For the process H₂O(l) → H₂O(g) at 373 K, Δ H = 40.79 kJ/mol, what is Δ Ssurr?
Δ Ssurr = -Δ H / T = -40.79 × 10³ / 373 = -109.4 J/K·mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
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Calculate Δ Ssurr for the combustion of 1 mol of CH₄(g) at 298 K, given Δ H = -890.00 kJ/mol .
For the surroundings, Δ Ssurr = -Δ H / T = -(-890.00 × 10³) / 298 ≈ 2986.6 J/K .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Calculate Δ Ssurr when 1 mol of H₂O(l) forms under standard conditions at 298 K, given Δf H° = -286 kJ/mol.
Δ Ssurr = -Δ H / T = -(-286 × 10³) / 298 = 959.7 J/K·mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Calculate Δ Ssurr when 2 mol of water freezes at 273 K if Δ Hfus = 6.01 kJ/mol.
For freezing, Δ H = -n × Δ Hfus = -2 × 6.01 = -12.02 kJ, Δ Ssurr = -Δ H / T = -(-12.02 × 10³) / 273 = 44.03 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Calculate Δ Ssurr when 3 mol of a liquid freezes at 250 K if Δ Hfus = 5.5 kJ/mol.
For freezing, Δ H = -n × Δ Hfus = -3 × 5.5 = -16.5 kJ, Δ Ssurr = -Δ H / T = -(-16.5 × 10³) / 250 = 66 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Calculate Δ Ssurr when 2 mol of a substance melts at 300 K if Δ Hfus = 8 kJ/mol.
Δ H = 2 × 8 = 16 kJ, Δ Ssurr = -Δ H / T = -16 × 10³ / 300 = -53.33 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
