State the equation for Gibbs free energy and explain what each of G 0 and G = 0 means for spontaneity. [3 marks]

§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 1: What happens when chemical reactions do not go through to completion?

Explore the relationship between entropy, enthalpy and Gibbs free energy in the context of a reaction's spontaneity, and evaluate given the equation delta-G = delta-H minus T delta-S why a reaction that has a negative delta-H is not always spontaneous, and why some reactions do not go to completion

A focused answer to the HSC Chemistry Module 5 dot point on entropy, enthalpy and Gibbs free energy. What entropy and enthalpy each measure, why delta-G = delta-H minus T delta-S decides spontaneity, why an exothermic reaction can still be non-spontaneous, and why most reactions stop at equilibrium rather than completion.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-28

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

In plain English

Think of a reaction as a ball rolling down a bumpy hillside made of two different slopes stacked together. One slope is enthalpy, the "downhill energy release" slope; the ball rolling down it releases heat, and a steeper downhill slope (a more negative delta-H) feels like it should always make the ball keep rolling. But the other slope is entropy, the "spreading out" slope, and its effect grows stronger the hotter the room gets, because temperature is literally multiplied onto it. If the entropy slope actually points uphill (disorder DECREASES, like gas molecules combining into fewer, more ordered molecules), then in a hot enough room, that uphill entropy penalty can outweigh the downhill enthalpy release, and the ball simply will not roll on its own anymore, even though releasing heat "feels" like it should be favourable. Gibbs free energy, delta-G, is just the net result of combining both slopes at a given temperature: negative overall means the ball keeps rolling (spontaneous), positive means it will not budge without a push, and exactly zero means the ball has settled at the bottom of a dip and stays there, which is exactly what chemists call equilibrium. That is also why most reactions do not run to 100 percent completion: as products pile up, the "reverse ball" starts finding its own downhill slope, and the system settles into a balanced dip somewhere in the middle rather than emptying out completely.

What this dot point is asking

NESA wants you to connect three thermodynamic quantities, enthalpy (delta-H), entropy (delta-S) and Gibbs free energy (delta-G), through the equation $\Delta G = \Delta H - T\Delta S$ , and use this to explain why a reaction proceeds spontaneously (or does not), and why an exothermic reaction is not automatically spontaneous. This directly answers Inquiry Question 1: reactions do not always go through to completion because the true criterion for spontaneity is delta-G, not delta-H alone, and most real reactions settle at an equilibrium position where delta-G for the overall system reaches zero rather than continuing until reactants are exhausted.

The answer

Enthalpy and entropy: two separate quantities

Enthalpy, delta-H, is the heat energy change of a reaction at constant pressure. A negative delta-H (exothermic) releases heat to the surroundings; a positive delta-H (endothermic) absorbs heat. For a long time, chemists assumed exothermic reactions were "favoured" and endothermic ones were not, but this is incomplete.

Entropy, S, measures the number of ways energy and particles can be dispersed or arranged in a system, informally its "disorder". Entropy tends to increase when:

a solid or liquid becomes a gas (large positive delta-S);
the number of moles of gas increases across a reaction (positive delta-S);
a solid or liquid dissolves into solution (positive delta-S);
particles become more free to move or occupy more microstates.

Entropy tends to decrease in the reverse of each of these changes, for example when gas moles decrease, such as in the Haber process, $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ , where 4 moles of gas become 2.

Gibbs free energy: the true test of spontaneity

Josiah Willard Gibbs combined enthalpy, entropy and temperature into a single quantity that correctly predicts spontaneity:

\Delta G = \Delta H - T\Delta S

where T is the absolute temperature in kelvin. The sign of delta-G determines the outcome:

$\Delta G < 0$ : the reaction is spontaneous in the forward direction at that temperature.
$\Delta G > 0$ : the reaction is non-spontaneous in the forward direction (the reverse reaction is spontaneous instead).
$\Delta G = 0$ : the system is at equilibrium; neither direction is favoured over the other.

