Second Law Thermo

8 MCQs1 revision card9-step worked example

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

Lesson

The second law of thermodynamics introduces a directional arrow to chemical processes that the first law cannot provide. The first law tells you energy is conserved; the second law tells you which way a process actually runs.

The trap that costs marks: judging spontaneity from ΔH alone. An endothermic reaction (ΔH > 0) can be spontaneous — ice melting at room temperature is the textbook example. NEET exploits this confusion regularly.

The concept. The second law states that in any spontaneous process, the total entropy of the system and surroundings increases (NCERT Class 11 Chemistry, Chapter 5, page 20). Entropy (S) is a measure of disorder or the number of accessible microstates. For a process at constant T and P, spontaneity is decided not by ΔH alone but by the Gibbs energy criterion:

ΔG = ΔH − TΔS

ΔG < 0 → spontaneous (forward-favoured)
ΔG = 0 → equilibrium
ΔG > 0 → non-spontaneous

The four-case table you must internalise:

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)

Bridge to NEET. Questions in this topic typically give you ΔH, ΔS, and a temperature, then ask whether the reaction is spontaneous. The distractor designed to trap you is the option that ignores temperature and judges from ΔH sign alone.

Watch out: when ΔH and ΔS have the same sign, temperature is the deciding factor. Calculate T = ΔH/ΔS to find the crossover — this is a common direct-application question.

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 1Easy RecallPractice

Which of the following correctly states the second law of thermodynamics?

MCQ 2Direct ApplicationPractice

For a reaction with ΔH = +50 kJ/mol and ΔS = +0.15 kJ/(mol·K), at what temperature does the reaction become spontaneous?

MCQ 3Concept TrapPractice

The dissolution of ammonium chloride in water is endothermic yet spontaneous at room temperature. This is because:

MCQ 4Easy RecallPractice

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

MCQ 5Direct ApplicationPractice

A reaction has ΔH = −400 kJ/mol and ΔS = −0.50 kJ/(mol·K). Above what temperature does the reaction become non-spontaneous?

MCQ 6Easy RecallPractice

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

MCQ 7Direct ApplicationPractice

For a certain reaction at 298 K: ΔH = −10.0 kJ/mol and ΔS = +0.040 kJ/(mol·K). What is ΔG at 298 K?

MCQ 8CalculationPractice

An endothermic reaction has ΔH = +80 kJ/mol and becomes spontaneous above 400 K. What is the minimum value of ΔS for the reaction?

Quick recall before you leave

Worked Example

Pattern: Predict spontaneity from ΔH and ΔS at given T using ΔG = ΔH − TΔS (pattern: spontaneity from Gibbs criterion, observed in NEET 2022, 2024).

1
Given
- ΔH = +125 kJ/mol (endothermic) - ΔS = +0.250 kJ/(mol·K) (entropy increases) - T₁ = 400 K, T₂ = 600 K
2
Required
Determine the sign of ΔG at each temperature to assess spontaneity.
3
Concept
At constant T and P, the Gibbs energy criterion decides spontaneity: ΔG = ΔH − TΔS. Since both ΔH and ΔS are positive, the reaction will be spontaneous only above the crossover temperature T = ΔH/ΔS (NCERT Class 11 Chemistry, Chapter 5).
4
Formula
ΔG = ΔH − TΔS
5
Substitution
Crossover: T = ΔH/ΔS = 125/0.250 = 500 K At 400 K: ΔG = 125 − (400 × 0.250) = 125 − 100 = +25 kJ/mol At 600 K: ΔG = 125 − (600 × 0.250) = 125 − 150 = −25 kJ/mol
6
Calculation
- Crossover temperature = 500 K - ΔG(400 K) = +25 kJ/mol → non-spontaneous - ΔG(600 K) = −25 kJ/mol → spontaneous Note: the integer values 125, 400, 600 in this problem are exact (problem-defined); they do not limit significant figures in the answer.
7
Final answer
(a) At 400 K: ΔG = +25 kJ/mol — **non-spontaneous.** (b) At 600 K: ΔG = −25 kJ/mol — **spontaneous.** The crossover temperature is 500 K. Below 500 K, the positive ΔH dominates; above 500 K, the TΔS term overwhelms ΔH.
8
Common trap
Judging from ΔH alone: "ΔH is positive, so the reaction is never spontaneous." This ignores the entropy term. When ΔS is also positive, a sufficiently high temperature makes TΔS > ΔH, driving ΔG negative.
9
Similar NEET-style question
A reaction has ΔH = +40 kJ/mol and ΔS = +0.100 kJ/(mol·K). At what temperature does this reaction reach equilibrium? *(Answer: T = 40/0.100 = 400 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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