26 public questions tagged with this topic.
Δ G = Δ H - TΔ S = -80.00 - 300 × (-0.2000) = -80.00 + 60.00 = -20.00 kJ/mol . The negative Δ G indicates spontaneity at 300 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ G = Δ H - TΔ S = -72.6 - 298 × 0.114 = -72.6 - 33.97 = -106.57 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
For Δ ng = 2 - (1 + 2.5) = -1.5 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -1299.50 - (-1.5 × 2.4776) = -1299.50 + 3.7164 = -1295.78 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
For Δ ng = 3 - 0 = 3 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = 128.10 - 3 × 2.4776 = 128.10 - 7.4328 = 120.67 ≈ 120.7 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
At constant pressure, Δ H = qp. Here, q = 1200 J (heat absorbed), so Δ H = 1200 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Δ G = Δ H - TΔ S, 30 = 90 - 300 Δ S, Δ S = (90 - 30) / 300 = 60 / 300 = 0.2 kJ/K = 200 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Moles of CO₂ = 22 / 44 = 0.5 mol . For C(s) + O₂(g) → CO₂(g) , Δ H = -393.5 kJ/mol . Heat released = 0.5 × 393.5 = 196.75 kJ (positive, as heat is released in exothermic reaction).
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For an exothermic reaction at constant pressure, heat evolved qp = Δ H, and since it’s exothermic, Δ H < 0, so qp = Δ H.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Using Hess’s law: Δ H = -393.5 - (-283) = -393.5 + 283 = -110.5 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
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
Δ H = Δ U + Δ ng RT. When Δ ng = 0 (no change in moles of gas), Δ H = Δ U, as in H₂(g) + Cl₂(g) → 2HCl(g).
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
The standard state of an element is its most stable form at 1 bar and 298 K. For carbon, this is graphite.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
