Hess's Law and Enthalpy Changes – Hess’s law of constant heat summation: NEET MCQs with Answers

15 free questions on Hess's Law and Enthalpy Changes – Hess’s law of constant heat summation for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Hess’s law of constant heat summation; Enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization, and solution.

Practise all 217 Hess's Law and Enthalpy Changes – Hess’s law of constant heat summation questions free Take a free timed test

Question 1

Previous-year question
For an endothermic reaction, consider the statements below and choose the correct answer. [Re-NEET 2024]
A. q_p is negative.
B. ΔrH is positive.
C. ΔrH is negative.
D. q_p is positive.
  1. B and D
  2. C and D
  3. A and B
  4. A and C
Show answer
Answer: (A) B and D
Explanation: In an endothermic reaction heat is absorbed by the system from the surroundings, so the heat at constant pressure q_p is positive and ΔrH (= q_p) is positive. Hence statements B and D are correct; A and C describe an exothermic reaction.

Question 2

Previous-year question
The standard enthalpy of neutralization of a strong acid by a strong base is −57.3 kJ mol⁻¹: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l), ΔH = −57.3 kJ mol⁻¹. What is the enthalpy change when 0.25 mol of HCl is neutralized by 0.25 mol of NaOH? [NEET 2024 (Sawai Madhopur)]
  1. −28.3 kJ
  2. −14.32 kJ
  3. +57.3 kJ
  4. −57.3 kJ
Show answer
Answer: (B) −14.32 kJ
Explanation: Neutralization of 1 mol H⁺ by 1 mol OH⁻ has ΔH = −57.3 kJ. Here only 0.25 mol of water is formed, so the enthalpy change for this amount is 0.25 × (−57.3) = −14.325 ≈ −14.32 kJ. The molar enthalpy of neutralization itself stays −57.3 kJ mol⁻¹; the −14.32 kJ value is the total heat change for 0.25 mol. Neutralization is exothermic, so a positive value is wrong.

Question 3

Previous-year question
Under standard conditions, the enthalpy change for the reaction H₂(g) + Br₂(g) → 2HBr(g) is −109 kJ mol⁻¹. The bond energies of H₂ and Br₂ are 435 kJ mol⁻¹ and 192 kJ mol⁻¹ respectively. What is the bond energy of HBr (in kJ mol⁻¹)? [NEET 2020 (COVID-19)]
  1. 368
  2. 736
  3. 518
  4. 259
Show answer
Answer: (A) 368
Explanation: ΔH = Σ(bond energies of reactants) − Σ(bond energies of products): −109 = (435 + 192) − 2x. So 2x = 627 + 109 = 736 and x = 368 kJ mol⁻¹. (736 is the energy for two H–Br bonds, not one.)

Question 4

Previous-year question
The bond dissociation energies of X₂, Y₂ and XY are in the ratio 1 : 0.5 : 1. ΔH for the formation of XY is −200 kJ mol⁻¹. The bond dissociation energy of X₂ will be [NEET 2018]
  1. 200 kJ mol⁻¹
  2. 100 kJ mol⁻¹
  3. 400 kJ mol⁻¹
  4. 800 kJ mol⁻¹
Show answer
Answer: (D) 800 kJ mol⁻¹
Explanation: Let the energies be a, 0.5a and a. For ½X₂ + ½Y₂ → XY, ΔH = ½a + ½(0.5a) − a = −0.25a = −200, so a = 800 kJ mol⁻¹.

Question 5

Previous-year question
The heat of combustion of carbon to CO₂ is −393.5 kJ/mol. The heat released upon formation of 35.2 g of CO₂ from carbon and oxygen gas is [AIPMT 2015]
  1. −315 kJ
  2. +315 kJ
  3. −630 kJ
  4. −3.15 kJ
Show answer
Answer: (A) −315 kJ
Explanation: 35.2 g of CO₂ is 35.2/44 = 0.8 mol. The heat released = 0.8 × (−393.5) = −315 kJ.

Question 6

Previous-year question
The bond dissociation energies of X₂, Y₂ and XY are in the ratio 1 : 0.5 : 1. ΔH for the formation of XY is –200 kJ mol⁻¹. The bond dissociation energy of X₂ will be: [NEET 2018]
  1. 800 kJ mol⁻¹
  2. 100 kJ mol⁻¹
  3. 200 kJ mol⁻¹
  4. 400 kJ mol⁻¹
Show answer
Answer: (A) 800 kJ mol⁻¹
Explanation: Let BE(X₂) = a, BE(Y₂) = 0.5a, BE(XY) = a. For ½X₂ + ½Y₂ → XY: ΔH = [½a + ½(0.5a)] – a = –0.25a = –200 → a = 800 kJ mol⁻¹.

Question 7

Previous-year question
The heat of combustion of carbon to CO₂ is –393.5 kJ/mol. The heat released upon formation of 35.2 g of CO₂ from carbon and oxygen gas is: [Re-AIPMT 2015]
  1. –630 kJ
  2. –3.15 kJ
  3. –315 kJ
  4. +315 kJ
Show answer
Answer: (C) –315 kJ
Explanation: 35.2 g CO₂ = 35.2/44 = 0.8 mol. Heat = 0.8 × (–393.5) = –314.8 ≈ –315 kJ.
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Question 8

Previous-year question
Consider the processes: ½A → B, ΔH = +150; 3B → 2C + D, ΔH = –125; E + A → 2D, ΔH = +350 (kJ/mol). For B + D → E + 2C, ΔH will be: [AIPMT Mains 2011]
  1. 325 kJ/mol
  2. 525 kJ/mol
  3. –175 kJ/mol
  4. –325 kJ/mol
Show answer
Answer: (C) –175 kJ/mol
Explanation: Taking (A → 2B, +300) + (3B → 2C + D, –125) – (E + A → 2D, +350) gives B + D → E + 2C with ΔH = 300 – 125 – 350 = –175 kJ/mol.

Question 9

Previous-year question
Enthalpy change for the reaction 4H(g) → 2H₂(g) is –869.6 kJ. The dissociation energy of the H–H bond is: [AIPMT Pre. 2011]
  1. –434.8 kJ
  2. 869.6 kJ
  3. +434.8 kJ
  4. +217.4 kJ
Show answer
Answer: (C) +434.8 kJ
Explanation: Forming 2 moles of H–H bonds releases 869.6 kJ, so bond energy per mole = 869.6/2 = 434.8 kJ. Dissociation energy is +434.8 kJ.

Question 10

Previous-year question
Given: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), ΔH = –26.8 kJ; FeO(s) + CO(g) → Fe(s) + CO₂(g), ΔH = –16.5 kJ. The correct target equation Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g) has ΔH equal to: [AIPMT 2010]
  1. –43.3 kJ
  2. –10.3 kJ
  3. +6.2 kJ
  4. +10.3 kJ
Show answer
Answer: (C) +6.2 kJ
Explanation: Target = (Reaction 1) – 2 × (Reaction 2): ΔH = –26.8 – 2(–16.5) = –26.8 + 33.0 = +6.2 kJ.

Question 11

Previous-year question
From the bond energies H–H = 431.37, C=C = 606.10, C–C = 336.49, C–H = 410.50 kJ mol⁻¹, the enthalpy for the reaction H₂C=CH₂ + H–H → H₃C–CH₃ will be: [AIPMT 2009]
  1. 553.0 kJ mol⁻¹
  2. 1523.6 kJ mol⁻¹
  3. –243.6 kJ mol⁻¹
  4. –120.0 kJ mol⁻¹
Show answer
Answer: (D) –120.0 kJ mol⁻¹
Explanation: Net: break one C=C and one H–H, form one C–C and two extra C–H. ΔH = [606.10 + 431.37] – [336.49 + 2(410.50)] = 1037.47 – 1157.49 = –120.0 kJ mol⁻¹.

Question 12

Previous-year question
Bond dissociation enthalpies of H₂, Cl₂ and HCl are 434, 242 and 431 kJ mol⁻¹ respectively. Enthalpy of formation of HCl is: [AIPMT 2008]
  1. –93 kJ mol⁻¹
  2. 245 kJ mol⁻¹
  3. 93 kJ mol⁻¹
  4. –245 kJ mol⁻¹
Show answer
Answer: (A) –93 kJ mol⁻¹
Explanation: ½H₂ + ½Cl₂ → HCl: Δ_fH = [½(434) + ½(242)] – 431 = 338 – 431 = –93 kJ mol⁻¹.

Question 13

Previous-year question
Given that bond energies of H–H and Cl–Cl are 430 kJ mol⁻¹ and 240 kJ mol⁻¹ respectively and Δ_fH for HCl is –90 kJ mol⁻¹. Bond enthalpy of HCl is: [AIPMT 2007]
  1. 245 kJ mol⁻¹
  2. 290 kJ mol⁻¹
  3. 380 kJ mol⁻¹
  4. 425 kJ mol⁻¹
Show answer
Answer: (D) 425 kJ mol⁻¹
Explanation: ½H₂ + ½Cl₂ → HCl: Δ_fH = [½(430) + ½(240)] – BE(HCl) = 335 – BE(HCl) = –90. So BE(HCl) = 335 + 90 = 425 kJ mol⁻¹.

Question 14

Previous-year question
Consider the reactions: (a) H⁺(aq) + OH⁻(aq) = H₂O(l), ΔH = –X₁; (b) H₂(g) + ½O₂(g) = H₂O(l), ΔH = –X₂; (c) CO₂(g) + H₂(g) = CO(g) + H₂O(l), –X₃; (d) C₂H₂(g) + 5/2 O₂(g) = 2CO₂(g) + H₂O(l), ΔH = X₄. Enthalpy of formation of H₂O(l) is: [AIPMT 2007]
  1. +X₁ kJ mol⁻¹
  2. –X₂ kJ mol⁻¹
  3. +X₃ kJ mol⁻¹
  4. –X₄ kJ mol⁻¹
Show answer
Answer: (B) –X₂ kJ mol⁻¹
Explanation: The formation of H₂O(l) from its elements is H₂(g) + ½O₂(g) → H₂O(l), which is reaction (b) with ΔH = –X₂ kJ mol⁻¹.

Question 15

Previous-year question
The enthalpy of hydrogenation of cyclohexene is –119.5 kJ mol⁻¹. If the resonance energy of benzene is –150.4 kJ mol⁻¹, its enthalpy of hydrogenation would be: [AIPMT 2006]
  1. –508.9 kJ mol⁻¹
  2. –208.1 kJ mol⁻¹
  3. –269.9 kJ mol⁻¹
  4. –358.5 kJ mol⁻¹
Show answer
Answer: (B) –208.1 kJ mol⁻¹
Explanation: Expected ΔH for three isolated double bonds = 3 × (–119.5) = –358.5 kJ. Benzene is stabilised by resonance (150.4 kJ), so actual ΔH = –358.5 + 150.4 = –208.1 kJ mol⁻¹.

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