What spontaneity means
All naturally occurring processes proceed in one direction only. Heat flows from a hot body to a cold one, a gas expands to fill its container, and carbon burns in dioxygen — but none of these reverses on its own. The first law permits the reverse just as readily as the forward change, so it cannot be the arbiter of direction. NCERT frames the central question of §5.6 precisely this way: what is the driving force of a spontaneously occurring change?
A common misreading is that a spontaneous reaction must happen the moment reactants meet. NCERT dismantles this with the hydrogen–oxygen mixture, which can sit for years at room temperature with no perceptible change yet is still called spontaneous. The reaction proceeds at an extremely slow rate, but it has the potential to proceed unaided.
A spontaneous process is one that has the potential to proceed without the assistance of any external agency. It is an irreversible process and may only be reversed by some external agency. Spontaneity says nothing about the rate.
Why enthalpy alone fails
By analogy with a stone falling or water flowing downhill — both losing potential energy — one might guess that reactions are spontaneous because energy decreases, as in exothermic reactions. Several exothermic reactions are indeed spontaneous:
$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$, $\;\Delta_r H^{\ominus} = -46.1\ \text{kJ mol}^{-1}$, and $\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$, $\;\Delta_r H^{\ominus} = -285.8\ \text{kJ mol}^{-1}$. The fall in enthalpy looks like a reasonable driving force.
The argument collapses against endothermic reactions that are nonetheless spontaneous. NCERT cites $\ce{N2(g) + O2(g) -> 2NO(g)}$ with $\Delta_r H = +33.2\ \text{kJ mol}^{-1}$ and $\ce{C(graphite,s) + 2S(l) -> CS2(l)}$ with $\Delta_r H = +128.5\ \text{kJ mol}^{-1}$. Both absorb heat, yet both proceed. A decrease in enthalpy is therefore a contributory factor for spontaneity, but it is not the whole story.
Entropy as a measure of disorder
Consider two gases A and B separated by a partition in an isolated container. Before mixing, a molecule drawn from the left is certainly A and one from the right is certainly B. Remove the partition and the gases diffuse; now a molecule picked at random could be either. The system has become less predictable — more chaotic. This drives diffusion even though $\Delta H = 0$.
NCERT names this new state function entropy, symbol $S$: a measure of the degree of randomness or disorder in a system. The greater the disorder in an isolated system, the higher its entropy. For a given substance the crystalline solid is the state of lowest entropy (most ordered) and the gaseous state is the state of highest entropy.
To make entropy quantitative, NCERT relates it to the heat exchanged. Heat has a randomising influence, but its effect depends on temperature: a system already hot is more chaotic, so the same heat added at a lower temperature produces a greater fractional rise in disorder. Entropy change is therefore inversely proportional to temperature, and for a reversible process:
$$ \Delta S = \frac{q_{\text{rev}}}{T} $$
Entropy is a state function, so $\Delta S$ depends only on the initial and final states, not the path. Its SI unit is $\text{J K}^{-1}\,\text{mol}^{-1}$.
The ΔS_total criterion and second law
The genuine test of spontaneity is not the entropy of the system alone but the entropy of the system plus its surroundings — effectively the universe. NCERT states it as:
$$ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 \quad \text{(spontaneous)} $$
At equilibrium the entropy of an isolated system is maximum and $\Delta S = 0$. Entropy rises during a spontaneous process until it reaches that maximum. This is the second law of thermodynamics: for an isolated system, the natural direction of a spontaneous change is the one that increases entropy.
The surroundings term is what rescues exothermic reactions. Heat released by the system is gained by the surroundings, so the heat absorbed by the surroundings equals $-\Delta H_{\text{system}}$, giving $\Delta S_{\text{surroundings}} = -\Delta H_{\text{system}}/T$. An exothermic reaction therefore raises the surroundings' entropy strongly, which is exactly why spontaneous exothermic reactions are so common even when the system's own entropy falls.
For the oxidation of iron, $\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$, the entropy change is $\Delta_r S = -549.4\ \text{J K}^{-1}\,\text{mol}^{-1}$ at 298 K and $\Delta_r H = -1648\times10^{3}\ \text{J mol}^{-1}$. Despite the negative entropy change, why is the reaction spontaneous?
