Definition
Enthalpy
H = U + PV. At constant pressure, ΔH = q_p (heat absorbed at constant P). For exothermic ΔH<0; endothermic ΔH>0.
-- NCERT Class 11 Chemistry, Ch. 5, p. 143Enthalpy Heat Capacity
8 MCQs2 revision cards9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2021, 2023Official key: NTA-verifiedLast updated: 21 Sep 2026
Enthalpy Heat Capacity, explained for NEET
NEET asks you to distinguish between enthalpy and internal energy, apply ΔH = q_p, and connect heat capacity to measurable heat flow. The confusion that costs marks here is not Hess's law or Gibbs energy — those belong to their own topics. The confusion is simpler and more fundamental: mixing up q_p and q_v, or forgetting that ΔH = q_p holds only at constant pressure.
Enthalpy defined. Enthalpy is a state function: H = U + PV, where U is internal energy, P is pressure, and V is volume (NCERT Class 11 Chemistry Chapter 5, page 143). You cannot measure H directly — you measure ΔH, the change.
The constant-pressure link. At constant pressure, the heat absorbed by the system equals the enthalpy change: ΔH = q_p. This is the operational definition that connects calorimetry to thermodynamic tables. At constant volume, it is internal energy that equals the heat: ΔU = q_v. Confusing these two conditions is a common source of wrong answers.
Heat capacity. Heat capacity C is the heat required to raise temperature by 1 K. Two versions matter:
- C_p (constant pressure): q_p = nC_pΔT, so ΔH = nC_pΔT.
- C_v (constant volume): q_v = nC_vΔT, so ΔU = nC_vΔT.
For an ideal gas, the relation C_p − C_v = R connects them (NCERT Class 11 Chemistry Chapter 5, page 145). This means C_p is always larger than C_v — the system at constant pressure must also do expansion work.
The relationship ΔH = ΔU + ΔnRT (for ideal-gas reactions at constant T and P) follows directly from H = U + PV. Here Δn = moles of gaseous products − moles of gaseous reactants. When Δn = 0, ΔH = ΔU.
Watch out: When a problem gives you C_p and asks for ΔU, you need C_v (or must convert via C_p − C_v = R). Don't substitute C_p into a constant-volume expression.
Can you answer these Enthalpy Heat Capacity 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 enthalpy of a system is defined as H = U + PV. Which statement about enthalpy is correct?
Show answer and why every option is right or wrong
Answer: C. By definition, at constant pressure the heat exchanged equals the enthalpy change: ΔH = q_p (NCERT Class 11 Chemistry Chapter 5, page 143). This is the operational meaning of enthalpy in calorimetry.
Why A is wrong: A is wrong because enthalpy is a state function (depends only on initial and final states, not on the path taken). Work and heat are path functions — enthalpy is not.
Why B is wrong: B is wrong because at constant volume, heat equals the change in internal energy (q_v = ΔU), not enthalpy. ΔH = q_p applies at constant pressure, not constant volume.
Why D is wrong: D is wrong because enthalpy is not directly measurable. A thermometer measures temperature. We measure ΔH indirectly through calorimetry (measuring q at constant P).
MCQ 2Easy RecallPractice
For an ideal gas, C_p − C_v equals:
Show answer and why every option is right or wrong
Answer: A. For an ideal gas, the relation C_p − C_v = R holds, where R = 8.314 J mol⁻¹ K⁻¹ (NCERT Class 11 Chemistry Chapter 5, page 145). The extra heat capacity at constant pressure accounts for expansion work.
Why B is wrong: B is wrong because R/2 appears in expressions for C_v of specific gas types (e.g., C_v = 3R/2 for monatomic), but the difference C_p − C_v is always exactly R for any ideal gas.
Why C is wrong: C is wrong because C_p > C_v for any ideal gas. At constant pressure the gas expands against external pressure, requiring additional energy beyond what heats the gas. The difference is never zero.
Why D is wrong: D is wrong because 2R has no basis in the derivation. From PV = nRT and H = U + PV, differentiating gives C_p − C_v = R exactly, not 2R.
MCQ 3Easy RecallPractice
A reaction in solution is carried out in a coffee-cup calorimeter (open to the atmosphere). The heat measured in this experiment directly gives:
Show answer and why every option is right or wrong
Answer: B. A coffee-cup calorimeter operates at constant (atmospheric) pressure. The heat measured under constant pressure equals q_p = ΔH (NCERT Class 11 Chemistry Chapter 5, page 143).
Why A is wrong: A is wrong because ΔU = q_v, the heat at constant volume. A coffee-cup calorimeter is open to the atmosphere (constant pressure, not constant volume). A bomb calorimeter gives q_v = ΔU.
