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. 161Spontaneity Entropy
8 MCQs3 revision cards9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2020, 2021, 2024, 2026Official key: NTA-verifiedLast updated: 25 Sep 2026
Spontaneity Entropy, explained for NEET
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 155 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.
Can you answer these Spontaneity Entropy 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?
Show answer and why every option is right or wrong
Answer: C. At the crossover, ΔG = 0 → T = ΔH/ΔS = 50000 J / 125 J/K = 400 K. Above 400 K, TΔS > ΔH, so ΔG < 0 (spontaneous). NCERT Class 11 Chemistry Chapter 5, Gibbs energy discussion.
Why A is wrong: A is wrong because 200 K comes from using ΔS in kJ/K without converting (50/0.125 = 400, not 200). The unit mismatch trap: ΔH is in kJ but ΔS in J/K — you must convert before dividing.
Why B is wrong: B is wrong because below 400 K the TΔS term is smaller than ΔH, making ΔG positive (non-spontaneous). Spontaneity requires T above the crossover, not below.
Why D is wrong: D is wrong because when both ΔH and ΔS are positive, spontaneity is temperature-dependent, not universal. Only ΔH < 0 with ΔS > 0 gives spontaneity at all temperatures.
MCQ 2Easy RecallPractice
Which quantity determines whether a process is spontaneous at constant temperature and pressure?
Show answer and why every option is right or wrong
Answer: B. At constant T and P, the Gibbs free energy change ΔG is the criterion: ΔG < 0 means spontaneous. NCERT Class 11 Chemistry Chapter 5, page 155.
Why A is wrong: A is wrong because ΔH alone cannot predict spontaneity — an endothermic reaction (ΔH > 0) can still be spontaneous if TΔS is large enough to make ΔG negative.
Why C is wrong: C is wrong because ΔS alone is not sufficient either — a reaction with positive ΔS can be non-spontaneous if ΔH is large and positive enough at that temperature.
Why D is wrong: D is wrong because ΔU is the criterion for constant-volume processes (Helmholtz energy context), not constant-pressure. NEET thermochemistry operates at constant P where ΔG applies.
MCQ 3Easy RecallPractice
For which combination of ΔH and ΔS is a reaction spontaneous at all temperatures?
Show answer and why every option is right or wrong
Answer: C. When ΔH is negative and ΔS is positive, ΔG = ΔH − TΔS is negative at every temperature (both terms favour spontaneity). NCERT Class 11 Chemistry Chapter 5.
Why A is wrong: A is wrong because with both ΔH and ΔS positive, ΔG = ΔH − TΔS is positive at low temperatures (where TΔS < ΔH). Spontaneity occurs only above T = ΔH/ΔS.
Why B is wrong: B is wrong because with ΔH < 0 and ΔS < 0, the −TΔS term becomes positive. At high temperatures, TΔS overwhelms ΔH, making ΔG > 0. Spontaneous only at low T.
Why D is wrong: D is wrong because with ΔH > 0 and ΔS < 0, both terms make ΔG positive at every temperature. This combination is never spontaneous.
MCQ 4Concept TrapPractice
The melting of ice at 280 K and 1 atm is spontaneous even though ΔH > 0. The correct explanation is:
Show answer and why every option is right or wrong
Answer: D. Ice melting is endothermic (ΔH > 0) but entropy increases (solid → liquid). At 280 K (above 273 K), TΔS exceeds ΔH, giving ΔG < 0. This is the textbook counter-example to judging by ΔH alone.
Why A is wrong: A is wrong because melting (fusion) is endothermic — you must supply heat to break the crystal lattice. ΔH_fusion is positive, not negative.
Why B is wrong: B is wrong because ice melting is spontaneous only above 273 K (0 °C at 1 atm). Below 273 K, TΔS < ΔH and ΔG > 0, so freezing is favoured instead.
Why C is wrong: C is wrong because melting increases entropy (more disorder in liquid than solid). ΔS > 0 for solid → liquid transitions.
MCQ 5Direct ApplicationPractice
A reaction has ΔH = −100 kJ/mol and ΔS = −200 J/(mol·K). Above what temperature does it become non-spontaneous?
Show answer and why every option is right or wrong
Answer: D. Crossover: T = ΔH/ΔS = 100000 J / 200 J/K = 500 K. Below 500 K, |ΔH| > TΔS and ΔG < 0. Above 500 K, TΔS dominates and ΔG > 0. NCERT Class 11 Chemistry Chapter 5.
Why A is wrong: A is wrong because 200 K results from misreading ΔS as −500 J/(mol·K): 100000 J ÷ 500 J/K = 200 K. With the given −200 J/(mol·K), T = 500 K.
Why B is wrong: B is wrong because when both ΔH and ΔS are negative, spontaneity is temperature-dependent. At high T the −TΔS term (which is positive here) overwhelms the negative ΔH.
Why C is wrong: C is wrong because 1000 K results from using ΔS = −100 J/K instead of −200 J/K (halving the denominator). The given ΔS is −200 J/(mol·K).
MCQ 6Easy RecallPractice
Entropy of a substance follows the order:
Show answer and why every option is right or wrong
Answer: B. Gases have maximum molecular disorder, liquids intermediate, solids minimum. S(gas) > S(liquid) > S(solid). NCERT Class 11 Chemistry Chapter 5.
Why A is wrong: A is wrong because this is the exact reverse of the correct order. Solids have the most ordered arrangement and therefore the lowest entropy.
Why C is wrong: C is wrong because it places solid above liquid. Liquids have more molecular freedom than solids, so S(liquid) > S(solid).
Why D is wrong: D is wrong because it places gas below liquid. Gas molecules have far greater translational freedom and occupy much larger volumes than liquids.
MCQ 7Direct ApplicationPractice
For a reaction at 300 K: ΔH = −30 kJ/mol and ΔS = +100 J/(mol·K). Calculate ΔG.
Show answer and why every option is right or wrong
Answer: A. ΔG = ΔH − TΔS = −30 kJ − (300 K × 0.100 kJ/K) = −30 − 30 = −60 kJ/mol. Convert ΔS to kJ/K before multiplying. NCERT Class 11 Chemistry Chapter 5.
Why B is wrong: B is wrong because +60 kJ results from using ΔG = ΔH + TΔS (adding instead of subtracting). The Gibbs equation has a minus sign: ΔG = ΔH − TΔS.
Why C is wrong: C is wrong because −30.1 kJ subtracts ΔS itself (0.100 kJ/K) instead of TΔS: the multiplication by T = 300 K was left out.
Why D is wrong: D is wrong because ΔG = 0 would mean equilibrium. Here both ΔH (negative) and TΔS (positive, subtracted) drive ΔG further negative, not to zero.
MCQ 8Concept TrapPractice
A student claims: "Since dissolving NH₄Cl in water is endothermic, it cannot be spontaneous." The error in this reasoning is:
Show answer and why every option is right or wrong
Answer: A. The student ignored the entropy term. NH₄Cl dissolution is endothermic but increases entropy significantly (ions dispersed in solvent). At room temperature TΔS > ΔH, so ΔG < 0. This is the classic mistake of predicting spontaneity from enthalpy alone.
Why B is wrong: B is wrong because NH₄Cl dissolution IS endothermic (the solution feels cold). The student's observation of ΔH > 0 is correct; it is the conclusion about spontaneity that is flawed.
Why C is wrong: C is wrong because this repeats the student's error. Endothermic reactions CAN be spontaneous when the entropy increase is large enough that TΔS > ΔH.
Why D is wrong: D is wrong because temperature is central to spontaneity — it appears directly in ΔG = ΔH − TΔS. The magnitude of TΔS changes with T, which can flip the sign of ΔG.
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Spontaneity Entropy: quick recall before you leave
How do you solve a Spontaneity Entropy question? A 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.
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What to remember before solving Spontaneity Entropy questions
Which Spontaneity Entropy formulas do you need for NEET?
Gibbs free energy
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
Where do students lose marks on Spontaneity Entropy?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
Correction
Spontaneity from ΔG = ΔH - TΔS. Endothermic reactions can be spontaneous if TΔS > ΔH (ice melting at room T).
More in Thermodynamics: 3 exam traps and mistakes · 5 formulas · 1 question pattern from its other lessons.
Spontaneity Entropy questions from past NEET papers
4 questions from NEET 2020, 2021, 2024, 2026. Answers verified against NTA official keys.
NEET 2026
1ΔS_system = 0; ΔS_surroundings = 0
2ΔS_system = 4.606 R; ΔS_surroundings = −4.606 R
3ΔS_system = 0; ΔS_surroundings = 4.606 R
4ΔS_system = 4.606 R; ΔS_surroundings = 0
NTA Answer: Option 4(final)
NEET 2024Revised key
1A and C
2A, B and D
3A, C and D
4C and D
NTA Answer: Option 3(revised_final)
NEET 2021
For irreversible expansion of an ideal gas under isothermal condition, the correct option is:
1ΔU ≠ 0, ΔS(total) = 0
2ΔU = 0, ΔS(total) = 0
3ΔU ≠ 0, ΔS(total) ≠ 0
4ΔU = 0, ΔS(total) ≠ 0
NTA Answer: Option 4(final)
NEET 2020
For the reaction, 2Cl(g) → Cl2 (g), the correct option is :
1∆rH < 0 and ∆rS < 0
2∆rH > 0 and ∆rS > 0
3∆rH > 0 and ∆rS < 0
4∆rH < 0 and ∆rS > 0
NTA Answer: Option 1(final)
How does NEET ask about Spontaneity Entropy?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
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.155
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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