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 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
Calculate Δ Ssurr when 1 mol of a substance melts at 300 K with Δ Hfus = 9 kJ/mol.
Δ Ssurr = -Δ H / T = -9 × 10³ / 300 = -30 J/K·mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
