#thermochemistry

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

View Details

Third Law of Thermodynamics and Applications

Calculate Δ H for C(s) + H₂O(g) → CO(g) + H₂(g) given: C(s) + O₂(g) → CO₂(g), Δ H = -393.5 kJ/mol; CO(g) + (1/2)O₂(g) →

Reverse second: Δ H = 283 kJ. Add third: Δ H = 241.8 kJ. Subtract first: Δ H = 283 + 241.8 - 393.5 = 131.3 kJ/mol.

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

View Details

Third Law of Thermodynamics and Applications

For the reaction C₂H₂(g) + (5/2)O₂(g) → 2CO₂(g) + H₂O(l) , Δ H = -1299.5 kJ/mol at 298 K. What is Δ U ? ( R = 8.314 J/mo

For Δ ng = 2 - (1 + 2.5) = -1.5 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -1299.50 - (-1.5 × 2.4776) = -1299.50 + 3.7164 = -1295.78 kJ/mol .

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

View Details

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

View Details

Heat Capacity and Calorimetry and Measurement of Enthalpy

Calculate Δ H for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) using: Δ Hf° (C₃H₈,g) = -103.8 kJ/mol, Δ Hf° (CO₂,g) = -393.5 kJ/

Δ H = [3(-393.5) + 4(-285.8)] - [-103.8 + 5(0)] = -1180.5 - 1143.2 + 103.8 = -2219.9 kJ/mol.

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

View Details

Heat Capacity and Calorimetry and Measurement of Enthalpy

Calculate the heat released when 22 g of CO₂ is formed from C(s) + O₂(g) → CO₂(g) . ( Δ H = -393.5 kJ/mol , molar mass o

Moles of CO₂ = 22 / 44 = 0.5 mol . For C(s) + O₂(g) → CO₂(g) , Δ H = -393.5 kJ/mol . Heat released = 0.5 × 393.5 = 196.75 kJ (positive, as heat is released in exothermic reaction).

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

View Details

Heat Capacity and Calorimetry and Measurement of Enthalpy

Using Hess’s law, calculate Δ H for 2S(s) + 3O₂(g) → 2SO₃(g) given: S(s) + O₂(g) → SO₂(g), Δ H = -296.8 kJ/mol; 2SO₂(g)

Multiply first by 2: 2S(s) + 2O₂(g) → 2SO₂(g), Δ H = -593.6 kJ. Add second: Δ H = -593.6 + (-197.8) = -791.4 kJ.

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

The enthalpy of combustion of methane is -890 kJ/mol. What is the heat released when 8 g of methane is burnt? (Molar mas

Moles of CH₄ = 8 / 16 = 0.5 mol. Heat released = 0.5 × 890 = 445 kJ.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

Calculate Δ H for 2C(s) + 3H₂(g) → C₂H₆(g) if Δ Hf° (C₂H₆,g) = -84.7 kJ/mol.

Δ H = Δ Hf° (C₂H₆) - [2 × Δ Hf° (C) + 3 × Δ Hf° (H₂)] = -84.7 - [0 + 0] = -84.7 kJ/mol.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

Calculate the heat released when 60 g of ethanol is combusted if Δ Hc = -1366.7 kJ/mol. (Molar mass = 46 g/mol)

Moles = 60 / 46 = 1.304 mol, heat released = 1.304 × 1366.7 = 1782.2 kJ.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

View Details

Enthalpy Changes - Reaction Formation Combustion and Hess's Law

The standard enthalpy change for 2C(s) + 2H₂(g) → C₂H₄(g) is 52.3 kJ/mol. What is Δ Hf° of C₂H₄(g)?

The reaction forms 1 mol of C₂H₄(g) from its elements in standard states, so Δ H = Δ Hf° (C₂H₄) = 52.3 kJ/mol, as Δ Hf° (C) = 0 and Δ Hf° (H₂) = 0.

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

View Details

Enthalpy Changes - Reaction Formation Combustion and Hess's Law

Calculate Δ H for C₂H₅OH(l) → C₂H₄(g) + H₂O(l) using: Δ Hf° (C₂H₅OH,l) = -277.7 kJ/mol, Δ Hf° (C₂H₄,g) = 52.3 kJ/mol, Δ

Δ H = [Δ Hf° (C₂H₄) + Δ Hf° (H₂O)] - Δ Hf° (C₂H₅OH) = [52.3 + (-285.8)] - (-277.7) = -233.5 + 277.7 = 44.2 kJ/mol.

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

View Details