Enthalpies Various

Answer: C. By the IUPAC convention (NCERT Class 11 Chemistry Chapter 5, page 150), the standard enthalpy of formation of an element in its reference state (most stable allotrope) is assigned as zero, since the "reactants" and "product" are the same substance.

Why A is wrong: A: Atomisation enthalpy is the energy to convert one mole of a substance into gaseous atoms — it is a positive, non-zero value even for elements in their standard state (e.g., O₂ → 2O). Formation enthalpy of the standard-state element is zero, not equal to atomisation enthalpy.

Why B is wrong: B: This is wrong because Δ_f H° of an element in its standard state is exactly zero, not 'always positive.' Some compounds do have positive Δ_f H° values, but the question asks about the element itself.

Why D is wrong: D: Sublimation enthalpy applies to solid → gas conversion and is a positive, non-zero quantity. It has no definitional connection to Δ_f H° of the standard-state element, which is zero.

Answer: C. Standard enthalpy of combustion (Δ_c H°) is defined per mole of the substance being burnt, with complete combustion in excess O₂ under standard conditions (1 bar, 298 K), as stated in NCERT Class 11 Chemistry Chapter 5.

Why A is wrong: A: The 'per mole' refers to one mole of the substance being combusted, not one mole of oxygen. This reversal of the reference species is a definitional trap.

Why B is wrong: B: Standard enthalpies are defined per mole, not per gram. Calorific value (energy per gram) is a related but different quantity.

Why D is wrong: D: 'Partial combustion' (e.g., carbon → CO instead of CO₂) violates the 'complete combustion' requirement. Incomplete combustion gives a different, smaller magnitude enthalpy value.

Answer: B. Average bond enthalpy = (502 + 427)/2 = 464.5 kJ/mol. For polyatomic molecules, tabulated "bond enthalpy" values are averages over all equivalent bonds in the molecule (NCERT Class 11 Chemistry Chapter 5).

Why A is wrong: A: 502 kJ/mol is the first O–H dissociation enthalpy only, not the average. Using the first bond energy alone overestimates the per-bond value.

Why C is wrong: C: 427 kJ/mol is the second O–H dissociation enthalpy, which is lower because the remaining OH radical is less stable than H₂O. It is not the average.

Why D is wrong: D: 929 kJ/mol is the total energy to break both O–H bonds (atomisation enthalpy of H₂O). The question asks for the average per bond, so this must be divided by 2.

Answer: B. The reaction shows one mole of C₂H₅OH(l) formed from its elements in their standard states — this is the definition of standard enthalpy of formation (NCERT Class 11 Chemistry Chapter 5, page 150). The value −277 kJ/mol belongs to the liquid product specifically.

Why A is wrong: A: The product is C₂H₅OH(l), not C₂H₅OH(g). The phase matters — the enthalpy of vaporisation separates these two values. Assigning this Δ_f H° to the gaseous form is a phase-mismatch error.

Why C is wrong: C: Combustion of ethanol would produce CO₂ and H₂O as products, with O₂ as a reactant. This reaction forms ethanol — it is a formation reaction, the reverse conceptual direction from combustion.

Why D is wrong: D: Graphite is an element in its standard state, so Δ_f H°(C, graphite) = 0 by convention, not −277 kJ/mol. The −277 value applies to the compound product.

Answer: A. Atomisation enthalpy is the enthalpy change to produce one mole of gaseous atoms from the substance in its standard state. For metallic sodium, this is the conversion Na(s) → Na(g), as per NCERT Class 11 Chemistry Chapter 5.

Why B is wrong: B: Na(s) → Na(l) is fusion (melting), not atomisation. Atomisation requires conversion all the way to gaseous atoms.

Why C is wrong: C: Na(g) → Na⁺(g) + e⁻ is the ionisation enthalpy, which starts from gaseous atoms and removes an electron. Atomisation stops at gaseous atoms without ionising them.

Why D is wrong: D: This involves ionisation AND solvation (aqueous phase), combining multiple enthalpy changes. Atomisation is strictly the solid-to-gaseous-atom step.

Answer: D. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Using Δ_rxn H° = Σ Δ_f H°(products) − Σ Δ_f H°(reactants): [−393.5 + 2(−285.8)] − [−74.8 + 0] = [−393.5 − 571.6] − [−74.8] = −965.1 + 74.8 = −890.3 kJ/mol.

Why A is wrong: A: This value (−679.3) makes two slips at once: one mole of H₂O instead of two, and no subtraction of Δ_f H° of CH₄: −393.5 − 285.8 = −679.3.

Why B is wrong: B: This value (−604.5) results from using only one mole of H₂O instead of two in the balanced equation. The stoichiometric coefficient of H₂O is 2, so the contribution is 2 × (−285.8) = −571.6, not −285.8.

Why C is wrong: C: Combustion reactions are exothermic by definition, so Δ_c H° must be negative. A positive value indicates the sign was flipped — likely from subtracting products from reactants instead of the correct order.

Answer: A. Enthalpy is a state function, so ΔH for the solid → gas conversion depends only on the initial and final states, not on the path taken to get there. The two-step path solid → liquid (fusion, +6.00 kJ/mol) → gas (vaporisation, +40.79 kJ/mol) must therefore add up to the same total enthalpy change as the direct solid → gas (sublimation) path: ΔsubH° = ΔfusH° + ΔvapH° = 6.00 + 40.79 = 46.79 kJ/mol (NCERT Class 11 Chemistry Chapter 5, pages 147–148).

Why B is wrong: B: This lands on the same number (46.79) but for the wrong reason — sublimation enthalpy is not an average of two dissimilar quantities; addition and averaging only coincidentally agree when both steps happen to be equal, which they are not here (trap: right number, wrong law — Hess's law/state-function additivity, not averaging, is what justifies the sum).

Why C is wrong: C: Fusion enthalpy is not 'spent' or subtracted from vaporisation enthalpy; both steps are endothermic and add together because both are physically necessary to reach the gaseous state from the solid, so 34.79 kJ/mol (a subtraction) has no thermodynamic basis (trap: confusing sequential steps with cancellation).

Why D is wrong: D: This ignores the fusion step entirely. Melting the solid into liquid is a necessary, energy-requiring step on the way to vapour, and its enthalpy contributes fully to the total sublimation enthalpy alongside vaporisation (trap: counting only the final leg of a two-step path).

Answer: A. Δ_rxn H° = Σ Δ_f H°(products) − Σ Δ_f H°(reactants) = [2 × (−46.1)] − [0 + 0] = −92.2 kJ. The stoichiometric coefficient of 2 for NH₃ must multiply the per-mole formation enthalpy. N₂ and H₂ are elements in standard states, so their Δ_f H° = 0.

Why B is wrong: B: −46.1 kJ is the formation enthalpy per mole of NH₃, but the balanced equation produces 2 moles. Forgetting to multiply by the stoichiometric coefficient is a common error in applying the standard enthalpy formula.

Why C is wrong: C: +46.1 kJ has the wrong sign and ignores the stoichiometric coefficient. The formation of NH₃ is exothermic (negative Δ_f H°), and the reaction produces 2 moles, making Δ_rxn H° = −92.2 kJ.

Why D is wrong: D: −138.3 kJ would correspond to multiplying by 3 instead of 2, likely confusing the coefficient of H₂ (3) with that of NH₃ (2). The relevant coefficient is that of the product whose Δ_f H° is being used.