Gibbs Free Energy and Thermodynamic Favorability吉布斯自由能与热力学有利性

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 9~50 min

Learning goals

  • Calculate the standard Gibbs free energy change with ΔG° = ΔH° − TΔS° or ΔG° = ΣnΔfG°(products) − ΣnΔfG°(reactants), and decide whether a reaction is thermodynamically favorable
  • Use the signs of ΔH° and ΔS° to decide at what temperatures a reaction is favorable, and find the temperature where ΔG° = 0
  • Relate the sign of ΔG° to how much product is present at equilibrium, and distinguish "thermodynamically favorable" from "fast"

Before you start

  • Enthalpy of formation (U6.8)
  • Absolute entropy and entropy change (U9.2)

Gibbs free energy

Whether a process is spontaneous depends on the total entropy, but the entropy of the surroundings is hard to calculate directly. At constant temperature and pressure, two quantities for the system alone, ΔH and ΔS, are enough; they combine into the Gibbs free energy change吉布斯自由能变 ΔG:

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ
  • ΔG° < 0: the reaction is thermodynamically favorable热力学有利; it proceeds forward spontaneously under standard conditions, and products are favored at equilibrium.
  • ΔG° > 0: the forward reaction is unfavorable (the reverse reaction is favorable).
  • Use ΔH° in kJ, convert ΔS° from J/K to kJ/K, and use T in kelvins.

ΔG° can also be found directly from standard free energies of formation, ΔfG°, the same way as ΔH° (elements have ΔfG° = 0):

ΔGrxn∘=∑n ΔfG∘(products)−∑n ΔfG∘(reactants)\begin{aligned} \Delta G^\circ_\text{rxn} = &\sum n\,\Delta_\text{f}G^\circ(\text{products}) \\ &- \sum n\,\Delta_\text{f}G^\circ(\text{reactants}) \end{aligned}

The role of temperature: four cases

ΔH° and ΔS° usually change little with temperature, so ΔG° is roughly a linear function of T with slope −ΔS°.

ΔH°ΔS°Favorable when
−+at all temperatures
−−at low T (T < ΔH°/ΔS°)
++at high T (T > ΔH°/ΔS°)
+−at no temperature

When ΔH° and ΔS° have the same sign, the temperature where ΔG° = 0, T = ΔH° ÷ ΔS°, divides favorable from unfavorable. Choose a reaction and drag the temperature (data: OpenStax Appendix G):

Try it: how ΔG° changes with temperature

Example 1: ΔG° two ways

Find ΔG° at 25 °C for 2NO(g) + O₂(g) → 2NO₂(g). Data: ΔfH° (kJ/mol) NO 90.25, NO₂ 33.2; S° (J·mol⁻¹·K⁻¹) NO 210.8, O₂ 205.2, NO₂ 240.1; ΔfG° (kJ/mol) NO 87.6, NO₂ 51.3.

Step-by-step solution

  1. Method 1, step 1: ΔH°. Products minus reactants; O₂ is an element, so ΔfH° = 0 (in kJ):

    ΔH∘=2(33.2)−2(90.25)=−114.1\begin{aligned} \Delta H^\circ &= 2(33.2) - 2(90.25) \\ &= -114.1 \end{aligned}

  2. Method 1, step 2: ΔS°. S° of O₂ is not 0 (in J/K):

    ΔS∘=2(240.1)−[2(210.8)+205.2]=−146.6\begin{aligned} \Delta S^\circ &= 2(240.1) \\ &\quad - [2(210.8) + 205.2] \\ &= -146.6 \end{aligned}

    In kJ/K: −0.1466 kJ/K (to match the units of ΔH°).

  3. Method 1, step 3: ΔG° = ΔH° − TΔS°. T = 298 K:

    ΔG∘=−114.1−298(−0.1466)=−114.1+43.7=−70.4 kJ\begin{aligned} \Delta G^\circ &= -114.1 - 298(-0.1466) \\ &= -114.1 + 43.7 \\ &= \boxed{-70.4\ \text{kJ}} \end{aligned}

  4. Method 2: free energies of formation. ΔfG° of O₂ is 0:

    ΔG∘=2(51.3)−2(87.6)=−72.6 kJ\begin{aligned} \Delta G^\circ &= 2(51.3) - 2(87.6) \\ &= \boxed{-72.6\ \text{kJ}} \end{aligned}

  5. Compare. −70.4 and −72.6 kJ are close; the difference comes from rounding and measurement error in the data. ΔG° < 0, so the reaction is thermodynamically favorable at 25 °C.

  6. Check the signs. Gas goes from 3 mol to 2 mol, so ΔS° < 0 and −TΔS° is positive (unfavorable); but ΔH° is more negative and wins.

Example 2: how hot must it be?

For PCl₅(g) → PCl₃(g) + Cl₂(g), ΔH° = +87.9 kJ and ΔS° = +170.5 J/K. Is the reaction favorable at 25 °C? Above what temperature does it become favorable?

Step-by-step solution

  1. At 25 °C: ΔG°. ΔS° in kJ/K is 0.1705 kJ/K, T = 298 K (in kJ):

    ΔG∘=87.9−298×0.1705=+37.1\begin{aligned} \Delta G^\circ &= 87.9 - 298 \times 0.1705 \\ &= +37.1 \end{aligned}

    ΔG° > 0, so the reaction is unfavorable at 25 °C.

