Coupling reactions — driving unfavorable reactions: Question bank

Before we start, five symbols we lean on constantly. Read these once — every trap uses them.

Definition The symbols this page assumes

Common intermediate — a single molecule made by one reaction and consumed by the other, so the two reactions become links in one chain. No such molecule ⇒ no chain ⇒ no coupling, only arithmetic on paper.
$Δ H$ = enthalpy change = the heat released or absorbed by a reaction (negative = gives out heat). $Δ S$ = entropy change = how much the disorder/spread of the system increases (positive = more disorder, e.g. a gas is made). If these two feel shaky, refresh with Entropy and Temperature dependence of ΔG before continuing.
$Δ G$ = Gibbs free energy change = the single quantity deciding direction at constant $T, P$ , built from the two above as $Δ G = Δ H - T Δ S$ . The $- T Δ S$ piece is just "temperature times the disorder gain, subtracted." See Gibbs Free Energy.
$K$ = the equilibrium constant: a single number saying how far a reaction goes before it stops. $K ≫ 1$ means "mostly products, reaction basically completes"; $K ≪ 1$ means "mostly reactants, reaction barely happens"; $K = 1$ means "sits half-and-half." It is tied to free energy by $Δ G^{\circ} = - R T ln K$ — see Relation between ΔG and K.
The little circle in $Δ G^{\circ}$ (read "delta-G-standard") means the free-energy change measured at a fixed reference standard state (1 bar, 1 mol/L, stated $T$ ). Same idea as plain $Δ G$ , just at agreed textbook conditions so numbers are comparable.

Now unpack the master picture. Figure s01 is an energy-level diagram — height = free energy $G$ , and you read a reaction by the drop or climb between levels.

Figure — Coupling reactions — driving unfavorable reactions

And here is why the intermediate is allowed to cancel — the state-function bookkeeping that makes $Δ G$ add. Every "add the $Δ G$ " trap traces back to this figure.

Spot the error

Δ G_{1} = + 20

Δ G_{2} = - 15

, sum

= + 5 > 0

but they share

O_{2}

, so add a bit more heat and it'll go."

The error is hoping heat fixes any deficit; heat only helps if it makes

Δ G_{2}

(or

Δ G_{1}

) more negative via the

- T Δ S

term — you must check the actual entropy change, not just "add heat." In figure s01 terms, the teal drop is still shorter than the orange climb.

"These two reactions both consume

H_{2} O

, so

H_{2} O

is their common intermediate — they're coupled."

A shared reactant is not an intermediate; an intermediate is produced by one and consumed by the other. Both merely using water does not link them into a single path.

K_{tot} = K_{1} + K_{2}

because we add the

Δ G^{\circ}

values."

Adding

Δ G^{\circ}

becomes multiplying

K

- R T ln K_{tot} = - R T ln K_{1} - R T ln K_{2} \Rightarrow ln K_{tot} = ln K_{1} + ln K_{2} \Rightarrow K_{tot} = K_{1} K_{2}

"Coupling makes an endergonic reaction exergonic, so the products of the uphill reaction now have lower energy than its reactants."

No — the uphill reaction's own energetics are unchanged; only the net combined reaction is exergonic. In figure s01 B still sits above A; only the endpoint C is low.

"The net

Δ G_{tot}

is the average of the two

Δ G

values."

It is the sum, not the average:

Δ G_{tot} = Δ G_{1} + Δ G_{2}

, because the two steps happen one after the other along one path (see figure s02).

"Since coupling drove the reaction, the shared intermediate accumulates over time."

The intermediate is produced then consumed at matched rates in the coupled path, so it stays low and roughly steady — it cancels in the overall equation and does not pile up.

Why questions

Why does

Δ G

add rather than combine some other way when reactions couple?

G

is a state function, so the change along a two-step path from start to finish equals the sum of the step changes; the shared intermediate cancels, leaving only the endpoints (figure s02).

Why must a real molecular intermediate exist — why isn't paper arithmetic enough?

Without a physical link, the two reactions occupy separate flasks of reality; reaction 1 still faces

Δ G_{1} > 0

and will not proceed no matter what number you write beside reaction 2.

