Dove's LibraryDove's Library

9.7 - Coupled Reactions

Overview

Plenty of useful reactions are thermodynamically unfavored: splitting water into hydrogen and oxygen, turning carbon dioxide into sugar, pulling copper metal out of its ore. They still happen. This topic covers the two ways to make them happen: supply energy from outside, or couple the unfavored reaction to a favored one so the overall ΔG∘\Delta G^\circ is negative.

Driving a reaction with outside energy

A process with ΔG∘>0\Delta G^\circ > 0 won't make much product on its own (K<1K < 1). An external source of energy can push it anyway. Under standard conditions, the minimum energy you have to supply is ΔG∘\Delta G^\circ.

Electrical energy

In an electrolytic cell (Topic 9.8), a power supply pushes electrons through a circuit to drive an unfavored redox reaction. The electrolysis of water is one example:
2H2O(l)→2H2(g)+O2(g)ΔG∘=+474.2 kJ/molrxn\mathrm{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \qquad \Delta G^\circ = +474.2\ \mathrm{kJ/mol_{rxn}}
Charging a battery works the same way. When a battery powers your phone, a favored redox reaction pushes electrons through the circuit. The charger forces that reaction to run backward, storing energy that the battery can release later.

Light

In photosynthesis, plants turn carbon dioxide and water into glucose:
6CO2(g)+6H2O(l)→C6H12O6(s)+6O2(g)ΔG∘≈+2880 kJ/molrxn\mathrm{6CO_2(g) + 6H_2O(l) \rightarrow C_6H_{12}O_6(s) + 6O_2(g)} \qquad \Delta G^\circ \approx +2880\ \mathrm{kJ/mol_{rxn}}
ΔG∘\Delta G^\circ is hugely positive. The energy comes from sunlight: chlorophyll absorbs light, and that energy drives a long chain of reactions that ends in glucose. The reverse, breaking glucose down to CO2\mathrm{CO_2} and water in respiration, is the favored direction.

Example 1: The minimum energy to make hydrogen

Hydrogen fuel can be made by electrolysis of water. What is the minimum energy needed to make 1.00 kg of H2(g)\mathrm{H_2(g)} at 25 °C? For H2O(l)\mathrm{H_2O(l)}, ΔGf∘=−237.1\Delta G_f^\circ = -237.1 kJ/mol.
Step 1: Energy per mole of H2\mathrm{H_2}. The reaction above has ΔG∘=+474.2 kJ/molrxn\Delta G^\circ = +474.2\ \mathrm{kJ/mol_{rxn}} and makes 2 mol H2\mathrm{H_2}, so each mole of H2\mathrm{H_2} needs 237.1 kJ.
Step 2: Moles of H2\mathrm{H_2}, with M=2.016\mathcal{M} = 2.016 g/mol:
n=1000 g2.016 g/mol=496.0 mol H2n = \frac{1000\ \mathrm{g}}{2.016\ \mathrm{g/mol}} = 496.0\ \mathrm{mol\ H_2}
Step 3: Multiply:
(496.0 mol)(237.1 kJ/mol)=1.18×105 kJ(496.0\ \mathrm{mol})(237.1\ \mathrm{kJ/mol}) = 1.18 \times 10^{5}\ \mathrm{kJ}
Answer: at least 1.18×1051.18 \times 10^{5} kJ (about 118 MJ, or 33 kWh). Real electrolyzers need more than this, because some of the electrical energy is always lost as heat.

Coupled reactions

The other strategy is to pair the unfavored reaction with a favored one. When reactions add, their ΔG∘\Delta G^\circ values add, the same way ΔH∘\Delta H^\circ values do in Hess's law (Topic 6.9). So the overall reaction can have ΔG∘<0\Delta G^\circ < 0 even though one of its pieces doesn't.
Coupled reactions share one or more common intermediates: species made by one reaction and used up by the other. When you add the equations, the common intermediates cancel, and what's left is the overall reaction you want. Since ΔG∘\Delta G^\circ values add, the equilibrium constants multiply (Topic 7.6):
ΔGoverall∘=ΔG1∘+ΔG2∘Koverall=K1×K2\Delta G^\circ_\mathrm{overall} = \Delta G^\circ_1 + \Delta G^\circ_2 \qquad\qquad K_\mathrm{overall} = K_1 \times K_2