Spontaneity is decided by $\Delta S_{\text{total}}$, not by $\Delta S_{\text{sys}}$ alone. The surroundings gain the heat released:
$$ \Delta S_{\text{surr}} = \frac{-\Delta_r H}{T} = \frac{1648\times10^{3}}{298} = 5530\ \text{J K}^{-1}\,\text{mol}^{-1} $$
$$ \Delta S_{\text{total}} = -549.4 + 5530 = 4980.6\ \text{J K}^{-1}\,\text{mol}^{-1} > 0 $$
Because $\Delta S_{\text{total}}$ is positive, the rusting of iron is spontaneous even though the solid product is far more ordered than the reactants.
Spontaneous ≠ fast, and don't forget the sign of ΔS_surroundings
Two errors recur. First, students equate "spontaneous" with "rapid" — but a spontaneous reaction can be immeasurably slow (the H₂/O₂ mixture). Rate belongs to kinetics; spontaneity belongs to thermodynamics. Second, the surroundings' entropy change carries a minus sign: $\Delta S_{\text{surr}} = -\Delta H_{\text{sys}}/T$. An exothermic system ($\Delta H_{\text{sys}} < 0$) gives a positive $\Delta S_{\text{surr}}$. Forgetting this sign flips the whole spontaneity verdict.
Rule: judge spontaneity by $\Delta S_{\text{total}} > 0$ (or $\Delta G < 0$). Never read the rate from the sign of $\Delta G$.
Gibbs energy: G = H − TS
Tracking the surroundings every time is awkward, because most chemical reactions occur in closed or open systems with changes in both enthalpy and entropy. NCERT defines a new state function, the Gibbs energy $G$, that bundles both effects into one quantity belonging to the system alone:
$$ G = H - TS $$
Gibbs energy is an extensive property and a state function. At constant temperature, the change for the system is:
$$ \Delta G = \Delta H - T\Delta S \qquad \text{(the Gibbs equation)} $$
This is one of the most important equations in chemistry. Starting from $\Delta S_{\text{total}} = \Delta S_{\text{sys}} - \Delta H_{\text{sys}}/T$ and multiplying by $T$ gives $T\Delta S_{\text{total}} = T\Delta S_{\text{sys}} - \Delta H_{\text{sys}}$. For a spontaneous process $\Delta S_{\text{total}} > 0$, so $T\Delta S_{\text{sys}} - \Delta H_{\text{sys}} > 0$, which is exactly $-\Delta G > 0$. Hence the Gibbs criterion at constant temperature and pressure:
| Sign of ΔG | Verdict | Reading |
|---|---|---|
ΔG < 0 | Spontaneous | Process proceeds in the forward direction |
ΔG = 0 | Equilibrium | No net change; free energy is at a minimum |
ΔG > 0 | Non-spontaneous | Forward process requires an external agency |
Physically, $\Delta H$ is the energy change of the reaction and $T\Delta S$ is the part not available to do useful work. $\Delta G$ is the net energy available to do useful work — which is why $G$ was historically called the free energy.
Once you can read the sign of $\Delta G$, the next step is linking it to $K$ through $\Delta_r G^{\ominus} = -RT\ln K$. See Gibbs Energy and Equilibrium.
The four ΔH/ΔS sign cases
Because $\Delta G = \Delta H - T\Delta S$ carries a temperature in the entropy term, the spontaneity verdict can flip as $T$ changes. NCERT Table 5.4 organises every possibility according to the signs of $\Delta H$ and $\Delta S$. This master table is the single most exam-relevant object in the topic.
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneity |
|---|---|---|---|
| − (exothermic) | + (disorder ↑) | Always − | Spontaneous at all temperatures |
| − (exothermic) | − (order ↑) | − at low T, + at high T | Spontaneous at low temperature |
| + (endothermic) | + (disorder ↑) | + at low T, − at high T | Spontaneous at high temperature |
| + (endothermic) | − (order ↑) | Always + | Non-spontaneous at all temperatures |
Two cases give a verdict independent of temperature: when the signs of $\Delta H$ and $\Delta S$ favour the same direction. The other two are temperature-dependent — and these are where NEET sets its crossover-temperature problems. The reasoning is purely the relative size of $\Delta H$ versus $T\Delta S$.
Finding the crossover temperature
The boundary between spontaneous and non-spontaneous behaviour is the temperature at which $\Delta G = 0$. Setting the Gibbs equation to zero and solving for $T$:
$$ \Delta G = \Delta H - T\Delta S = 0 \;\;\Longrightarrow\;\; T = \frac{\Delta H}{\Delta S} $$
This assumes $\Delta H$ and $\Delta S$ do not vary with temperature — the standard NEET assumption. For an endothermic, entropy-increasing reaction the reaction is spontaneous above this temperature; for an exothermic, entropy-decreasing reaction it is spontaneous below it. A unit caution is essential: $\Delta H$ is usually quoted in $\text{kJ mol}^{-1}$ while $\Delta S$ is in $\text{J K}^{-1}\,\text{mol}^{-1}$, so convert before dividing.
For a reaction $\Delta H = 35.5\ \text{kJ mol}^{-1}$ and $\Delta S = 83.6\ \text{J K}^{-1}\,\text{mol}^{-1}$, with both assumed independent of temperature. At what temperatures is the reaction spontaneous?
At the boundary $\Delta G = 0$, so:
$$ T = \frac{\Delta H}{\Delta S} = \frac{35.5\times10^{3}\ \text{J mol}^{-1}}{83.6\ \text{J K}^{-1}\,\text{mol}^{-1}} \approx 424.6\ \text{K} $$
Both $\Delta H$ and $\Delta S$ are positive (case 3), so $T\Delta S$ outgrows $\Delta H$ as temperature rises and $\Delta G$ turns negative above the crossover. The reaction is spontaneous for $T > 425\ \text{K}$. This is NEET 2017 Q.36 in full.
Entropy change in phase transitions
A phase change carried out at its equilibrium temperature — the melting point for fusion or the boiling point for vaporisation — is a reversible process at constant temperature and pressure. At that temperature the two phases coexist in equilibrium, so $\Delta G = 0$. Substituting into the Gibbs equation gives a clean route to the entropy of transition:
$$ \Delta G = \Delta H - T\Delta S = 0 \;\;\Longrightarrow\;\; \Delta S = \frac{\Delta H}{T} $$
Here $\Delta H$ is the enthalpy of fusion or vaporisation and $T$ is the transition temperature in kelvin. Both melting and boiling increase disorder, so $\Delta S$ is positive in the forward (solid→liquid→gas) direction.
Water boils at $373\ \text{K}$ with an enthalpy of vaporisation $\Delta_{\text{vap}}H = 40.79\ \text{kJ mol}^{-1}$. Find the entropy of vaporisation.
At the boiling point liquid and vapour are in equilibrium ($\Delta G = 0$), so $\Delta S = \Delta H / T$:
$$ \Delta_{\text{vap}}S = \frac{40.79\times10^{3}\ \text{J mol}^{-1}}{373\ \text{K}} \approx 109.4\ \text{J K}^{-1}\,\text{mol}^{-1} $$
The positive value confirms that vaporisation raises entropy: a mole of gas is far more disordered than a mole of liquid.
Spontaneity, entropy and Gibbs energy at a glance
- Spontaneous = potential to proceed unaided and irreversibly; it is silent about the rate.
- Neither $\Delta H < 0$ alone nor $\Delta S > 0$ alone is a reliable criterion.
- True criterion (second law): $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$, with $\Delta S_{\text{surr}} = -\Delta H_{\text{sys}}/T$.
- For a system at constant $T,p$: $\Delta G = \Delta H - T\Delta S$; spontaneous when $\Delta G < 0$, equilibrium when $\Delta G = 0$.
- Master table: ($-$,$+$) spontaneous always; ($-$,$-$) low T; ($+$,$+$) high T; ($+$,$-$) never.
- Crossover temperature $T = \Delta H/\Delta S$; phase-transition entropy $\Delta S = \Delta H/T$ at the transition point.