Coupling reactions — driving unfavorable reactions: Visual walkthrough

We build everything from one single idea: a quantity called free energy that only ever wants to go downhill. Let's meet it.

Definition Two flavours of

$Δ G$ — read this before Step 1 Throughout this page you will meet two closely related symbols, and it matters which is which:

$Δ G$ (no circle) — the free-energy change of a reaction as it is actually running, at the real amounts present. This is the quantity that decides direction moment-to-moment (Steps 1–5, and the metallurgy numbers in Step 7, which are quoted at a stated temperature).
$Δ G^{\circ}$ (with a circle $^{\circ}$ , read "standard") — the free-energy change under a fixed reference condition (defined fully in Step 6). This is a fixed constant for a reaction and is the one that maps onto the equilibrium constant $K$ (Steps 6, and the ATP/glucose numbers in Step 7).

Both add for coupled reactions for the same state-function reason. We will flag which flavour is in play at each step so you never lose track.

Step 1 — What "downhill" even means

WHAT. Every chemical mixture has a number attached to it called its Gibbs free energy, written $G$ . Read it as "how much useful push is stored, given the current temperature and pressure." A reaction converts a starting mixture into a finished mixture, so it moves $G$ from one value to another. The change is what matters:

$Δ G = G_{after} - G_{before}$

$Δ G$ (the triangle $Δ$ means "change in") — the whole quantity we track. No circle here: this is the actual free-energy change.
$G_{after}$ — free energy of the products (the finish line).
$G_{before}$ — free energy of the reactants (the start line).

WHY. We need a single number that decides which way a reaction runs. Nature's rule (proved in Gibbs Free Energy) is beautifully simple: a reaction runs forward only if it goes downhill in $G$ , i.e.

$Δ G < 0 (spontaneous, "wants to go")$

PICTURE. Think of $G$ as height on a landscape. Reactants sit at one height, products at another. A ball rolls only downhill.

Figure — Coupling reactions — driving unfavorable reactions

Step 2 — The two players: an uphill reaction and a downhill one

WHAT. We take two reactions.

$(1) A \to B, Δ G_{1} > 0 (uphill, won’t go alone)$ $(2) B \to C, Δ G_{2} < 0 (downhill, wants to go)$

$A, B, C$ — three chemical states (real molecules).
$Δ G_{1} > 0$ — reaction (1) climbs; its products are higher than its reactants.
$Δ G_{2} < 0$ — reaction (2) drops; its products are lower.

WHY. This is exactly the "weak person + falling rock" from the parent note. Reaction (1) is the box we can't lift. Reaction (2) is the heavy weight that wants to fall. Notice the letter $B$ appears in both — it is the finish of (1) and the start of (2).

PICTURE. Two separate slopes drawn on the same height axis: one going up (red), one going down (green). They touch at $B$ .

Step 3 — Why we are allowed to ADD (state-function magic)

WHAT. Line up the two reactions so $B$ appears on both sides and cancel it, exactly like cancelling in Hess's Law:

A B A \to \to \to B C C

The $B$ produced by (1) is the same $B$ consumed by (2), so it disappears, leaving one net reaction $A \to C$ .

WHY. Because $G$ is a state function — its value depends only on where you are, never on how you got there. So the total drop from $A$ to $C$ is just the sum of the two drops:

$Δ G_{total} = A \to B Δ G_{1} + B \to C Δ G_{2}$

$Δ G_{total}$ — the height change for the whole journey $A \to C$ .
$Δ G_{1}$ — the uphill leg $A \to B$ (a positive contribution, works against us).
$Δ G_{2}$ — the downhill leg $B \to C$ (a negative contribution, works for us).

PICTURE. Stack the two legs end-to-end. The height of $A$ and the height of $C$ are all that matter; $B$ 's height is just a waypoint the path passes through. (This is the logic of Hess's Law.)

Step 4 — When does the net path actually run downhill?

First, the $∣ \dots ∣$ symbol. The two vertical bars around a number mean its absolute value — its plain size, throwing away any minus sign. So $∣ - 360∣ = 360$ and $∣ + 7 ∣ = 7$ . We use it to talk about "how big is the drop" without the sign getting in the way.

WHAT. Apply the one rule from Step 1 to the net reaction:

$Δ G_{total} < 0 ⟺ Δ G_{1} + Δ G_{2} < 0$

Rearrange. Since $Δ G_{1} > 0$ and $Δ G_{2} < 0$ , write $Δ G_{2} = - ∣Δ G_{2} ∣$ — that is, " $Δ G_{2}$ is a negative number, so it equals minus its own size":

$Δ G_{1} - ∣Δ G_{2} ∣ < 0 ⟺ ∣Δ G_{2} ∣ > Δ G_{1}$

$∣Δ G_{2} ∣$ — the size of the downhill drop (how heavy the falling weight is), a positive number.
$Δ G_{1}$ — the size of the uphill climb (how heavy the box is), already positive.

WHY. The falling weight must release more energy than the box demands. Only the surplus is what actually drags the box up. If they were exactly equal, the system sits still (net $Δ G = 0$ , equilibrium).

PICTURE. Two vertical bars side by side — a red "cost" bar of height $Δ G_{1}$ and a green "payout" bar of height $∣Δ G_{2} ∣$ . The net is the green minus the red. Three cases: green taller (goes), equal (stuck), red taller (fails).

Step 5 — The degenerate cases (never leave a gap)

WHAT. We must show every outcome, not just the winning one.

Case	Relationship	Net $Δ G_{total}$	Result
Win	$	\Delta G_2	> \Delta G_1$
Tie	$	\Delta G_2	= \Delta G_1$
Lose	$	\Delta G_2	< \Delta G_1$
No link	$B$ not shared	has no chemical meaning	you may not couple

WHY. The last row is the trap — and it needs care. You can physically write down two numbers and add them; arithmetic never stops you. But if the two reactions share no real molecule, there is no single path $A \to C$ , nothing cancels in Step 3, and the sum $Δ G_{1} + Δ G_{2}$ describes no actual transformation. It is a number without a reaction attached — chemically meaningless. Each reaction simply keeps its own separate $Δ G$ and its own separate fate. Coupling requires a genuine shared intermediate; without one, the "total" is bookkeeping, not chemistry.

PICTURE. Four mini-landscapes side by side: falling weight wins, ties, loses — and the fourth where the rope is cut (no pulley), so the box stays put no matter how heavy the weight.

Step 6 — From adding $Δ G$ to multiplying $K$

WHAT. Every reaction also has an equilibrium constant $K$ — a number saying how far it goes (big $K$ = lots of product, tiny $K$ = barely any product). It links to $Δ G^{\circ}$ by the bridge from Relation between ΔG and K:

$Δ G^{\circ} = - R T ln K$

$Δ G^{\circ}$ — the standard free-energy change. The little circle $^{\circ}$ means "measured under a fixed reference condition": every gas at $1 bar$ pressure, every dissolved species at $1 M$ concentration, pure solids/liquids in their normal state, at a stated temperature. It is a fixed constant for a reaction, unlike the general $Δ G$ (no circle) from Steps 1–5 which changes as the actual amounts shift during the reaction. We use $Δ G^{\circ}$ here precisely because it is fixed, so it maps cleanly onto the constant $K$ .
$R$ — the gas constant, $T$ — absolute temperature; together $R T$ is just a positive scale factor.
$ln K$ — the natural logarithm of $K$ .

Substitute the additive law (standard flavour) into this bridge:

$Δ G_{tot}^{\circ} - R T ln K_{tot} = Δ G_{1}^{\circ} - R T ln K_{1} + Δ G_{2}^{\circ} - R T ln K_{2}$

Divide every term by $- R T$ :

$ln K_{tot} = ln K_{1} + ln K_{2} = ln (K_{1} K_{2})$

WHY. A logarithm turns adding into multiplying: $ln a + ln b = ln (ab)$ . So the additive $Δ G^{\circ}$ becomes a multiplicative $K$ . This is why a hopeless $K_{1} = 1 0^{- 4}$ can be rescued by a giant $K_{2} = 1 0^{7}$ .

PICTURE. A number line in powers of ten (a "log ruler"). Adding the exponents $- 4$ and $+ 7$ lands you at $+ 3$ — i.e. $K_{tot} = 1 0^{3}$ , safely above $1$ .

Recall Check yourself: the

$K$ example $K_{1} = 1 0^{- 4}$ , $K_{2} = 1 0^{7}$ . What is $K_{tot}$ , and does the net run? Multiply: $1 0^{- 4} \times 1 0^{7} = 1 0^{3}$ ::: $K_{tot} = 1 0^{3} ≫ 1$ , so yes, the net reaction is favorable.

Step 7 — Two real couplings on the same picture

WHAT. The identical diagram explains metallurgy and biology. Both sets of numbers below are per mole of reaction as written, in $kJ mol^{- 1}$ . The metallurgy pair uses actual $Δ G$ evaluated at the same temperature $T \approx 1400 K$ (both legs must be quoted at one temperature or you cannot add them); the biology pair uses standard $Δ G^{\circ}$ at $298 K$ .

Metallurgy (see Ellingham Diagram): shared intermediate is oxygen. Both legs are read off the Ellingham diagram at the same temperature $T \approx 1400 K$ : $ZnO \to Zn + \frac{1}{2} O_{2}, Δ G_{1} \approx + 318 kJ mol^{- 1} (at T \approx 1400 K)$ $C + \frac{1}{2} O_{2} \to C O, Δ G_{2} \approx - 236 kJ mol^{- 1} (at T \approx 1400 K)$ Here $Δ G_{2}$ is estimated from $Δ G_{2} \approx Δ H_{2} - T Δ S_{2}$ with $Δ H_{2} \approx - 111 kJ mol^{- 1}$ and $Δ S_{2} \approx + 89 J K^{- 1} mol^{- 1}$ , giving $- 111 - 1400 (0.089) \approx - 236 kJ mol^{- 1}$ . $Δ G_{tot} = + 318 + (- 236) = + 82 kJ mol^{- 1} > 0$ At $1400 K$ this pair has not yet tipped — carbon's payout ( $236$ ) is still smaller than zinc's cost ( $318$ ). The reduction of $ZnO$ by carbon becomes spontaneous only at a higher temperature (near $1200^{\circ} C \approx 1500 K$ ), where the carbon line has fallen far enough that $∣Δ G_{2} ∣ > 318$ . This is exactly why the Ellingham Diagram is read at the crossover temperature, not at a single fixed value.

Biology (see ATP and Bioenergetics): shared intermediate is the phosphate group $P_{i}$ , and both legs are standard values at $298 K$ : $Glucose + P_{i} \to G6P + H_{2} O, Δ G^{\circ} = + 13.8 kJ mol^{- 1}$ $ATP + H_{2} O \to ADP + P_{i}, Δ G^{\circ} = - 30.5 kJ mol^{- 1}$ $Δ G_{net}^{\circ} = + 13.8 + (- 30.5) = - 16.7 kJ mol^{- 1} < 0 ✓$

WHY. Both are just Step 4 with numbers, and a genuine shared molecule ( $O_{2}$ or $P_{i}$ ) lets us add. The biology case wins right away; the metallurgy case shows the honest reality that a coupling can still lose at one temperature and only win once conditions change.

The metallurgy case has a temperature twist worth spelling out. The carbon reaction $C + \frac{1}{2} O_{2} \to C O$ increases the number of gas molecules (roughly $\frac{1}{2}$ mole of $O_{2}$ becomes $1$ mole of $C O$ ), so its entropy change $Δ S_{2}$ is positive. Recall $Δ G = Δ H - T Δ S$ : with $Δ S_{2} > 0$ , the term $- T Δ S_{2}$ grows more negative as $T$ rises, so $Δ G_{2}$ slides steadily downward with temperature (this is exactly the downward-sloping carbon line on the Ellingham Diagram). At $1400 K$ the carbon payout is still too small and coupling loses; only above the crossover temperature does $∣Δ G_{2} ∣$ finally exceed $Δ G_{1}$ and the net tip to negative. The full account of why $Δ G_{2}$ walks downhill with $T$ lives in Entropy and Temperature dependence of ΔG.

PICTURE. Two payout-vs-cost bar pairs side by side — metallurgy (net still positive at $1400 K$ ) and biology (net negative), each with its shared intermediate labelled.

The one-picture summary

Everything above compressed into a single landscape: the box climbs ( $Δ G_{1} > 0$ , red), the weight falls ( $Δ G_{2} < 0$ , green), they are tied at the shared intermediate $B$ , and — when the fall beats the climb — the net drop from $A$ to $C$ is negative and the whole thing runs.

Recall Feynman retelling of the whole walkthrough

Imagine a hilly landscape where "downhill" means a reaction happens by itself. One reaction ( $A \to B$ ) is a hill going up — it can't happen alone. Another ( $B \to C$ ) is a steep drop. Crucially they meet at the same spot $B$ — that shared spot is a rope tying them together. Before you tie them, make sure the same amount of the shared molecule is on both ends (if one reaction needs two, run the other one twice — and double its energy too). Because free energy only cares about where you start and stop (not the wiggles in between), the total trip $A \to C$ just adds the up-part and the down-part. If the drop is deeper than the climb is tall, the whole trip is downhill and everything moves — the falling half literally drags the climbing half along. But if the drop is not deep enough (like carbon reducing zinc oxide at only 1400 K), the trip is still uphill and nothing happens until you change conditions — heat it more and the falling half gets heavier. Switch language from heights to "how far a reaction goes" ( $K$ ) and the adding of heights becomes multiplying of those numbers, because logarithms turn plus into times. That single picture is metallurgy (oxygen is the rope, and heating makes the carbon weight heavier) and biology (ATP's phosphate is the rope). Cut the rope — remove the shared molecule — and the sum means nothing at all, no matter how heavy the weight.

Connections

Parent topic — the full concept note.
Gibbs Free Energy — where the $Δ G < 0$ rule (Step 1) comes from.
Hess's Law — why we may add and cancel the intermediate (Step 3).
Relation between ΔG and K — the $Δ G^{\circ} = - R T ln K$ bridge (Step 6).
Ellingham Diagram & Entropy and Temperature dependence of ΔG — the metallurgy case and its temperature dependence (Step 7).
ATP and Bioenergetics — the biology case (Step 7).