Why C is wrong: C is wrong because ΔG (Gibbs energy) is not directly measurable as heat in any calorimeter. ΔG = ΔH − TΔS requires separate knowledge of entropy change.
Why D is wrong: D is wrong because entropy change ΔS is not directly given by heat measured in a simple calorimeter. ΔS = q_rev/T applies only for reversible processes and requires additional analysis.
MCQ 4Direct ApplicationPractice
For the gaseous reaction N₂(g) + 3H₂(g) → 2NH₃(g), the relationship between ΔH and ΔU at temperature T is:
Show answer and why every option is right or wrong
Answer: A. Δn_g = 2 − (1 + 3) = −2. Using ΔH = ΔU + Δn_g RT gives ΔH = ΔU + (−2)RT = ΔU − 2RT. The negative Δn_g reflects a decrease in gas moles.
Why B is wrong: B is wrong because it uses Δn_g = +2 instead of −2. The correct calculation is Δn_g = moles of gaseous products − moles of gaseous reactants = 2 − 4 = −2, giving a negative correction term.
Why C is wrong: C is wrong because Δn_g = +4 has no basis here. A common error is using the total moles (1 + 3 + 2 = 6 or 4) instead of the difference (products − reactants = 2 − 4 = −2).
Why D is wrong: D is wrong because ΔH = ΔU only when Δn_g = 0 (equal moles of gas on both sides). Here Δn_g = −2, so ΔH ≠ ΔU.
MCQ 5Direct ApplicationPractice
2 moles of an ideal gas are heated from 300 K to 500 K at constant pressure. Given C_p = 29.1 J mol⁻¹ K⁻¹, the enthalpy change ΔH is:
Show answer and why every option is right or wrong
Answer: B. ΔH = nC_pΔT = 2 × 29.1 × (500 − 300) = 2 × 29.1 × 200 = 11640 J = 11.64 kJ. At constant pressure, enthalpy change is calculated using C_p.
Why A is wrong: A is wrong because 5.82 kJ results from using n = 1 instead of n = 2. The problem states 2 moles; the factor n must match: ΔH = 2 × 29.1 × 200 = 11640 J.
Why C is wrong: C is wrong because 8.31 kJ likely arises from substituting R (8.314) or using C_v instead of C_p. At constant pressure, ΔH = nC_pΔT — C_v would give ΔU, not ΔH.
Why D is wrong: D is wrong because 29.1 kJ comes from using only C_p × ΔT without the mole factor (29.1 × 200 = 5820, not 29100), or from a unit error. Careful: ΔH = nC_pΔT, all three factors required.
MCQ 6Direct ApplicationPractice
In a bomb calorimeter experiment, the heat measured gives q_v. For a gaseous reaction with Δn_g = +1, which relationship is correct at temperature T?
Show answer and why every option is right or wrong
Answer: D. In a bomb calorimeter, q_v = ΔU. Then ΔH = ΔU + Δn_g RT = q_v + (1)RT = q_v + RT. The correction converts constant-volume data to the constant-pressure enthalpy change.
Why A is wrong: A is wrong because it uses −RT, implying Δn_g = −1. The problem states Δn_g = +1, so the correction term is +RT, not −RT.
Why B is wrong: B is wrong because ΔH = q_v holds only when Δn_g = 0. With Δn_g = +1, the PV work term contributes an additional RT to the enthalpy.
Why C is wrong: C is wrong because +2RT implies Δn_g = +2. The problem explicitly gives Δn_g = +1, so the correction is exactly RT.
MCQ 7CalculationPractice
For a reaction where ΔH = −100 kJ and Δn_g = −3, at T = 300 K (R = 8.314 J mol⁻¹ K⁻¹), the value of ΔU is closest to:
Show answer and why every option is right or wrong
Answer: C. ΔH = ΔU + Δn_g RT → ΔU = ΔH − Δn_g RT = −100 − (−3)(8.314 × 10⁻³)(300) = −100 + 7.48 = −92.52 kJ ≈ −92.5 kJ. Rearranging correctly and watching the sign of Δn_g is essential.
Why A is wrong: A is wrong because −107.5 kJ results from using ΔU = ΔH + Δn_g RT without flipping the sign correctly. ΔU = ΔH − Δn_g RT: subtracting (−3)(2.494) = subtracting −7.48, which means adding 7.48 to −100.
Why B is wrong: B is wrong because ΔU = ΔH only when Δn_g = 0. Here Δn_g = −3, so the Δn_g RT correction (≈ 7.5 kJ) is significant and cannot be ignored.
Why D is wrong: D is wrong because −75.1 kJ implies a correction of about +25 kJ, which would require Δn_g ≈ −10 at 300 K. The actual correction for Δn_g = −3 is only about 7.5 kJ.
