Third Law of Thermodynamics and Applications

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

Calculate Δ H for C(s) + H₂(g) + O₂(g) → CH₃OH(l) given: CH₃OH(l) + (3/2)O₂(g) → CO₂(g) + 2H₂O(l), Δ H = -726.5 kJ/mol;

Reverse first: Δ H = 726.5 kJ. Second: Δ H = -393.5 kJ. Third (×2): Δ H = -571.6 kJ. Add: 726.5 - 393.5 - 571.6 = -238.6 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

Calculate the work done when 2 mol of an ideal gas expands isothermally and reversibly from 8 L to 24 L at 400 K. (R = 8

w = -nRT ln(V₂/V₁) = -2 × 8.314 × 400 × ln(24/8) = -6651.2 × ln(3) = -6651.2 × 1.099 = -7309.7 J.

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 CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g) , Δ H = -103.00 kJ/mol at 298 K. What is Δ U ? ( R = 8.314 J/mol·K )

For Δ ng = 2 - 2 = 0 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -103.00 - 0 = -103.00 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

Calculate Δ H for 2C(s) + H₂(g) → C₂H₂(g) given: C₂H₂(g) + (5/2)O₂(g) → 2CO₂(g) + H₂O(l), Δ H = -1299.5 kJ/mol; C(s) + O

Reverse first: Δ H = 1299.5 kJ. Second (×2): Δ H = -787 kJ. Third: Δ H = -285.8 kJ. Add: 1299.5 - 787 - 285.8 = 226.7 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

Calculate Δ H for C(s) + 2S(s) → CS₂(l) given: CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g), Δ H = -1076.8 kJ/mol; C(s) + O₂(g) →

Reverse first: Δ H = 1076.8 kJ. Second: Δ H = -393.5 kJ. Third (×2): Δ H = -593.6 kJ. Add: 1076.8 - 393.5 - 593.6 = 89.7 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 an ideal gas (γ = 1.33) compressed adiabatically from 18 L to 6 L at 300 K, what is the final temperature?

T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (18/6)⁰.³³ = 300 × 3⁰.³³ = 300 × 1.44 = 432 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 the reaction C(s) + O₂(g) → CO₂(g), Δ H = -393.5 kJ/mol at 298 K. What is Δ U? (R = 8.314 J/mol·K)

Δ ng = 1 - 1 = 0, RT = 2.48 kJ, Δ H = Δ U + Δ ng RT, Δ U = -393.5 - 0 = -393.5 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 the reaction N₂(g) + 2O₂(g) → 2NO₂(g), Δ H = 66.4 kJ/mol at 298 K, what is Δ U? (R = 8.314 J/mol·K)

Δ ng = 2 - 3 = -1, RT = 2.48 kJ, Δ H = Δ U + Δ ng RT, Δ U = 66.4 - (-1 × 2.48) = 66.4 + 2.48 = 68.88 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

Calculate Δ Stotal when 2 mol of water vapor condenses at 373 K if Δ Hvap = 40.79 kJ/mol.

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

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Third Law of Thermodynamics and Applications

For a reaction with Δ H = -80.00 kJ/mol and Δ S = -200.0 J/K·mol , what is Δ G at 300 K?

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

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Third Law of Thermodynamics and Applications

For H₂(g) + Br₂(g) → 2HBr(g), Δ H = -72.6 kJ/mol, Δ S = 114 J/K at 298 K. What is Δ G?

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

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