In the notes below, material marked A‑level only is not required for AS students but is essential for the full A‑level syllabus.
This short section supplies the background that A‑level topics build on.
Combustion of propane (simplified, ΔH_f° values only):
\[ \mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)} \] \[ \Delta H^{\circ}_{\text{rxn}} = \big[3(-393.5) + 4(-285.8)\big] - \big[-104.7 + 5(0)\big] = -2\,221\ \text{kJ mol}^{-1} \]The large negative ΔH tells us the reaction is exothermic; no entropy or ΔG is needed at AS level.
Entropy measures the dispersal of energy and matter in a system. Statistically it reflects the number of microscopic arrangements (microstates) compatible with a macroscopic state.
\[ S = k_{\mathrm{B}} \ln W \]For any spontaneous process at constant temperature and pressure:
\[ \Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 \]When the process occurs at constant pressure (the usual case in chemistry), the heat exchanged with the surroundings is $q_{\text{p}} = \Delta H_{\text{system}}$. Therefore:
\[ \Delta S_{\text{surroundings}} = -\frac{\Delta H_{\text{system}}}{T} \]Tabulated at 298 K and 1 bar. By the Third Law, a perfect crystal at 0 K has $S = 0$, so all $S^{\circ}$ values are positive.
| Substance | $S^{\circ}$ (J mol‑1K‑1) | State (298 K) |
|---|---|---|
| H₂(g) | 130.7 | Gas |
| O₂(g) | 205.0 | Gas |
| H₂O(l) | 69.9 | Liquid |
| CO₂(g) | 213.7 | Gas |
Dissolution of ammonium chloride, $\mathrm{NH_4Cl(s) \rightarrow NH_4^{+}(aq) + Cl^{-}(aq)}$.
ΔG° values are tabulated for many reactions; otherwise calculate from ΔH° and ΔS°.
ΔH° ΔS° ──► ΔG° = ΔH° – TΔS° ──► Sign of ΔG° → Spontaneity?
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ΔG° = –RT ln K
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Equilibrium constant K
From the Zn/Cu cell example (see §5.3) we obtained $\Delta G^{\circ} = -2.12\times10^{5}\ \text{J mol}^{-1}$.
\[ E^{\circ} = -\frac{\Delta G^{\circ}}{nF} = -\frac{-2.12\times10^{5}}{2 \times 96\,485} = +1.10\ \text{V} \]The value matches the tabulated cell potential, confirming the consistency of the thermodynamic relationships.
A plot of $\ln K$ versus $1/T$ is linear:
Equilibrium: $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$
| T (K) | K |
|---|---|
| 500 | 1.0×10⁻³ |
| 600 | 4.0×10⁻³ |
| 700 | 1.5×10⁻² |
Calculate $\ln K$ and $1/T$, fit a straight line (by hand or calculator). The slope obtained is $-1.20\times10^{4}\ \text{K}$, giving
\[ \Delta H^{\circ} = -\text{slope}\times R = (1.20\times10^{4})(8.314) \approx 1.0\times10^{5}\ \text{J mol}^{-1} \]The positive ΔH° shows the synthesis of ammonia is endothermic under these conditions.
When ΔH° and ΔS° have the same sign, the sign of ΔG can change with temperature. The temperature at which the reaction becomes spontaneous (ΔG = 0) is:
\[ T_{\text{eq}} = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \]A negative ΔG tells us that a reaction is thermodynamically favourable, but it says nothing about how fast it proceeds. The activation energy $E_a$ governs the rate; catalysts lower $E_a$ without changing ΔG.
| Species | $S^{\circ}$ (J mol‑1K‑1) |
|---|---|
| N₂(g) | 191.5 |
| O₂(g) | 205.0 |
| NO(g) | 210.8 |
Positive ΔS° favours spontaneity at higher temperatures (see §3.8).
Reaction: $\displaystyle \mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$
| Species | ΔH_f° (kJ mol⁻¹) | S° (J mol⁻¹K⁻¹) |
|---|---|---|
| CH₄(g) | -74.8 | 186.3 |
| O₂(g) | 0.0 | 205.0 |
| CO₂(g) | -393.5 | 213.7 |
| H₂O(l) | -285.8 | 69.9 |
Such a huge $K$ confirms the reaction proceeds essentially to completion.
Cell potential:
\[ E^{\circ}_{\text{cell}} = 0.34 - (-0.76) = +1.10\ \text{V} \]Corresponding ΔG° (using the ΔG° ↔ E° link):
\[ \Delta G^{\circ} = -nF E^{\circ}_{\text{cell}} = -(2)(96\,485)(1.10) = -2.12\times10^{5}\ \text{J mol}^{-1} \]From ΔG° to equilibrium constant:
\[ K = \exp\!\Bigl(\frac{2.12\times10^{5}}{8.314\times298}\Bigr) \approx 1.6\times10^{37} \]Consider the reaction $\mathrm{A(g) \rightarrow B(g)}$ with $\Delta H^{\circ}=+40\ \text{kJ mol}^{-1}$ and $\Delta S^{\circ}=+120\ \text{J mol}^{-1}\text{K}^{-1}$.
\[ T_{\text{eq}} = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{40\,000}{120} \approx 333\ \text{K} \]Below 333 K, ΔG > 0 (non‑spontaneous); above 333 K, ΔG < 0 (spontaneous).
| Quantity | Expression | Typical Use |
|---|---|---|
| Statistical entropy | $S = k_{\mathrm{B}}\ln W$ | Conceptual basis of disorder. |
| Standard entropy change | $\Delta S^{\circ}_{\text{rxn}} = \sumu_i S^{\circ}_{\text{products}} - \sumu_i S^{\circ}_{\text{reactants}}$ | Calculate ΔS° for a reaction. |
| Enthalpy change | $\Delta H = \sum n\Delta H_f^{\circ}(\text{products}) - \sum n\Delta H_f^{\circ}(\text{reactants})$ | Hess’s law, bond‑energy calculations. |
| Gibbs free energy | $\Delta G = \Delta H - T\Delta S$ | Assess spontaneity. |
| Standard Gibbs free energy | $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$ | Link to equilibrium constant & cell potential. |
| ΔG° ↔ K | $\Delta G^{\circ} = -RT\ln K$ | Calculate K or ΔG°. |
| ΔG° ↔ E° | $\Delta G^{\circ} = -nF E^{\circ}$ | Electrochemical cell calculations. |
| van ’t Hoff equation | $\ln K = -\dfrac{\Delta H^{\circ}}{R}\dfrac{1}{T} + \dfrac{\Delta S^{\circ}}{R}$ | Determine ΔH° and ΔS° from temperature‑dependence of K. |
| Temperature of spontaneity | $T_{\text{eq}} = \dfrac{\Delta H^{\circ}}{\Delta S^{\circ}}$ | Find the temperature at which ΔG changes sign. |
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