Latest questions in this category.
For adiabatic expansion, T₂ = 500 × (6/12)⁰.⁴ ≈ 500 × 0.7579 = 378.9 K . Then, Δ T = 378.9 - 500 = -121.1 K . Work done: w = nCvΔ T = 1 × 20.785 × (-121.1) ≈ -2516 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
The standard enthalpy of formation of an element in its standard state (e.g., O₂(g) at 1 bar, 298 K) is zero by definition.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
Δ G = Δ H - TΔ S = 0, so -50 × 10³ + T × 100 = 0, T = 50000 / 100 = 500 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
For isothermal process, Δ U = 0, so q = -w. Compression means w > 0 (4988 J), so q = -4988 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
Δ H = 2 × Δ Hf° (NO₂) - [Δ Hf° (N₂) + 2 × Δ Hf° (O₂)] = 2 × 33.2 - [0 + 0] = 66.4 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
The enthalpy change for melting is the enthalpy of fusion (Δ Hfus), which is positive as it’s an endothermic process.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
Δ G° = -RT ln K, -20 × 10³ = -8.314 × 298 × ln K, ln K = 8.07, K = e⁸.⁰⁷ ≈ 3160.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
w = -Pext Δ V = -2 × (10 - 5) × 101.3 = -2 × 5 × 101.3 = -1013 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
In a reversible adiabatic process, PV^γ = constant, where γ = Cp/Cv.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
For Δ ng = -1.5 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -286.00 - (-1.5 × 2.4776) = -286.00 + 3.7164 = -282.28 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
Δ H = [2 × Δ Hf° (SO₃)] - [2 × Δ Hf° (SO₂) + Δ Hf° (O₂)] = [2 × (-395.7)] - [2 × (-296.8) + 0] = -791.4 + 593.6 = -197.8 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
The standard enthalpy of formation is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states, i.e., C(s) + O₂(g) → CO₂(g).
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