The four sign combinations of delta-H and delta-S give four distinct behaviours as temperature changes:

$\Delta H$	$\Delta S$	$\Delta G = \Delta H - T\Delta S$	Spontaneity
negative	positive	always negative	spontaneous at all temperatures
positive	negative	always positive	non-spontaneous at all temperatures
negative	negative	negative at low T, positive at high T	spontaneous only when cold
positive	positive	positive at low T, negative at high T	spontaneous only when hot

Why a negative delta-H does not guarantee spontaneity

The classic HSC misconception is assuming exothermic always means spontaneous. Looking at the equation $\Delta G = \Delta H - T\Delta S$ , a negative delta-H only produces a negative delta-G if the $-T\Delta S$ term does not outweigh it. If delta-S is negative, then $-T\Delta S$ is a POSITIVE quantity that grows in magnitude as T increases. At a high enough temperature, this positive term can exceed the magnitude of the negative delta-H, flipping the overall sign of delta-G to positive, so the reaction becomes non-spontaneous despite releasing heat.

Why reactions do not always go to completion

Delta-G as calculated from standard enthalpies and entropies describes the reaction under a defined set of starting conditions, typically pure reactants converting to pure products. As a real reaction proceeds, reactant concentrations fall and product concentrations rise, and the effective driving force for the forward reaction weakens while the driving force for the reverse reaction strengthens. The reaction continues until the forward and reverse processes are equally favoured, at which point the overall system sits at $\Delta G = 0$ : this is dynamic equilibrium, not the reactants being fully used up.

The magnitude of delta-G-standard (not just its sign) sets how far toward completion the equilibrium lies, through:

\Delta G^{\circ} = -RT\ln K

A very large negative delta-G-standard corresponds to an enormous equilibrium constant K, so the equilibrium mixture is, for practical purposes, almost all product (the reaction appears to "go to completion"). A delta-G-standard that is only modestly negative corresponds to a K only a little above 1, giving a genuine equilibrium mixture with significant amounts of both reactants and products, exactly the reversible reactions studied throughout Module 5.

Worked calculations

Worked example 1: the Haber process at 298 K.

N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}, \quad \Delta H = -92\ \text{kJ mol}^{-1}, \quad \Delta S = -199\ \text{J K}^{-1}\text{mol}^{-1}

Convert delta-S: $-199\ \text{J K}^{-1}\text{mol}^{-1} = -0.199\ \text{kJ K}^{-1}\text{mol}^{-1}$ .

\Delta G = \Delta H - T\Delta S = -92 - (298)(-0.199) = -92 + 59.302 = -32.7\ \text{kJ mol}^{-1}

Delta-G is negative, so the forward reaction is spontaneous at 298 K, even though delta-S is negative here, because the exothermic delta-H term still dominates at this temperature.

Worked example 2: calcium carbonate decomposition, finding the crossover temperature.

CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}, \quad \Delta H = +178\ \text{kJ mol}^{-1}, \quad \Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}

At 298 K: $\Delta G = 178 - (298)(0.161) = 178 - 47.978 = 130\ \text{kJ mol}^{-1}$ (positive, non-spontaneous at room temperature, matching everyday experience).

Setting $\Delta G = 0$ to find the minimum spontaneous temperature:

T = \frac{\Delta H}{\Delta S} = \frac{178\ \text{kJ mol}^{-1}}{0.161\ \text{kJ K}^{-1}\text{mol}^{-1}} = 1105.6\ \text{K} \approx 833\ ^{\circ}\text{C}

Above about 833 degrees C, delta-G turns negative and decomposition becomes spontaneous, which is why industrial lime kilns run at roughly 900 to 1000 degrees C.

Marker's note: full marks require showing the delta-S unit conversion explicitly as its own step, since submitting delta-S in J directly against delta-H in kJ is the single most common HSC arithmetic error in this topic, producing an answer that is out by a factor of 1000.

Exam technique for delta-G questions

Always show four labelled steps: (1) convert delta-S to kJ K⁻¹ mol⁻¹ if delta-H is in kJ mol⁻¹, (2) substitute into $\Delta G = \Delta H - T\Delta S$ with T in kelvin, (3) evaluate and give the correct sign, (4) state the conclusion in words ("spontaneous", "non-spontaneous" or "at equilibrium") rather than leaving a bare number. For a "why is a reaction not spontaneous despite negative delta-H" question, always name the sign of delta-S and explicitly describe how the $-T\Delta S$ term behaves as temperature changes, rather than only stating the final sign of delta-G. For "why does a reaction not go to completion", link back to delta-G becoming zero at equilibrium, and if asked quantitatively, bring in $\Delta G^{\circ} = -RT\ln K$ to connect the SIZE of delta-G to the SIZE of K.

