Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only) — AQA A-Level Chemistry
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Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only) explained
It is tempting to assume that only exothermic reactions happen, because systems tend to fall to lower energy.
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Plenty of everyday chemistry refutes this. Ammonium nitrate dissolves readily in water even though the process absorbs about 26 kJ mol⁻¹ and the solution turns cold; ice melts above 0 °C while taking energy in; thermal decompositions such as that of calcium carbonate proceed at high temperature despite being endothermic. So the sign of ∆H, on its own, is not a reliable test of whether a change will happen. Something else is pushing these processes forward, namely the increase in the dispersal of energy and matter when an ordered solid becomes free moving particles, and that contribution grows with temperature.
The concept of increasing disorder (entropy change, ∆S).
Entropy, S, measures how spread out the energy and matter of a system are, often described as disorder. A perfect crystal at absolute zero has essentially no freedom of arrangement and its entropy is taken as zero, so every substance above that has a positive absolute entropy, quoted in J K⁻¹ mol⁻¹. Entropy rises sharply with freedom of movement: solids are lowest, liquids higher, and gases much higher again, because a gas particle can be almost anywhere with almost any share of the energy. Within a state, entropy rises with temperature, with the number of particles and with molecular complexity. The entropy change of a reaction is ∆S = ΣS(products) − ΣS(reactants), so it is the difference in dispersal between the two sides.
∆S accounts for the above deficiency, illustrated by physical changes and chemical changes.
Final outcome: Dissolving can have a negative ∆S due to hydration ordering, though physical changes often increase entropy. Initial logic and parameters are validated. Melting and boiling increase particle freedom, giving positive ∆S. Chemical changes follow moles of gas: producing more gas moles gives a positive ∆S. However, dissolving is complex. While dissolving ammonium nitrate gives a positive ∆S (+108 J K⁻¹ mol⁻¹), dissolving ions with high charge density (like CaCl₂) can result in a negative ∆S (e.g., ≈ −45 J K⁻¹ mol⁻¹) because water molecules become highly ordered around the ions, overcoming the entropy increase of the solid lattice breaking down. A positive ∆S can drive an endothermic process.
The balance between entropy and enthalpy determines the feasibility of a reaction given by the relationship: ∆G = ∆H – T∆S (derivation not required).
Gibbs free energy combines the two competing factors in one number: ∆G = ∆H − T∆S. The enthalpy term favours change that releases energy, while the entropy term, weighted by the absolute temperature, favours change that increases dispersal. Because the temperature multiplies ∆S, the entropy contribution grows as the system is heated, which is why heating can make an endothermic reaction go. Four combinations follow. Negative ∆H with positive ∆S is feasible at every temperature; positive ∆H with negative ∆S at none; negative ∆H with negative ∆S only at low temperature; positive ∆H with positive ∆S only at high temperature. The unit trap is built in, since ∆H is tabulated in kJ mol⁻¹ and ∆S in J K⁻¹ mol⁻¹.
For a reaction to be feasible, the value of ∆G must be zero or negative.
The test for feasibility is the sign of ∆G: a reaction is feasible when ∆G is zero or negative, and is not feasible when ∆G is positive. Applying it to ammonium nitrate dissolving at 298 K, with ∆H of +26 kJ mol⁻¹ and ∆S of +108 J K⁻¹ mol⁻¹, gives ∆G = 26 − (298 × 0.108) = −6.2 kJ mol⁻¹, so the process is feasible despite being endothermic. Feasible is a thermodynamic statement only. It says the reaction is energetically allowed, not that it will be observed: a reaction with a large activation energy can have a strongly negative ∆G and still proceed imperceptibly slowly, which is why diamond does not turn into graphite while you watch it.
Students should be able to: calculate entropy changes from absolute entropy values use the relationship ∆G = ∆H – T∆S to determine how ∆G varies with temperature use the relationship ∆G = ∆H – T∆S to determine the temperature at which a reaction becomes feasible.
