Coupling reactions — driving unfavorable reactions: Worked examples

Before anything: three symbols, in plain words.

$Δ G$ ("delta G") = the free-energy change of a reaction. Negative = the reaction wants to run (downhill). Positive = it does not want to run (uphill). See Gibbs Free Energy.
$Δ G^{\circ}$ (with the little circle) = the same thing measured under standard conditions (1 bar, stated concentrations). We use it because it is a fixed number you can look up.
$K$ = the equilibrium constant. $K > 1$ means products are favored; $K < 1$ means reactants are. Linked to $Δ G^{\circ}$ by $Δ G^{\circ} = - R T ln K$ — see Relation between ΔG and K.

Here $R = 8.314 J K^{- 1} mol^{- 1}$ is the gas constant, $T$ the temperature in kelvin, and $ln$ is the natural logarithm (the "how many $e$ -foldings" function).

The scenario matrix

Every coupling problem falls into exactly one of these cells. The worked examples below are tagged with the cell they cover.

Cell	Situation	What decides the outcome	Example
A	Favorable step wins ( $∣ Δ G_{2} ∣ > Δ G_{1}$ )	$Δ G_{tot} < 0$ → net runs	Ex 1
B	Favorable step loses ( $∣ Δ G_{2} ∣ < Δ G_{1}$ )	$Δ G_{tot} > 0$ → net stalls	Ex 2
C	Exactly balanced ( $∣ Δ G_{2} ∣ = Δ G_{1}$ )	$Δ G_{tot} = 0$ → equilibrium	Ex 3
D	Temperature flips a losing case into a winning one	entropy term $- T Δ S$ grows with $T$	Ex 4
E	Work in $K$ -space, not $Δ G$ -space	$K_{tot} = K_{1} K_{2}$ crosses 1	Ex 5
F	Fake coupling — no shared intermediate	arithmetic is meaningless; net stalls	Ex 6
G	Real-world word problem (biology, ATP)	must find the shared species first	Ex 7
H	Exam twist — reverse a reaction, sign of $Δ G$ flips	reversing negates $Δ G$	Ex 8
I	Non-standard conditions — $Q$ tips a borderline case	$Δ G = Δ G^{\circ} + R T ln Q$	Ex 9

Figure — Coupling reactions — driving unfavorable reactions

Reading this figure (Cell A shown). The horizontal line is a $Δ G$ axis in kJ, ticked every $10$ from $- 30$ to $+ 30$ ; the tall black bar at the centre marks $Δ G = 0$ . We draw two arrows head-to-tail: a black up-arrow of length $+ 25$ (the unfavorable step $Δ G_{1}$ ) starting at $0$ , then a red down-arrow of length $- 40$ (the favorable step $Δ G_{2}$ ) starting where the first ended. The red dot marks the final tip. It lands at $+ 25 - 40 = - 15$ , left of zero in the red "net RUNS" region. That single picture is the whole topic; every example below is one placement of these two arrows.

Example 1 — Cell A: favorable step wins

Worked example The clean success

$(1) A \to B, Δ G_{1} = + 25 kJ$ $(2) B \to C, Δ G_{2} = - 40 kJ$ $B$ is the shared intermediate. Does $A \to C$ run?

Forecast: The down-arrow ( $40$ ) is longer than the up-arrow ( $25$ ). Guess the sign of the sum.

Step 1 — Confirm the intermediate cancels. $B$ is made by (1) and eaten by (2), so it disappears from the overall equation, leaving $A \to C$ . Why this step? Only a genuine shared species lets us add — this is Hess's Law in action.

Step 2 — Add the free energies. $Δ G_{tot} = (+ 25) + (- 40) = - 15 kJ .$ Why this step? $G$ is a state function, so along the path $A \to B \to C$ the $Δ G$ values sum.

Step 3 — Read the sign. $- 15 < 0$ , so the net reaction is spontaneous. On the figure, the arrow-tip lands in the red (left) region. Why this step? Spontaneity at constant $T, P$ is decided solely by $Δ G < 0$ .

Verify: Sanity — the favorable release ( $40$ ) must exceed the uphill demand ( $25$ ), surplus $= 15$ kJ, which is exactly $- Δ G_{tot}$ . Units: kJ throughout. ✓

Example 2 — Cell B: favorable step loses

Worked example When coupling isn't enough

$(1) X \to Y, Δ G_{1} = + 50 kJ$ $(2) Y \to Z, Δ G_{2} = - 30 kJ$ Shared intermediate $Y$ . Does $X \to Z$ run at this temperature?

Forecast: The up-arrow ( $50$ ) now dwarfs the down-arrow ( $30$ ). Where does the tip land?

