ΔS_univ by −T gives a quantity that depends only on the system: the Gibbs free energy change, ΔG = ΔH − TΔS. At constant temperature and pressure, a process is spontaneous when ΔG is negative, at equilibrium when ΔG is zero, and nonspontaneous when ΔG is positive.

Worked example 2. For the ammonia synthesis at 298.15 K, ΔG° = −91.8 kJ − (298.15 K)(−0.1981 kJ/K) = −32.7 kJ. Note the conversion of ΔS from J/K to kJ/K. Appendix G also lists free energies of formation, and 2 × ΔG°f(NH₃) = 2(−16.5) = −33.0 kJ. The two routes agree to within the rounding of the tabulated values. The same −32.7 kJ also equals −T × ΔS_univ from example 1.

Because T is always positive, the signs of ΔH and ΔS set the pattern:

• ΔH negative, ΔS positive: spontaneous at every temperature.
• ΔH positive, ΔS negative: nonspontaneous at every temperature.
• ΔH negative, ΔS negative: spontaneous below the crossover temperature T = ΔH/ΔS. The ammonia synthesis is in this group, with T = −91 800 J ÷ −198.1 J/K = 463 K, about 190 °C.
• ΔH positive, ΔS positive: spontaneous above the crossover temperature. Melting and boiling are in this group.

Worked example: why water boils near 100 °C

For H₂O(l) → H₂O(g), Appendix G gives ΔH° = −241.82 − (−285.83) = +44.01 kJ and ΔS° = 188.8 − 70.0 = +118.8 J/K. The crossover temperature is 44 010 J ÷ 118.8 J/K = 370 K, or about 97 °C. The measured normal boiling point is 373 K. The small gap arises because ΔH and ΔS both change with temperature, while this estimate uses their 25 °C values. Even so, data measured at 25 °C predict the boiling point to within about 3 K.

At 25 °C, ΔG° = −228.59 − (−237.1) = +8.51 kJ, which is positive. Yet water left in an open dish evaporates. The resolution is that ΔG° refers to standard states, here water vapor at 1 bar. The actual free energy change depends on the actual vapor pressure through ΔG = ΔG° + RT ln Q. If the air holds water vapor at 0.0100 bar, Q = 0.0100 and ΔG = 8.51 kJ + (8.314 × 10⁻³ kJ/(mol·K))(298.15 K) ln(0.0100) = 8.51 − 11.42 = −2.91 kJ/mol. Evaporation is spontaneous under those conditions.

From ΔG° to an equilibrium constant

At equilibrium ΔG = 0 and Q = K, so ΔG° = −RT ln K, or K = e^(−ΔG°/RT). A negative ΔG° gives K greater than 1, meaning products are favored at equilibrium. A positive ΔG° gives K less than 1. Because the relationship is exponential, a change of about 5.7 kJ/mol in ΔG° changes K tenfold at 25 °C.

Worked example 3. For the vaporization of water at 25 °C, K = exp[−8510 ÷ (8.314 × 298.15)] = 0.0323. Here K is the equilibrium vapor pressure in bar, which is 24.2 torr. OpenStax Table 9.2 lists a measured 23.8 torr at 25 °C. Getting this close from tabulated thermodynamic data alone is a strong check that the method works.

Worked example 4. For the ammonia synthesis, K = exp[−(−33 000) ÷ (8.314 × 298.15)] ≈ 6 × 10⁵ at 25 °C. Report K to only one or two significant figures. An uncertainty of 0.1 kJ in ΔG° already changes K by about 4 percent.

Sign, unit and standard-state slips

Most errors in free-energy problems come from units and from confusing standard and actual conditions. Write units on every term and check that the sign of each result matches your prediction.

• Combining ΔH in kJ with ΔS in J/K without converting. Divide ΔS by 1000 first.
• Using Celsius temperature. T in ΔG = ΔH − TΔS and in RT ln K must be in kelvins.
• Setting S° of an element to zero. Only ΔH°f and ΔG°f are zero for elements in their reference forms.
• Treating ΔG° and ΔG as the same thing. ΔG° tells you about standard conditions and fixes K; ΔG decides what happens under the actual conditions.
• Concluding that a reaction with negative ΔG must be fast.
• Using R = 0.08206 L·atm/(mol·K) in energy equations. Use 8.314 J/(mol·K).