Part 1: Entropy and the balance of energy dispersal

Explain entropy changes from physical states and particle numbers, then calculate ΔS using a balanced equation.

Energy can be distributed in different ways

Entropy describes the dispersal of energy in a system. A system with more accessible ways to distribute energy has greater entropy; “disorder” is a useful shorthand only when connected to the actual particles and their freedom of arrangement and motion.

For the same substance, a gas normally has higher entropy than its liquid and solid at comparable conditions because its particles can occupy far more positions and distribute energy more widely. Melting and vaporisation therefore have positive entropy changes for the substance. Heating within one phase also generally increases entropy.

Do not confuse S, the entropy of a specified substance, with ΔS, the change for a process. Standard entropy of an element at room temperature is not zero merely because its standard enthalpy of formation is zero. Those are different reference conventions and different physical quantities.

Use gaseous particles as a strong clue, not a substitute for data

For N₂(g) + 3H₂(g) → 2NH₃(g), four moles of gaseous molecules become two. The reduced opportunities for dispersal suggest a negative entropy change. For CaCO₃(s) → CaO(s) + CO₂(g), gas is produced from solids and entropy increases.

Count stoichiometric gaseous amounts rather than the number of distinct chemical formulae. A change from one gaseous species to two does not establish a sign without looking at coefficients. Where competing effects are not clear, calculate using the supplied entropies instead of claiming that a simple count settles every reaction.

Worked example: coefficients and units matter

For original illustrative entropy data in J K⁻¹ mol⁻¹, take S(N₂) = 190, S(H₂) = 130 and S(NH₃) = 195. For N₂ + 3H₂ → 2NH₃, ΔS = 2(195) − [190 + 3(130)] = 390 −580 = −190 J K⁻¹ mol⁻¹ of reaction as written.

The negative sign matches the gas-amount argument. If the balanced equation is halved, the entropy change for that new reaction is −95 J K⁻¹ mol⁻¹. Reversing it changes the sign. Do not multiply the entire answer again after already using the coefficients.

Why an entropy decrease can still accompany a feasible reaction

A reaction may make its chemical system more ordered while releasing heat that increases energy dispersal in the surroundings. This is why “ΔS of the reacting chemicals is negative, so the reaction cannot happen” is incomplete. The enthalpy and temperature must also be considered.

At constant temperature and pressure, the Gibbs equation combines these effects in a practical criterion. First calculate the system entropy change correctly; then combine it with the enthalpy change using consistent energy units. The next part shows that an exothermic process with negative ΔS can be favourable at low temperatures.

Quick checks

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

Sources