#entropy change

Third Law of Thermodynamics and Applications

Calculate Δ Ssys when 2 mol of an ideal gas expands isothermally from 1 atm to 0.25 atm at 300 K. ( R = 8.314 J/mol·K )

For isothermal expansion, Δ Ssys = nR ln(P₁/P₂) = 2 × 8.314 × ln(1/0.25) = 16.628 × 1.3863 ≈ 23.0 J/K .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

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Third Law of Thermodynamics and Applications

For a reaction with Δ H = 60 kJ/mol and Δ S = 150 J/K, what is Δ G at 400 K?

Δ G = Δ H - TΔ S = 60 - 400 × 0.15 = 60 - 60 = 0 kJ/mol.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

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Third Law of Thermodynamics and Applications

For a reaction with Δ H = 120 kJ/mol and Δ S = 300 J/K, what is Δ G at 400 K?

Δ G = Δ H - TΔ S = 120 - 400 × 0.3 = 120 - 120 = 0 kJ/mol.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

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Thermodynamic Terms - System Surroundings and Types of Systems

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

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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

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Heat Capacity and Calorimetry and Measurement of Enthalpy

For a reaction with Δ H = 90 kJ/mol and Δ G = 30 kJ/mol at 300 K, what is Δ S?

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

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Heat Capacity and Calorimetry and Measurement of Enthalpy

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

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Heat Capacity and Calorimetry and Measurement of Enthalpy

For a reaction with Δ H = -120 kJ and Δ G = -130 kJ at 298 K, what is Δ S?

Δ G = Δ H - TΔ S, -130 = -120 - 298 Δ S, Δ S = (-130 + 120) / -298 = 0.0336 kJ/K = 33.6 J/K.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy

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Gibbs Energy and Gibbs Energy Change and Equilibrium

For the vaporization of 72 g of water at 373 K ( Δ Hvap = 40.79 kJ/mol , molar mass = 18 g/mol), what is Δ Ssys ?

Moles = 72 / 18 = 4 mol . For vaporization, Δ Ssys = n × Δ Hvap / T = 4 × 40790 / 373 ≈ 437.48 J/K ≈ 437.5 J/K .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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Gibbs Energy and Gibbs Energy Change and Equilibrium

For a process with Δ H = 50 kJ and Δ G = 20 kJ at 300 K, what is Δ S?

Δ G = Δ H - TΔ S, so 20 = 50 - 300 Δ S, 300 Δ S = 30, Δ S = 0.1 kJ/K = 100 J/K.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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First Law of Thermodynamics and Enthalpy and Internal Energy

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

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Enthalpy Changes - Reaction Formation Combustion and Hess's Law

Calculate Δ Ssys for the vaporization of 36 g of ethanol at 351 K if Δ Hvap = 38.6 kJ/mol . (Molar mass of C₂H₅OH = 46.0

Moles = 36 / 46.07 ≈ 0.7812 mol . For vaporization, Δ Ssys = n × Δ Hvap / T = 0.7812 × 38.6 × 10³ / 351 ≈ 86.0 J/K .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law

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