Spontaneity Entropy

8 MCQs1 revision card9-step worked example

Source: NCERT ThermodynamicsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

A reaction can be exothermic yet non-spontaneous, or endothermic yet spontaneous. The trap that costs marks: judging spontaneity from ΔH alone while ignoring entropy and temperature.

Entropy (S) is the thermodynamic measure of disorder or randomness of a system. NCERT Class 11 Chemistry Chapter 5, page 20 defines it as a state function whose change for a reversible process is ΔS = q_rev / T. Higher disorder → higher entropy. For any substance, S(gas) > S(liquid) > S(solid).

The Second Law connection: in an isolated system, entropy of the universe never decreases for a spontaneous process: ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0. But real NEET problems don't ask about isolated systems — they give you ΔH and ΔS at a temperature and expect you to use the Gibbs equation.

The Gibbs criterion: ΔG = ΔH − TΔS. A process is spontaneous when ΔG < 0, at equilibrium when ΔG = 0, and non-spontaneous when ΔG > 0. Four combinations arise:

ΔH	ΔS	Spontaneity
−	+	Always spontaneous (ΔG < 0 at all T)
+	−	Never spontaneous (ΔG > 0 at all T)
−	−	Spontaneous at low T (below T = ΔH/ΔS)
+	+	Spontaneous at high T (above T = ΔH/ΔS)

The classic example: ice melting at room temperature is endothermic (ΔH > 0) but spontaneous because TΔS > ΔH, making ΔG < 0.

Watch-out: when ΔH and ΔS have the same sign, spontaneity is temperature-dependent. The crossover temperature is T = ΔH/ΔS. NEET distractors exploit students who skip this calculation and pick based on the sign of ΔH alone.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Direct ApplicationPractice

For a reaction with ΔH = +50 kJ/mol and ΔS = +125 J/(mol·K), at what temperature does the reaction become spontaneous?

MCQ 2Easy RecallPractice

Which quantity determines whether a process is spontaneous at constant temperature and pressure?

MCQ 3Easy RecallPractice

For which combination of ΔH and ΔS is a reaction spontaneous at all temperatures?

MCQ 4Concept TrapPractice

The melting of ice at 280 K and 1 atm is spontaneous even though ΔH > 0. The correct explanation is:

MCQ 5Direct ApplicationPractice

A reaction has ΔH = −100 kJ/mol and ΔS = −200 J/(mol·K). Above what temperature does it become non-spontaneous?

MCQ 6Easy RecallPractice

Entropy of a substance follows the order:

MCQ 7Direct ApplicationPractice

For a reaction at 300 K: ΔH = −30 kJ/mol and ΔS = +100 J/(mol·K). Calculate ΔG.

MCQ 8Concept TrapPractice

A student claims: "Since dissolving NH₄Cl in water is endothermic, it cannot be spontaneous." The error in this reasoning is:

Quick recall before you leave

Worked Example

Pattern: P.CHE.U04.SPONTANEITY_GIBBS — Predict spontaneity from ΔH and ΔS at given T using ΔG = ΔH − TΔS.

1
Given
A reaction has ΔH = +170 kJ/mol and ΔS = +170 J/(mol·K). Determine whether it is spontaneous at (a) 500 K and (b) 1200 K.
2
Required
ΔG at each temperature; spontaneity verdict for each.
3
Concept
Spontaneity at constant T and P is determined by ΔG = ΔH − TΔS. Both ΔH and ΔS are positive, so this is the "spontaneous at high T" case. The crossover temperature where ΔG = 0 is T = ΔH/ΔS.
4
Formula
ΔG = ΔH − TΔS
5
Substitution
First convert units: ΔS = 170 J/(mol·K) = 0.170 kJ/(mol·K). (a) At T = 500 K: ΔG = 170 kJ − (500 K × 0.170 kJ/K) = 170 − 85 = +85 kJ/mol (b) At T = 1200 K: ΔG = 170 kJ − (1200 K × 0.170 kJ/K) = 170 − 204 = −34 kJ/mol
6
Calculation
Crossover temperature: T = ΔH/ΔS = 170000 J / 170 J/K = 1000 K (exact division — the numbers 170000 and 170 are given problem values, not measurements requiring sig-fig treatment).
7
Final answer
(a) At 500 K: ΔG = +85 kJ/mol → **non-spontaneous**. (b) At 1200 K: ΔG = −34 kJ/mol → **spontaneous**. The crossover temperature is 1000 K. Note: the temperature values (500 K, 1200 K) and the given ΔH, ΔS values are treated as exact problem-defined quantities; they do not limit significant figures in this context.
8
Common trap
Predicting spontaneity from ΔH alone. A student seeing ΔH = +170 kJ might immediately mark "non-spontaneous at all T" — ignoring that the positive ΔS drives spontaneity above 1000 K. Always compute ΔG. A second common error: forgetting to convert ΔS from J to kJ before substituting. At 1200 K, using 170 instead of 0.170 gives TΔS = 204000 kJ — an absurd value that should trigger a sanity check.
9
Similar NEET-style question
A reaction has ΔH = −80 kJ/mol and ΔS = −160 J/(mol·K). Is the reaction spontaneous at 600 K? Find the crossover temperature. *Approach:* T_crossover = 80000/160 = 500 K. At 600 K (above crossover), ΔG = −80 − (600)(−0.160) = −80 + 96 = +16 kJ → non-spontaneous. Both negative: spontaneous only below 500 K. ---

Before solving, remember these

Definition

Entropy

S = measure of disorder. ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0 (second law). Spontaneous process: ΔS_universe > 0.

-- NCERT Class 11 Chemistry, Ch. 5, p. 20

Formulas

Enthalpy and ΔH at constant P

H = U + P V; Δ H = q_{p}

Enthalpy = internal energy + PV. At constant pressure, heat absorbed equals enthalpy change.

Symbol	Quantity	SI Unit
H	enthalpy	J
U	internal energy	J
P	pressure	Pa
V	volume	m^3

Valid when

At constant pressure
Closed system

First law of thermodynamics (chemistry)

Δ U = q + w

Internal energy change = heat added to system + work done ON system. Note: chemistry uses w as work done ON system; physics uses w as work done BY.

Symbol	Quantity	SI Unit
ΔU	internal energy change	J
q	heat	J
w	work on system	J

Valid when

Closed system
Sign convention chosen consistently

Gibbs free energy

Δ G = Δ H - T Δ S

Spontaneity criterion. ΔG<0: spontaneous. ΔG=0: equilibrium. ΔG>0: non-spontaneous.

Symbol	Quantity	SI Unit
ΔG	Gibbs energy change	kJ
ΔH	enthalpy	kJ
ΔS	entropy	kJ/K
T	temperature	K

Valid when

Constant T and P

ΔG° and K

Δ G^{\circ} = - R T ln K

Standard free energy change relates to equilibrium constant. K>1: ΔG°<0; K<1: ΔG°>0.

Symbol	Quantity	SI Unit
ΔG°	standard free energy	J/mol
R	gas constant 8.314	J/mol/K
T	temp	K
K	equilibrium constant	-

Valid when

Standard state
Equilibrium

Hess's law

Δ H_{net} = \sum Δ H_{steps}

Total enthalpy change is independent of path. Useful for computing ΔH of reactions not directly measurable.

Symbol	Quantity	SI Unit
ΔH	enthalpy change	J or kJ

Valid when

State function (path-independent)
Same initial/final states

Standard enthalpy of reaction

Δ H_{rxn}^{\circ} = \sum Δ H_{f}^{\circ} (prod) - \sum Δ H_{f}^{\circ} (react)

From standard formation enthalpies. ΔH°_f of element in standard state = 0.

Symbol	Quantity	SI Unit
ΔH°_f	standard formation enthalpy	kJ/mol

Valid when

Standard state (1 bar, 298 K)
Stoichiometric coefficients applied

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Hess's law: ΔH sign on reversal

Category: Sign Convention

When reversing a reaction, ΔH changes sign. Multiply: same as multiply ΔH.

When it triggers

Hess's law problem with combination of multiple reactions.

How to avoid

If reaction is reversed: ΔH → -ΔH. If multiplied by factor n: ΔH → n×ΔH. Apply systematically when combining.

Mistake

Wrong sign of ΔG° from K. ΔG° = -RT ln K (negative ln means K<1, ΔG°>0).

