Lecture on Heat of Neutralization

Introduction

Topic: Heat of Neutralization
Focus: Calculating the standard heat of neutralization (ΔH⁰) and solving related problems

Standard Heat of Neutralization (ΔH⁰):
- [ΔH⁰ = -\frac{Q}{n}]
- Q = Amount of heat released
- n = Number of moles of acid or base
- Negative sign indicates exothermic reaction
Heat (Q):
- [Q = m_s \times C_s \times ΔT]
- m_s = Mass of the solution
- C_s = Specific heat capacity of the solution
- ΔT = Change in temperature (T₂ - T₁)

Mass of the Solution (m_s):
- Volume of solution = Volume of acid + Volume of base
- Density of solution (usually 1 g/cm³)
- Mass = Volume × Density
Temperature Change (ΔT):
- T₁: Average of initial temperatures of acid and base
- T₂: Final temperature of the mixture

Given:
- 75 cm³ of 0.2 M NaOH
- 75 cm³ of 0.2 M HCl
- Initial temperature: 14.7°C
- Final temperature: 16.0°C
- Specific heat capacity: 4.2 J/g·°C
Steps:
1. Calculate volume and mass of the solution:
  - Volume = 75 cm³ + 75 cm³ = 150 cm³
  - Mass = 150 cm³ × 1 g/cm³ = 150 g
2. Calculate Q:
  - ΔT = 16.0 - 14.7 = 1.3°C
  - Q = 150 g × 4.2 J/g·°C × 1.3°C
  - Q = 819 J
3. Calculate number of moles:
  - Moles of acid = 0.2 mol/dm³ × 75 cm³ / 1000 = 0.015 mol
4. Calculate ΔH⁰:
  - ΔH⁰ = -\frac{819 J}{0.015 mol} = -54.6 kJ/mol

Given:
- 50 cm³ of 0.33 M KOH
- 50 cm³ of 0.33 M HNO₃
- Initial temperatures: 31°C (KOH), 18°C (HNO₃)
- Final temperature: 34.1°C
- Heat capacity of calorimeter: 0.488 J/K
- Specific heat capacity: 4.208 J/g·K
Steps:
1. Calculate initial temperature (T₁):
  - T₁ = (31 + 33) / 2 = 32°C
2. Calculate volume and mass of the solution:
  - Volume = 50 cm³ + 50 cm³ = 100 cm³
  - Mass = 100 cm³ × 1 g/cm³ = 100 g
3. Calculate Q:
  - ΔT = 34.1 - 32 = 2.1°C
  - Q = (100 g × 4.208 J/g·°C + 0.488 J/°C) × 2.1°C
  - Q = 884.705 J
4. Calculate number of moles:
  - Moles of base = 0.33 mol/dm³ × 50 cm³ / 1000 = 0.0165 mol
5. Calculate ΔH⁰:
  - ΔH⁰ = -\frac{884.705 J}{0.0165 mol} = -53.618 kJ/mol