  2. Find the dividing temperature. ΔG° = 0 when ΔH° = TΔS°:

    T=ΔH∘ΔS∘=87.90.1705=516 K\begin{aligned} T &= \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{87.9}{0.1705} \\ &= 516\ \text{K} \end{aligned}

  3. Which side is favorable? ΔH° and ΔS° are both positive, so the higher the temperature, the more negative −TΔS° and the smaller ΔG°. The reaction is favorable for T>516 K\boxed{T > 516\ \text{K}} (about 243 °C).

  4. Check. At T = 600 K: ΔG° = 87.9 − 600 × 0.1705 = −14.4 kJ < 0, favorable as expected.

  5. State the assumption. ΔH° and ΔS° are assumed constant with temperature, so 516 K is an estimate.

Practice: at what temperatures is the reaction thermodynamically favorable?
  1. ΔH° = −90 kJ, ΔS° = +40 J/K
  2. ΔH° = +120 kJ, ΔS° = +250 J/K
  3. ΔH° = −150 kJ, ΔS° = −200 J/K
  4. ΔH° = +60 kJ, ΔS° = −30 J/K
  5. Ice melting (endothermic, ΔS > 0)
  6. NH₃(g) + HCl(g) → NH₄Cl(s) (exothermic, gases become a solid)
  7. 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g) (exothermic, more gas)
  8. N₂(g) + 2O₂(g) → 2NO₂(g) (endothermic, less gas)

Common mistakes

Mistake 1: “The more negative ΔG°, the faster the reaction.” ΔG° only says which side the equilibrium favors; it says nothing about rate. The rate depends on the activation energy (U9.4).

Mistake 2: “Subtract ΔS° in J/K straight from ΔH° in kJ.” Divide ΔS° by 1000 to convert it to kJ/K first, or the answer will be off by a factor of a thousand.

Mistake 3: “A reaction with ΔG° > 0 forms no product at all.” ΔG° > 0 means K < 1, so there is little product at equilibrium, but not none. Such a reaction can also be driven by coupling or by electrical energy (U9.6, U9.10).

Quick check

For Br₂(l) → Br₂(g), ΔH° = +30.9 kJ and ΔS° = +93.3 J/K. Estimate the boiling point of bromine (the temperature where ΔG° = 0):

ΔfG° (kJ/mol): TiCl₄(l) −737.2, H₂O(l) −237.1, TiO₂(s) −888.8, HCl(g) −95.3. What is ΔG° for TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 4HCl(g)?

A reaction has ΔH° > 0 and ΔS° > 0. Which statement is correct?

A reaction has ΔH° = −100 kJ and ΔS° = −150 J/K. What is ΔG° at 500 K?

At 25 °C a reaction has ΔG° = +15 kJ. Which statement is correct?

What does “thermodynamically favorable” mean?

AP-style free response

AP-style free response

Heating dark brown silver oxide gives silvery silver metal and oxygen: 2Ag₂O(s) → 4Ag(s) + O₂(g). Data: ΔfH°(Ag₂O) = −31.05 kJ/mol; S° (J·mol⁻¹·K⁻¹) Ag₂O 121.3, Ag 42.55, O₂ 205.2 (OpenStax Appendix G).

(a) Without calculating, predict the sign of ΔS°, and justify your prediction.

(b) Calculate ΔH° and ΔS°.

(c) Calculate ΔG° at 25 °C, and decide whether the reaction is thermodynamically favorable at that temperature.

(d) Assuming ΔH° and ΔS° don’t change with temperature, calculate the temperature above which the reaction becomes favorable.

(e) Use ΔG° = ΔH° − TΔS° to explain why heating decomposes silver oxide.

Model answer and scoring

(a) ΔS° > 0: a solid decomposes and forms O₂ gas.

(b) ΔH° = 0 − 2 × (−31.05) = +62.1 kJ; ΔS° = (4 × 42.55 + 205.2) − 2 × 121.3 = 375.4 − 242.6 = +132.8 J/K.

(c) ΔG° = 62.1 − 298 × 0.1328 = +22.5 kJ > 0, so the reaction is unfavorable at 25 °C (ΔfG° values give +22.4 kJ too).

(d) T = ΔH° ÷ ΔS° = 62.1 ÷ 0.1328 = 468 K (about 195 °C).

(e) ΔH° > 0 is unfavorable and ΔS° > 0 is favorable. As the temperature rises, −TΔS° becomes more and more negative; above 468 K its magnitude exceeds ΔH°, so ΔG° < 0 and the decomposition becomes favorable.

Scoring: (a) gas formed; (b) both calculations; (c) ΔG° and judgment; (d) dividing temperature; (e) TΔS° grows with temperature.

Summary

  • ΔG° = ΔH° − TΔS°, or ΔG° = ΣnΔfG°(products) − ΣnΔfG°(reactants); ΔG° < 0 means thermodynamically favorable.
  • The signs of ΔH° and ΔS° decide the role of temperature; when they share a sign, T = ΔH° ÷ ΔS° is the dividing temperature.
  • Units: convert ΔS° to kJ/K and use T in K. Thermodynamically favorable doesn’t mean fast.

AP tip: When explaining why a reaction is favorable, mention both terms, for example: “ΔH° < 0 is favorable and ΔS° < 0 is unfavorable; at low temperature the ΔH° term dominates.” Show the unit conversion when calculating ΔG°.

Practice this chapter: Unit 9 · Thermodynamics and Electrochemistry · 12 questionsPractice this topic: U9.3 · Gibbs Free Energy and Thermodynamic Favorability · 7 questions

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