Step 1 — Confirm the intermediate cancels. $Y$ is made by (1) and consumed by (2), so it cancels, leaving the net reaction $X \to Z$ . Why this step? We must verify a genuine shared species exists before adding — otherwise the sum is meaningless (see Cell F).

Step 2 — Add. $Δ G_{tot} = (+ 50) + (- 30) = + 20 kJ .$ Why this step? Same addition rule — the sign of the result is what matters.

Step 3 — Interpret. $+ 20 > 0$ : the net reaction does not run forward. The falling weight is too light to lift the box. Why this step? A positive total means the surroundings would have to supply $20$ kJ; nothing here does.

Verify: Deficit $= 50 - 30 = 20$ kJ, matching $Δ G_{tot}$ . On the figure the tip is right of zero (black region). This is the honest counterexample to "just couple anything." ✓

Example 3 — Cell C: exactly balanced (equilibrium)

Worked example The knife-edge case

$(1) P \to Q, Δ G_{1} = + 18 kJ$ $(2) Q \to S, Δ G_{2} = - 18 kJ$ Shared intermediate $Q$ . What is the state of $P \to S$ ?

Forecast: Two arrows of equal length, opposite directions. Where does the tip sit?

Step 1 — Confirm the intermediate cancels, then add. $Q$ is made by (1) and consumed by (2), so it cancels, leaving $P \to S$ , and we may sum the free energies: $Δ G_{tot} = (+ 18) + (- 18) = 0 kJ .$ Why this step? The cancellation of the shared $Q$ licenses the addition; $G$ is a state function, so only the endpoints $P$ and $S$ survive.

Step 2 — Interpret $Δ G = 0$ . The system sits at equilibrium: no net drive in either direction. Products and reactants coexist with no tendency to change. Why this step? $Δ G = 0$ is precisely the definition of a system at equilibrium.

Step 3 — Translate to $K$ . Using $Δ G^{\circ} = - R T ln K$ , if $Δ G_{tot}^{\circ} = 0$ then $ln K_{tot} = 0$ , so $K_{tot} = 1$ . Why this step? $K = 1$ means neither side is favored — the algebraic mirror of $Δ G = 0$ .

Verify: $- R T ln (1) = - R T \cdot 0 = 0 = Δ G_{tot}$ . Consistent. ✓

Example 4 — Cell D: temperature flips the sign

Worked example Metallurgy: heat rescues a losing coupling

Unfavorable: $ZnO \to Zn + \frac{1}{2} O_{2}, Δ G_{1} = + 318 kJ$ Favorable: $C + \frac{1}{2} O_{2} \to C O, Δ G_{2} (T)$ . Shared intermediate: oxygen. At low $T$ , $Δ G_{2} = - 137$ kJ (loses). At high $T$ , $Δ G_{2} = - 360$ kJ. Show heating turns Cell B into Cell A.

Forecast: Low $T$ gives sum $= 318 - 137$ . Positive or negative? Then high $T$ : $318 - 360$ ?

Step 1 — Low-temperature total. $Δ G_{tot} = 318 + (- 137) = + 181 kJ > 0 (Cell B, stalls) .$ Why this step? At low $T$ carbon's oxidation isn't yet negative enough; we add because the shared $O_{2}$ (released by ZnO, grabbed by carbon) cancels.

Step 2 — Why does $Δ G_{2}$ get more negative as $T$ rises? Because $Δ G = Δ H - T Δ S$ , and the reaction $C + \frac{1}{2} O_{2} \to C O$ increases gas moles ( $\frac{1}{2}$ mol $O_{2}$ → 1 mol $C O$ ), so $Δ S > 0$ . A positive $Δ S$ makes $- T Δ S$ ever more negative as $T$ grows. See Entropy and Temperature dependence of ΔG. Why this step? This is why the Ellingham lines slope — the tool that decides which metal reduces which.

Step 3 — High-temperature total. $Δ G_{tot} = 318 + (- 360) = - 42 kJ < 0 (Cell A, runs) .$ Why this step? We add the two $Δ G$ values again — the shared $O_{2}$ still cancels, so only the magnitude of the carbon term has changed with temperature; the addition rule is unchanged.

Verify: The sign literally flipped, $+ 181 \to - 42$ , purely from temperature. This is the whole logic of the Ellingham Diagram: raise $T$ until the carbon line drops below the metal-oxide line. ✓

The figure shows the two carbon-arrows: short at low $T$ (tip lands right of zero, black) and long at high $T$ (tip crosses into red). Same up-arrow both times — only the down-arrow grows.

