Practice questions focused on bond enthalpy, entropy, and spontaneity. Helps students assess reaction feasibility and prepare for chemistry exams.
Using Δ G = Δ H - TΔ S , Δ G = 75.00 - 350 × 0.2500 = 75.00 - 87.50 = -12.50 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For H₂(g) → 2H(g), Δ H = 2 × Δ Hf (H) - Δ Hf (H₂) = 2 × 218 - 0 = 436 kJ/mol, which is the bond enthalpy.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For CO(g) → C(g) + O(g) , bond enthalpy is Δ H = Δ Ha(C) + Δ Ha(O) - Δ Hf(CO) = 715.0 + 249.0 - (-110.5) = 1074.5 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
Δ S = nR ln(V₂/V₁) = 1 × 8.314 × ln(16/4) = 8.314 × ln(4) = 8.314 × 1.386 = 11.52 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For C₂H₆(g) → 2C(g) + 6H(g), Δ H = 2 × 715 + 6 × 218 - (-84.7) = 1430 + 1308 + 84.7 = 2822.7 kJ/mol. Bonds: 1(C-C) + 6(C-H), C-C = 2822.7 - 6 × 413 = 2822.7 - 2478 = 344.7 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For isothermal expansion, Δ Ssys = nR ln(V₂/V₁) = 1 × 8.314 × ln(4/2) = 8.314 × 0.6931 ≈ 5.76 J/K .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For an ideal gas, Cp - Cv = R, where R is the gas constant.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
Bond breaking is endothermic (Δ H > 0), and entropy increases as one molecule becomes two (Δ S > 0).
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
Δ Ssys = Δ Hvap / T = 35 × 10³ / 400 = 87.5 J/K, Δ Ssurr = -Δ H / T = -35 × 10³ / 400 = -87.5 J/K, Δ Stotal = 87.5 - 87.5 = 0 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
Δ S = nR ln(V₂/V₁) = 1 × 8.314 × ln(5/20) = 8.314 × ln(0.25) = 8.314 × (-1.386) = -11.52 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
Δ S = nR ln(V₂/V₁) = 1 × 8.314 × ln(5/10) = 8.314 × (-0.693) = -5.76 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
Moles of C = 12 / 12 = 1 mol. Heat released = 1 × 393.5 = 393.5 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