Common traps

Using Celsius instead of kelvin for T: T must always be converted to kelvin before substituting into $\Delta G = \Delta H - T\Delta S$ ; a Celsius value can even be negative, which breaks the equation entirely.
Forgetting to convert delta-S units: Delta-H is usually given in kJ mol⁻¹ and delta-S in J K⁻¹ mol⁻¹; substituting without converting delta-S to kJ introduces a 1000-fold error.
Assuming exothermic always means spontaneous: A negative delta-H is necessary-looking but not sufficient; a sufficiently negative delta-S at high temperature can still make delta-G positive.
Confusing spontaneity with speed: A negative delta-G means a reaction is thermodynamically favoured, not that it happens quickly; reaction rate is governed by activation energy and kinetics, a completely separate concept from delta-G.
Treating delta-G = 0 as "the reaction stops": At delta-G = 0 the system is at dynamic equilibrium; both the forward and reverse reactions are still occurring, just at equal rates, so concentrations stop changing rather than the reaction itself stopping.

Examples in context

Example 1. Lime kilns at Port Kembla and regional NSW cement works. Industrial lime kilns thermally decompose calcium carbonate, $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$ , an endothermic reaction with a strongly positive delta-S because a gas is released from two solids. Operators must heat the kiln well above the thermodynamic crossover temperature of around 833 degrees C, typically to 900 to 1000 degrees C, both to ensure delta-G is comfortably negative (favouring the forward reaction) and to achieve a commercially useful reaction rate, illustrating that spontaneity (delta-G) and rate are related but separate considerations in real plant design.

Example 2. Why the Haber process is not run at very high temperature. The Haber process, $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ , is exothermic with a decrease in entropy (delta-S negative, since 4 moles of gas become 2). This means the $-T\Delta S$ term is positive and grows with temperature, so as temperature rises the reaction becomes progressively less spontaneous (delta-G becomes less negative), consistent with the Le Chatelier prediction that heating an exothermic equilibrium shifts it toward reactants. This thermodynamic argument, not just the Le Chatelier shift, is part of why the industrial reactor uses a compromise temperature of about 400 degrees C rather than running hotter to speed up the reaction.

Try this

Q1. State the equation for Gibbs free energy and explain what each of $\Delta G < 0$ , $\Delta G > 0$ and $\Delta G = 0$ means for spontaneity. [3 marks]

Cue. $\Delta G = \Delta H - T\Delta S$ ; negative means spontaneous forward, positive means non-spontaneous forward (reverse spontaneous), zero means equilibrium.

Q2. A reaction has $\Delta H = -58\ \text{kJ mol}^{-1}$ and $\Delta S = -176\ \text{J K}^{-1}\text{mol}^{-1}$ . Calculate delta-G at 400 K and state whether the reaction is spontaneous at this temperature. [3 marks]

Cue. Convert delta-S to $-0.176\ \text{kJ K}^{-1}\text{mol}^{-1}$ ; $\Delta G = -58 - (400)(-0.176) = -58 + 70.4 = +12.4\ \text{kJ mol}^{-1}$ ; positive, so non-spontaneous at 400 K (despite being exothermic).

Q3. Explain why a reaction that is spontaneous (delta-G negative) does not necessarily go to completion. [3 marks]

Cue. As product builds up, the reverse reaction becomes more favoured until delta-G for the overall system reaches zero (equilibrium); the extent of reaction at that point depends on the size of delta-G-standard via $\Delta G^{\circ} = -RT\ln K$ , not just its sign.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor each reaction, state the expected sign of delta-S (positive, negative or approximately zero) and justify it in one sentence: (a)

2NO_{2(g)} \rightarrow N_2O_{4(g)}

; (b)

H_2O_{(l)} \rightarrow H_2O_{(g)}

; (c)

H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}

Show worked solution →

A 3-mark identify-and-justify needs the correct sign AND a reason based on particle dispersal for each part.

