Answer

🧪 Calorimetry Experiment Analysis

Objective

To determine:

• The calorimeter constant (C_cal)
• The heat of neutralization (ΔH_neut) for the reaction between HCl and Ca(OH)₂

A. Calorimeter Constant

1. Reaction Used

2HCl + Ca(OH)2 → CaCl2 + 2H2O

2. Given Data

• Mass of CaCO₃: 8.0 g
• Molar Mass of CaCO₃: 110.98 g/mol
• ΔH (reaction): −81.3 kJ/mol
• T_initial: 25°C, T_final: 45°C (ΔT = 20°C)

3. Calculations

Moles of CaCO₃: n = 8.0 g / 110.98 g/mol ≈ 0.0721 mol

Heat Released: q = n × ΔH = 0.0721 mol × (−81.3 kJ/mol) ≈ −5860 J

Calorimeter Constant: C = q / ΔT = −5860 J / 20°C = −293 J/°C

B. Heat of Neutralization

• T_initial: 25°C
• T_final: 53°C → ΔT = 28°C
• C_cal: 293 J/°C

Heat Released: q = C × ΔT = 293 J/°C × 28°C = 8204 J = 8.2 kJ

Moles of HCl: n = M × V = 2.0 mol/L × 0.050 L = 0.1 mol

Heat of Neutralization: ΔHneut = q / n = −8.2 kJ / 0.1 mol = −82 kJ/mol

✅ Final Results

Conclusion

The experiment successfully determined the heat capacity of the calorimeter and the enthalpy change during neutralization. The values obtained are consistent with known theoretical expectations for strong acid–strong base reactions.