#Gibbs free energy

Variation of Conductivity with Concentration and Measurement

A cell reaction has E°cell = 0.74 V and Δr G° = -142610 J/mol at 298 K. How many electrons are transferred? (F = 96500 C

Δr G° = -n F E°cell , -142610 = -n × 96500 × 0.74 . n = (142610/96500 × 0.74) = (142610/71410) ≈ 2 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

View Details

Variation of Conductivity with Concentration and Measurement

A cell reaction has Δr G° = -285900 J/mol and involves 3 electrons at 298 K. What is the standard emf? (F = 96500 C/mol)

Δr G° = -n F E°cell , -285900 = -3 × 96500 × E°cell . E°cell = (285900/3 × 96500) = 0.987 V ≈ 0.99 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

View Details

Nernst Equation and Gibbs Energy and Equilibrium Constant

A cell has E°cell = 1.10 V and Δr G° = -424600 J/mol at 298 K. What is the number of electrons transferred? (F = 96500 C

Δr G° = -n F E°cell , -424600 = -n × 96500 × 1.10 . n = (424600/96500 × 1.10) = (424600/106150) ≈ 4 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant

View Details

Nernst Equation and Gibbs Energy and Equilibrium Constant

A cell has E°cell = 0.88 V and Δr G° = -169840 J/mol at 298 K. How many electrons are transferred? (F = 96500 C/mol)

Δr G° = -n F E°cell , -169840 = -n × 96500 × 0.88 . n = (169840/96500 × 0.88) = (169840/84920) ≈ 2 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant

View Details

Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the standard Gibbs energy change for a cell with E°cell = 0.77 V and 2 electrons transferred? (F = 96500 C/mol)

Δr G° = -n F E°cell = -2 × 96500 × 0.77 = -148610 J/mol = -148.61 kJ/mol .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

View Details

Electrochemical Cells - Galvanic Cells and Electrode Potential

A cell reaction has Δr G° = -212300 J/mol and involves 2 electrons at 298 K. What is the equilibrium constant?

Δr G° = -n F E°cell , E°cell = (212300/2 × 96500) = 1.10 V . E°cell = (0.059/n) log Kc , 1.10 = (0.059/2) log Kc , log Kc = 37.29, Kc ≈ 10³⁷ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

View Details

Electrochemical Cells - Galvanic Cells and Electrode Potential

A cell reaction has Δr G° = -212300 J/mol and involves 2 electrons at 298 K. What is the equilibrium constant?

Δr G° = -n F E°cell , E°cell = (212300/2 × 96500) = 1.10 V . E°cell = (0.059/n) log Kc , 1.10 = (0.059/2) log Kc , log Kc = 37.29, Kc ≈ 10³⁷ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

View Details

Third Law of Thermodynamics and Applications

For a reaction with Δ H = -80.00 kJ/mol and Δ S = -200.0 J/K·mol , what is Δ G at 300 K?

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

View Details

Third Law of Thermodynamics and Applications

For H₂(g) + Br₂(g) → 2HBr(g), Δ H = -72.6 kJ/mol, Δ S = 114 J/K at 298 K. What is Δ G?

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

View Details

Third Law of Thermodynamics and Applications

For the reaction 2A(g) → B(g) , Δ H = -40.00 kJ/mol , Δ S = -100.0 J/K·mol at 300 K. What is Δ G ?

Δ G = Δ H - TΔ S = -40.00 - 300 × (-0.1000) = -40.00 + 30.00 = -10.00 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

View Details

Third Law of Thermodynamics and Applications

For a reaction with Δ G° = 13.8 kJ/mol at 298 K, what is the equilibrium constant K ? ( R = 8.314 J/mol·K )

Using Δ G° = -RT ln K , we get 13.8 × 10³ = -8.314 × 298 × ln K . Thus, ln K = -(13.8 × 10³/8.314 × 298) = -5.57 , and K = e⁻⁵.⁵⁷ ≈ 0.0038 .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

View Details

Third Law of Thermodynamics and Applications

For a reaction with Δ H = 60 kJ/mol and Δ S = 150 J/K, what is Δ G at 400 K?

Δ G = Δ H - TΔ S = 60 - 400 × 0.15 = 60 - 60 = 0 kJ/mol.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

View Details