Latest questions in this category.
For a spontaneous process under constant T and p, Gibbs free energy decreases (Δ G < 0).
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Δ S = nR ln(V₂/V₁) = 1 × 8.314 × ln(5/25) = 8.314 × ln(0.2) = 8.314 × (-1.609) = -13.38 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
For irreversible expansion against constant pext, w = -pext Δ V, where Δ V is the volume change.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
For reversible isothermal expansion, w = -nRT ln(P₁/P₂) = -1 × 8.314 × 300 × ln(10/2) = -2494.2 × 1.6094 ≈ -4013 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Δ G = Δ H - TΔ S < 0, -150 - T(-0.5) < 0, 0.5T < 150, T < 300 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Moles = 72 / 18 = 4 mol . For vaporization, Δ Ssys = n × Δ Hvap / T = 4 × 40790 / 373 ≈ 437.48 J/K ≈ 437.5 J/K .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ/mol . Using Δ H = Δ U + Δ ng RT , Δ U = -1124.00 - (-3 × 2.4776) = -1124.00 + 7.4328 = -1116.57 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Δ G° = -RT ln K = -8.314 × 298 × ln(2 × 10⁶) = -2477.6 × 14.509 = -35962 J/mol = -35.96 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
For an isothermal process (Δ U = 0), Δ U = q + w = 0, so q = -w. For reversible compression, w = -nRT ln(V₂/V₁) (positive), so q = nRT ln(V₂/V₁), which is negative since V₂ < V₁.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) , Δ ng = 1 - 3 = -2 . Then, RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -890.30 - (-2 × 2.4776) = -890.30 + 4.9552 = -885.34 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Δ G = 0 at equilibrium, Δ H - TΔ S = 0, -200 × 10³ - T(-250) = 0, T = 200000 / 250 = 800 K. Spontaneous below 800 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
Moles = 36 / 18 = 2 mol, Δ S = n × Δ Hfus / T = 2 × 6.01 × 10³ / 273 = 44 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
