Total enthalpy change of a reaction is independent of the path between initial and final states. Allows calculation of ΔH for unknown reactions from sum of known ΔH for elementary reactions.
-- NCERT Class 11 Chemistry, Ch. 5, p. 14Hess Law
8 MCQs2 revision cards9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap that costs marks in Hess's law problems is forgetting to reverse the sign of ΔH when you reverse a thermochemical equation. You see two or three given reactions, you need a target reaction, and somewhere in the manipulation you flip an equation without flipping its ΔH. The answer you get has the wrong sign — and it sits right there among the options.
What Hess's law actually says. Enthalpy is a state function: the total enthalpy change for converting reactants to products depends only on the initial and final states, not on the path taken (NCERT Class 11 Chemistry Chapter 5, page 14). In practice, this means you can combine known thermochemical equations — reversing, multiplying, and adding them — to compute the ΔH of a reaction that cannot be measured directly.
The operational formula:
ΔH(net) = Σ ΔH(steps)
The two manipulation rules you must apply without exception:
- Reverse a reaction → negate ΔH. If the given reaction has ΔH = −393.5 kJ/mol, the reversed reaction has ΔH = +393.5 kJ/mol.
- Multiply a reaction by factor n → multiply ΔH by n. Half the reaction means half the ΔH. Double the reaction means double the ΔH.
NEET application. Questions give you 2–3 thermochemical equations and ask for ΔH of a target reaction. The standard approach: identify which given equations need reversing or scaling to cancel intermediates, adjust each ΔH accordingly, then sum. The wrong-sign-on-reversal distractor appears frequently as an option — it catches students who manipulate the chemical equation correctly but leave ΔH untouched.
Watch-out. Before summing, verify that every intermediate species cancels. If something does not cancel, you have either reversed the wrong equation or applied the wrong stoichiometric multiplier.
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
Hess's law is a direct consequence of which thermodynamic property of enthalpy?
MCQ 2Easy RecallPractice
When a thermochemical equation is reversed, what happens to the sign and magnitude of ΔH?
MCQ 3Easy RecallPractice
If a thermochemical equation is multiplied by a factor of 3, how is the enthalpy change affected?
MCQ 4Direct ApplicationPractice
Given:
MCQ 5Direct ApplicationPractice
Given:
MCQ 6Direct ApplicationPractice
For the reaction 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566.0 kJ. What is ΔH for CO₂(g) → CO(g) + ½O₂(g)?
MCQ 7CalculationPractice
Given:
MCQ 8CalculationPractice
Given:
Quick recall before you leave
Worked Example
- 1
Given
(i) C(graphite) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ/mol (ii) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ/mol
- 2
Required
Find ΔH for: C(graphite) + ½O₂(g) → CO(g)
- 3
Concept
Hess's law: the target ΔH can be found by combining the given equations so that intermediates cancel and the net equation matches the target. Since enthalpy is a state function, the path taken does not affect the total ΔH (NCERT Class 11 Chemistry Chapter 5, page 14).
- 4
Formula
ΔH(target) = Σ ΔH(adjusted steps)
- 5
Substitution
Keep equation (i) as-is: C(graphite) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ/mol Reverse equation (ii): CO₂(g) → CO(g) + ½O₂(g), ΔH = −(−283.0) = +283.0 kJ/mol Add the two equations. CO₂ cancels on both sides. ½O₂ on the right partially cancels O₂ on the left, leaving ½O₂ on the left. Net: C(graphite) + ½O₂(g) → CO(g)
- 6
Calculation
ΔH = ΔH₁ + (−ΔH₂) = (−393.5) + (+283.0) = −110.5 kJ/mol
- 7
Final answer
ΔH = −110.5 kJ/mol This is the standard enthalpy of formation of CO(g). The negative value confirms the reaction is exothermic — consistent with the known stability of CO formation from graphite.
- 8
Common trap
Forgetting to reverse the sign of ΔH₂ when reversing equation (ii). If you use −283.0 instead of +283.0, you get: (−393.5) + (−283.0) = −676.5 kJ/mol — a plausible-looking but incorrect answer that typically appears as a distractor option.
- 9
Similar NEET-style question
Given: (i) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), ΔH₁ = −26.8 kJ (ii) C(graphite) + O₂(g) → CO₂(g), ΔH₂ = −393.5 kJ (iii) C(graphite) + ½O₂(g) → CO(g), ΔH₃ = −110.5 kJ Find ΔH for: 2Fe(s) + 3/2 O₂(g) → Fe₂O₃(s). *Approach:* Reverse equation (i), then use equations (ii) and (iii) with appropriate multipliers to cancel all intermediates. Remember to negate ΔH₁ on reversal. ---
Before solving, remember these
Formulas
Enthalpy and ΔH at constant 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)
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
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
Hess's law
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
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.
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.
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: sign error
Correction
Reversing a reaction: ΔH → -ΔH. Multiplying by n: ΔH → n·ΔH. Apply consistently before summing.
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
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
1A and C
2A, B and D
3A, C and D
4C and D
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
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
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–CHCH
2H CCH–CCH 2 2 2 CH 3 |
3H C C –CHCH
4H CCH–CHCH 2 2 2 2
NTA Answer: Option 1(final)
NEET 2023
1–137.26 cal
2–1381.80 cal
3–13.73 cal
41372.60 cal
NTA Answer: Option 2(final)
NEET 2022
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
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.14
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