Entropy, enthalpy and Gibbs free energy explained: HSC Chemistry Module 5

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.

(a) Calculate delta-G at 298 K.

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

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

(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.

(a) Delta-G at 298 K.

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

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 $\Delta G = 0$:

Convert to degrees Celsius:

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).

(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".

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.

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".