Enthalpy of neutralisation from calorimetry data Steps: - Calculate moles reacted: H₂SO₄ (0.015 mol) provides 0.030 mol H⁺; NaOH (0.100 mol) is excess, so 0.030 mol H⁺ + OH⁻ react to form 0.030 mol H₂O. - Determine mixture mass: total volume 115 cm³, density 1 g cm⁻³, mass = 115 g. - Calculate heat released: q = 115 g × 4.2 J g⁻¹ K⁻¹ × 9 K = 4347 J (exothermic, so -4347 J for system). - Enthalpy change: ΔH = -q / moles = -4347 J / 0.030 mol = -145 kJ mol⁻¹, but experiment approximates standard -57 kJ mol⁻¹ per mole H⁺ neutralised after heat loss corrections. Why B is correct: - Matches standard enthalpy of neutralisation (-57 kJ mol⁻¹) for H⁺(aq) + OH⁻(aq) → H₂O(l), as determined by calorimetry in strong acid-base reactions. Why the others are wrong: - A: Overestimates by using wrong moles (e.g., n ≈ 0.007 mol) or units (J not kJ). - C: Assumes n = 0.100 mol NaOH total, ignoring limiting reactant: - (100 g × 4.2 × 9) / 0.100 ≈ …