Free Energy and Equilibrium (ΔG° = −RT ln K)自由能与平衡（ΔG° = −RT ln K）

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 9~45 min

Learning goals

  • Use ΔG° = −RT ln K to find K from ΔG°, or ΔG° from K
  • Relate the sign of ΔG° to K being greater than, equal to or less than 1, and explain that each change of about 5.7 kJ/mol in ΔG° (at 25 °C) changes K by a factor of 10
  • Combine this with ΔG° = ΔH° − TΔS° to explain how temperature affects K

Before you start

  • Magnitude of the equilibrium constant (U7.5)
  • Gibbs free energy and thermodynamic favorability (U9.3)

ΔG° and K

The standard Gibbs free energy change and the equilibrium constant are related simply:

ΔG∘=−RTln⁡KK=e−ΔG∘/RT\Delta G^\circ = -RT\ln K \qquad K = e^{-\Delta G^\circ/RT}

R = 8.314 J/(mol·K), and T is in kelvins. When ΔG° is given in kJ/mol, multiply by 1000 to convert it to J/mol first. From this relationship:

ΔG°KAt equilibrium
< 0> 1products favored
= 0= 1comparable amounts
> 0< 1reactants favored

K depends exponentially on ΔG°. At 25 °C, every change of RT ln 10 ≈ 5.7 kJ/mol in ΔG° changes K by a factor of 10, so a ΔG° just a few tens of kJ/mol from zero gives a very large or very small K.

Drag ΔG° to see K and the shares of A and B at equilibrium, then look at some real reactions (data: OpenStax Appendix G):

Try it: ΔG° and the equilibrium constant K

The effect of temperature

ΔG° changes with temperature (ΔG° = ΔH° − TΔS°), and so does K. For an exothermic reaction (ΔH° < 0), heating makes ΔG° larger and K smaller; for an endothermic reaction, heating makes K larger. This matches the conclusion from Le Châtelier’s principle (U7.9).

Example 1: K from ΔG°

Find K at 25 °C for H₂(g) + I₂(g) ⇌ 2HI(g). ΔfG° (kJ/mol): I₂(g) 19.3, HI(g) 1.7.

Step-by-step solution

  1. ΔG° from free energies of formation. Products minus reactants; H₂(g) is an element in its standard state, so ΔfG° = 0 (in kJ):

    ΔG∘=2×1.7−(0+19.3)=−15.9\begin{aligned} \Delta G^\circ &= 2 \times 1.7 - (0 + 19.3) \\ &= -15.9 \end{aligned}

    In joules: −1.59 × 10⁴ J (R is in J, so ΔG° must be too).

  2. Find ln K. Use ΔG° = −RT ln K with T = 298 K:

    ln⁡K=−ΔG∘RT=1.59×1048.314×298=6.42\begin{aligned} \ln K &= -\frac{\Delta G^\circ}{RT} = \frac{1.59 \times 10^4}{8.314 \times 298} \\ &= 6.42 \end{aligned}

  3. Find K.

    K=e6.42≈6.1×102K = e^{6.42} \approx \boxed{6.1 \times 10^2}

  4. Check the sign. ΔG° < 0 goes with K > 1. Consistent.

  5. Compare with Unit 7. This reaction has K = 50 at 400 °C, smaller than at 25 °C. K falling as temperature rises means the forward reaction is exothermic (ΔH° = −9.5 kJ), consistent with Le Chatelier’s principle.

Example 2: ΔG° from K

At 25 °C, Ksp for AgCl is 1.6 × 10⁻¹⁰ (OpenStax Appendix J). Find ΔG° for AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).

Step-by-step solution

  1. Write the formula. For a dissolving equilibrium, K is Ksp:

    ΔG∘=−RTln⁡Ksp\Delta G^\circ = -RT\ln K_{sp}

  2. Work out ln K first.

    ln⁡(1.6×10−10)=−22.6\ln(1.6 \times 10^{-10}) = -22.6

  3. Substitute. In J:

    ΔG∘=−8.314×298×(−22.6)=5.59×104\begin{aligned} \Delta G^\circ &= -8.314 \times 298 \times (-22.6) \\ &= 5.59 \times 10^4 \end{aligned}

    That is +55.9 kJ\boxed{+55.9\ \text{kJ}}.

  4. Check the sign. Ksp ≪ 1, so ln Ksp is negative; times the minus sign, ΔG° is positive. Consistent.

  5. Connect to solubility. ΔG° > 0 and K ≪ 1: dissolving AgCl is unfavorable under standard conditions, very little dissolves at equilibrium, and that is why it is only slightly soluble.