(a) $2NO_{2(g)} \rightarrow N_2O_{4(g)}$: Delta-S is negative. Two moles of gas combine into one mole of gas, reducing the number of independent gas particles and hence the number of ways to disperse energy.
(b) $H_2O_{(l)} \rightarrow H_2O_{(g)}$: Delta-S is positive. Converting a liquid to a gas greatly increases the volume and disorder available to the molecules; vaporisation is the classic large-positive-entropy change.
(c) $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$: Delta-S is approximately zero. There are 2 moles of gas on each side of the equation, so the number of ways to disperse energy among gas particles is essentially unchanged.

Marking criteria: 1 mark per part for the correct sign linked explicitly to a change (or lack of change) in the number of gas moles or physical state.

foundation4 marksA reaction has

\Delta H = -92\ \text{kJ mol}^{-1}

and

\Delta S = -199\ \text{J K}^{-1}\text{mol}^{-1}

(the Haber process,

N_2 + 3H_2 \rightleftharpoons 2NH_3

). (a) Calculate delta-G at 298 K. (b) State whether the reaction is spontaneous at this temperature. (c) Explain, without recalculating, whether raising the temperature makes the reaction more or less spontaneous.

Show worked solution →

(a) Calculate delta-G at 298 K.

Convert delta-S to kJ K⁻¹ mol⁻¹ so units match delta-H:

\Delta S = -199\ \text{J K}^{-1}\text{mol}^{-1} = -0.199\ \text{kJ K}^{-1}\text{mol}^{-1}

Substitute into $\Delta G = \Delta H - T\Delta S$ :

\Delta G = -92\ \text{kJ mol}^{-1} - (298\ \text{K})(-0.199\ \text{kJ K}^{-1}\text{mol}^{-1})

\Delta G = -92\ \text{kJ mol}^{-1} + 59.302\ \text{kJ mol}^{-1}

\Delta G = -32.7\ \text{kJ mol}^{-1} \ \text{(3 s.f.)}

(b) Spontaneity. Since $\Delta G < 0$ at 298 K, the reaction is spontaneous at this temperature (thermodynamically favoured to proceed toward ammonia), which is consistent with the Haber process operating at all.

(c) Effect of raising temperature. Both delta-H and delta-S are negative, so the $-T\Delta S$ term is positive and GROWS larger (more positive) as T increases. This makes delta-G progressively less negative (and eventually positive at high enough T), so raising the temperature makes the forward reaction less spontaneous. This is consistent with Le Chatelier's principle, which predicts that heating an exothermic equilibrium shifts it toward reactants.

Marking criteria: (a) 1 mark for correctly converting delta-S units, 1 mark for correct substitution, 1 mark for the correct numerical answer with sign and 3 significant figures. (b) 1 mark for correctly concluding "spontaneous" directly from the sign of delta-G. (c) full marks require reasoning from the growing magnitude of the $-T\Delta S$ term, not just restating Le Chatelier without the thermodynamic link.

core5 marksThe thermal decomposition of calcium carbonate,

CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}

, has

\Delta H = +178\ \text{kJ mol}^{-1}

and

\Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}

. (a) Calculate delta-G at 298 K and state whether the reaction is spontaneous at room temperature. (b) Calculate the minimum kiln temperature (in degrees Celsius) at which the decomposition becomes spontaneous.

Show worked solution →

(a) Delta-G at 298 K.

Convert delta-S: $161\ \text{J K}^{-1}\text{mol}^{-1} = 0.161\ \text{kJ K}^{-1}\text{mol}^{-1}$ .

\Delta G = \Delta H - T\Delta S = 178\ \text{kJ mol}^{-1} - (298\ \text{K})(0.161\ \text{kJ K}^{-1}\text{mol}^{-1})

\Delta G = 178\ \text{kJ mol}^{-1} - 47.978\ \text{kJ mol}^{-1} = 130\ \text{kJ mol}^{-1} \ \text{(3 s.f.)}

Since $\Delta G > 0$ at 298 K, the decomposition is not spontaneous at room temperature, which matches everyday observation: limestone does not spontaneously decompose on a bench.