Why does heating help ZnO reduction but the reaction still needs carbon?

Heating makes the carbon-oxidation

Δ G_{2}

far more negative through the

- T Δ S

term (a gas is made, so

Δ S > 0

), but only carbon provides the favorable oxygen-grabbing partner that supplies the surplus. This tilting-with-temperature is exactly what an Ellingham Diagram plots:

Δ G^{\circ}

versus

T

as sloped lines, and reduction becomes possible where the carbon line dips below the metal-oxide line.

Why is ATP hydrolysis called the cell's "universal falling weight"?

Its

Δ G^{\circ} \approx - 30.5

kJ/mol is large and negative enough to drive many otherwise-uphill biosyntheses, and its phosphate transfers directly, giving a true shared intermediate. See ATP and Bioenergetics.

Why can a very small

K_{1}

be "rescued" by a huge

K_{2}

, yet a coupled pair with

Δ G_{1} + Δ G_{2} > 0

still cannot run?

Multiplying

K

and adding

Δ G^{\circ}

are the same fact (

ln K

tracks

- Δ G^{\circ}

): a huge

K_{2}

just means

Δ G_{2}^{\circ}

is very negative, so it can overpower a positive

Δ G_{1}^{\circ}

. But if even so the total

Δ G

stays positive, then

K_{tot} < 1

and the net won't run — the rescue works only when the sum actually goes negative.

Why does temperature change

Δ G

but not automatically enable coupling?

Δ G = Δ H - T Δ S

shifts with

T

, but coupling still requires a shared intermediate; temperature only tunes the magnitudes (heights in figure s01), not the existence of the molecular link. See Entropy and Temperature dependence of ΔG.

Why do we use

Δ G

and not

Δ H

to decide if coupling works?

Spontaneity at constant

T, P

is set by

Δ G

, which folds in both the heat term

Δ H

and the disorder term

- T Δ S

; an endothermic step (

Δ H > 0

) can still be favorable if its entropy gain

Δ S

is large enough. See Gibbs Free Energy.

Edge cases

What if

Δ G_{1} = 0

(reaction 1 already at equilibrium) and it couples to

Δ G_{2} < 0

The net becomes

Δ G_{2} < 0

, so the combined process runs forward, pulling reaction 1 off its equilibrium in the product direction.

What if both reactions are favorable (

Δ G_{1}, Δ G_{2} < 0

They still add to something more negative, so the net runs even harder — coupling isn't reserved for rescuing uphill steps, though that's its dramatic use.

What if the "favorable" step is only slightly negative, e.g.

Δ G_{2} = - 1

kJ against

Δ G_{1} = + 20

kJ?

The sum is

+ 19 > 0

, so the net stays non-spontaneous; the falling weight is too light to lift the bucket, and no shared intermediate can force it.

What happens to

K_{tot}

if one reaction has

K = 1

exactly (

Δ G^{\circ} = 0

Multiplying by

1

leaves

K_{tot} =

the other reaction's

K

; that step neither helps nor hinders the overall equilibrium.

What if the shared intermediate is also a bulk reactant present in huge excess (like water in a cell)?

Then its concentration barely changes and it does not behave as a low-level linking intermediate; genuine coupling needs the intermediate to be produced and consumed by the paired steps, not swamped. Recall that Hess's Law adds reactions on paper, but real coupling also demands the physical shared species.

Degenerate case: two reactions written to "share" an intermediate that actually cancels to zero net change — is that coupling?

No — if the intermediate cancels and nothing else combines, you've written a trivial identity; real coupling produces a new overall reaction

A \to C

with distinct endpoints.

Limiting case: as

Δ G_{2} \to - \infty

, what happens to the net?

The net

Δ G_{tot} \to - \infty

and

K_{tot} \to \infty

, meaning the uphill reaction is driven essentially to completion — the ideal "infinitely heavy weight."

Connections

Relation between ΔG and K — why additive $Δ G^{\circ}$ becomes multiplicative $K$ .
Gibbs Free Energy — the $Δ G < 0$ criterion underneath every trap here.

Question bank — Coupling reactions — driving unfavorable reactions

True or false — justify

Spot the error

Why questions

Edge cases

Connections