Root cause: sign error

Correction

K > 1: ln K > 0 → ΔG° < 0 (forward favoured). K < 1: ln K < 0 → ΔG° > 0 (reverse favoured). At equilibrium K=1, ΔG°=0.

Mistake

Forgets to negate ΔH when reversing a reaction in Hess's law combination.

Root cause: sign error

Correction

Reversing a reaction: ΔH → -ΔH. Multiplying by n: ΔH → n·ΔH. Apply consistently before summing.

Mistake

Predicts spontaneity from ΔH alone (ignoring T and ΔS).

Root cause: concept gap

Correction

Spontaneity from ΔG = ΔH - TΔS. Endothermic reactions can be spontaneous if TΔS > ΔH (ice melting at room T).

Past Year Questions

8 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Phosphoric acid ionizes in three steps with their ionization constant values - 35 - K a 1 , K a 2 a n d K a 3 , respectively, while K is the overall ionization constant. Which of the following statements are true? A. lo g K = lo g K a1 + lo g K a 2 + lo g K a 3 B. H PO is a stronger acid than 3 4 H 2 P O −4 and H P O 24 − C. K a1 > K a 2 > K a 3 D. K a 1 = K a 3 + 2 K a 2 Choose the correct answer from the options given below :

1A, B and C only

2A and B only

3A and C only

4B, C and D only

NTA Answer: Option 1(final)

NEET 2024Revised key

In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2NaHCO → Na CO + CO + H O 3(s) 2 3(s) 2(g) 2 (g) D. Cl → 2Cl 2(g) (g) Choose the correct answer from the options given below:

1A and C

2A, B and D

3A, C and D

4C and D

NTA Answer: Option 3(revised_final)

NEET 2024Revised key

Match List I with List II. List-I List-II (Process) (Conditions) A. Isothermal process I. No heat exchange B. Isochoric process II. Carried out at constant temperature C. Isobaric process III. Carried out at constant volume D. Adiabatic process IV. Carried out at constant pressure Choose the correct answer from the options given below:

1A-IV, B-III, C-II, D-I

2A-IV, B-II, C-III, D-I

3A-I, B-II, C-III, D-IV

4A-II, B-III, C-IV, D-I

NTA Answer: Option 4(revised_final)

NEET 2024Revised key

The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is (Given R = 2.0 cal K–1 mol–1)

10 calorie

2–413.14 calories

3413.14 calories

4100 calories

NTA Answer: Option 2(revised_final)

NEET 2023

Which amongst the following molecules on polymerization produces neoprene? Cl |

1H C C–CHCH

2H CCH–CCH 2 2 2 CH 3 |

3H C C –CHCH

4H CCH–CHCH 2 2 2 2

NTA Answer: Option 1(final)

NEET 2023

The equilibrium concentrations of the species in the reaction AB CDare 2, 3, 10 and 6 mol L–1,  respectively at 300 K. Gº for the reaction is (R = 2 cal/mol K)

1–137.26 cal

2–1381.80 cal

3–13.73 cal

41372.60 cal

NTA Answer: Option 2(final)

NEET 2022

The given graph is a representation of kinetics of a reaction. The y and x axes for zero and first order reactions, respectively are

1zero order (y = rate and x = concentration), first order (y = rate and x = t ) ½

2zero order (y = concentration and x = time), first order (y = t and x = concentration) ½

3zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)

4zero order (y = rate and x = concentration), first order (y = t and x = concentration) ½

NTA Answer: Option 4(final)

NEET 2021

For irreversible expansion of an ideal gas under isothermal condition, the correct option is: ΔU ≠ ΔS ΔU ΔS

10, = 0

2= 0, = 0 total total ΔU ≠ ΔS ≠ ΔU ΔS ≠

30, 0

4= 0, 0 total total

NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Combine multiple thermochemical equations using Hess's law to find ΔH of target reaction.

Multi StepMedium

Common distractors

wrong sign when reversing

Forgets to negate ΔH when reversing equation

Predict spontaneity from ΔH and ΔS at given T using ΔG = ΔH - TΔS.

InterpretationMedium

Common distractors

ignores temperature

Uses ΔH alone for spontaneity

Sources

NCERT refs: Class 11 Chemistry Chapter 5, p.20

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