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. 151Hess's Law
8 MCQs6 revision cards9-step worked example
Source: NCERT ThermodynamicsOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Hess's Law, explained for NEET
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 151). 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.
Can you answer these Hess's Law 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?
Show answer and why every option is right or wrong
Answer: A. Hess's law holds because enthalpy is a state function — its change depends only on the initial and final states, not the path. This is stated directly in NCERT Class 11 Chemistry Chapter 5, page 151.
Why B is wrong: B — Enthalpy is extensive, not intensive (it scales with amount of substance). Being intensive would not by itself guarantee path-independence.
Why C is wrong: C — While enthalpy is indeed extensive, that describes its dependence on quantity of matter, not its path-independence. Hess's law follows specifically from the state-function property.
Why D is wrong: D — This is the opposite of what Hess's law requires. Enthalpy change is path-independent; if it were path-dependent, combining reaction steps would give different results depending on the route.
MCQ 2Easy RecallPractice
When a thermochemical equation is reversed, what happens to the sign and magnitude of ΔH?
Show answer and why every option is right or wrong
Answer: D. Reversing a reaction reverses the direction of heat flow. The magnitude of ΔH stays the same; only the sign flips. This is a direct corollary of the state-function nature of enthalpy (NCERT Class 11 Chemistry Chapter 5, page 151).
Why A is wrong: A — The sign does change on reversal, but the magnitude does not double. Doubling occurs only when you multiply the entire equation by 2.
Why B is wrong: B — The sign must change when the reaction direction reverses. If an exothermic forward reaction becomes endothermic on reversal, the sign cannot stay the same.
Why C is wrong: C — If the sign stayed the same after reversal, you could create energy from nothing by running a reaction forward then backward — this would violate the first law.
MCQ 3Easy RecallPractice
If a thermochemical equation is multiplied by a factor of 3, how is the enthalpy change affected?
Show answer and why every option is right or wrong
Answer: C. Enthalpy is an extensive property — it scales linearly with the amount of substance. Multiplying the balanced equation by 3 means tripling the moles involved, so ΔH is also tripled. This follows directly from the Hess's law manipulation rules (NCERT Class 11 Chemistry Chapter 5, page 151).
Why A is wrong: A — ΔH scales with stoichiometric coefficients because enthalpy is extensive. If it remained unchanged on scaling, Hess's law calculations using fractional or multiple equations would be impossible.
Why B is wrong: B — Dividing ΔH by 3 would apply if you divided the equation by 3, not multiplied. The scaling is proportional: multiply equation by n, multiply ΔH by the same n.
Why D is wrong: D — Sign changes occur when you reverse a reaction, not when you multiply it. Multiplying by a positive factor preserves the sign.
MCQ 4Direct ApplicationPractice
Given:
(i) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ/mol
(ii) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ/mol
Using Hess's law, find ΔH for: C(s) + ½O₂(g) → CO(g)
Show answer and why every option is right or wrong
Answer: C. Target = equation (i) minus equation (ii). ΔH = ΔH₁ − ΔH₂ = (−393.5) − (−283.0) = −110.5 kJ/mol. This is the standard enthalpy of formation of CO(g). (NCERT Class 11 Chemistry Chapter 5, page 151.)
Why A is wrong: A — This comes from adding the two ΔH values instead of subtracting: (−393.5) + (−283.0) = −676.5. Addition would be correct only if both equations are used as-is in the same direction, but here equation (ii) must effectively be reversed to cancel CO₂.
Why B is wrong: B — This is the sign-reversal trap. A student who reverses equation (ii) correctly but forgets to also account for the subtraction arithmetic gets +110.5. The correct subtraction gives (−393.5) − (−283.0) = −110.5. (Trap: sign reversal on Hess's law combination.)
Why D is wrong: D — This combines both errors: adding instead of subtracting, and then flipping the sign. No valid Hess's law pathway yields this value.
MCQ 5Direct ApplicationPractice
Given:
(i) H₂(g) + ½O₂(g) → H₂O(l), ΔH = −286 kJ/mol
(ii) H₂(g) + ½O₂(g) → H₂O(g), ΔH = −242 kJ/mol
Using Hess's law, find ΔH for: H₂O(l) → H₂O(g)
Show answer and why every option is right or wrong
Answer: B. Reverse equation (i): H₂O(l) → H₂(g) + ½O₂(g), ΔH = +286 kJ/mol. Add equation (ii) as-is: ΔH = −242 kJ/mol. Net ΔH = +286 + (−242) = +44 kJ/mol. Vaporisation is endothermic, so the positive sign is physically expected. (NCERT Class 11 Chemistry Chapter 5, page 151.)
Why A is wrong: A — Adding both ΔH values without reversing either equation gives −528 kJ/mol. But to get the target H₂O(l) → H₂O(g), equation (i) must be reversed.
Why C is wrong: C — This comes from subtracting in the wrong order: (−286) − (−242) = −44. The student reversed equation (ii) instead of equation (i). Since vaporisation absorbs heat, a negative ΔH for liquid-to-gas is a physical red flag. (Trap: sign reversal applied to the wrong equation.)
Why D is wrong: D — This is +286 + 242 = +528, where the student reversed equation (i) correctly but also reversed equation (ii) unnecessarily, flipping its sign to +242. Only one equation needs reversal.
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)?
Show answer and why every option is right or wrong
Answer: B. First, divide by 2: CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. Then reverse: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Both manipulation rules applied: halve the ΔH, then negate it. (NCERT Class 11 Chemistry Chapter 5, page 151.)
Why A is wrong: A — This applies the halving step correctly (−566.0 ÷ 2 = −283.0) but forgets to reverse the sign for the reversed reaction. The decomposition of CO₂ is endothermic, so ΔH must be positive. (Trap: sign reversal forgotten.)
Why C is wrong: C — This simply keeps the original ΔH without halving or reversing. The target reaction has different stoichiometry and opposite direction — both adjustments are required.
Why D is wrong: D — This reverses the sign (+566.0) but forgets to halve. The target reaction involves 1 mol CO₂, not 2, so the magnitude must be divided by 2 before sign reversal.
MCQ 7CalculationPractice
Given:
(i) 2H₂(g) + O₂(g) → 2H₂O(g), ΔH₁ = −484 kJ
(ii) 3H₂(g) + O₂(g) → 2H₂O(g) + H₂(g) is a trivially rearranged form of equation (i).
A student claims that for the reaction H₂O(g) → H₂(g) + ½O₂(g), ΔH = −242 kJ. Identify the error.
Show answer and why every option is right or wrong
Answer: D. Halving equation (i): H₂(g) + ½O₂(g) → H₂O(g), ΔH = −242 kJ. Reversing: H₂O(g) → H₂(g) + ½O₂(g), ΔH = +242 kJ. The student halved correctly but kept the negative sign instead of reversing it. The decomposition of water is endothermic, confirming the sign must be positive. (NCERT Class 11 Chemistry Chapter 5, page 151.)
Why A is wrong: A — Decomposing water into its elements is endothermic. A negative ΔH would mean the decomposition releases heat, which contradicts the known thermochemistry.
Why B is wrong: B — The student did halve correctly: −484 ÷ 2 = −242. The error is not in scaling but in failing to negate the sign upon reversal.
Why C is wrong: C — Equation (i) is the correct starting point. The error is purely in the sign manipulation, not in equation selection.
MCQ 8CalculationPractice
Given:
(i) A → B, ΔH₁ = −100 kJ
(ii) B → C, ΔH₂ = −200 kJ
(iii) C → D, ΔH₃ = +150 kJ
Using Hess's law, find ΔH for D → A.
Show answer and why every option is right or wrong
Answer: A. First find A → D by summing all three: ΔH(A→D) = (−100) + (−200) + (+150) = −150 kJ. Then reverse for D → A: ΔH = +150 kJ. Two steps: sum the forward path, then negate for the reverse direction. (NCERT Class 11 Chemistry Chapter 5, page 151.)
Why B is wrong: B — This is ΔH for A → D, not D → A. The question asks for the reverse direction. Forgetting to negate after computing the forward-path sum is the sign-reversal trap. (Trap: sign reversal forgotten on the final answer.)
Why C is wrong: C — This appears to come from taking the absolute values and summing only two of them (100 + 150 = 250). All three equations must be included, and signs must be preserved throughout.
Why D is wrong: D — This results from summing all three absolute values (100 + 200 − 150 = 250, then negating incorrectly). Algebraic summation with correct signs gives −150 for A → D, then +150 for D → A.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Hess's Law: quick recall before you leave
How do you solve a Hess's Law question? A 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 151).
- 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.
---
What to remember before solving Hess's Law questions
Which Hess's Law formulas do you need for NEET?
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
Where do students lose marks on Hess's Law?
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
Reversing a reaction: ΔH → -ΔH. Multiplying by n: ΔH → n·ΔH. Apply consistently before summing.
More in Thermodynamics: 2 exam traps and mistakes · 5 formulas · 1 question pattern from its other lessons.
Hess's Law questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Hess's Law?
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
Sources
NCERT refs: Class 11 Chemistry Chapter 5, p.151
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 →