Formula
Gibbs free energy
G = H - TS. ΔG = ΔH - TΔS at constant T,P. ΔG < 0: spontaneous; ΔG = 0: equilibrium; ΔG > 0: non-spontaneous.
-- NCERT Class 11 Chemistry, Ch. 5, p. 160Gibbs Free Energy
8 MCQs2 revision cards9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2020, 2023Official key: NTA-verifiedLast updated: 25 Sep 2026
Gibbs Free Energy, explained for NEET
The trap: You see ΔH is negative and mark the reaction as spontaneous. You just lost a mark. Spontaneity is decided by ΔG, not ΔH alone — and NEET exploits this confusion repeatedly.
The concept: Gibbs free energy combines enthalpy and entropy into a single spontaneity criterion at constant temperature and pressure (NCERT Class 11 Chemistry Chapter 5, page 159):
ΔG = ΔH − TΔS
- ΔG < 0 → spontaneous (forward-favoured)
- ΔG = 0 → equilibrium
- ΔG > 0 → non-spontaneous
This means an endothermic reaction (ΔH > 0) CAN be spontaneous if TΔS is large enough — ice melting at room temperature is the textbook example.
The equilibrium link: Standard Gibbs energy connects to the equilibrium constant (NCERT Class 11 Chemistry Chapter 5, page 162):
ΔG° = −RT ln K
The sign logic: K > 1 means ln K > 0, so ΔG° < 0 (products favoured). K < 1 means ln K < 0, so ΔG° > 0 (reactants favoured). At K = 1, ΔG° = 0 exactly.
Bridge to NEET: Questions test two skills — (1) predicting spontaneity from given ΔH and ΔS at a stated temperature, and (2) inferring the sign of ΔG° from a given K value. Both are direct-application level but carry medium negative-marking risk because the sign errors feel trivial until you're under time pressure.
Watch-out: Temperature must be in kelvin. ΔS is often given in J/K while ΔH is in kJ — unit mismatch before subtraction is a common silent error.
Can you answer these Gibbs Free Energy 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 = +40 kJ/mol and ΔS = +120 J/(mol·K), at what temperature does the reaction become spontaneous?
Show answer and why every option is right or wrong
Answer: A. ΔG = ΔH − TΔS < 0 requires T > ΔH/ΔS = 40000 J / 120 J/K = 333.3 K. Above 333 K, TΔS exceeds ΔH, making ΔG negative (NCERT Class 11 Chemistry Chapter 5, page 159).
Why B is wrong: Confuses the inequality direction — TΔS must EXCEED ΔH, so T must be ABOVE the threshold, not below (trap: judging spontaneity from ΔH sign alone).
Why C is wrong: Ignores temperature dependence — when both ΔH and ΔS are positive, spontaneity depends on T (trap: assuming positive ΔS guarantees spontaneity at all T).
Why D is wrong: Ignores that TΔS can overcome positive ΔH at sufficiently high T (trap: assuming endothermic means never spontaneous).
MCQ 2Direct ApplicationPractice
For a reaction at equilibrium (ΔG = 0) with ΔH = −50 kJ/mol and ΔS = −100 J/(mol·K), what is the equilibrium temperature?
Show answer and why every option is right or wrong
Answer: D. At equilibrium ΔG = 0, so ΔH = TΔS. T = ΔH/ΔS = −50000 J / −100 J/K = 500 K (NCERT Class 11 Chemistry Chapter 5, page 159).
Why A is wrong: A: 200 K does not follow from the data. Check it: at 200 K, TΔS = 200 × (−0.100 kJ/K) = −20 kJ, not the −50 kJ needed to cancel ΔH.
Why B is wrong: B: at 50 K, TΔS = −5 kJ, one-tenth of ΔH. 50 K comes from dropping a factor of 10 in the kJ → J conversion (−5000 J / −100 J/K).
Why C is wrong: Multiplies instead of dividing, or uses unconverted values incorrectly (trap: arithmetic error from unit confusion).
MCQ 3Direct ApplicationPractice
If the equilibrium constant K for a reaction is 10⁻⁴ at 298 K, the sign of ΔG° is:
Show answer and why every option is right or wrong
Answer: B. ΔG° = −RT ln K. Since K = 10⁻⁴ < 1, ln K < 0, so −RT ln K > 0. ΔG° is positive, meaning reactants are favoured at equilibrium (NCERT Class 11 Chemistry Chapter 5, page 162).
Why A is wrong: Confuses the double-negative: the formula has a negative sign AND ln K is negative, so the product is positive, not negative (trap: sign error in ΔG° = −RT ln K).
