Unit 9 · Topic 9.7 Beta

Coupled Reactions

An unfavored reaction can be made to proceed by coupling it to a strongly favored one through a shared intermediate.

Practice 4: Model AnalysisPractice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

Your muscles build proteins, pump ions and contract, and almost every one of those jobs is thermodynamically unfavored on its own. The cell gets them done by pairing each one with a reaction that is strongly favored, most often the breakdown of ATP. Smelters use the same trick to get copper out of ore.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. When two reactions are added, what happens to their K values?

  1. They multiply
  2. They add
  3. They subtract
  4. The larger one is kept
Show the answer

Adding reactions multiplies equilibrium constants (topic 7.6).

  • Correct: They multiply:
  • They add:
  • They subtract:
  • The larger one is kept:

2. A reaction has ΔG° = +40 kJ/mol. Is it favored?

  1. No, because ΔG° > 0
  2. Yes, because ΔG° is large
  3. Yes, at every temperature
  4. It cannot be known
Show the answer

A process is favored when ΔG° < 0.

  • Correct: No, because ΔG° > 0:
  • Yes, because ΔG° is large:
  • Yes, at every temperature:
  • It cannot be known:

Part 4 · See it

See it first

Bar chart of ΔG° in kJ/mol: the unfavored step rises to +86, the favored step falls to −300, and their sum, the overall coupled reaction, falls to −214, so the coupled process is favored.
An unfavored step (+86 kJ/mol) and a favored step (−300 kJ/mol) add to an overall ΔG° of −214 kJ/mol. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. An unfavored reaction has ΔG° > 0it does not proceed to a useful extent on its own
  2. A second reaction consumes a product of the first (a shared intermediate)the two reactions add into one overall process
  3. ΔG° is a state functionthe ΔG° values add, and the K values multiply
  4. The driving reaction is more favored than the driven one is unfavoredthe overall ΔG° is negative and the coupled process is favored

Part 6 · Key ideas

Key ideas

  • A coupled reaction pairs an unfavored reaction (ΔG° > 0) with a favored one (ΔG° < 0) through a shared intermediate.
  • When reactions add, their ΔG° values add and their K values multiply.
  • The overall process is favored if ΔG°(total) < 0. Neither step’s own ΔG° changes.
  • Examples: Cu₂S + O₂ (smelting), ATP hydrolysis driving cell reactions.
  • External energy (electric current, light) can also drive a process with ΔG° > 0.

Part 7 · Misconception

A common mistake

The wrong idea: Coupling a favored reaction to an unfavored one changes the unfavored reaction’s ΔG° to a negative value.

What actually happens: Each reaction keeps its own ΔG°. Coupling makes them act as one overall process, and it is the sum, ΔG°₁ + ΔG°₂, that is negative.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Getting copper from its ore

Copper can be obtained from the mineral chalcocite, Cu₂S. Data at 298 K:

Two reactions and their standard free energy changes at 298 K
ReactionEquationΔG° (kJ/mol)
1Cu₂S(s) → 2 Cu(s) + S(s)+86.2
2S(s) + O₂(g) → SO₂(g)−300.1

1. Is Reaction 1 thermodynamically favored at 298 K on its own?

  1. Yes: it makes three moles of product from one mole of reactant.
  2. No: ΔG° > 0, so heating Cu₂S by itself does not give copper.
  3. Yes: decompositions like this are favored at room temperature.
  4. No: it is favored, but too slow to observe.
Show the answer

ΔG° = +86.2 kJ/mol > 0: not favored on its own.

  • Yes: it makes three moles of product from one mole of reactant.: Counting moles of solid does not decide favorability; ΔG° does, and it is positive.
  • Correct: No: ΔG° > 0, so heating Cu₂S by itself does not give copper.: Right: ΔG° = +86.2 kJ/mol, so the decomposition is not favored.
  • Yes: decompositions like this are favored at room temperature.: Many decompositions are not favored (this one, or water into H₂ and O₂).
  • No: it is favored, but too slow to observe.: That would describe kinetic control, which needs ΔG° < 0. Here ΔG° > 0.

2. Calculate ΔG° for the overall reaction Cu₂S(s) + O₂(g) → 2 Cu(s) + SO₂(g).

Type a number and its unit.

Show the answer

Adding Reaction 1 and Reaction 2 cancels S(s), the shared intermediate. ΔG° adds: +86.2 + (−300.1) = −213.9 kJ/mol.

  • Answer: -213.9 kJ/mol

3. Why does coupling Reaction 1 with Reaction 2 allow copper to be produced?

  1. Reaction 2 releases heat that changes ΔG° of Reaction 1 to a negative value.
  2. Reaction 2 acts as a catalyst for Reaction 1, lowering its activation energy.
  3. Oxygen makes Cu₂S less stable, so Reaction 1 becomes favored without Reaction 2.
  4. They share S(s): Reaction 2 uses the S that Reaction 1 makes; total ΔG° < 0.
Show the answer

Coupling requires a common intermediate. When the steps add, ΔG° values add, and the large negative ΔG° of Reaction 2 carries Reaction 1.

  • Reaction 2 releases heat that changes ΔG° of Reaction 1 to a negative value.: Coupling does not change ΔG° of either reaction; the ΔG° values add.
  • Reaction 2 acts as a catalyst for Reaction 1, lowering its activation energy.: A catalyst cannot make an unfavored reaction favored, and Reaction 2 is not a catalyst.
  • Oxygen makes Cu₂S less stable, so Reaction 1 becomes favored without Reaction 2.: Reaction 1 has ΔG° = +86.2 kJ/mol whatever else is present; it is the overall reaction that is favored.
  • Correct: They share S(s): Reaction 2 uses the S that Reaction 1 makes; total ΔG° < 0.: Right: a shared intermediate links them, and the sum of ΔG° values is −213.9 kJ/mol.

Model

Using ATP in a cell

Cells store energy in ATP. The first step of breaking down glucose attaches a phosphate group to glucose. Standard free energy changes at pH 7 (the standard state biochemists use), taken here for 37 °C (310 K):

Step A: glucose + phosphate → glucose-6-phosphate + H₂O; ΔG° = +13.8 kJ/mol
Step B: ATP + H₂O → ADP + phosphate; ΔG° = −30.5 kJ/mol
Overall: glucose + ATP → glucose-6-phosphate + ADP

glucose + phosphateglucose-6-phosphateA: +13.8ATPADP + phosphateB: −30.5phosphate passed on

Key: arrows show each step; the number on each is ΔG° in kJ/mol.

4. Calculate ΔG° for the overall coupled reaction, glucose + ATP → glucose-6-phosphate + ADP.

Type a number and its unit.

Show the answer

ΔG°(overall) = ΔG° of Step A + ΔG° of Step B = +13.8 + (−30.5) = −16.7 kJ/mol. The water and phosphate cancel when the steps are added.

  • Answer: -16.7 kJ/mol

5. Which statement best explains how the cell makes Step A happen?

  1. It adds an enzyme that makes ΔG° for Step A negative.
  2. It raises the temperature to 37 °C, which makes Step A favored on its own.
  3. It couples Step A to ATP hydrolysis, so the overall ΔG° is negative.
  4. It removes glucose-6-phosphate, so ΔG for Step A becomes zero.
Show the answer

An unfavored step is driven by pairing it with a strongly favored one through a common intermediate (here, the phosphate group).

  • It adds an enzyme that makes ΔG° for Step A negative.: Enzymes speed reactions but cannot change ΔG°. The coupling is what makes the overall process favored.
  • It raises the temperature to 37 °C, which makes Step A favored on its own.: Step A has ΔG° = +13.8 kJ/mol at 37 °C; it is still not favored alone.
  • Correct: It couples Step A to ATP hydrolysis, so the overall ΔG° is negative.: Right: the coupled process has ΔG° = −16.7 kJ/mol, so it is favored.
  • It removes glucose-6-phosphate, so ΔG for Step A becomes zero.: Lowering the product concentration helps a little, but the main driver here is coupling to ATP.

6. If a cell paired Step A with a reaction that had ΔG° = −10.0 kJ/mol instead of ATP hydrolysis (with a suitable shared species), would the overall process be favored?

  1. Yes: any favored reaction can drive any unfavored one.
  2. Yes: the two ΔG° values multiply to a negative number.
  3. No: +13.8 + (−10.0) = +3.8 kJ/mol, still positive.
  4. No: coupled reactions need ΔG values of exactly equal size.
Show the answer

Coupling works only if ΔG°(driving) + ΔG°(driven) < 0.

  • Yes: any favored reaction can drive any unfavored one.: The sum of ΔG° values must be negative. Here it is positive.
  • Yes: the two ΔG° values multiply to a negative number.: ΔG° values add when reactions are coupled; K values multiply.
  • Correct: No: +13.8 + (−10.0) = +3.8 kJ/mol, still positive.: Right: the driving reaction must be more favored than the driven one is unfavored.
  • No: coupled reactions need ΔG values of exactly equal size.: They need not be equal; the driving reaction just has to outweigh the driven one.

7. Reaction 1: X → Y, ΔG° = +25.0 kJ/mol. Reaction 2: Y → Z, ΔG° = −40.0 kJ/mol. Calculate ΔG° for X → Z.

Type a number and its unit.

Show the answer

Y is made in reaction 1 and used in reaction 2, so the reactions add to X → Z. ΔG° = +25.0 + (−40.0) = −15.0 kJ/mol.

  • Answer: -15.0 kJ/mol

Part 9 · Summary

Summary

An unfavored reaction can be made to proceed by coupling it to a strongly favored one through a shared intermediate. The ΔG° values add (and K values multiply), so the overall process is favored if the total ΔG° is negative. External energy such as electricity or light can also drive an unfavored process.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections