Class 11 Chemistry Notes Chapter 6 (Chapter 6) – Lab Manual (English) Book

Detailed Notes with MCQs of Chapter 6 from your Chemistry Lab Manual. This chapter deals with a fundamental concept in thermochemistry: Determining the Enthalpy of Dissolution. While it's a practical experiment, the underlying principles are crucial for understanding energy changes in chemical processes, which often appear in various government examinations.

Chapter 6: Determination of Enthalpy of Dissolution of Copper Sulphate or Potassium Nitrate

1. Introduction: What is Enthalpy of Dissolution (ΔsolH)?

• Definition: The enthalpy of dissolution (or enthalpy change of solution) is the enthalpy change observed when one mole of a substance dissolves completely in a large excess of a solvent (usually water) such that further dilution causes no detectable heat change.
• Symbol: ΔsolH
• Units: Typically expressed in kilojoules per mole (kJ/mol).
• Nature of Process: Dissolution can be either:
  • Exothermic: Heat is released during dissolution (ΔsolH < 0). The temperature of the solution increases. Example: Dissolving anhydrous CuSO₄, NaOH, or concentrated H₂SO₄ in water. The dissolution of hydrated copper sulphate (CuSO₄·5H₂O) is slightly exothermic.
  • Endothermic: Heat is absorbed from the surroundings during dissolution (ΔsolH > 0). The temperature of the solution decreases. Example: Dissolving KNO₃, NH₄NO₃, NaCl, or KCl in water.

2. Theoretical Basis: Lattice Enthalpy and Hydration Enthalpy

The process of dissolving an ionic solid in water can be thought of as occurring in two hypothetical steps:

• Step 1: Breaking the Crystal Lattice: Energy is required to overcome the electrostatic forces holding the ions together in the crystal lattice. This energy is the Lattice Enthalpy (ΔlatticeH). It is always positive (endothermic).

  • MX(s) → M⁺(g) + X⁻(g) ; ΔH = ΔlatticeH > 0

• Step 2: Hydration of Ions: Energy is released when the separated gaseous ions are surrounded and stabilized by solvent (water) molecules. This energy is the Hydration Enthalpy (ΔhydrationH). It is always negative (exothermic).

  • M⁺(g) + X⁻(g) + aq → M⁺(aq) + X⁻(aq) ; ΔH = ΔhydrationH < 0

• Overall Enthalpy of Dissolution: The net enthalpy change (ΔsolH) is the sum of these two steps:

  • ΔsolH = ΔlatticeH + ΔhydrationH

• Predicting Exo/Endothermic Nature:

  • If |ΔhydrationH| > |ΔlatticeH|, the overall process is exothermic (ΔsolH is negative). More energy is released during hydration than is consumed to break the lattice.
  • If |ΔlatticeH| > |ΔhydrationH|, the overall process is endothermic (ΔsolH is positive). More energy is required to break the lattice than is released during hydration.

3. Principle of Calorimetry

This experiment uses a simple calorimeter (like a thermos flask or an insulated beaker) to measure the heat change.

• Assumption: The calorimeter is perfectly insulated, meaning no heat is exchanged with the surroundings.
• Heat Exchange: Any heat released or absorbed by the dissolution process causes a change in the temperature of the solvent (water) and the dissolved solute.
• Calculation: The heat change (q) associated with the temperature change of the solution is calculated using:
  • q = m × c × ΔT
  • Where:
    • q = heat absorbed or released by the solution (in Joules, J)
    • m = mass of the solution (in grams, g). Often approximated as the mass of the solvent (water) if the amount of solute is small. m = m_water + m_salt.
    • c = specific heat capacity of the solution (in J g⁻¹ K⁻¹ or J g⁻¹ °C⁻¹). Often approximated as the specific heat capacity of water (c_water ≈ 4.184 J g⁻¹ K⁻¹).
    • ΔT = change in temperature (in K or °C). ΔT = T_final - T_initial.

4. Relating Heat Change (q) to Enthalpy of Dissolution (ΔsolH)

• The heat change q calculated above is the heat absorbed or released by the solution (the surroundings from the perspective of the dissolving salt).
• The enthalpy change of the system (the dissolving salt) is equal in magnitude but opposite in sign to the heat change of the surroundings (the solution).
• ΔH_dissolution = -q
• To get the molar enthalpy of dissolution (ΔsolH), we divide by the number of moles (n) of the solute dissolved:
  • ΔsolH = -q / n
  • Where n = mass of solute / molar mass of solute.
• Sign Convention Recap:
  • If dissolution is exothermic, T_final > T_initial, ΔT is positive, q is positive (heat absorbed by water), so ΔsolH = -q/n is negative.
  • If dissolution is endothermic, T_final < T_initial, ΔT is negative, q is negative (heat lost by water), so ΔsolH = -q/n is positive.

5. Experimental Considerations (Focus on Concepts)

• Calorimeter: Use an insulating container (thermos flask is ideal) to minimize heat exchange with the environment. A beaker placed in a larger beaker packed with cotton wool is a common lab setup.
• Measurements: Accurate measurement of the mass/volume of water and the mass of the salt is crucial. Accurate temperature readings before mixing and the maximum/minimum temperature after dissolution are essential. Use a thermometer with good precision (e.g., 0.1 °C).
• Stirring: Continuous stirring ensures the salt dissolves completely and the heat is distributed uniformly throughout the solution, giving an accurate final temperature reading.
• Complete Dissolution: Ensure all the added salt dissolves to measure the enthalpy change for the specified amount.