Why C is wrong: ΔG° = 0 only when K = 1 exactly; K = 10⁻⁴ ≠ 1 (trap: confusing ΔG = 0 at equilibrium with ΔG° = 0).
Why D is wrong: All information needed is present — K and T fully determine ΔG° through the formula (trap: not recognising ΔG° = −RT ln K as sufficient).
MCQ 4Direct ApplicationPractice
For a reaction at 300 K, ΔH = −40 kJ mol⁻¹ and ΔS = −100 J K⁻¹ mol⁻¹. The Gibbs energy change ΔG is:
Show answer and why every option is right or wrong
Answer: B. B is correct. ΔG = ΔH − TΔS, with ΔS converted to kJ: ΔS = −0.100 kJ K⁻¹ mol⁻¹. ΔG = −40 − 300 × (−0.100) = −40 + 30 = −10 kJ mol⁻¹. The reaction is spontaneous at 300 K, but only because the favourable ΔH still outweighs the unfavourable −TΔS; at a high enough temperature it would stop being so.
Why A is wrong: A is wrong because −70 kJ mol⁻¹ adds TΔS instead of subtracting it: −40 + (−30). With ΔS negative, the −TΔS term is POSITIVE and works against spontaneity.
Why C is wrong: C is wrong because +2.996 × 10⁴ kJ mol⁻¹ leaves ΔS in J while ΔH is in kJ: −40 − 300 × (−100). The units must match before subtracting.
Why D is wrong: D is wrong because +10 kJ mol⁻¹ gets the sign of the result wrong. ΔH (−40) is larger in magnitude than −TΔS (+30), so ΔG is negative.
MCQ 5CalculationPractice
The standard Gibbs energy of a reaction is −13.6 kJ/mol at 298 K. The equilibrium constant K is approximately: (R = 8.314 J/(mol·K))
Show answer and why every option is right or wrong
Answer: A. ΔG° = −RT ln K → ln K = −ΔG°/(RT) = 13600/(8.314 × 298) = 13600/2477.6 = 5.49. K = e⁵·⁴⁹ ≈ 242 ≈ 2.4 × 10² (NCERT Class 11 Chemistry Chapter 5, page 162).
Why B is wrong: B reports ln K itself instead of K: ln K = 13600/(8.314 × 298) = 5.49 ≈ 5.5, and the final step K = e⁵·⁴⁹ was never taken.
Why C is wrong: Uses ln K ≈ 5.49 but rounds down aggressively or confuses e⁵ ≈ 148 with the answer (trap: arithmetic rounding error in exponential).
Why D is wrong: Uses log₁₀ instead of ln, or miscalculates the exponential as e⁶·⁵⁷ ≈ 712 (trap: confusing ln with log₁₀ or arithmetic slip in division).
MCQ 6Direct ApplicationPractice
For a reaction with ΔH = −10 kJ/mol and ΔS = −30 J/(mol·K), the reaction is:
Show answer and why every option is right or wrong
Answer: C. ΔG = ΔH − TΔS = (−10000) − T(−30) = −10000 + 30T. For ΔG < 0: 30T < 10000, i.e. T < 333 K. Spontaneous only below 333 K (NCERT Class 11 Chemistry Chapter 5, page 159).
Why A is wrong: ΔH < 0 does not guarantee spontaneity at all T when ΔS is also negative — at high T the +30T term overwhelms the negative ΔH (trap: predicting spontaneity from ΔH alone).
Why B is wrong: High-T spontaneity requires ΔS > 0 so that −TΔS becomes more negative with increasing T; here ΔS < 0 so higher T makes ΔG more positive (trap: reversing the temperature-dependence logic).
Why D is wrong: Both ΔH < 0 and ΔS < 0 means there IS a temperature range of spontaneity (low T); it's not non-spontaneous everywhere (trap: confusing negative ΔS with 'never spontaneous').
MCQ 7Easy RecallPractice
At equilibrium, which of the following is true?
Show answer and why every option is right or wrong
Answer: C. At equilibrium, ΔG = 0 always. However, ΔG° = −RT ln K, which is zero only if K = 1. For most reactions K ≠ 1, so ΔG° ≠ 0 at equilibrium (NCERT Class 11 Chemistry Chapter 5, pages 159 and 27).
Why A is wrong: ΔG° = 0 requires K = 1 exactly; equilibrium does not require K = 1, only that Q = K (trap: conflating ΔG = 0 with ΔG° = 0).
Why B is wrong: ΔG (not ΔG°) is the criterion that equals zero at equilibrium; ΔG° is a fixed property of the reaction at standard conditions (trap: swapping the roles of ΔG and ΔG°).
