#bond enthalpies

3 public questions tagged with this topic.

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

View Details

Gibbs Energy and Gibbs Energy Change and Equilibrium

Using bond enthalpies (C-H = 413 kJ/mol, Cl-Cl = 243 kJ/mol, C-Cl = 328 kJ/mol, H-Cl = 431 kJ/mol), calculate Δ H for CH

Bonds broken: C-H (413) + Cl-Cl (243) = 656 kJ. Bonds formed: C-Cl (328) + H-Cl (431) = 759 kJ. Δ H = 656 - 759 = -103 kJ/mol.

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

Using bond enthalpies (C-H = 413 kJ/mol, O=O = 498 kJ/mol, C=O = 803 kJ/mol, O-H = 467 kJ/mol), calculate Δ H for CH₄(g)

Bonds broken: 4(C-H) + 2(O=O) = 4 × 413 + 2 × 498 = 1652 + 996 = 2648 kJ. Bonds formed: 2(C=O) + 4(O-H) = 2 × 803 + 4 × 467 = 1606 + 1868 = 3474 kJ. Δ H = 2648 - 3474 = -826 kJ/mol.

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

View Details

#bond enthalpies

No questions with this tag

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,

Using bond enthalpies (C-H = 413 kJ/mol, Cl-Cl = 243 kJ/mol, C-Cl = 328 kJ/mol, H-Cl = 431 kJ/mol), calculate Δ H for CH

Using bond enthalpies (C-H = 413 kJ/mol, O=O = 498 kJ/mol, C=O = 803 kJ/mol, O-H = 467 kJ/mol), calculate Δ H for CH₄(g)

Please allow advertisements on this page