Part 3: Entropy of the system, surroundings and total

Explain spontaneous endothermic change by tracking both particle dispersal and energy transfer. Calculate the total, not just one contribution.

Enthalpy alone does not decide direction

Some changes absorb heat and still proceed at room temperature, including dissolution of ammonium nitrate under suitable conditions. The reacting chemicals gain enthalpy, so “systems always change toward lower enthalpy” cannot be a complete rule.

Entropy describes the number of possible arrangements of particles and energy; at this level it is often described as disorder. More accessible arrangements mean greater entropy. The natural thermodynamic direction of a change increases total entropy, including both system and surroundings. This does not guarantee that the change is fast.

Define the system as the reacting substances and the surroundings as everything outside that boundary. A system may become more ordered if it releases enough energy to increase the entropy of its surroundings by a greater amount. “Entropy must increase” is incomplete unless it specifies total entropy.

Standard molar entropy S° has units J mol⁻¹ K⁻¹. Unlike standard formation enthalpy, an element’s standard entropy at 298 K is not defined to be zero. The zero-entropy reference applies to a perfect crystal at absolute zero in the ideal third-law limit, not all matter at ordinary temperatures.

Use states and gas amounts, then consider competing effects

For the same substance, melting usually increases entropy and vaporisation causes a larger increase because gas particles have much greater freedom of arrangement. Heating within one state also increases entropy by making more energy arrangements accessible.

A reaction producing more moles of gas usually increases system entropy; consuming gas to make condensed phases usually decreases it. Count gaseous stoichiometric amounts and state their physical significance. Total number of formula units without states is a weak guide.

Dissolving an ionic lattice disperses ions from fixed lattice positions, tending to increase entropy. However, hydration can organise surrounding water, especially around small highly charged ions. The overall system entropy change is a balance, so do not assume every dissolution has positive ΔSsystem. Use supplied entropy data where qualitative effects compete.

In a reaction with a gas increase and solid formation, explain both. Pearson 9CH0/03 June 2023 Q10(c) required recognition that forming more gaseous CO outweighed the opposing effect of forming a solid; a generic statement of “more particles” did not fully address that system.

Apply the reasoning to four contrasting changes

Dissolving NH₄NO₃(s) to NH₄⁺(aq) and NO₃⁻(aq) can cool the surroundings: the dissolution is endothermic. Ion dispersal contributes a positive system entropy change sufficiently large for total entropy to increase under the conditions. Cooling alone therefore does not establish infeasibility.

When ethanoic acid reacts with ammonium carbonate, carbon dioxide is evolved and a solution containing ammonium ethanoate forms. Formation of gas and dispersal of the solid contribute to increasing system entropy. In a cooling demonstration the surroundings entropy decreases, so the positive system contribution must outweigh it for the observed thermodynamically favourable change.

Burning magnesium consumes oxygen gas and forms magnesium oxide solid, so system entropy decreases. The strongly exothermic reaction transfers much energy to the surroundings, whose entropy increases enough to give positive total entropy. Ignition is still needed because there is an activation barrier.

Mixing solid barium hydroxide octahydrate with solid ammonium chloride produces ammonia gas and a liquid mixture. Gas production and the change from crystalline reactants to mobile particles give a large positive system entropy contribution; a substantial temperature fall shows an endothermic reaction. This is a supervised demonstration: toxic soluble barium compounds and ammonia require the appropriate controls and disposal.

Worked example: keep the mole-of-reaction basis

For N₂(g) + 3H₂(g) → 2NH₃(g), use these rounded illustrative molar entropies: N₂ 192, H₂ 131 and NH₃ 193 J mol⁻¹ K⁻¹. ΔSsystem = ΣνS(products) − ΣνS(reactants) = 2(193) − [192 + 3(131)] = −199 J mol⁻¹ K⁻¹ for the equation as written.

The negative sign agrees with reducing four moles of gas to two. Each coefficient multiplies its molar entropy. The result refers to one mole of reaction, which produces two moles NH₃; do not divide or multiply the enthalpy by two unless the whole equation basis changes.

Suppose ΔH = −92.0 kJ mol⁻¹ of reaction at 298 K. At constant pressure with surroundings effectively at temperature T, ΔSsurroundings = −ΔH/T = −(−92000)/298 = +308.7 J mol⁻¹ K⁻¹. Convert kJ to J before combining this with the system value.

ΔStotal = −199 + 308.7 = +109.7 J mol⁻¹ K⁻¹, or +110 J mol⁻¹ K⁻¹ to three significant figures. The overall process is thermodynamically favourable under the stated conditions even though the system becomes less disordered. A calculation stopping at −199 has answered a different question.

Changing T changes the balance

For an endothermic reaction with ΔSsystem > 0, the surroundings contribution −ΔH/T is negative but becomes smaller in magnitude as temperature increases. Above a suitable threshold, the positive system term can dominate.

For an exothermic reaction with ΔSsystem < 0, the positive surroundings contribution becomes smaller as temperature rises. At sufficiently high temperature it may no longer outweigh the negative system term. These statements assume the supplied ΔH and ΔS remain approximately constant and the same states persist.

Use kelvin because temperature appears as a thermodynamic ratio, not as a temperature difference. A calculation at 25 °C needs approximately 298 K. A positive total entropy change identifies the favourable direction, not a reaction rate, percentage yield or time to completion.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources