This entire sub-topic is additional higher level. It does not exist in the SL course,
and SL candidates are not examined on any of it. If you are HL, it is five hours of teaching time and
it answers a question the rest of Reactivity 1 leaves open: why do some endothermic reactions happen
at all?
🎯What you need to be able to do
- Predict whether a physical or chemical change will increase or decrease the entropy of a system.
- Calculate standard entropy changes \( \Delta S^{\ominus} \) from standard entropy values.
- Apply \( \Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} \) to calculate any unknown term, taking care over units.
- Interpret the sign of \( \Delta G \) and determine the temperature at which a reaction becomes spontaneous.
- Perform calculations using \( \Delta G = \Delta G^{\ominus} + RT\ln Q \) and, at equilibrium, \( \Delta G^{\ominus} = -RT\ln K \).
📚The chemistry
Entropy
Entropy, \( S \), is a measure of the dispersal or distribution of matter and/or energy in a
system. The more ways the energy can be distributed among the particles, the higher the
entropy. The older description — “a measure of disorder” — is a serviceable
intuition but the syllabus wording is about dispersal, and that is what you should write.
Under the same conditions:
\[ S(\text{gas}) \;>\; S(\text{liquid}) \;>\; S(\text{solid}) \]
To predict the sign of \( \Delta S \) for a reaction, look for these in order:
- A change in the number of moles of gas dominates everything else. More gas
molecules on the right means \( \Delta S \) is positive; fewer means negative. If the moles of gas
are unchanged, look further.
- A change of state — melting, boiling and dissolving a solid all increase
entropy; freezing and condensing decrease it.
- An increase in the number of particles — one molecule becoming two
increases entropy even without a change of state.
- Increasing temperature increases entropy, since energy can be spread over more
accessible levels.
Standard entropy values \( S^{\ominus} \) are in the data booklet, in
J K−1 mol−1 — note the joules, not kilojoules.
Unlike enthalpies of formation, elements do not have zero standard entropy: only a
perfect crystal at 0 K has zero entropy, because that is the only state in which there is exactly one
way to arrange the energy.
\[ \Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants}) \]
Gibbs energy
Enthalpy alone does not decide whether a reaction happens: ammonium nitrate dissolving is endothermic
and happens readily, and ice melts at room temperature although it absorbs energy. What decides is
Gibbs energy, which combines the enthalpy and entropy contributions:
\[ \Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} \]
At constant pressure, a change is spontaneous if \( \Delta G \) is negative.
“Spontaneous” means thermodynamically feasible — it says nothing whatever about
rate. The conversion of diamond into graphite has a negative \( \Delta G \) at room
temperature and is not something anyone worries about, because the activation energy is enormous.
Units. This is the single biggest source of lost marks in R1.4.
\( \Delta H \) is in kJ mol−1 but \( \Delta S \) is in
J K−1 mol−1. Before substituting into
\( \Delta G = \Delta H - T\Delta S \) you must convert one of them — almost always by
dividing \( \Delta S \) by 1000. Forget, and your answer is out by a factor of a
thousand and will usually still look like a plausible number. And \( T \) is in
kelvin, always.
The four cases
Since \( \Delta G = \Delta H - T\Delta S \), the signs of \( \Delta H \) and \( \Delta S \) between
them determine when a reaction is spontaneous:
\( \Delta H \) negative, \( \Delta S \) positive
spontaneous at all temperatures
both terms drive \( \Delta G \) negative
\( \Delta H \) positive, \( \Delta S \) negative
never spontaneous
both terms drive \( \Delta G \) positive
\( \Delta H \) negative, \( \Delta S \) negative
spontaneous at low temperature
the \( -T\Delta S \) term is positive and grows with \( T \), so heating eventually stops it
\( \Delta H \) positive, \( \Delta S \) positive
spontaneous at high temperature
the \( -T\Delta S \) term is negative and grows with \( T \), so heating eventually starts it
In the last two cases there is a crossover temperature at which
\( \Delta G = 0 \) and the reaction changes from non-spontaneous to spontaneous. Set
\( \Delta G = 0 \) and rearrange:
\[ T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} \]
This is why limestone decomposes in a kiln but not on a wall: \( \Delta H \) is +178 kJ
mol−1 and \( \Delta S \) is strongly positive (a gas is produced), so there is a
temperature above which it goes.
Gibbs energy and equilibrium
As a reaction proceeds towards equilibrium, \( \Delta G \) becomes less negative
and finally reaches zero at equilibrium — which is the thermodynamic definition
of equilibrium. Away from standard conditions:
\[ \Delta G = \Delta G^{\ominus} + RT\ln Q \]
where \( Q \) is the reaction quotient (R2.3). At
equilibrium \( \Delta G = 0 \) and \( Q = K \), so:
\[ \Delta G^{\ominus} = -RT\ln K \]
Both are in the data booklet, with \( R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \) — joules again,
so \( \Delta G^{\ominus} \) comes out in J mol−1 unless you convert.
The relationship is worth reading qualitatively, because that is how it is most often examined:
- \( \Delta G^{\ominus} \) negative ⇒ \( \ln K \) positive ⇒
\( K > 1 \): products are favoured at equilibrium.
- \( \Delta G^{\ominus} = 0 \) ⇒ \( K = 1 \): comparable amounts of both.
- \( \Delta G^{\ominus} \) positive ⇒ \( K < 1 \): reactants are favoured,
though the reaction still proceeds to some extent — a positive
\( \Delta G^{\ominus} \) does not mean nothing happens.
Because the relationship is logarithmic, a modest change in
\( \Delta G^{\ominus} \) produces an enormous change in \( K \). Around room temperature, every
−5.7 kJ mol−1 multiplies \( K \) by roughly ten.
✏️Worked example
For the thermal decomposition of calcium carbonate,
\( \mathrm{CaCO_3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO_2}(g) \):
\( \Delta H^{\ominus} = +178\ \mathrm{kJ\,mol^{-1}} \);
\( S^{\ominus} \) values are \( \mathrm{CaCO_3} \) 92.9, \( \mathrm{CaO} \) 39.7 and
\( \mathrm{CO_2} \) 213.8 J K−1 mol−1.
(a) Predict the sign of \( \Delta S^{\ominus} \) without calculating, then calculate it.
(b) Calculate \( \Delta G^{\ominus} \) at 298 K and state whether the reaction is spontaneous
at that temperature.
(c) Determine the temperature above which the reaction becomes spontaneous.
(d) Calculate \( K \) at 298 K, taking \( R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \), and
comment on the value.
(a) A solid produces a solid and a gas, so the number of moles of gas
rises from 0 to 1. Gases have by far the highest entropy, so \( \Delta S^{\ominus} \) must be
positive. Calculating:
\[ \Delta S^{\ominus} = (39.7 + 213.8) - (92.9) = 253.5 - 92.9 = +160.6\ \mathrm{J\,K^{-1}\,mol^{-1}} \]
(b) Convert \( \Delta S^{\ominus} \) to kJ before substituting:
\( +0.1606\ \mathrm{kJ\,K^{-1}\,mol^{-1}} \).
\[ \Delta G^{\ominus} = 178 - (298)(0.1606) = 178 - 47.9 = +130\ \mathrm{kJ\,mol^{-1}} \]
\( \Delta G^{\ominus} \) is positive, so the reaction is not
spontaneous at 298 K — limestone does not decompose at room temperature, which is
fortunate for every building made of it.
(c) Set \( \Delta G^{\ominus} = 0 \):
\[ T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \frac{178}{0.1606} = 1108\ \mathrm{K} \quad (\approx 835\ ^{\circ}\mathrm{C}) \]
Above about 1110 K the \( -T\Delta S^{\ominus} \) term outweighs the positive
\( \Delta H^{\ominus} \), \( \Delta G^{\ominus} \) turns negative, and the decomposition becomes
spontaneous. This is why a lime kiln is run at around 1200 K.
(d) Use \( \Delta G^{\ominus} = -RT\ln K \), with
\( \Delta G^{\ominus} \) in joules to match \( R \):
\[ \ln K = -\frac{\Delta G^{\ominus}}{RT} = -\frac{130\,000}{(8.31)(298)} = -\frac{130\,000}{2476} = -52.5 \]
\[ K = e^{-52.5} = 1.6 \times 10^{-23} \]
An extraordinarily small value: at room temperature the equilibrium lies overwhelmingly to
the left and the partial pressure of carbon dioxide above solid limestone is negligible. Note
how a \( \Delta G^{\ominus} \) of only +130 kJ mol−1 — not a large number by
the standards of this course — corresponds to \( K \) being twenty-three orders of magnitude
below one. That is the logarithm at work.
Check it. Four independent checks are available here, and using them will catch
almost every error. Sign of \( \Delta S \): gas produced, so positive — and it came
out positive. Magnitude of \( T\Delta S \): at 298 K it should be a few tens of kJ, and 47.9
is; a value of 47 900 means the J-to-kJ conversion was missed. Consistency of (b) and (c):
the reaction is not spontaneous at 298 K, so the crossover temperature must be above
298 K — and 1108 K is. Consistency of (b) and (d): \( \Delta G^{\ominus} \) is
positive, so \( K \) must be less than 1, and \( 10^{-23} \) certainly is.
Reading “spontaneous” as “fast”. A negative
\( \Delta G \) says only that a reaction is
thermodynamically feasible — that it can
proceed without a continuous input of energy. It says nothing about how long it takes, because rate
is decided by the
activation energy
(
R2.2), which does not appear in the Gibbs equation
at all. The combustion of petrol in air is spontaneous at room temperature and does not happen
without a spark; the conversion of diamond to graphite is spontaneous and takes geological time.
Whenever a question offers “the reaction will be fast” as a conclusion from
\( \Delta G \), it is wrong.
📝Practise
Work through these on paper, then reveal the answer. All are HL only.
1. Predict, with reasons, the sign of \( \Delta S \) for: (a) \( \mathrm{H_2O}(l) \rightarrow \mathrm{H_2O}(g) \), (b) \( \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightarrow 2\mathrm{NH_3}(g) \), (c) \( \mathrm{NH_4NO_3}(s) \rightarrow \mathrm{NH_4^{+}}(aq) + \mathrm{NO_3^{-}}(aq) \), (d) \( 2\mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow 2\mathrm{H_2O}(l) \).
(a) Positive. A liquid becomes a gas: the particles become far more widely dispersed and the energy can be distributed over many more arrangements. (b) Negative. Four moles of gas become two, so the number of gas particles halves; fewer particles means fewer ways of distributing the energy. (c) Positive. An ordered solid lattice breaks up into ions dispersed throughout the solution — a large increase in the dispersal of matter. (This is precisely why the process is spontaneous even though it is endothermic.) (d) Strongly negative. Three moles of gas become two moles of liquid: both the reduction in particle number and the change of state to a much more ordered phase decrease entropy. Note that (d) is nevertheless spontaneous, because \( \Delta H \) is very large and negative — enthalpy wins.
2. For \( 2\mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow 2\mathrm{H_2O}(l) \), \( \Delta H^{\ominus} = -572\ \mathrm{kJ\,mol^{-1}} \) and \( \Delta S^{\ominus} = -327\ \mathrm{J\,K^{-1}\,mol^{-1}} \). Calculate \( \Delta G^{\ominus} \) at 298 K and state whether there is a temperature above which the reaction ceases to be spontaneous.
Convert the entropy: \( \Delta S^{\ominus} = -0.327\ \mathrm{kJ\,K^{-1}\,mol^{-1}} \). \( \Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = -572 - (298)(-0.327) = -572 + 97.4 = \mathbf{-475\ kJ\,mol^{-1}} \). Strongly negative, so the reaction is spontaneous at 298 K. Yes, in principle there is such a temperature. Both \( \Delta H \) and \( \Delta S \) are negative, which is the “spontaneous at low temperature” case: as \( T \) rises, the \( -T\Delta S \) term becomes an increasingly large positive contribution, and at \( T = \dfrac{\Delta H}{\Delta S} = \dfrac{-572}{-0.327} = 1749\ \mathrm{K} \) it exactly cancels the enthalpy term. Above about 1750 K, \( \Delta G \) would be positive. (In practice the water is a gas long before then, so the \( \Delta H \) and \( \Delta S \) values used no longer apply — a good reminder that these calculations assume the values are temperature-independent, which is only approximately true.)
3. Explain, using Gibbs energy, why ammonium nitrate dissolving in water is spontaneous even though it is endothermic.
Spontaneity is decided by \( \Delta G = \Delta H - T\Delta S \), not by \( \Delta H \) alone. For this process \( \Delta H \) is positive (about +25 kJ mol−1) — the lattice enthalpy required to separate the ions exceeds the energy released on hydrating them, which is why the solution gets cold. But \( \Delta S \) is large and positive: a highly ordered crystalline lattice is replaced by ions dispersed randomly throughout the solvent, an enormous increase in the dispersal of matter. At 298 K the \( -T\Delta S \) term is therefore a substantial negative contribution, and it outweighs the positive \( \Delta H \). The result is \( \Delta G < 0 \) and the process is spontaneous. This is the general answer to “how can an endothermic process happen?” — it happens when the entropy increase is large enough, and it becomes more favourable as the temperature rises.
4. For \( \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightarrow 2\mathrm{NH_3}(g) \), \( \Delta H^{\ominus} = -92\ \mathrm{kJ\,mol^{-1}} \) and \( \Delta S^{\ominus} = -199\ \mathrm{J\,K^{-1}\,mol^{-1}} \). (a) Calculate \( \Delta G^{\ominus} \) at 298 K. (b) Find the temperature above which the reaction is no longer spontaneous. (c) Comment on the industrial implication.
(a) \( \Delta S^{\ominus} = -0.199\ \mathrm{kJ\,K^{-1}\,mol^{-1}} \). \( \Delta G^{\ominus} = -92 - (298)(-0.199) = -92 + 59.3 = \mathbf{-32.7\ kJ\,mol^{-1}} \) — negative, so spontaneous at 298 K. (b) Setting \( \Delta G^{\ominus} = 0 \): \( T = \dfrac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \dfrac{-92}{-0.199} = \mathbf{462\ K} \) (about 189 °C). Above this temperature \( \Delta G^{\ominus} \) becomes positive and the forward reaction is no longer thermodynamically favoured. (c) This is the central dilemma of the
Haber process. Thermodynamics says run it
cold, since the yield of ammonia is greatest at low temperature; kinetics says run it
hot, because at 462 K the reaction is impossibly slow. Industry therefore uses a
compromise temperature of roughly 700 K together with an
iron catalyst and
high pressure, accepting a lower equilibrium yield in exchange for reaching it in a usable time and recycling the unreacted gases. See
R2.3.
5. A reaction has \( \Delta G^{\ominus} = -22.8\ \mathrm{kJ\,mol^{-1}} \) at 298 K. Calculate \( K \), and state what happens to \( K \) if \( \Delta G^{\ominus} \) becomes \( -11.4\ \mathrm{kJ\,mol^{-1}} \).
Use \( \Delta G^{\ominus} = -RT\ln K \), converting to joules: \( \ln K = -\dfrac{\Delta G^{\ominus}}{RT} = -\dfrac{-22\,800}{(8.31)(298)} = \dfrac{22\,800}{2476} = 9.21 \). So \( K = e^{9.21} = \mathbf{1.0 \times 10^{4}} \) — a large value, so the equilibrium lies well to the right and products dominate. If \( \Delta G^{\ominus} \) is halved to −11.4 kJ mol−1, then \( \ln K = 4.60 \) and \( K = e^{4.60} = \mathbf{1.0 \times 10^{2}} \). Halving \( \Delta G^{\ominus} \) does not halve \( K \); it takes its square root. That is the practical meaning of the logarithmic relationship, and it is why quite modest changes in \( \Delta G^{\ominus} \) — from a catalyst-free change of temperature, say — can move an equilibrium constant by many orders of magnitude.
6. Explain what \( \Delta G = 0 \) means physically, and why a reaction with a positive \( \Delta G^{\ominus} \) still proceeds to some extent.
\( \Delta G = 0 \) is the condition for equilibrium. As a reaction proceeds, \( \Delta G \) becomes progressively less negative, because the reaction quotient \( Q \) rises and \( \Delta G = \Delta G^{\ominus} + RT\ln Q \). When \( \Delta G \) reaches zero there is no longer any thermodynamic driving force in either direction: the forward and reverse processes are equally favoured, the composition stops changing, and \( Q = K \). Physically, the system has reached the composition of minimum Gibbs energy, and any movement away from it in either direction would raise \( G \). Why a positive \( \Delta G^{\ominus} \) does not mean “no reaction”: \( \Delta G^{\ominus} \) refers to standard conditions, meaning all species present at unit concentration or pressure. At the start of a real reaction there are no products at all, so \( Q \) is very small, \( \ln Q \) is a large negative number, and \( \Delta G = \Delta G^{\ominus} + RT\ln Q \) can easily be negative even when \( \Delta G^{\ominus} \) is positive. The reaction therefore proceeds until \( Q \) has risen enough to bring \( \Delta G \) to zero. A positive \( \Delta G^{\ominus} \) means \( K < 1 \) — the equilibrium favours reactants — not that nothing happens.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and
everything above it on this page still stands.
- Your data booklet — the standard entropy table, the Gibbs equation, \( \Delta G = \Delta
G^{\ominus} + RT\ln Q \), \( \Delta G^{\ominus} = -RT\ln K \) and the value of \( R \). Everything
on this page except the reasoning is provided in the examination.
- Chemistry LibreTexts — a careful treatment of why entropy is about the dispersal of energy
rather than “disorder”, which is worth reading once if the definition feels
hand-waving.
- Khan Academy — worked Gibbs energy and equilibrium-constant problems, useful for drilling
the unit conversions until they stop costing marks.