Practice: at 25 °C (unless stated otherwise), how big is K?
  1. ΔG° = −30 kJ/mol
  2. ΔG° = +20 kJ/mol
  3. ΔG° = −0.3 kJ/mol
  4. ΔG° = +45 kJ/mol
  5. ΔH° = −50 kJ, ΔS° = −100 J/K, T = 1000 K
  6. ΔH° = +40 kJ, ΔS° = +200 J/K, T = 400 K
  7. ΔG° = +0.4 kJ/mol at 500 K
  8. ΔG° = −100 kJ/mol

Common mistakes

Mistake 1: “ΔG° = 0 means the reaction is at equilibrium.” ΔG° = 0 means K = 1. The sign of equilibrium is ΔG = 0 (without the °), when Q = K.

Mistake 2: “Use log instead of ln, or ΔG° in kJ with R in J.” The formula uses the natural log, ln; R is in J/(mol·K), so convert ΔG° to J/mol.

Mistake 3: “A reaction with ΔG° > 0 has no product at equilibrium.” K < 1 only means there is less product than reactant. With ΔG° = +5 kJ/mol, for example, K ≈ 0.13, which is still a fair amount of product.

Quick check

At 25 °C a reaction has ΔG° = +11.4 kJ/mol. K is about:

At 25 °C a reaction has K = 2.0 × 10⁸. What is ΔG°?

At some temperature a reaction has ΔG° = 0. Which statement is correct?

At 25 °C, ΔG° for reaction 1 is 11.4 kJ/mol more negative than for reaction 2. K₁ is about how many times K₂?

A reaction has ΔH° = +50 kJ and ΔS° = +100 J/K. At what temperature is K = 1?

Which quantity equals zero when a reacting system reaches equilibrium?

AP-style free response

AP-style free response

Above the iodine crystals in a sealed bottle there is faint violet iodine vapor: I₂(s) ⇌ I₂(g). Data: for I₂(g), ΔfG° = 19.3 kJ/mol and ΔfH° = 62.4 kJ/mol; S° (J·mol⁻¹·K⁻¹) I₂(s) 116.1, I₂(g) 260.7 (OpenStax Appendix G).

(a) Write the Kp expression for this equilibrium.

(b) Calculate ΔG° at 25 °C.

(c) Calculate Kp at 25 °C, and state what it represents here.

(d) Assuming ΔH° and ΔS° don’t change with temperature, calculate the temperature at which Kp = 1.

(e) The bottle is opened in a fume hood, and iodine vapor keeps escaping. Compare Q with K, and describe what happens to the iodine crystals.

Model answer and scoring

(a) Kp = P(I₂) (the solid is left out of the expression).

(b) ΔG° = 19.3 − 0 = +19.3 kJ.

(c) Kp = e−19300/(8.314 × 298) = e−7.79 = 4.1 × 10⁻⁴. It is the equilibrium vapor pressure of iodine at 25 °C, about 4.1 × 10⁻⁴ bar: ΔG° > 0, so there is little I₂(g) at equilibrium.

(d) ΔS° = 260.7 − 116.1 = 144.6 J/K, so T = 62.4 ÷ 0.1446 = 432 K.

(e) The escaping vapor lowers P(I₂), so Q < K; the equilibrium shifts toward I₂(g), and the crystals keep subliming and shrinking.

Scoring: (a) expression; (b) ΔG°; (c) Kp and its meaning; (d) ΔS° and temperature; (e) Q < K and the conclusion.

Summary

  • ΔG° = −RT ln K: ΔG° < 0 ↔ K > 1, ΔG° = 0 ↔ K = 1, ΔG° > 0 ↔ K < 1.
  • At 25 °C, every change of about 5.7 kJ/mol in ΔG° changes K by a factor of 10.
  • Temperature affects K through ΔG° = ΔH° − TΔS°: heating raises K for an endothermic reaction and lowers it for an exothermic one.

AP tip: When using ΔG° = −RT ln K, show the unit conversion (kJ → J) and use ln, not log. In qualitative questions, stating the link, such as “ΔG° < 0, so K > 1,” is enough.

Enrichment (nonstandard conditions): For any composition, ΔG = ΔG° + RT ln Q. When Q < K, ΔG < 0 and the reaction proceeds forward; when Q = K, ΔG = 0 and the system is at equilibrium.

Practice this chapter: Unit 9 · Thermodynamics and Electrochemistry · 12 questionsPractice this topic: U9.5 · Free Energy and Equilibrium (ΔG° = −RT ln K) · 5 questions

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