Q. Consider the following reactions: (i) H⁺(aq) + OH⁻(aq) = H₂O(l), ΔH = −X₁ kJ mol⁻¹ (ii) H₂(g) + ½O₂(g) = H₂O(l), ΔH = −X₂ kJ mol⁻¹ (iii) CO₂(g) + H₂(g) = CO(g) + H₂O, ΔH = −X₃ kJ mol⁻¹ (iv) C₂H₂(g) + (5/2)O₂(g) = 2CO₂(g) + H₂O(l), ΔH = −X₄ kJ mol⁻¹ Enthalpy of formation of H₂O(l) is
A) +X₃ kJ mol⁻¹
B) −X₄ kJ mol⁻¹
C) +X₁ kJ mol⁻¹
D) −X₂ kJ mol⁻¹
Explanation
[Solution from source] Enthalpy of formation is for forming 1 mol of the compound from elements in their standard states: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −X₂. The neutralisation enthalpy −X₁ does not start from elements. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.