6. Specific Cases:

• Potassium Nitrate (KNO₃):
  • Molar Mass ≈ 101.1 g/mol
  • Dissolution is endothermic. Expect the temperature to decrease (T_final < T_initial). ΔsolH will be positive.
• Copper(II) Sulphate Pentahydrate (CuSO₄·5H₂O):
  • Molar Mass ≈ 249.68 g/mol (Remember to include water of hydration!)
  • Dissolution is slightly exothermic. Expect the temperature to increase (T_final > T_initial). ΔsolH will be slightly negative. (Note: Dissolving anhydrous CuSO₄ is highly exothermic).

7. Precautions and Sources of Error

• Heat Loss/Gain: The primary source of error. No calorimeter is perfectly insulated.
• Incomplete Dissolution: If the salt doesn't dissolve completely, the measured heat change will be lower than expected.
• Measurement Errors: Inaccuracies in measuring mass (salt, water) or temperature.
• Heat Capacity Approximation: Assuming the specific heat capacity of the solution is the same as water introduces a small error. Assuming the calorimeter absorbs negligible heat is another approximation.
• Thermometer Lag: The thermometer bulb takes time to respond to temperature changes.
• Splashing: Loss of material during transfer or dissolution.

8. Key Takeaways for Exams

• Know the definition of ΔsolH and its units.
• Understand the difference between exothermic and endothermic dissolution and relate it to temperature change (ΔT) and the sign of ΔsolH.
• Be able to explain ΔsolH in terms of ΔlatticeH and ΔhydrationH.
• Understand the principle of calorimetry (q = mcΔT).
• Know the formula ΔsolH = -q / n and how to use it (including sign convention).
• Recognize common examples of exothermic (NaOH, H₂SO₄) and endothermic (KNO₃, NH₄NO₃) dissolutions.
• Be aware of major sources of error in calorimetric experiments.

Multiple Choice Questions (MCQs)

1. The enthalpy change when one mole of a substance dissolves completely in a large excess of solvent is called:
   a) Enthalpy of formation
   b) Enthalpy of hydration
   c) Enthalpy of dissolution
   d) Lattice enthalpy

2. If the dissolution of a salt in water is an endothermic process, it means that:
   a) The temperature of the solution increases.
   b) Heat is released to the surroundings.
   c) ΔsolH is negative.
   d) |ΔlatticeH| > |ΔhydrationH|.

3. The dissolution of potassium nitrate (KNO₃) in water causes the temperature of the solution to decrease. This indicates that the dissolution process is:
   a) Exothermic, ΔsolH < 0
   b) Endothermic, ΔsolH > 0
   c) Isothermal, ΔsolH = 0
   d) Adiabatic, q = 0

4. The relationship between enthalpy of dissolution (ΔsolH), lattice enthalpy (ΔlatticeH), and hydration enthalpy (ΔhydrationH) is given by:
   a) ΔsolH = ΔlatticeH - ΔhydrationH
   b) ΔsolH = ΔhydrationH - ΔlatticeH
   c) ΔsolH = ΔlatticeH + ΔhydrationH
   d) ΔsolH = -(ΔlatticeH + ΔhydrationH)

5. In a calorimetric experiment, 5g of a salt is dissolved in 100g of water. The temperature changes from 25°C to 22°C. Assuming the specific heat capacity of the solution is 4.2 J g⁻¹ °C⁻¹ and the mass of the solution is (100+5)g, the heat change 'q' for the solution is:
   a) q = (105) × (4.2) × (25 - 22) J
   b) q = (100) × (4.2) × (22 - 25) J
   c) q = (105) × (4.2) × (22 - 25) J
   d) q = (5) × (4.2) × (25 - 22) J

6. If the heat absorbed by the water (q_water) during an endothermic dissolution process is calculated to be 1500 J when 0.1 moles of salt are dissolved, the molar enthalpy of dissolution (ΔsolH) is: (Note: If water temperature drops, q_water is negative, heat lost by water) Let's rephrase: If the temperature of water drops, indicating heat was absorbed by the salt from the water, and the magnitude of heat transferred is 1500 J for 0.1 moles.
   a) -15.0 kJ/mol
   b) +15.0 kJ/mol
   c) -1.50 kJ/mol
   d) +1.50 kJ/mol

7. The standard unit for molar enthalpy of dissolution is:
   a) Joules (J)
   b) Kilojoules (kJ)
   c) Joules per mole (J/mol)
   d) Kilojoules per mole (kJ/mol)

8. Which of the following is a major source of error in determining the enthalpy of dissolution using a simple calorimeter?
   a) Using distilled water
   b) Stirring the solution
   c) Heat exchange with the surroundings
   d) Accurately weighing the salt

9. Dissolving 4.0 g of NaOH (Molar Mass = 40 g/mol) in 100 mL of water causes the temperature to rise significantly. The enthalpy of dissolution (ΔsolH) for NaOH is therefore:
   a) Positive
   b) Negative
   c) Zero
   d) Cannot be determined from the information

10. When calculating the enthalpy of dissolution of CuSO₄·5H₂O, the molar mass used should be:
    a) The molar mass of CuSO₄ only
    b) The molar mass of CuSO₄ plus the mass of 5 water molecules
    c) The molar mass of water only
    d) The molar mass of Cu only

Answer Key for MCQs:

1. c
2. d
3. b
4. c
5. c (q = m × c × ΔT = 105g × 4.2 J/g°C × (22-25)°C)
6. b (System absorbs heat, ΔH is positive. q_system = +1500 J. ΔsolH = q_system / n = +1500 J / 0.1 mol = +15000 J/mol = +15.0 kJ/mol)
7. d
8. c
9. b (Temperature rise indicates exothermic process, ΔsolH < 0)
10. b

Study these notes carefully, focusing on the definitions, relationships, formulas, and sign conventions. Good luck with your preparation!