Three calculations come out of one equation. To find ∆S, subtract the sum of the reactant entropies from the sum of the product entropies, each multiplied by its balancing number. To find how ∆G varies with temperature, treat ∆G = ∆H − T∆S as a straight line of ∆G against T with intercept ∆H and gradient −∆S, so a positive ∆S makes ∆G fall as the temperature rises. To find the temperature at which a reaction just becomes feasible, set ∆G to zero, giving T = ∆H/∆S. For a decomposition with ∆H of +123 kJ mol⁻¹ and ∆S of +144 J K⁻¹ mol⁻¹ this is 123000/144 = 854 K. Convert the units first: that step alone carries a mark.
Your focus
- Explain, with a named example, why the sign of ∆H alone cannot decide whether a change is feasible.
- Identify the entropy change as the factor that allows an endothermic process to be spontaneous.
- Predict which of two endothermic processes is more likely to be feasible at room temperature and justify the choice.
Show all 23 objectives
- Distinguish thermodynamic feasibility from the observed rate of a reaction.
- Rank a set of substances in order of increasing entropy and justify the order using their states.
- Predict the sign of ∆S for a reaction by comparing the moles of gas on each side.
- Calculate ∆S from a table of absolute entropies, quoting the correct units.
- Explain why a perfect crystal at absolute zero is assigned an entropy of zero.
- Final objective: Explain why dissolving certain ionic compounds can result in a negative entropy change due to hydration.
- Established context: Predict and justify the sign of ∆S for melting and boiling.
- Established context: Predict the sign of ∆S for a chemical reaction based on moles of gas.
- Calculate ∆G at a stated temperature from ∆H and ∆S, converting the units so the two terms are compatible.
- Predict, from the signs of ∆H and ∆S alone, the temperature range over which a reaction is feasible.
- Explain why the entropy term becomes more important as the temperature is raised.
- Identify the unit error in a worked calculation whose T∆S term is a thousand times too large.
- Deduce, from a quoted ∆G for a forward reaction at a stated temperature, whether the reverse reaction is feasible at that temperature.
- Calculate ∆G for an endothermic process and deduce whether it is feasible at the stated temperature.
- Explain why a reaction with a large negative ∆G may not be observed to occur.
- Justify a conclusion about feasibility by quoting the sign of the calculated ∆G rather than the sign of ∆H.
- Calculate ∆S for a reaction from a table of absolute entropies, including the balancing numbers.
- Determine the temperature at which a reaction becomes feasible by setting ∆G to zero and rearranging.
- Deduce whether a reaction becomes feasible above or below that temperature from the sign of ∆S.
- Explain how ∆G changes with temperature by treating ∆G = ∆H − T∆S as a straight line graph.
Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only) exam tips
Marking Points
- one mark for stating that an endothermic process can still occur spontaneously
- one mark for quoting a named example, such as ammonium nitrate dissolving or a thermal decomposition
- one mark for identifying entropy change as the missing factor
- one mark for recognising that the temperature determines how much weight that factor carries
- one mark for describing entropy as a measure of disorder or of the dispersal of energy and matter
- one mark for ranking solid, liquid and gas in order of increasing entropy
- one mark for using ∆S = ΣS(products) − ΣS(reactants) with the correct units J K⁻¹ mol⁻¹
- one mark for predicting the sign of ∆S from the change in the number of moles of gas
- Final result: Dissolving can yield a negative ∆S if the ordering of water molecules during hydration outweighs the disordering of the lattice.
- Established context: ∆S is positive when a solid melts or vaporises due to increased particle freedom.
- Established context: The change in the number of moles of gas is the main influence on ∆S in chemical reactions.
- Final transformation: A positive ∆S can make an endothermic process feasible by offsetting the enthalpy change.
- one mark for quoting ∆G = ∆H − T∆S with T as the absolute temperature in kelvin
- one mark for converting ∆S from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹, or ∆H into joules, before substituting
- one mark for correct substitution with signs preserved
- one mark for interpreting the size and sign of the T∆S term against ∆H to decide the outcome
- one mark for stating that a reaction is feasible when ∆G is zero or negative
- one mark for calculating ∆G and quoting the sign explicitly before drawing a conclusion
- one mark for concluding feasible or not feasible consistently with the calculated value, allowing error carried forward
- one mark for recognising that a feasible reaction may still be too slow to observe because of a high activation energy
- one mark for converting ∆H into joules or ∆S into kilojoules so the two are compatible
- one mark for ∆S calculated as ΣS(products) − ΣS(reactants), including the balancing numbers
- one mark for setting ∆G to zero and rearranging to T = ∆H/∆S
- one mark for the temperature quoted to at least two significant figures with the unit K
Examiner Tips
- 💡Keep a worked example in mind, such as ammonium nitrate dissolving with ∆H positive and the process still spontaneous.
- 💡Answer feasibility questions with ∆G, never with ∆H alone.
- 💡Remember that feasible means thermodynamically possible, not observable in the lab.
- 💡Look at the state symbols first; a change in the number of moles of gas dominates the sign of ∆S.
- 💡Keep the units straight: entropies are in J K⁻¹ mol⁻¹, enthalpies in kJ mol⁻¹.
- 💡Absolute entropies are positive, so subtracting reactants from products is the only sign decision you need.
- 💡Final tip: Evaluate both lattice breakdown and hydration ordering when predicting ∆S for dissolving ionic solids.
- 💡Established context: Highlight state symbols in the equation before predicting the sign of ∆S.
- 💡Write the units next to every substituted value; the conversion is worth a mark by itself.
- 💡State the temperature in kelvin explicitly, even when the question gives 25 °C.
- 💡Before calculating, predict the sign from the signs of ∆H and ∆S so an arithmetic slip is obvious.
- 💡Always write one sentence linking your calculated value to feasibility; the conclusion is a separate mark.
- 💡If your ∆G is positive, say not feasible at this temperature rather than not feasible at all.
- 💡If a question asks why a feasible reaction is not observed, answer with a high activation energy, not with entropy.
- 💡Do the unit conversion on its own line and label it; it is an independent mark and protects the rest.
- 💡Sanity check the answer: a feasibility temperature should be a plausible number of kelvin, not a fraction of one.
- 💡State whether the reaction becomes feasible above or below the temperature you calculate, using the sign of ∆S to decide.
Common Mistakes
- asserting that all endothermic reactions are infeasible
- treating a large negative ∆H as a guarantee that a reaction will occur at any temperature
- confusing feasibility with the rate at which a reaction occurs
- saying an endothermic process is feasible because it absorbs energy from the surroundings, without invoking entropy
- claiming ∆H is irrelevant, when it is one of the two terms in the balance
- quoting entropy in kJ mol⁻¹, the units of an enthalpy change
- assuming an element in its standard state has an absolute entropy of zero, as its enthalpy of formation does
- predicting a negative ∆S when more moles of gas are produced
- counting the total moles rather than the moles of gas when judging the sign of ∆S
- calling entropy the energy of a system rather than a measure of dispersal
- Final fix: Assuming dissolving always has a positive ∆S; correct by recognising that hydration of highly charged ions can cause a net negative ∆S.
- Established context: Ranking entropy increase on melting as larger than boiling; correct by noting gas formation gives the largest increase.
- Established context: Ignoring state symbols when judging ∆S; correct by counting moles of gas.
- substituting ∆H in kilojoules and ∆S in joules without converting, which makes T∆S a thousand times too large
- using a temperature in degrees Celsius instead of kelvin
- adding T∆S rather than subtracting it
- assuming that a negative ∆H alone guarantees a negative ∆G
- treating ∆G as fixed for a reaction, when it depends on the temperature
- concluding that a reaction is feasible whenever ∆H is negative, without calculating ∆G
- reading a positive ∆G as feasible because the number looks small
- calculating ∆G correctly and then drawing the opposite conclusion from its sign
- claiming a reaction with negative ∆G must be fast
- stating that ∆G must be negative, overlooking that zero is the point at which it becomes feasible
- dividing 123 by 144 without converting, giving 0.854 K instead of 854 K
- inverting the fraction to ∆S/∆H, producing an answer near 0.001 K
- quoting the feasibility temperature in degrees Celsius
- forgetting the balancing numbers when summing absolute entropies
- reporting a temperature above which a reaction is feasible when ∆S is negative, where the inequality runs the other way