39 public questions tagged with this topic.
Δ H = -2 × 40.79 = -81.58 kJ, Δ Ssys = -Δ Hvap × n / T = -40.79 × 10³ × 2 / 373 = -218.8 J/K, Δ Ssurr = -Δ H / T = 81.58 × 10³ / 373 = 218.8 J/K, Δ Stotal = -218.8 + 218.8 = 0 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ G = Δ H - TΔ S = -80.00 - 300 × (-0.2000) = -80.00 + 60.00 = -20.00 kJ/mol . The negative Δ G indicates spontaneity at 300 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ G = Δ H - TΔ S = -72.6 - 298 × 0.114 = -72.6 - 33.97 = -106.57 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
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
Δ Ssys = Δ Hfus / Tm = 6.01 × 10³ / 273 = 22.01 J/K, Δ Ssurr = -Δ H / T = -6.01 × 10³ / 263 = -22.85 J/K, Δ Stotal = 22.01 - 22.85 = -0.84 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
In an isolated system (q = 0), spontaneity requires Δ Stotal = Δ Ssys > 0, as Δ Ssurr = 0.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
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
For an isolated system, a spontaneous process increases entropy (Δ S > 0) until equilibrium, where Δ S = 0. Since Δ U = 0, Δ S > 0 for spontaneity.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Δ G = Δ H - TΔ S, 30 = 90 - 300 Δ S, Δ S = (90 - 30) / 300 = 60 / 300 = 0.2 kJ/K = 200 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Entropy decreases when disorder decreases, e.g., H₂O(l) → H₂O(s), as liquid to solid reduces randomness.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Δ S = nR ln(V₂/V₁) = 1 × 8.314 × ln(5/25) = 8.314 × ln(0.2) = 8.314 × (-1.609) = -13.38 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Δ G = Δ H - TΔ S < 0, -150 - T(-0.5) < 0, 0.5T < 150, T < 300 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