ATP in living cells

Cells run on coupled reactions. The favored partner is usually the breakdown of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and inorganic phosphate, written Pi\mathrm{P_i}:
ATP+H2O→ADP+PiΔG∘=−30.5 kJ/molrxn\text{ATP} + \mathrm{H_2O} \rightarrow \text{ADP} + \mathrm{P_i} \qquad \Delta G^\circ = -30.5\ \mathrm{kJ/mol_{rxn}}
The first step in using glucose for energy attaches a phosphate group to it, making glucose 6-phosphate. On its own, that step is unfavored:
glucose+Pi→glucose 6-phosphate+H2OΔG∘=+13.8 kJ/molrxn\text{glucose} + \mathrm{P_i} \rightarrow \text{glucose 6-phosphate} + \mathrm{H_2O} \qquad \Delta G^\circ = +13.8\ \mathrm{kJ/mol_{rxn}}
Add the two reactions. Pi\mathrm{P_i} and H2O\mathrm{H_2O} show up on both sides, so they cancel:
glucose+ATP→glucose 6-phosphate+ADPΔG∘=+13.8+(−30.5)=−16.7 kJ/molrxn\text{glucose} + \text{ATP} \rightarrow \text{glucose 6-phosphate} + \text{ADP} \qquad \Delta G^\circ = +13.8 + (-30.5) = -16.7\ \mathrm{kJ/mol_{rxn}}
(These ΔG∘\Delta G^\circ values are the ones biochemists use, for a standard state at pH 7.)
9.7 - Coupled Reactions
The diagram shows the bookkeeping: up 13.8 kJ/molrxn13.8\ \mathrm{kJ/mol_{rxn}} for reaction 1, down 30.5 kJ/molrxn30.5\ \mathrm{kJ/mol_{rxn}} for reaction 2, a net drop of 16.7 kJ/molrxn16.7\ \mathrm{kJ/mol_{rxn}}. In the cell, an enzyme moves the phosphate group straight from ATP to glucose, so free phosphate never actually forms. Adding the two equations is how you find ΔG∘\Delta G^\circ for that one-step reaction.
The equilibrium constants at 25 °C show what coupling does. On its own, reaction 1 has K=3.8×10−3K = 3.8 \times 10^{-3}, so reactants are strongly favored. ATP breakdown has K=2.2×105K = 2.2 \times 10^{5}. The coupled reaction has K=(3.8×10−3)(2.2×105)=8.4×102K = (3.8 \times 10^{-3})(2.2 \times 10^{5}) = 8.4 \times 10^{2}, so products are favored.

Example 2: Copper from its ore

Copper(I) sulfide, Cu2S\mathrm{Cu_2S}, is a common copper ore. Breaking it down is unfavored:
Cu2S(s)→2Cu(s)+S(s)ΔG∘=+86.2 kJ/molrxn\mathrm{Cu_2S(s) \rightarrow 2Cu(s) + S(s)} \qquad \Delta G^\circ = +86.2\ \mathrm{kJ/mol_{rxn}}
Smelters blow air through the hot ore, which couples that step to the burning of sulfur:
S(s)+O2(g)→SO2(g)ΔG∘=−300.2 kJ/molrxn\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)} \qquad \Delta G^\circ = -300.2\ \mathrm{kJ/mol_{rxn}}
(a) Write the overall reaction and find its ΔG∘\Delta G^\circ. (b) Identify the common intermediate.
(a) Add the two reactions. S(s)\mathrm{S(s)} is made in the first and used in the second, so it cancels:
Cu2S(s)+O2(g)→2Cu(s)+SO2(g)ΔG∘=+86.2+(−300.2)=−214.0 kJ/molrxn\mathrm{Cu_2S(s) + O_2(g) \rightarrow 2Cu(s) + SO_2(g)} \qquad \Delta G^\circ = +86.2 + (-300.2) = -214.0\ \mathrm{kJ/mol_{rxn}}
(b) Sulfur. The overall reaction is strongly favored, which is why heating sulfide ores in air works (and why smelters have to capture the SO2\mathrm{SO_2} they make).