(b) Minimum temperature for spontaneity. The crossover occurs where $\Delta G = 0$ :

0 = \Delta H - T\Delta S \quad \Rightarrow \quad T = \frac{\Delta H}{\Delta S}

T = \frac{178\ \text{kJ mol}^{-1}}{0.161\ \text{kJ K}^{-1}\text{mol}^{-1}} = 1105.59\ \text{K}

Convert to degrees Celsius:

T = 1105.59 - 273 = 832.59\ ^{\circ}\text{C} \approx 833\ ^{\circ}\text{C} \ \text{(3 s.f.)}

Above approximately 833 degrees C, delta-G becomes negative and the decomposition becomes spontaneous, consistent with industrial lime kilns operating at around 900 to 1000 degrees C to ensure a comfortable margin above this threshold.

Marking criteria: (a) 1 mark for correct unit conversion, 1 mark for correct substitution and evaluation, 1 mark for correctly concluding "not spontaneous" from the positive sign. (b) 1 mark for rearranging to $T = \Delta H / \Delta S$ from setting delta-G to zero, 1 mark for the correct numerical temperature (accept 832 to 833 degrees C, or 1105 to 1106 K before conversion).

core4 marksThe graph below is an owned illustrative plot of delta-G against temperature for a reaction with

\Delta H = +178\ \text{kJ mol}^{-1}

and

\Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}

(as in the calcium carbonate decomposition). (a) Describe the shape of the line and explain why it has that shape in terms of

\Delta G = \Delta H - T\Delta S

. (b) Identify, from the graph, the approximate temperature at which the reaction becomes spontaneous, and state how this point is defined thermodynamically.

Show worked solution →

(a) Shape and explanation. The graph is a straight line with a positive delta-G intercept at $T = 0$ K (equal to delta-H) and a NEGATIVE gradient, so delta-G falls steadily as temperature rises, crossing from positive to negative. This matches the equation $\Delta G = \Delta H - T\Delta S$ rearranged as $\Delta G = (-\Delta S)T + \Delta H$ , a straight line in T with gradient $-\Delta S$ and y-intercept $\Delta H$ ; because delta-S is positive here, the gradient $-\Delta S$ is negative, so delta-G decreases linearly as T increases.

(b) Spontaneity crossover. Reading the graph, the line crosses delta-G = 0 at approximately $T \approx 1106$ K (about 833 degrees C), matching the calculated value in part (a) of the calculation question above. This point is defined as the temperature at which the system is exactly at equilibrium between the forward and reverse processes ( $\Delta G = 0$ ); below this temperature the reaction is non-spontaneous (delta-G positive) and above it the reaction is spontaneous (delta-G negative).

Marking criteria: (a) 1 mark for describing the straight-line, negative-gradient shape, 1 mark for linking the shape correctly to the equation (gradient = negative delta-S, intercept = delta-H). (b) 1 mark for reading an approximate crossover temperature consistent with the graph, 1 mark for correctly defining that point as delta-G = 0 (equilibrium), not "reaction stops".

exam4 marksExplain why a reaction with a negative delta-H is not always spontaneous, using the equation

\Delta G = \Delta H - T\Delta S

and a named example.

Show worked solution →

A negative delta-H only guarantees a negative delta-G if the $-T\Delta S$ term does not outweigh it. If delta-S is negative (the reaction decreases disorder, for example by reducing the moles of gas), then $-T\Delta S$ is POSITIVE, and at sufficiently high temperature this positive term can become larger in magnitude than the negative delta-H, making the overall delta-G positive despite the exothermic delta-H.

Named example. The Haber process, $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ , has $\Delta H = -92\ \text{kJ mol}^{-1}$ (exothermic) but $\Delta S = -199\ \text{J K}^{-1}\text{mol}^{-1}$ (entropy decreases, since 4 moles of gas become 2). At 298 K, delta-G is still negative (about $-32.7\ \text{kJ mol}^{-1}$ ) and the reaction is spontaneous, but as temperature rises the growing positive $-T\Delta S$ term progressively cancels the negative delta-H, and above a sufficiently high temperature delta-G becomes positive, so the forward reaction is no longer spontaneous. This is exactly why the industrial reactor cannot simply be run at a very high temperature to speed up ammonia production; doing so would work against the thermodynamic favourability of the forward reaction as well as shift the equilibrium position via Le Chatelier's principle.

