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Enthalpies of bond dissociation, combustion, formation and physical changes: NEET MCQs with answers

89 questions on this topic of Chemical Thermodynamics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following compounds has a standard molar enthalpy of formation, ΔfH°, equal to zero at 298 K and 1 bar?

  1. Br₂(g)
  2. O₃(g)
  3. C(diamond)
  4. I₂(s)
Show answer and explanation

Answer: (D) I₂(s)

By convention, the standard molar enthalpy of formation of an element in its most stable state of aggregation at 298 K and 1 bar is zero. At 298 K, iodine exists stably as a solid (I₂(s)), so its ΔfH° = 0. Bromine's standard state is liquid (Br₂(l)), carbon's is graphite (C(graphite)), and oxygen's is O₂(g).

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2. The enthalpies of combustion of carbon(graphite), hydrogen gas, and liquid ethanoic acid (CH₃COOH) are -393.5 kJ mol⁻¹, -285.8 kJ mol⁻¹, and -874.0 kJ mol⁻¹ respectively. What is the standard enthalpy of formation of ethanoic acid?

  1. +194.7 kJ mol⁻¹
  2. +484.6 kJ mol⁻¹
  3. -194.7 kJ mol⁻¹
  4. -484.6 kJ mol⁻¹
Show answer and explanation

Answer: (D) -484.6 kJ mol⁻¹

Formation of ethanoic acid: 2C(graphite) + 2H₂(g) + O₂(g) → CH₃COOH(l). ΔfH° = 2ΔcH°[C] + 2ΔcH°[H₂] - ΔcH°[CH₃COOH] = 2(-393.5) + 2(-285.8) - (-874.0) = -787.0 - 571.6 + 874.0 = -484.6 kJ mol⁻¹.

hard

3. Given the following thermochemical data: (1) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g); ΔrH° = -26.8 kJ (2) FeO(s) + CO(g) → Fe(s) + CO₂(g); ΔrH° = -16.5 kJ What is the standard enthalpy change for the reaction: Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g)?

  1. +43.3 kJ mol⁻¹
  2. -10.3 kJ mol⁻¹
  3. -43.3 kJ mol⁻¹
  4. +6.2 kJ mol⁻¹
Show answer and explanation

Answer: (D) +6.2 kJ mol⁻¹

Target equation: Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g). This can be obtained by taking reaction (1) and subtracting 2 × reaction (2): ΔrH° = ΔrH°(1) - 2 × ΔrH°(2) = -26.8 - 2(-16.5) = -26.8 + 33.0 = +6.2 kJ mol⁻¹.

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4. The standard enthalpy of atomisation of phosphorus, P₄(g), is 1260 kJ mol⁻¹. What is the average P-P single bond dissociation enthalpy in a tetrahedral P₄ molecule?

  1. 315 kJ mol⁻¹
  2. 105 kJ mol⁻¹
  3. 420 kJ mol⁻¹
  4. 210 kJ mol⁻¹
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Answer: (D) 210 kJ mol⁻¹

The tetrahedral P₄ molecule contains 6 P-P single bonds (one along each of the six edges of the tetrahedron). The atomisation reaction is P₄(g) → 4P(g), which requires breaking all 6 P-P bonds. Hence, the bond enthalpy of P-P = 1260 / 6 = 210 kJ mol⁻¹.

easy

5. For a pure compound, the enthalpy of fusion at its normal melting point (200 K) is 6.0 kJ mol⁻¹, and the enthalpy of vaporisation at its normal boiling point (300 K) is 30.0 kJ mol⁻¹. Assuming heat capacities of solid, liquid, and gas phases are negligible over this temperature range, what is the estimated enthalpy of sublimation of the compound at 200 K?

  1. 36.0 kJ mol⁻¹
  2. 24.0 kJ mol⁻¹
  3. 5.0 kJ mol⁻¹
  4. 18.0 kJ mol⁻¹
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Answer: (A) 36.0 kJ mol⁻¹

Sublimation converts solid directly into vapour. According to Hess's law, ΔsubH = ΔfusH + ΔvapH. Therefore, ΔsubH ≈ 6.0 + 30.0 = 36.0 kJ mol⁻¹.

hard

6. Calculate the resonance stabilisation energy of benzene from the following data: ΔfH° of C₆H₆(g) (experimental) = +83 kJ mol⁻¹ ΔsubH° of C(graphite) = 717 kJ mol⁻¹ Bond dissociation enthalpy of H-H = 436 kJ mol⁻¹ Bond enthalpy of C-H = 413 kJ mol⁻¹ Bond enthalpy of C-C single bond = 347 kJ mol⁻¹ Bond enthalpy of C=C double bond = 611 kJ mol⁻¹

  1. -152 kJ mol⁻¹
  2. -358 kJ mol⁻¹
  3. -83 kJ mol⁻¹
  4. -235 kJ mol⁻¹
Show answer and explanation

Answer: (A) -152 kJ mol⁻¹

For the Kekulé structure of benzene, atomisation: 6C(g) + 6H(g) → C₆H₆(g) involves 3 C-C, 3 C=C, and 6 C-H bonds. Energy released in forming bonds = 3(347) + 3(611) + 6(413) = 1041 + 1833 + 2478 = 5352 kJ mol⁻¹. Energy required to produce gaseous atoms: 6ΔsubH°[C] + 3 B-E(H-H) = 6(717) + 3(436) = 4302 + 1308 = 5610 kJ mol⁻¹. Thus, calculated ΔfH° = 5610 - 5352 = +235 kJ mol⁻¹. Resonance energy = Experimental ΔfH° - Calculated ΔfH° = 83 - 235 = -152 kJ mol⁻¹.

