Coupled Reactions偶联反应

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 9~40 min

Learning goals

  • Explain coupled reactions, in which a thermodynamically unfavorable reaction combines with a favorable one through a common intermediate so that the overall reaction is favorable
  • Add ΔG° values (or multiply K values) for coupled reactions, and decide whether the overall reaction is favorable
  • Decide whether two reactions can be coupled; they must share a common intermediate

Before you start

  • Properties of the equilibrium constant (U7.6)
  • Free energy and equilibrium (ΔG° = −RT ln K) (U9.5)

Using a favorable reaction to drive an unfavorable one

A reaction with ΔG° > 0 is unfavorable under standard conditions. One way to obtain its products is to couple偶联 it with a reaction whose ΔG° is very negative. The two reactions share a common intermediate共同的中间物, a product of one reaction that is a reactant in the other. Adding the two equations cancels the intermediate and gives a new overall reaction:

ΔGoverall∘=ΔG1∘+ΔG2∘Koverall=K1×K2\begin{gathered} \Delta G^\circ_\text{overall} = \Delta G^\circ_1 + \Delta G^\circ_2 \\ K_\text{overall} = K_1 \times K_2 \end{gathered}

This is the same idea as Hess’s law and the properties of K (U6.9, U7.6). As long as ΔG°₁ + ΔG°₂ < 0, the overall reaction is favorable. One way to picture it: the driving reaction keeps using up the intermediate, which keeps pulling the target reaction’s equilibrium forward.

The common intermediate is the key. If two reactions share no substance, putting them in the same container just lets each proceed on its own; the free energy released by one can’t be passed to the other.

Choose a target reaction and a driving reaction, and see whether they can be coupled and whether the result is favorable (ΔfG° values from OpenStax Appendix G; Ksp and Kf from Appendices J and K):

Try it: driving an unfavorable reaction by coupling

Besides coupling to another reaction, an unfavorable reaction can be driven by an outside energy source: light (as in photosynthesis) or electrical energy (electrolysis, U9.10). Living things couple many reactions to the hydrolysis of ATP.

Example 1: reducing copper(I) oxide with hydrogen

At 25 °C, Cu₂O(s) → 2Cu(s) + ½O₂(g) has ΔG° = +146.0 kJ, and H₂(g) + ½O₂(g) → H₂O(l) has ΔG° = −237.1 kJ. Find ΔG° and K for Cu₂O(s) + H₂(g) → 2Cu(s) + H₂O(l), and explain why it is favorable.

Step-by-step solution

  1. Find the common intermediate. O₂ is formed in the first reaction and used up in the second, so it cancels when they are added:

    • ① Cu₂O(s) → 2Cu(s) + ½O₂(g), ΔG° = +146.0
    • ② H₂(g) + ½O₂(g) → H₂O(l), ΔG° = −237.1
    • ① + ②: Cu₂O(s) + H₂(g) → 2Cu(s) + H₂O(l)

    All values in kJ.

  2. Add the ΔG° values. When equations add, so do their ΔG° values:

    ΔG∘=+146.0+(−237.1)=−91.1 kJ\begin{aligned} \Delta G^\circ &= +146.0 + (-237.1) \\ &= \boxed{-91.1\ \text{kJ}} \end{aligned}

  3. Find K. ΔG° in J, T = 298 K:

    ln⁡K=9.11×1048.314×298=36.8K=e36.8≈9×1015\begin{aligned} \ln K &= \frac{9.11 \times 10^4}{8.314 \times 298} = 36.8 \\ K &= e^{36.8} \approx \boxed{9 \times 10^{15}} \end{aligned}

  4. Check. ΔG° < 0 goes with K ≫ 1. Consistent.

  5. Explain why it is favorable. Decomposing Cu₂O on its own is unfavorable (ΔG° > 0); H₂ keeps using up the O₂ that the decomposition produces, linking a very favorable reaction to the unfavorable one, so the overall ΔG° is negative and Cu₂O is reduced to copper.

Example 2: no common intermediate, no coupling

A student wants to dissolve AgCl (Ksp = 1.6 × 10⁻¹⁰). The student puts AgCl in a beaker and adds equal amounts of HCl and NaOH solutions, reasoning that the very favorable neutralization H⁺ + OH⁻ → H₂O (K = 1.0 × 10¹⁴) will “drive” the AgCl to dissolve. Is this reasoning correct? How could it be improved?

Step-by-step solution

  1. Identify: what coupling needs. The two reactions must share an intermediate: one forms it and the other uses it up, so the favorable reaction can “pull” the unfavorable one.

  2. Check for a shared substance.

    • Dissolving AgCl: forms Ag⁺ and Cl⁻
    • Neutralization: uses H⁺ and OH⁻

    They share no substance.

  3. Judge the reasoning. The two reactions proceed separately, and the free energy from neutralization can’t drive AgCl to dissolve: the reasoning is wrong. In fact the Cl⁻ from the HCl (a common ion) makes AgCl even less soluble.

  4. Improvement: pick a reaction that uses up Ag⁺. For example, add aqueous ammonia:

    AgX++2 NHX3⇌[Ag(NHX3)X2]X+\ce{Ag+ + 2NH3 <=> [Ag(NH3)2]+}

    This reaction and the dissolving of AgCl share Ag⁺ as a common intermediate.

  5. Connect to the effect. NH₃ keeps binding Ag⁺, so [Ag⁺] stays very small and the dissolving equilibrium keeps shifting forward; AgCl dissolves far more in aqueous ammonia than in water.

Practice: what happens when these two reactions are coupled?
  1. A → B (ΔG° = +30 kJ); B → C (ΔG° = −50 kJ)
  2. A → B (ΔG° = +30 kJ); B → C (ΔG° = −10 kJ)
  3. A → B (ΔG° = +30 kJ); X → Y (ΔG° = −100 kJ)
  4. CuO(s) → Cu(s) + ½O₂(g) (+129.7 kJ); H₂(g) + ½O₂(g) → H₂O(l) (−237.1 kJ)
  5. Fe₂O₃(s) → 2Fe(s) + ³⁄₂O₂(g) (+742.2 kJ); 3H₂(g) + ³⁄₂O₂(g) → 3H₂O(l) (−711.3 kJ)
  6. N₂(g) + O₂(g) → 2NO(g) (+175.2 kJ); 2NO(g) + O₂(g) → 2NO₂(g) (−72.6 kJ)
  7. AgCl dissolving; H⁺ + OH⁻ → H₂O
  8. K₁ = 1 × 10⁻⁵; K₂ = 1 × 10³ (common intermediate)

Common mistakes

Mistake 1: “Any very favorable reaction in the same container can drive an unfavorable one.” The reactions must share a common intermediate for free energy to be “passed along.”

Mistake 2: “When reactions are coupled, ΔG° values add, so K values add too.” Adding ΔG° values corresponds to multiplying K values, because ΔG° = −RT ln K and ln K₁ + ln K₂ = ln(K₁K₂).

Mistake 3: “Coupling makes the target reaction’s own ΔG° negative.” The target reaction’s ΔG° doesn’t change; what is favorable is a new overall reaction, whose products include those of the driving reaction.

Quick check

Reaction 1 has ΔG° = +25 kJ and reaction 2 has ΔG° = −40 kJ, and they share a common intermediate. What is ΔG° for the overall coupled reaction?

Two reactions that can be coupled have K₁ = 2.0 × 10⁻⁴ and K₂ = 5.0 × 10⁶. What is K for the overall reaction?

Which pair of reactions can be coupled?

AgCl is nearly insoluble in water but dissolves in aqueous ammonia, mainly because:

A target reaction has ΔG° = +60 kJ. For coupling to make the overall reaction favorable, ΔG° for the driving reaction must be:

Many unfavorable syntheses in cells happen because they are coupled to ATP hydrolysis. What is required for this coupling to work?

AP-style free response

AP-style free response

Chromium metal can be made by the aluminothermic reaction Cr₂O₃(s) + 2Al(s) → 2Cr(s) + Al₂O₃(s). Data (25 °C, OpenStax Appendix G): ΔfG° (kJ/mol) Cr₂O₃(s) −1058.1, Al₂O₃(s) −1582, H₂O(l) −237.1.

(a) Treat this reaction as two coupled reactions. Write the two reactions and identify the common intermediate.

(b) Calculate ΔG° for each of the two reactions in (a) and for the overall reaction.

(c) A student suggests using hydrogen instead of aluminum: Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l). Calculate ΔG° for this reaction at 25 °C, and explain why aluminum can reduce Cr₂O₃ but hydrogen can’t.

(d) Estimate the order of magnitude of K for the aluminothermic reaction at 25 °C.

(e) A mixture of Cr₂O₃ and aluminum powder can sit at room temperature for a long time without reacting; it has to be ignited with a burning magnesium ribbon. With such a negative ΔG°, why is ignition needed?

Model answer and scoring

(a) Cr₂O₃(s) → 2Cr(s) + ³⁄₂O₂(g); 2Al(s) + ³⁄₂O₂(g) → Al₂O₃(s). The common intermediate is O₂.

(b) First: +1058.1 kJ; second: −1582 kJ; overall: +1058.1 + (−1582) = −523.9 kJ.

(c) For 3H₂(g) + ³⁄₂O₂(g) → 3H₂O(l), ΔG° = 3 × (−237.1) = −711.3 kJ, so the overall ΔG° = +1058.1 − 711.3 = +346.8 kJ > 0, which is unfavorable. The driving reaction’s ΔG° must be more negative than −1058.1 kJ; forming Al₂O₃ (−1582 kJ) meets this, but forming water (−711.3 kJ) doesn’t.

(d) log K = 523 900 ÷ (2.303 × 8.314 × 298) ≈ 92, so K is about 10⁹².

(e) The reaction is under kinetic control: reactions between solids have very high activation energies, so at room temperature the rate is extremely slow. The burning magnesium provides the high temperature to start the reaction, and the large amount of heat the reaction releases keeps it going (U9.4).

Scoring: (a) both reactions and the intermediate; (b) all three ΔG° values; (c) calculation and comparison; (d) order of magnitude of K; (e) high activation energy, ignition needed.

Summary

  • Coupling: an unfavorable reaction and a favorable one are added through a common intermediate; the overall ΔG° = ΔG°₁ + ΔG°₂ and K = K₁ × K₂.
  • If the overall ΔG° < 0, the driving reaction can drive the target reaction; two reactions with no shared substance can’t be coupled.
  • Light and electrical energy can also drive reactions with ΔG° > 0.

AP tip: To explain coupling, state that the two reactions add, the common intermediate cancels, the ΔG° values add (or K values multiply), and the overall ΔG° < 0 (or K > 1). Saying only “there is a favorable reaction” isn’t enough.

Practice this chapter: Unit 9 · Thermodynamics and Electrochemistry · 12 questionsPractice this topic: U9.6 · Coupled Reactions · 5 questions

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