Marking criteria: 1 mark for stating that delta-G, not delta-H alone, determines spontaneity, 1 mark for explaining the role of a negative delta-S making $-T\Delta S$ positive and temperature-dependent, 1 mark for a named, thermodynamically consistent example, 1 mark for explicitly linking the explanation back to the sign of delta-G rather than delta-H.

exam7 marksEvaluate the claim that 'a reaction with a negative delta-G always goes to completion.' In your answer, refer to the relationship between delta-G, equilibrium and the equilibrium constant K, and use a specific chemical example.

Show worked solution →

This is a 7-mark EVALUATE: markers reward a judgement backed by the delta-G to K relationship and a worked example, not just a restatement of the claim.

Band 6 plan.

State the claim is FALSE and give the thesis: a negative delta-G means a reaction is spontaneous and thermodynamically favoured, but "spontaneous" describes direction, not extent; almost all reactions are reversible to some degree and settle at an equilibrium position, not 100 percent conversion, unless K is extremely large.
Introduce the quantitative link: $\Delta G^{\circ} = -RT\ln K$ . A very negative delta-G-standard corresponds to an extremely large K (reaction lies almost entirely toward products, appearing to "go to completion" in practice), while a delta-G-standard that is only slightly negative corresponds to a K only modestly greater than 1, meaning a genuine equilibrium mixture with significant amounts of both reactants and products remains.
Explain the mechanism at the particle level: as a reaction with negative delta-G proceeds, product concentration rises and reactant concentration falls; the reverse reaction becomes increasingly favourable (its own effective delta-G rises toward zero) until the forward and reverse rates (and delta-G values) balance at delta-G = 0, which is the equilibrium point, not full conversion.
Worked example: the esterification of ethanoic acid with ethanol, $CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O$ , has a modestly negative delta-G-standard corresponding to $K \approx 4$ ; this gives a typical yield of around 67 percent at equilibrium, clearly demonstrating a spontaneous (delta-G negative) reaction that does NOT go to completion. Contrast this with a reaction like the combustion of methane, $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(g)}$ , which has an extremely large negative delta-G-standard and correspondingly enormous K, so it proceeds to what is, for all practical purposes, completion.
Judgement: the claim is only true in the limiting case of an extremely large K; in general a negative delta-G establishes the DIRECTION and favourability of a reaction but the actual extent of reaction at equilibrium depends on the MAGNITUDE of delta-G (via K), so the claim as stated is an overgeneralisation.

Model paragraph (excerpt). The claim that a negative delta-G always drives a reaction to completion conflates spontaneity with extent of reaction. A negative delta-G, by $\Delta G = \Delta H - T\Delta S$ , establishes only that the forward process is thermodynamically favoured over the reverse at a given temperature; it says nothing about how far the reaction proceeds before the forward and reverse rates balance. This extent is captured instead by the equilibrium constant K, related to delta-G-standard by $\Delta G^{\circ} = -RT\ln K$ . Esterification reactions such as ethanoic acid with ethanol have $K \approx 4$ , a value only modestly favouring products, and correspondingly plateau at around 67 percent yield despite delta-G-standard being negative; by contrast, the combustion of methane has such an overwhelmingly negative delta-G-standard, and hence such an enormous K, that essentially no measurable reactant remains at equilibrium. The claim is therefore true only in the limiting case of a very large K, and is a meaningful overgeneralisation for the majority of moderate, reversible reactions studied in Module 5.

Marker's note: top-band answers (1) correctly reject the claim rather than agreeing with it, (2) explicitly bring in $\Delta G^{\circ} = -RT\ln K$ to connect delta-G's sign/magnitude to K, (3) use at least one concrete named reaction with an approximate K or yield figure, and (4) end with an explicit, qualified judgement (true only for very large K) rather than a vague "it depends".