JEE Mains Chemistry · Chemical Thermodynamics

Entropy, Gibbs Energy and Spontaneity

Whether a change goes on its own: ΔG = ΔH − TΔS, the four sign cases, the entropy change of a reaction or a phase change, and the temperature at which ΔG crosses zero.

All Chemical Thermodynamics notes JEE Mains Chemistry strategy

Why this matters

Twenty PYQs, twelve of them numerical, and two from 2026. Five read the signs of ΔH and ΔS, five compute an entropy or Gibbs energy change, and ten find the temperature at which ΔG changes sign — the largest single cluster in the chapter.

Concept 1 of 3: Spontaneity from the signs of ΔH and ΔS

ΔG = ΔH − TΔS. The enthalpy term hardly changes with temperature, but the entropy term grows with T. So the sign of ΔS decides which way a rise in temperature pushes ΔG, and the two signs together decide whether a change is always, never or only sometimes spontaneous.

Definition

At constant $T$ and $p$ : $\Delta G < 0$ spontaneous; $\Delta G = 0$ equilibrium (a reversible change); $\Delta G > 0$ non-spontaneous, and the reverse change is spontaneous.
When $\Delta H$ and $\Delta S$ share a sign, $\Delta G$ changes sign at $T = \frac{\Delta H}{\Delta S}$ .
Second law: $\Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{surr}} > 0$ for a spontaneous change, and $\frac{\Delta G_{\mathrm{sys}}}{\Delta S_{\mathrm{total}}} = -T$ at constant pressure.
Partial derivatives: $\left(\frac{\partial G}{\partial T}\right)_p = -S$ , $\left(\frac{\partial G}{\partial p}\right)_T = V$ , $\left(\frac{\partial H}{\partial T}\right)_p = C_p$ , $\left(\frac{\partial U}{\partial T}\right)_V = C_V$ .

ΔH	ΔS	Sign of ΔG	Spontaneous
Negative	Positive	Negative at every temperature	At all temperatures
Positive	Negative	Positive at every temperature	At no temperature
Positive	Positive	Negative above ΔH/ΔS	At high temperature An endothermic change that goes at 373 K but not at 273 K belongs in this row.
Negative	Negative	Negative below ΔH/ΔS	At low temperature

The sign of ΔS decides which way a rise in temperature pushes ΔG.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q33Moderate

Example 1 · Chemical Thermodynamics · Entropy, Gibbs Energy and Spontaneity

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

Exothermic is not enough

An exothermic change with

\Delta S < 0

stops being spontaneous above

\Delta H/\Delta S

. Only

\Delta H < 0

with

\Delta S > 0

is spontaneous at every temperature.

Swapping the two derivatives of G

G falls as temperature rises, at the rate S, and rises with pressure, at the rate V:

\left(\partial G/\partial T\right)_p = -S

and

\left(\partial G/\partial p\right)_T = V

. Matching lists swap them.

Drill 4 more on spontaneity from the signs of δh and δs

Concept 2 of 3: Entropy change and Gibbs energy of a reaction

Entropy measures how spread out the energy and the particles are. Making gas raises it; freezing, losing gas molecules or sticking molecules to a surface lowers it. For a reaction, use tabulated entropies just like formation enthalpies — except that no element has zero entropy.

Definition

$\Delta_r S^\circ = \sum\nu S^\circ(\text{products}) - \sum\nu S^\circ(\text{reactants})$ . Elements have non-zero $S^\circ$ .
$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$ . Put $\Delta S$ in kJ K⁻¹ (divide by 1000) first.
Phase change at its transition temperature: $\Delta S = \frac{\Delta H_{\mathrm{trans}}}{T_{\mathrm{trans}}}$ .
Heating with no phase change: $\Delta S = \int \frac{C_p}{T}\,dT = C_p\ln\frac{T_2}{T_1}$ for constant $C_p$ . A path through phase changes adds one term per step.
Surroundings: $\Delta S_{\mathrm{surr}} = -\frac{q_{\mathrm{sys}}}{T}$ .
$\Delta S < 0$ : freezing (at any temperature), $\mathrm{N_2 + 3H_2 \to 2NH_3}$ , adsorption. $\Delta S > 0$ : melting, vaporising, dissolving NaCl.

Entropy and Gibbs energy of reaction

\Delta_r S^\circ = \sum \nu S^\circ_{\mathrm{products}} - \sum \nu S^\circ_{\mathrm{reactants}},\qquad \Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ

Worked example

For

\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}

\Delta_r H^\circ = +178

kJ mol⁻¹.

S^\circ

(J K⁻¹ mol⁻¹):

\mathrm{CaCO_3}

92.9, CaO 39.8,

\mathrm{CO_2}

213.7. Find

\Delta_r S^\circ

and

\Delta_r G^\circ

at 298 K.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q29Moderate

Example 2 · Chemical Thermodynamics · Entropy, Gibbs Energy and Spontaneity

Consider the following data for the reaction

X_{2}(\text{ }g) +Y_{2}(\text{ }g) \rightleftharpoons 2XY(g)

at 600 K. The

\Delta_{r}G^{\ominus}

(in

kJ\,mol^{-1}

) for the reaction is:

Compound	$\Delta_{f}H_{600\text{ }K}^{\Theta}$ $\left( kJmol^{- 1} \right)$	$S_{600\text{ }K}^{\Theta}$ $\left( Jmol^{- 1}{\text{ }K}^{- 1} \right)$
$XY(g)$	42	200
$X_{2}(\text{ }g)$	8	140
$Y_{2}(\text{ }g)$	80	250

ΔS in joules, ΔH in kilojoules

This is the most common slip in the chapter. With

\Delta H = 50

kJ,

\Delta S = 100

J K⁻¹ and T = 400 K,

50 - 400 \times 100

is nonsense;

50 - 400 \times 0.100 = 10

kJ is right.