MCQ 8CalculationPractice
A monatomic ideal gas (C_v = 3R/2) is heated from 200 K to 400 K. The ratio ΔH/ΔU for this process is:
Show answer and why every option is right or wrong
Answer: D. For a monatomic ideal gas, C_v = 3R/2 and C_p = C_v + R = 5R/2. ΔH = nC_pΔT and ΔU = nC_vΔT, so ΔH/ΔU = C_p/C_v = (5R/2)/(3R/2) = 5/3. The temperature values and moles cancel in the ratio.
Why A is wrong: A is wrong because ΔH/ΔU = 1 only if C_p = C_v, which never holds for an ideal gas (C_p − C_v = R > 0). The ratio equals γ = C_p/C_v, which is always > 1.
Why B is wrong: B is wrong because 3/2 is the value of C_v/R for a monatomic gas, not the ratio C_p/C_v. The ratio ΔH/ΔU = C_p/C_v = (5R/2)/(3R/2) = 5/3.
Why C is wrong: C is wrong because ΔH/ΔU = 2 would require C_p = 2C_v, i.e., C_p = 3R. But C_p = C_v + R = 3R/2 + R = 5R/2, and (5R/2)/(3R/2) = 5/3, not 2.
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Enthalpy Heat Capacity: quick recall before you leave
How do you solve a Enthalpy Heat Capacity question? A worked example
- 1
Given
For the combustion reaction:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
ΔH = −890 kJ mol⁻¹ at T = 298 K.
R = 8.314 J mol⁻¹ K⁻¹. - 2
Required
Find ΔU for this reaction.
- 3
Concept
Enthalpy and internal energy are related by ΔH = ΔU + Δn_g RT, where Δn_g is the change in moles of gaseous species only. Liquids and solids are excluded from Δn_g.
- 4
Formula
ΔU = ΔH − Δn_g RT
- 5
Substitution
Gaseous products: CO₂ = 1 mol (H₂O is liquid, excluded).
Gaseous reactants: CH₄ + 2O₂ = 1 + 2 = 3 mol.
Δn_g = 1 − 3 = −2.
ΔU = −890 − (−2)(8.314 × 10⁻³)(298) - 6
Calculation
Δn_g RT = (−2)(8.314 × 10⁻³)(298) = −4.955 kJ
ΔU = −890 − (−4.955) = −890 + 4.955 = −885.045 kJ
Note on exact values: The coefficient 2 in Δn_g and the stoichiometric coefficients are exact counting numbers. R = 8.314 J mol⁻¹ K⁻¹ is a defined constant. These do not limit significant figures. The precision is governed by ΔH (3 significant figures). - 7
Final answer
ΔU ≈ −885 kJ mol⁻¹
Reported to 3 significant figures, consistent with the given ΔH = −890 kJ mol⁻¹ (which has 3 significant figures per convention, as the trailing zero after 89 in the context of thermochemical data is significant). - 8
Common trap
Forgetting that H₂O(l) is a liquid and including it in Δn_g. If you mistakenly count H₂O as gaseous: Δn_g = (1 + 2) − (1 + 2) = 0, giving ΔU = ΔH = −890 kJ — which is wrong. Always check state symbols.
Another common error: using the wrong sign for Δn_g (reactants − products instead of products − reactants), which flips the correction term. - 9
Similar NEET-style question
For the reaction 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −198 kJ mol⁻¹ at 298 K. Calculate ΔU. (Answer: Δn_g = 2 − 3 = −1; ΔU = −198 − (−1)(8.314 × 10⁻³)(298) = −198 + 2.478 ≈ −195.5 kJ mol⁻¹.)
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What to remember before solving Enthalpy Heat Capacity questions
Formula
Cp - Cv = nR
For ideal gas: C_p - C_v = nR. Per mole: c_p - c_v = R. C_p > C_v because at constant P, heat also does expansion work.
-- NCERT Class 11 Chemistry, Ch. 5, p. 145Which Enthalpy Heat Capacity formulas do you need for NEET?
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
More in Thermodynamics: 4 exam traps and mistakes · 5 formulas · 2 question patterns from its other lessons.
Enthalpy Heat Capacity questions from past NEET papers
2 questions from NEET 2021, 2023. Answers verified against NTA official keys.
NEET 2023
1ΔH = ΔU + Δn_gRT
2ΔH – ΔU = –ΔnRT
3ΔH + ΔU = ΔnR
4ΔH = ΔU – Δn_gRT
NTA Answer: Option 1(final)
NEET 2021
1C_V = RC_P
2C_P + C_V = R
3C_P − C_V = R
4C_P = RC_V
NTA Answer: Option 3(final)
Sources
NCERT refs: Class 11 Chemistry Chapter 5, p.143 | Class 11 Chemistry Chapter 5, p.145
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.
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