Gibbs Free Energy

8 MCQs9-step worked example

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

Lesson

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 24):

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 27):

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.

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 = +40 kJ/mol and ΔS = +120 J/(mol·K), at what temperature does the reaction become 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?

MCQ 3Direct ApplicationPractice

If the equilibrium constant K for a reaction is 10⁻⁴ at 298 K, the sign of ΔG° is:

MCQ 4Easy RecallPractice

Which combination of ΔH and ΔS makes a reaction spontaneous at ALL temperatures?

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))

MCQ 6Direct ApplicationPractice

For a reaction with ΔH = −10 kJ/mol and ΔS = −30 J/(mol·K), the reaction is:

MCQ 7Easy RecallPractice

At equilibrium, which of the following is true?

MCQ 8CalculationPractice

A reaction has ΔH = +80 kJ/mol and ΔS = +200 J/(mol·K). At 500 K, calculate ΔG and determine spontaneity.

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.) ---

Before solving, remember these

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. 24

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. 27

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.24 | Class 11 Chemistry Chapter 5, p.27

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