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7. Calculate the standard enthalpy of combustion of gaseous propene, C₃H₆(g), from the following standard enthalpies of formation: ΔfH°[C₃H₆(g)] = +20.4 kJ mol⁻¹ ΔfH°[CO₂(g)] = -393.5 kJ mol⁻¹ ΔfH°[H₂O(l)] = -285.8 kJ mol⁻¹

  1. -2058.3 kJ mol⁻¹
  2. -2017.5 kJ mol⁻¹
  3. +2058.3 kJ mol⁻¹
  4. -2099.1 kJ mol⁻¹
Show answer and explanation

Answer: (A) -2058.3 kJ mol⁻¹

The combustion reaction is: C₃H₆(g) + 9/2 O₂(g) → 3CO₂(g) + 3H₂O(l). ΔcH° = [3 × ΔfH°(CO₂) + 3 × ΔfH°(H₂O)] - [ΔfH°(C₃H₆) + 9/2 ΔfH°(O₂)]. ΔcH° = [3(-393.5) + 3(-285.8)] - [20.4 + 0] = [-1180.5 - 857.4] - 20.4 = -2037.9 - 20.4 = -2058.3 kJ mol⁻¹.

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8. The enthalpy of combustion of carbon(graphite) to CO₂(g) is -393.5 kJ mol⁻¹, and that of carbon monoxide (CO) to CO₂(g) is -283.0 kJ mol⁻¹. What is the standard enthalpy of formation of carbon monoxide, CO(g)?

  1. -676.5 kJ mol⁻¹
  2. +110.5 kJ mol⁻¹
  3. -110.5 kJ mol⁻¹
  4. +676.5 kJ mol⁻¹
Show answer and explanation

Answer: (C) -110.5 kJ mol⁻¹

Reaction (1): C(graphite) + O₂(g) → CO₂(g); ΔH₁ = -393.5 kJ mol⁻¹. Reaction (2): CO(g) + 1/2 O₂(g) → CO₂(g); ΔH₂ = -283.0 kJ mol⁻¹. The formation of CO is: C(graphite) + 1/2 O₂(g) → CO(g). ΔfH°[CO(g)] = ΔH₁ - ΔH₂ = -393.5 - (-283.0) = -110.5 kJ mol⁻¹.

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9. The enthalpy change for the complete combustion of 1 mole of methane at 298 K is -890.3 kJ mol⁻¹ according to: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) What is the value of ΔU° for this reaction at 298 K? (Take R = 8.314 J K⁻¹ mol⁻¹)

  1. -895.25 kJ mol⁻¹
  2. -880.45 kJ mol⁻¹
  3. -890.30 kJ mol⁻¹
  4. -885.35 kJ mol⁻¹
Show answer and explanation

Answer: (D) -885.35 kJ mol⁻¹

For the reaction, Δng = n(products, gas) - n(reactants, gas) = 1 - (1 + 2) = -2 mol. We know ΔH° = ΔU° + ΔngRT, so ΔU° = ΔH° - ΔngRT. ΔU° = -890.3 kJ - [(-2 mol) × (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹) × 298 K] = -890.3 - (-4.955) = -885.35 kJ mol⁻¹.

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10. The bond dissociation enthalpies of H-H, Cl-Cl, and H-Cl are 435 kJ mol⁻¹, 243 kJ mol⁻¹, and 431 kJ mol⁻¹ respectively. What is the standard enthalpy of formation of HCl(g)?

  1. -184.0 kJ mol⁻¹
  2. -92.0 kJ mol⁻¹
  3. +92.0 kJ mol⁻¹
  4. +184.0 kJ mol⁻¹
Show answer and explanation

Answer: (B) -92.0 kJ mol⁻¹

For the reaction 1/2 H₂(g) + 1/2 Cl₂(g) → HCl(g), ΔrH° = 1/2 B-E(H-H) + 1/2 B-E(Cl-Cl) - B-E(H-Cl). Thus, ΔfH°[HCl] = 1/2(435) + 1/2(243) - 431 = 217.5 + 121.5 - 431 = -92.0 kJ mol⁻¹.

easy

11. The enthalpy of atomisation of an element or compound is defined as the enthalpy change accompanying the:

  1. Dissolution of one mole of a solid substance into an infinite volume of water
  2. Total dissociation of one mole of a substance into gaseous ions
  3. Breaking of exactly one mole of a specific bond in a polyatomic molecule
  4. Formation of one mole of gaseous atoms from one mole of substance in its standard state
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Answer: (D) Formation of one mole of gaseous atoms from one mole of substance in its standard state

Enthalpy of atomisation (ΔaH°) is defined as the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase, or producing 1 mole of gaseous atoms from the substance in its standard state. Ionic dissociation corresponds to lattice enthalpy or ionization.

easy

12. Given that the enthalpy of sublimation of dry ice (solid CO₂) is 25.2 kJ mol⁻¹ and its enthalpy of fusion is 8.4 kJ mol⁻¹ at its triple point, what is the standard enthalpy of vaporisation (ΔvapH°) of liquid CO₂?

  1. 33.6 kJ mol⁻¹
  2. -16.8 kJ mol⁻¹
  3. 16.8 kJ mol⁻¹
  4. 2.0 kJ mol⁻¹
Show answer and explanation

Answer: (C) 16.8 kJ mol⁻¹

According to Hess's law, the phase transitions are related by ΔsubH = ΔfusH + ΔvapH. Therefore, ΔvapH = ΔsubH - ΔfusH = 25.2 - 8.4 = 16.8 kJ mol⁻¹.

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