Example 5 — Cell E: work entirely in K-space

Worked example The multiplication rescue

Reaction (1): $K_{1} = 1 0^{- 5}$ (barely happens). Reaction (2): $K_{2} = 1 0^{8}$ . They share an intermediate. Is the coupled process favorable, and what is $Δ G_{tot}^{\circ}$ at $T = 298$ K?

Forecast: A tiny $K_{1}$ times a huge $K_{2}$ . Guess whether $K_{tot}$ ends up above or below 1.

Step 1 — Multiply the constants. $K_{tot} = K_{1} \times K_{2} = 1 0^{- 5} \times 1 0^{8} = 1 0^{3} .$ Why this step? Additive $Δ G$ becomes multiplicative $K$ because $ln (K_{1} K_{2}) = ln K_{1} + ln K_{2}$ .

Step 2 — $K_{tot} = 1 0^{3} ≫ 1$ , so the net reaction strongly favors products. Why this step? $K > 1$ ⇔ products dominate ⇔ $Δ G^{\circ} < 0$ .

Step 3 — Convert to free energy. $Δ G_{tot}^{\circ} = - R T ln K_{tot} = - (8.314) (298) ln (1 0^{3}) J \approx - 17, 100 J = - 17.1 kJ .$ Why this step? Confirms Cell E lands in Cell A once translated back to $Δ G$ .

Verify: $ln (1 0^{3}) = 3 ln 10 \approx 6.908$ ; $8.314 \times 298 \times 6.908 \approx 17, 114$ J. Negative sign ⇒ spontaneous. ✓

Example 6 — Cell F: fake coupling (no shared intermediate)

Worked example The trap

$(1) M \to N, Δ G_{1} = + 20 kJ$ $(2) P \to Q, Δ G_{2} = - 35 kJ$ These reactions share no molecule. A student writes $Δ G_{tot} = + 20 - 35 = - 15$ and claims reaction (1) now runs. Are they right?

Forecast: Trust your gut before the algebra: is $M \to N$ affected by an unrelated reaction?

Step 1 — Check for a shared intermediate. $N$ from (1) is not consumed by (2); $P$ of (2) has nothing to do with (1). No common species. Why this step? Addition of $Δ G$ is licensed only by a genuine path linking the two. No link, no license.

Step 2 — Correct conclusion. Reaction (1) keeps its own $Δ G_{1} = + 20 kJ > 0$ and does not run. The " $- 15$ " on paper is meaningless arithmetic. Why this step? Coupling is chemistry, not bookkeeping — see the parent note's steel-manned mistake.

Verify: With no intermediate, each reaction's fate is set by its own $Δ G$ : (1) stalls ( $+ 20$ ), (2) runs ( $- 35$ ), independently. The sum $- 15$ predicts nothing about either. ✓

Example 7 — Cell G: biology word problem

Worked example Phosphorylating glucose with ATP

A cell needs to attach a phosphate to glucose, an uphill step: $Glucose + P_{i} \to Glucose-6-P + H_{2} O, Δ G^{\circ} = + 13.8 kJ/mol .$ It couples this to ATP hydrolysis: $ATP + H_{2} O \to ADP + P_{i}, Δ G^{\circ} = - 30.5 kJ/mol .$ Does the coupled reaction proceed, and by how much surplus energy? See ATP and Bioenergetics.

Forecast: Down-arrow $30.5$ , up-arrow $13.8$ . Which is longer, and by how much?

Step 1 — Identify the shared intermediate. The phosphate group $P_{i}$ — released by ATP hydrolysis and consumed in phosphorylating glucose (in the cell it is transferred directly, never released to water). Why this step? Without confirming $P_{i}$ is genuinely shared, we cannot add — this is the biology version of Cell F's warning.

Step 2 — Add the two standard free energies. $Δ G_{net}^{\circ} = (+ 13.8) + (- 30.5) = - 16.7 kJ/mol .$ Why this step? The shared $P_{i}$ cancels between the two equations, so the state-function property of $G$ lets us sum the two $Δ G^{\circ}$ into the net transformation Glucose → Glucose-6-P.

Step 3 — Interpret. $- 16.7 < 0$ : phosphorylation runs. Surplus $= 16.7$ kJ/mol is dissipated, guaranteeing the reaction goes essentially to completion. Why this step? $∣ - 30.5 ∣ = 30.5 > 13.8$ is the Cell-A win condition.

