More practice questions

4. Q4. The enthalpies of all elements in their standard states are (i) unity; (ii) zero; (iii) < 0; (iv) different for each element (NCERT Exercise 5.3, page 165). Which statement is true, and what does it really mean?

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   (ii), as a convention: the standard enthalpy of formation of an element in its reference state is taken as zero. Absolute enthalpies cannot be measured, so this sets a common zero from which every compound’s Δ_fH^⊖ is counted.

5. Q5. Which of these has Δ_rH^⊖ equal to a standard enthalpy of formation? (a) CO(g) + ½O₂(g) → CO₂(g); (b) ½N₂(g) + O₂(g) → NO₂(g); (c) C(diamond) + O₂(g) → CO₂(g); (d) 2H₂(g) + O₂(g) → 2H₂O(l).

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   Only (b): one mole of NO₂ from its elements in their reference states. (a) starts from a compound, CO; (c) uses diamond, not graphite; (d) forms 2 mol of water.

6. Q6. Using NCERT Table 5.2, find Δ_rH^⊖ for 2NO(g) + O₂(g) → 2NO₂(g).

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   Δ_rH^⊖ = 2(33.18) − [2(90.25) + 0] = 66.36 − 180.50, about −114.1 kJ mol⁻¹.

7. Q7. Given S(rhombic) + O₂(g) → SO₂(g), Δ_rH^⊖ = −296.8 kJ mol⁻¹, and SO₂(g) + ½O₂(g) → SO₃(g), Δ_rH^⊖ = −98.9 kJ mol⁻¹, find Δ_fH^⊖ of SO₃(g).

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   Add the two steps (Hess’s law): −296.8 + (−98.9) = −395.7 kJ mol⁻¹, in agreement with NCERT Table 5.2 (−395.72).

8. Q8. For Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l), Δ_rH^⊖ = −33.3 kJ mol⁻¹. What is Δ_rH^⊖ for 4Fe(s) + 6H₂O(l) → 2Fe₂O₃(s) + 6H₂(g)?

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   The equation is reversed (sign changes) and doubled (value doubles): Δ_rH^⊖ = −2 × (−33.3) = +66.6 kJ mol⁻¹.