Enthalpy and entropy — OCR A-Level Chemistry

This row is a course heading, not an assessed answer target. It introduces the entropy content of section 5.2.2, where you learn that entropy measures how energy is dispersed within a system and that dispersal increases as a system becomes more disordered. Read the heading as a map: the following statements define entropy, then apply it to predicting the direction of change and to calculating entropy changes. Build your understanding by comparing a solid, a liquid and a gas of the same substance: the gas has far more ways to arrange its particles and its energy, so its entropy is highest. Practise explaining everyday spontaneous changes, such as a gas expanding to fill a flask, in terms of increased dispersal of energy rather than in terms of 'randomness' alone. Check that you can move between the qualitative idea of dispersal and quantitative entropy values measured in J K⁻¹ mol⁻¹.

Entropy is a measure of how widely energy is dispersed within a system. When energy is spread over many available arrangements, entropy is high; when it is concentrated in few arrangements, entropy is low. A more disordered system has more ways to arrange its particles and their energy, so its entropy is greater. Compare the same substance as a solid, a liquid and a gas: in the solid, particles vibrate about fixed positions and energy is localised, giving low entropy; in the gas, particles move freely throughout the container and energy is dispersed over many arrangements, giving high entropy. A useful test is to ask how many ways the energy could be shared out: more ways means greater dispersal and greater entropy. Entropy is measured in J K⁻¹ mol⁻¹, so keep the kelvin and per-mole units when values are quoted.

Entropy S measures the dispersal of energy among the accessible microstates of a system, so a larger number of arrangements at a given energy gives a larger entropy. To explain differences in magnitude, compare the freedom of motion and the number of ways particles can be arranged. A gas has far more accessible positions and momenta than a liquid, and a liquid more than a well-ordered solid, so S(gas) > S(liquid) > S(solid) for the same substance. Raising temperature increases the spread of populated energy levels and so increases entropy. Increasing the number of particles, for example in a reaction that produces more moles of gas, also increases entropy because there are more independent ways to distribute the energy. Dissolving an ionic solid increases entropy because ordered ions become free to move in solution.

This item tests whether you can rank the entropy of the three physical states of the same substance and justify the ranking. For a given substance at comparable conditions, the gas has the greatest entropy, the liquid has an intermediate value and the solid has the lowest. The reason is the number of accessible microstates: particles in a gas move freely through a large volume and can occupy many positions and momenta, particles in a liquid can slide past one another but remain close together, and particles in a solid vibrate about fixed lattice positions. When you answer a multiple-choice question, read whether it asks for the order from lowest to highest or the reverse, and check whether the substance and conditions are the same for each state. Eliminate options that place a solid above a gas or that treat entropy as increasing as particles become more ordered.

This point asks you to reason qualitatively about how the entropy of the system changes when the number of gaseous molecules changes during a reaction. Gases have far more freedom of motion and far more accessible microstates than liquids or solids, so a reaction that increases the number of gas molecules usually increases the entropy of the system, and one that decreases the number of gas molecules usually decreases it. Compare the total moles of gas on each side of the equation. For example, N₂O₄(g) → 2NO₂(g) converts one gas molecule into two, so the system entropy rises; the reverse change lowers it. Judge only the sign and direction of the change, and keep the reasoning tied to the gaseous species counted in the balanced equation.

This point requires you to calculate the entropy change of the system for a reaction from tabulated standard entropies. Use ∆S = ΣS(products) − ΣS(reactants), multiplying each standard entropy by its stoichiometric coefficient from the balanced equation. Entropies are usually quoted in J K⁻¹ mol⁻¹, so the result is in J K⁻¹ mol⁻¹; divide by 1000 to express it in kJ K⁻¹ mol⁻¹ when a related quantity needs that unit. A positive ∆S means the system entropy increases, which is typical when gaseous moles increase. Related quantities include the value per mole of a named reactant and unit conversions, so keep track of coefficients and units throughout.

This row is a course-guidance heading within the enthalpy and entropy topic, not an assessed answer target. Read it as a signpost to the Gibbs free-energy relationship ΔG = ΔH − TΔS, which combines the enthalpy change of the system, the entropy change of the system and the temperature in kelvin. Work through your specification and textbook to see how ΔG is calculated from ΔH and ΔS at a stated temperature, and how the sign of ΔG links to whether a process is feasible. Practise converting temperatures to kelvin by adding 273, keeping entropy in J K⁻¹ mol⁻¹ and enthalpy in kJ mol⁻¹ consistent before combining them, and interpreting negative and positive ΔG values. Use past questions to check you can select the correct data and units rather than memorising a fixed answer.

Feasibility is judged by the total entropy change of the universe, which combines two system quantities: the enthalpy change ∆H and the entropy change ∆S, with temperature T in kelvin. The entropy term is written T∆S, so its sign follows ∆S: a positive ∆S gives a positive T∆S, and a negative ∆S gives a negative T∆S. The Gibbs relation ∆G = ∆H − T∆S links them, and a process is feasible when ∆G is negative, equivalently when ∆H − T∆S < 0. A negative ∆H favours feasibility; a positive ∆S also favours it. Temperature scales the T∆S term, so a reaction with positive ∆H and positive ∆S becomes feasible only above a threshold temperature, while one with negative ∆H and negative ∆S is feasible only below a threshold.

The Gibbs equation combines enthalpy and entropy into one feasibility test: ∆G = ∆H – T∆S. Here ∆H is the enthalpy change of the process, ∆S its entropy change and T the temperature in kelvin (K = °C + 273). Both ∆H and ∆S must be expressed in consistent energy units, so convert ∆S from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000 before subtracting. A process is feasible when ∆G is negative. Worked example: ∆H = +50 kJ mol⁻¹, ∆S = +100 J K⁻¹ mol⁻¹ at 298 K gives T∆S = 298 × 0.100 = 29.8 kJ mol⁻¹, so ∆G = 50 − 29.8 = +20.2 kJ mol⁻¹, not feasible. At 600 K, T∆S = 60 kJ mol⁻¹ and ∆G = −10 kJ mol⁻¹, feasible. The sign of ∆G also shows how temperature shifts feasibility: when ∆H and ∆S have the same sign, one temperature range is feasible and the other is not.

A negative ∆G tells you a process is thermodynamically feasible, but it says nothing about how fast the process happens. Feasibility is a thermodynamic judgement about the relative free energy of reactants and products; the rate is a kinetic matter governed by activation energy and the reaction pathway. A reaction with a large negative ∆G can still be immeasurably slow at room temperature if its activation energy is high, because few particles have enough energy to overcome the barrier. For example, the combustion of methane is strongly feasible yet methane is stable in air until ignited. Conversely, a process with a positive ∆G is not feasible overall, though local or coupled steps may occur. So ∆G predicts whether a process can happen, not whether it will happen at an observable rate; catalysts change rate by lowering activation energy without altering ∆G.