Practice question

Given: What is Δ H for C₆H₆(l) + 15/2O₂(g) -> 6CO₂(g) + 3H₂O(l) if Δ_f H°: C₆H₆(l) = 49 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(l) = -285.8 kJ/mol? Formula: Δ_r H = [6(-393.5) + 3(-285.8)] - [49 + 0] = -2361 - 857.4 - 49 = -3267.4 kJ/mol.. Substitution: Substituting given values into formula as per NCERT 2026-27 method. Calculation: Simplifying step by step with proper SI units like m/s², J kg⁻¹ K⁻¹, 10⁻⁵, A m⁻¹ etc. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.