Why D is wrong: At equilibrium ΔG = 0, not negative; negative ΔG means the reaction still proceeds forward, which contradicts equilibrium (trap: confusing 'spontaneous' with 'at equilibrium').
MCQ 8CalculationPractice
A reaction has ΔH = +80 kJ/mol and ΔS = +200 J/(mol·K). At 500 K, calculate ΔG and determine spontaneity.
Show answer and why every option is right or wrong
Answer: A. ΔG = ΔH − TΔS = 80 kJ − (500 K × 0.200 kJ/K) = 80 − 100 = −20 kJ/mol. Negative ΔG means spontaneous at 500 K (NCERT Class 11 Chemistry Chapter 5, page 159).
Why B is wrong: Uses ΔS = 200 without converting J to kJ: 80 − (500 × 200) = 80 − 100000 → then wrongly divides somewhere to get +20, or adds instead of subtracting (trap: unit mismatch J vs kJ for ΔS).
Why C is wrong: Uses ΔG = ΔH + TΔS instead of ΔH − TΔS: 80 + 100 = 180, then assigns wrong sign (trap: sign error in the Gibbs formula).
Why D is wrong: Same addition error as option C but retains positive sign: 80 + 100 = +180 (trap: using + instead of − in ΔG = ΔH − TΔS).
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Gibbs Free Energy: quick recall before you leave
How do you solve a Gibbs Free Energy question? A worked example
Pattern: Predict spontaneity from ΔH and ΔS at given T using ΔG = ΔH − TΔS (P.CHE.U04.SPONTANEITY_GIBBS).
- 1
Given
• ΔH = −40 kJ/mol• ΔS = −120 J/(mol·K)• T₁ = 298 K, T₂ = 400 K
- 2
Required
ΔG at each temperature; spontaneity verdict.
- 3
Concept
ΔG = ΔH − TΔS. Negative ΔG → spontaneous. The threshold temperature where ΔG = 0 is T = ΔH/ΔS.
- 4
Formula
ΔG = ΔH − TΔS
- 5
Substitution
Convert ΔS to kJ: −120 J/(mol·K) = −0.120 kJ/(mol·K)
(a) ΔG₂₉₈ = −40 − (298)(−0.120) = −40 + 35.76
(b) ΔG₄₀₀ = −40 − (400)(−0.120) = −40 + 48.0 - 6
Calculation
(a) ΔG₂₉₈ = −4.24 kJ/mol
(b) ΔG₄₀₀ = +8.0 kJ/mol - 7
Final answer
(a) At 298 K: ΔG = −4.24 kJ/mol → spontaneous
(b) At 400 K: ΔG = +8.0 kJ/mol → non-spontaneous
Threshold: T = 40/0.120 = 333 K. Below 333 K, spontaneous; above 333 K, non-spontaneous.
Note on exact values: T₁ = 298 K and T₂ = 400 K are problem-defined exact values and do not limit significant figures. - 8
Common trap
Predicting spontaneity from ΔH alone: "ΔH is negative so it must be spontaneous." This ignores the −TΔS term. Here, at 400 K the entropy penalty (+48 kJ) overwhelms the enthalpy drive (−40 kJ), flipping ΔG positive.
- 9
Similar NEET-style question
"For a reaction with ΔH = +30 kJ/mol and ΔS = +100 J/(mol·K), calculate the minimum temperature for spontaneity." (Answer: T > 300 K.)
---
What to remember before solving Gibbs Free Energy questions
Formula
ΔG° and equilibrium constant
ΔG° = -RT ln(K). At equilibrium ΔG = 0. K > 1: forward favoured; K < 1: reverse favoured.
-- NCERT Class 11 Chemistry, Ch. 5, p. 162Which Gibbs Free Energy 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
ΔG° and 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
Where do students lose marks on Gibbs Free Energy?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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.
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: 2 exam traps and mistakes · 4 formulas · 1 question pattern from its other lessons.
Gibbs Free Energy questions from past NEET papers
2 questions from NEET 2020, 2023. Answers verified against NTA official keys.
NEET 2023
1–137.26 cal
2–1381.80 cal
3–13.73 cal
41372.60 cal
NTA Answer: Option 2(final)
NEET 2020
1−8.314 J mol⁻¹K⁻¹ × 300 K × ln(4×10¹³)
2−8.314 J mol⁻¹K⁻¹ × 300 K × ln(2×10¹³)
38.314 J mol⁻¹K⁻¹ × 300 K × ln(2×10¹³)
48.314 J mol⁻¹K⁻¹ × 300 K × ln(3×10¹³)
NTA Answer: Option 2(final)
How does NEET ask about Gibbs Free Energy?
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.159 | Class 11 Chemistry Chapter 5, p.162
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →