#enthalpy change

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

Calculate Δ H for the reaction C(s) + 2H₂(g) → CH₄(g) given: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δ H = -890.3 kJ/mol; C(

Reverse first: Δ H = 890.3 kJ. Second: Δ H = -393.5 kJ. Third (×2): Δ H = -571.6 kJ. Add: 890.3 - 393.5 - 571.6 = -74.8 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 = 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

What is the relationship between Δ H and Δ U for the reaction 2SO₂(g) + O₂(g) → 2SO₃(g) at 298 K? (R = 8.314 J/mol·K)

Δ H = Δ U + Δ ng RT. Here, Δ ng = 2 - 3 = -1, RT = 8.314 × 298 × 10⁻³ = 2.48 kJ, so Δ H = Δ U - 2.48 kJ.

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

Which of the following is always true for an exothermic process at constant pressure?

For an exothermic process, heat is released (qp < 0), and at constant pressure, Δ H = qp, so Δ H < 0.

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 H for C(s) + CO₂(g) → 2CO(g) using: C(s) + O₂(g) → CO₂(g), Δ H = -393.5 kJ/mol; CO(g) + (1/2)O₂(g) → CO₂(g),

The Δ Hf for CO can be found from the second reaction: Δ Hf (CO) = Δ Hf (CO₂) + 283 = -393.5 + 283 = -110.5 kJ/mol.

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 H for C₂H₄(g) + H₂(g) → C₂H₆(g) using bond enthalpies: C=C = 610 kJ/mol, H-H = 436 kJ/mol, C-H = 413 kJ/mol,

Bonds broken: C=C (610) + H-H (436) = 1046 kJ. Bonds formed: C-C (346) + 2 × C-H (2 × 413 = 826) = 1172 kJ. Δ H = 1046 - 1172 = -126 kJ/mol.

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

Calculate Δ H for 2C(s) + O₂(g) → 2CO(g) given: C(s) + O₂(g) → CO₂(g), Δ H = -393.5 kJ/mol; 2CO(g) + O₂(g) → 2CO₂(g), Δ

Multiply first equation by 2: 2C(s) + 2O₂(g) → 2CO₂(g), Δ H = -787 kJ. Reverse second: 2CO₂(g) → 2CO(g) + O₂(g), Δ H = 566 kJ. Add: Δ H = -787 + 566 = -221 kJ.

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 CO(g) + H₂O(g) → CO₂(g) + H₂(g) , Δ H = -41.20 kJ/mol at 298 K, what is Δ U ? ( R = 8.314 J/mol·K )

For Δ ng = 2 - 2 = 0 , Δ H = Δ U + Δ ng RT . Since Δ ng = 0 , Δ U = Δ H = -41.20 kJ/mol .

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 the reaction H₂(g) + I₂(g) → 2HI(g), Δ H = 52.96 kJ/mol at 298 K. What is Δ U? (R = 8.314 J/mol·K)

Δ ng = 2 - 2 = 0, so Δ H = Δ U + Δ ng RT = Δ U. Thus, Δ U = 52.96 kJ/mol.

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

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