The enthalpy of dissolution represents a fundamental thermodynamic property that measures the heat change when one mole of a solute dissolves in a solvent to form an infinitely dilute solution. This experiment provides students with hands-on experience in calorimetry and thermodynamics while determining the heat of solution for common salts like copper sulphate (CuSO₄) and potassium nitrate (KNO₃).
Understanding dissolution enthalpy is crucial for various industrial applications, including pharmaceutical formulation, chemical processing, and environmental studies. This comprehensive guide will walk you through the complete experimental procedure for determining the enthalpy of dissolution in laboratory conditions.
To determine the enthalpy of dissolution of copper sulphate (CuSO₄) or potassium nitrate (KNO₃) using calorimetric methods in a controlled laboratory environment.
Keywords: enthalpy of dissolution, heat of solution, calorimetry experiment, copper sulphate enthalpy, potassium nitrate dissolution
The enthalpy of dissolution (ΔH_solution) is defined as the amount of heat absorbed or released when one mole of a solute dissolves completely in a solvent under constant pressure and temperature conditions.
When a solute dissolves in a solvent, two primary processes occur:
The net enthalpy change is expressed as: ΔH_solution = ΔH_lattice + ΔH_hydration
ΔH_solution = (m × s × ΔT) / n
Where:
Step 1: Initial Temperature Measurement
Step 2: Addition of Solute
Step 3: Temperature Monitoring
Step 4: Data Collection
Time (min) | Temperature (°C) | Observations |
---|---|---|
0.0 | T₁_initial | Initial stable temp |
0.5 | ||
1.0 | ||
1.5 | ||
2.0 | ||
2.5 | First solute addition | |
3.0 | Peak temperature | |
3.5 | ||
4.0 | ||
4.5 | ||
5.0 | T₂_final |
Parameter | Value | Unit |
---|---|---|
Mass of water | 100 | g |
Mass of solute | 5.00 | g |
Initial temperature (T₁) | 25.0 | °C |
Final temperature (T₂) | 29.5 | °C |
Temperature change (ΔT) | 4.5 | °C |
Moles of solute | 0.0313 (for CuSO₄) | mol |
Heat absorbed (q) | 1881 | J |
Enthalpy of dissolution | 60.1 | kJ/mol |
Given:
Calculation: q = m × s × ΔT = 105 × 4.18 × 4.5 = 1975.05 J ΔH = q/n = 1975.05 J ÷ 0.0313 mol = 63.1 kJ/mol
Copper Sulphate (CuSO₄):
Potassium Nitrate (KNO₃):
Note: The negative sign indicates exothermic process, positive indicates endothermic.
A1: Enthalpy of dissolution is the heat change when one mole of a solute dissolves in a solvent to form an infinitely dilute solution under standard conditions.
A2: Polystyrene is a good thermal insulator with low heat capacity, minimizing heat loss to surroundings and providing accurate results.
A3: Factors include:
A4: Quick addition prevents heat loss to the surroundings and ensures accurate measurement of temperature change during dissolution.
A5: Stirring ensures uniform temperature distribution throughout the solution and promotes complete dissolution of the solute.
A6: Results would be inaccurate as the actual amount of solute dissolved would be less than the measured amount, leading to incorrect enthalpy calculation.
A7: Endothermic dissolution absorbs heat (ΔH positive) while exothermic dissolution releases heat (ΔH negative).
A8: Distilled water is free from impurities that could affect the dissolution process and introduce errors in heat measurements.
This experiment demonstrates fundamental principles of:
The enthalpy of dissolution experiment provides valuable insights into the thermodynamic behavior of solutes in solution. Through careful measurement and analysis, students can understand the energy relationships in dissolution processes and develop essential laboratory skills in calorimetry.
Whether studying the exothermic dissolution of copper sulphate or the endothermic process of potassium nitrate, this experiment reinforces theoretical concepts while introducing practical applications in chemistry and related fields.
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