Verify: $30.5 - 13.8 = 16.7$ kJ/mol surplus, matching $- Δ G_{net}^{\circ}$ . Units kJ/mol throughout. ✓

Example 8 — Cell H: exam twist (reverse a reaction)

Worked example The sign-flip you must catch

You are given: $(1) D \to E, Δ G_{1} = - 60 kJ, (2) E \to F, Δ G_{2} = + 45 kJ .$ Here $E$ is the shared intermediate. Find $Δ G$ for $D \to F$ and for the reverse $F \to D$ .

Forecast: Guess $Δ G$ for the forward path, then predict the sign of the reverse.

Step 1 — Forward path $D \to F$ (intermediate $E$ cancels): $Δ G_{D \to F} = (- 60) + (+ 45) = - 15 kJ .$ Why this step? Straight Hess addition; $E$ is made then consumed, so it cancels.

Step 2 — Reverse the whole reaction. Reversing a reaction negates its $Δ G$ : $Δ G_{F \to D} = - (Δ G_{D \to F}) = - (- 15) = + 15 kJ .$ Why this step? $Δ G$ is a difference of state functions ( $G_{final} - G_{initial}$ ); swapping endpoints flips the sign. This is the classic exam booby-trap.

Step 3 — Read outcomes. $D \to F$ runs ( $- 15 < 0$ ); $F \to D$ does not ( $+ 15 > 0$ ). Why this step? Spontaneity is decided solely by the sign of $Δ G$ : negative runs, positive stalls — the same core criterion we applied in every cell above.

Verify: Forward and reverse totals are exact negatives ( $- 15$ and $+ 15$ ), as required for any state function. ✓

Example 9 — Cell I: non-standard conditions ( $Q$ tips a borderline coupling)

Worked example How the cell rescues a barely-positive net reaction

Suppose a coupled net reaction $A ⇌ B$ has $Δ G_{net}^{\circ} = + 4.0 kJ/mol$ — positive, so under standard conditions it stalls (a borderline Cell B). But the cell holds the product low and the reactant high, giving a reaction quotient $Q = [B] / [A] = 0.01$ . Does the reaction now run, at $T = 310 K$ (body temperature)?

Forecast: Products are $100 \times$ scarcer than reactants ( $Q = 0.01 < 1$ ). Guess whether that pulls $Δ G$ up or down, and whether $4.0$ kJ/mol can be overcome.

Step 1 — Write the non-standard drive. $Δ G = Δ G^{\circ} + R T ln Q .$ Why this step? $Δ G^{\circ}$ only describes standard conditions ( $Q = 1$ ). Real cells are far from that, and the $R T ln Q$ term is exactly the correction — see Relation between ΔG and K.

Step 2 — Evaluate the correction. With $R = 8.314 J K^{- 1} mol^{- 1}$ , $T = 310$ K, $Q = 0.01$ : $R T ln Q = (8.314) (310) ln (0.01) J/mol \approx (2577) (- 4.605) \approx - 11, 870 J/mol = - 11.9 kJ/mol .$ Why this step? $ln Q < 0$ because $Q < 1$ : scarce product pulls the drive downward, the exact mechanism promised in the intro callout.

Step 3 — Add and read the sign. $Δ G = (+ 4.0) + (- 11.9) = - 7.9 kJ/mol < 0.$ Why this step? The concentration term outweighs the standard uphill term, so the actual reaction runs even though the standard one would not.

Verify: $ln (0.01) = - 2 ln 10 \approx - 4.605$ ; $8.314 \times 310 \times (- 4.605) \approx - 11, 867$ J/mol $= - 11.87$ kJ/mol; total $4.0 - 11.87 = - 7.87 \approx - 7.9$ kJ/mol. Negative ⇒ runs. This is how living cells drive borderline couplings by managing concentrations, not just enthalpy. ✓

Coverage check (Mermaid)

Recall Which cell was hardest?

The fake-coupling Cell F, because the arithmetic looks valid. Always ask "what molecule do they share?" first. Reversing a reaction does what to its ΔG? ::: Negates it (Cell H). A coupled total ΔG = 0 means what? ::: Equilibrium, and $K_{tot} = 1$ (Cell C). Two ways to flip a borderline positive net ΔG negative? ::: Raise $T$ (Cell D) or lower $Q$ by keeping products scarce (Cell I).

Connections

Gibbs Free Energy — the $Δ G < 0$ criterion behind every cell.
Hess's Law — licenses adding the two $Δ G$ values.
Relation between ΔG and K — turns Cell E's $K$ -multiplication into $Δ G$ , and supplies the $R T ln Q$ term of Cell I.
Ellingham Diagram — the visual home of Cell D (temperature flip).
Entropy and Temperature dependence of ΔG — why heating changes $Δ G_{2}$ .
ATP and Bioenergetics — Cell G's real-world engine.