Example 3: Iron from iron ore (AP-style)

Two reactions at 25 °C:
(1)2Fe2O3(s)→4Fe(s)+3O2(g)ΔG∘=+1484.4 kJ/molrxn\text{(1)}\quad \mathrm{2Fe_2O_3(s) \rightarrow 4Fe(s) + 3O_2(g)} \qquad \Delta G^\circ = +1484.4\ \mathrm{kJ/mol_{rxn}}(2)2CO(g)+O2(g)→2CO2(g)ΔG∘=−514.4 kJ/molrxn\text{(2)}\quad \mathrm{2CO(g) + O_2(g) \rightarrow 2CO_2(g)} \qquad \Delta G^\circ = -514.4\ \mathrm{kJ/mol_{rxn}}
In a blast furnace, carbon monoxide turns iron(III) oxide into iron:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)\mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)}
(a) Show how to combine reactions (1) and (2) to get this reaction, and calculate its ΔG∘\Delta G^\circ. (b) Identify the common intermediate. (c) Calculate KK at 25 °C. (d) A student says iron can't be made from Fe2O3\mathrm{Fe_2O_3} because reaction (1) is not thermodynamically favored. Explain the flaw in this reasoning.
(a) The target has one Fe2O3\mathrm{Fe_2O_3}, so take 12\tfrac{1}{2} of reaction (1). It has three CO, so take 32\tfrac{3}{2} of reaction (2). Multiplying a reaction multiplies its ΔG∘\Delta G^\circ by the same number:
Fe2O3(s)→2Fe(s)+32O2(g)ΔG∘=12(+1484.4)=+742.2 kJ/molrxn\mathrm{Fe_2O_3(s) \rightarrow 2Fe(s) + \tfrac{3}{2}O_2(g)} \qquad \Delta G^\circ = \tfrac{1}{2}(+1484.4) = +742.2\ \mathrm{kJ/mol_{rxn}}3CO(g)+32O2(g)→3CO2(g)ΔG∘=32(−514.4)=−771.6 kJ/molrxn\mathrm{3CO(g) + \tfrac{3}{2}O_2(g) \rightarrow 3CO_2(g)} \qquad \Delta G^\circ = \tfrac{3}{2}(-514.4) = -771.6\ \mathrm{kJ/mol_{rxn}}ΔGoverall∘=+742.2+(−771.6)=−29.4 kJ/molrxn\Delta G^\circ_\mathrm{overall} = +742.2 + (-771.6) = -29.4\ \mathrm{kJ/mol_{rxn}}
(b) O2\mathrm{O_2} is made in the first step and used up in the second, so it cancels. Oxygen is the common intermediate.
(c)
K=e−ΔG∘/RT=e−(−29,400)/[(8.314)(298)]=e11.87=1.4×105K = e^{-\Delta G^\circ/RT} = e^{-(-29{,}400)/[(8.314)(298)]} = e^{11.87} = 1.4 \times 10^{5}
(d) Reaction (1) never has to happen by itself. Coupled with the oxidation of CO, which is strongly favored, the overall reaction has ΔG∘<0\Delta G^\circ < 0 and K>1K > 1, so products are favored. (Blast furnaces run hot so the reaction goes at a useful rate, which is a kinetics issue, Topic 9.4.)

Common mistakes

  • Saying an unfavored reaction "can't happen." It can, with outside energy or by coupling.
  • Adding ΔG∘\Delta G^\circ values without scaling them. Multiply a reaction by a number, and multiply its ΔG∘\Delta G^\circ by the same number. Reverse it, and flip the sign.
  • Adding K values. When reactions add, ΔG∘\Delta G^\circ values add but KK values multiply.
  • Leaving the common intermediate in the overall equation. It should cancel.
  • Saying energy is stored in ATP's phosphate bond and released when it breaks. Breaking any bond takes energy (Topic 6.7). ATP breakdown is favored because of the whole reaction: the products are lower in free energy than ATP and water.

AP® is a registered trademark of the College Board, which is not affiliated with Dove's Library and was not involved in the production of, and does not endorse, this website.

Providing free, high-quality AP notes to help students succeed, earn college credit, and reach their educational goals, ensuring everyone has access to the resources they need.

© 2026 Standard Learning Group, Inc., which operates Dove's Library. All rights reserved.