Heating entropy needs the 1/T

The entropy of warming is

\int C_p\,dT/T

, not

\int C_p\,dT

(that is the enthalpy). Each phase change on the way adds its own

\Delta H/T

, at its own temperature.

Drill 4 more on entropy change and gibbs energy of a reaction

Concept 3 of 3: Temperature at which ΔG changes sign

When ΔH and ΔS have the same sign, the two terms of ΔG pull against each other, and at one temperature they balance. There ΔG = 0. That temperature is a boiling point, a melting point, a transition point, or the lowest temperature at which a reduction starts to work.

Definition

$T = \frac{\Delta H}{\Delta S}$ , with $\Delta H$ in J (or $\Delta S$ in kJ K⁻¹).
$\Delta H > 0$ , $\Delta S > 0$ : spontaneous ABOVE this temperature. $\Delta H < 0$ , $\Delta S < 0$ : spontaneous BELOW it.
Boiling and melting points: $T_b = \frac{\Delta_{\mathrm{vap}}H}{\Delta_{\mathrm{vap}}S}$ , $T_f = \frac{\Delta_{\mathrm{fus}}H}{\Delta_{\mathrm{fus}}S}$ .
If $\Delta G^\circ$ is given as a function of T, set it to zero and solve. If $\Delta S$ itself depends on T, solve $\Delta H = T\,\Delta S(T)$ .
From a table: find $\Delta_r H^\circ$ and $\Delta_r S^\circ$ first, then divide.

Temperature at which ΔG changes sign

\Delta G = 0\ \Rightarrow\ T = \frac{\Delta H}{\Delta S}

Worked example

For

\mathrm{Br_2(l) \to Br_2(g)}

\Delta H = 30.9

kJ mol⁻¹ and

\Delta S = 93.0

J K⁻¹ mol⁻¹. Find the boiling point of bromine.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 21 · Q60Moderate

Example 3 · Chemical Thermodynamics · Entropy, Gibbs Energy and Spontaneity

Data given for the following reaction is as follows:

FeO_{(s)} + C_{\text{(graphite)}} \longrightarrow Fe_{(s)} + CO_{(g)}

Substance	$\Delta H^{\circ}\ \left( kJ\ mol^{- 1} \right)$	$\Delta S^{\circ}\ \left( J\ mol^{- 1}\ K^{- 1} \right)$
$FeO_{(s)}$	-266.3	57.49
$C_{\text{(graphite)}}$	0	5.74
$Fe_{(s)}$	0	27.28
$CO_{(g)}$	-110.5	197.6

The minimum temperature in

K

at which the reaction becomes spontaneous is (Integer answer)

A factor of a thousand

With

\Delta H = 60

kJ and

\Delta S = 150

J K⁻¹,

T = 60\,000/150 = 400

K, not 0.4 K. Convert before dividing.

Below, not above

For an exothermic change with falling entropy,

\Delta H/\Delta S

is the temperature to stay BELOW. Quoting it as a minimum temperature reverses the answer.

Drill 9 more on temperature at which δg changes sign

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Entropy change and Gibbs energy of a reaction
Entropy and Gibbs energy of reaction
$\Delta_r S^\circ = \sum \nu S^\circ_{\mathrm{products}} - \sum \nu S^\circ_{\mathrm{reactants}},\qquad \Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$
Temperature at which ΔG changes sign
Temperature at which ΔG changes sign
$\Delta G = 0\ \Rightarrow\ T = \frac{\Delta H}{\Delta S}$

Reference tables (1)

Spontaneity from the signs of ΔH and ΔS4 rows

ΔH	ΔS	Sign of ΔG	Spontaneous
Negative	Positive	Negative at every temperature	At all temperatures
Positive	Negative	Positive at every temperature	At no temperature
Positive	Positive	Negative above ΔH/ΔS	At high temperature An endothermic change that goes at 373 K but not at 273 K belongs in this row.
Negative	Negative	Negative below ΔH/ΔS	At low temperature

The sign of ΔS decides which way a rise in temperature pushes ΔG.

Watch out for (6)

Exothermic is not enough
→ Spontaneity from the signs of ΔH and ΔS
Swapping the two derivatives of G
→ Spontaneity from the signs of ΔH and ΔS
ΔS in joules, ΔH in kilojoules
→ Entropy change and Gibbs energy of a reaction
Heating entropy needs the 1/T
→ Entropy change and Gibbs energy of a reaction
A factor of a thousand
→ Temperature at which ΔG changes sign
Below, not above
→ Temperature at which ΔG changes sign

Test yourself on Chemical Thermodynamics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Thermodynamics chapter test Drill all 20 questions