#reaction enthalpy

CHEMISTRY

For the reaction NO(g) + 1/2O₂(g) -> NO₂(g), if Δ_f H°: NO(g) = 90.25 kJ/mol, NO₂(g) = 33.18 kJ/mol, what is Δ_

Given: For the reaction NO(g) + 1/2O₂(g) -> NO₂(g), if Δ_f H°: NO(g) = 90.25 kJ/mol, NO₂(g) = 33.18 kJ/mol, what is Δ_r H? These values define the system as per NCERT data. Formula: Δ_r H = [33.18] - [90.25 + 0] = 33.18 - 90.25 = -57.07 kJ/mol.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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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 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

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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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Gibbs Energy and Gibbs Energy Change and Equilibrium

Calculate Δ H for C(s) + 2H₂O(g) → CO₂(g) + 2H₂(g) using: Δ Hf° (H₂O,g) = -241.8 kJ/mol, Δ Hf° (CO₂,g) = -393.5 kJ/mol.

Δ H = [Δ Hf° (CO₂) + 2 × Δ Hf° (H₂)] - [Δ Hf° (C) + 2 × Δ Hf° (H₂O)] = [-393.5 + 0] - [0 + 2(-241.8)] = -393.5 + 483.6 = 90.1 kJ/mol.

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

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Bond Enthalpy and Entropy and Spontaneity

For the process 2NO₂(g) → N₂O₄(g) , Δ H = -57.20 kJ/mol at 298 K, what is Δ U ? ( R = 8.314 J/mol·K )

For Δ ng = 1 - 2 = -1 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -57.20 - (-1 × 2.4776) = -57.20 + 2.4776 = -54.72 kJ/mol .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity

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