Enthalpies Various

8 MCQs1 revision card9-step worked example

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

Lesson

This topic is about the different types of standard enthalpy changes — formation, combustion, dissociation, atomisation, sublimation, ionisation, electron gain, solution, and hydration — not the Hess's law arithmetic of combining them. The NEET trap here is definitional precision: confusing the conditions under which each enthalpy type is defined, and misapplying the "per mole" convention.

Standard enthalpy of formation (Δ_f H°) is the enthalpy change when one mole of a compound is formed from its elements in their standard states at 1 bar and 298 K (NCERT Class 11 Chemistry Chapter 5, page 16). The critical clause: elements must be in their most stable allotropic form. So Δ_f H° uses O₂(g), not O(g); C(graphite), not C(diamond). By convention, Δ_f H° of any element in its standard state is zero.

Standard enthalpy of combustion (Δ_c H°) is the enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen under standard conditions. "Complete" means carbon → CO₂, hydrogen → H₂O(l). Writing H₂O(g) instead of H₂O(l) is a common error that changes the numerical value by the enthalpy of vaporisation.

Bond dissociation enthalpy (Δ_bond H°) is the energy required to break one mole of a specific bond in a gaseous molecule homolytically. It is always positive (endothermic). For polyatomic molecules like H₂O, the first O–H bond dissociation enthalpy differs from the second; the "bond enthalpy" quoted in tables is the average.

Other standard enthalpies — atomisation (one mole of gaseous atoms from any phase), sublimation (solid → gas, one mole), ionisation enthalpy (removal of electron from gaseous atom), electron gain enthalpy (addition of electron to gaseous atom), and lattice enthalpy (separation of ionic solid into gaseous ions) — each have strict "per mole" and "phase" requirements. Mixing up the required phases (e.g., starting from liquid instead of solid for sublimation) directly produces wrong numerical answers.

The master formula linking these types is the standard enthalpy of reaction from formation enthalpies:

Δ_rxn H° = Σ Δ_f H°(products) − Σ Δ_f H°(reactants)

This is the workhorse equation for computing reaction enthalpies from tabulated data.

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

The standard enthalpy of formation of an element in its most stable allotropic form at 298 K and 1 bar is:

MCQ 2Easy RecallPractice

Which of the following correctly defines the standard enthalpy of combustion?

MCQ 3Direct ApplicationPractice

The bond dissociation enthalpy of the first O–H bond in H₂O(g) is 502 kJ/mol, and that of the second O–H bond is 427 kJ/mol. What is the average O–H bond enthalpy in water?

MCQ 4Direct ApplicationPractice

For the reaction: 2C(graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l), Δ_f H° = −277 kJ/mol. Which statement is correct?

MCQ 5Easy RecallPractice

Standard enthalpy of atomisation of Na(s) is 108.4 kJ/mol. This represents:

MCQ 6CalculationPractice

Given: Δ_f H°[CO₂(g)] = −393.5 kJ/mol, Δ_f H°[H₂O(l)] = −285.8 kJ/mol, Δ_f H°[CH₄(g)] = −74.8 kJ/mol. Calculate Δ_c H° for CH₄(g).

MCQ 7Easy RecallPractice

The enthalpy of sublimation of a substance is defined as the enthalpy change for:

MCQ 8Direct ApplicationPractice

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), Δ_f H° of NH₃(g) = −46.1 kJ/mol. What is Δ_rxn H° for this reaction?

Quick recall before you leave

Worked Example

1
Given
- Δ_f H°[C₂H₄(g)] = +52.3 kJ/mol - Δ_f H°[CO₂(g)] = −393.5 kJ/mol - Δ_f H°[H₂O(l)] = −285.8 kJ/mol
2
Required
Standard enthalpy of combustion of C₂H₄(g), i.e., Δ_c H°.
3
Concept
Combustion means complete reaction with O₂: all carbon → CO₂(g), all hydrogen → H₂O(l). Δ_rxn H° is computed from standard formation enthalpies using Hess's law in its formation-enthalpy form.
4
Formula
Δ_rxn H° = Σ Δ_f H°(products) − Σ Δ_f H°(reactants)
5
Balanced equation and substitution
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) Δ_c H° = [2 × Δ_f H°(CO₂) + 2 × Δ_f H°(H₂O)] − [Δ_f H°(C₂H₄) + 3 × Δ_f H°(O₂)] Δ_f H°[O₂(g)] = 0 (element in standard state) Δ_c H° = [2(−393.5) + 2(−285.8)] − [52.3 + 0]
6
Calculation
Products: 2(−393.5) + 2(−285.8) = −787.0 + (−571.6) = −1358.6 kJ Reactants: 52.3 + 0 = 52.3 kJ Δ_c H° = −1358.6 − 52.3 = −1410.9 kJ/mol **Note on exact values:** The stoichiometric coefficients (2, 3) are exact counting numbers and do not affect significant-figure count. The result is reported to one decimal place, consistent with the given data.
7
Final answer
**Δ_c H° of C₂H₄(g) = −1410.9 kJ/mol** The negative sign confirms the reaction is exothermic, as expected for combustion.
8
Common trap
Using H₂O(g) instead of H₂O(l) for combustion enthalpy calculations. The standard enthalpy of combustion convention requires liquid water as the product. Using the gaseous-phase formation enthalpy of water (−241.8 kJ/mol instead of −285.8 kJ/mol) gives a numerically smaller magnitude answer, which is a common wrong option in NEET.
9
Similar NEET-style question
Calculate the standard enthalpy of combustion of C₂H₆(g) given: Δ_f H°[C₂H₆(g)] = −84.7 kJ/mol, Δ_f H°[CO₂(g)] = −393.5 kJ/mol, Δ_f H°[H₂O(l)] = −285.8 kJ/mol. *(Balanced equation: C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O; answer: −1559.7 kJ/mol.)* ---

Before solving, remember these

Definition

Standard enthalpies of formation

ΔH_f° = enthalpy change when 1 mole of compound forms from its elements in standard states (1 bar, 298 K). ΔH°_reaction = ΣΔH_f°(products) - ΣΔH_f°(reactants).

-- NCERT Class 11 Chemistry, Ch. 5, p. 16

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.16

Test yourself on this topic with real past-paper questions:

Practice this topic →

Free NEET study resources

Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.

Enthalpies Various

Lesson

Practice MCQs

Quick recall before you leave

Standard enthalpy of reaction from formation enthalpies

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

Correction

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Sources

Free NEET study resources