NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice
Calculate Δ r G for the reaction at 27 o C H 2 g + 2 Ag + aq ⇌ 2 Ag s + 2 H + aq Given : P H 2 = 0.5 bar ; Ag + = 10 - 5 M ; H + = 1 0 - 3 M ; Δ f G ∘ Ag + aq = 77.1 kJ/mol
Options
- A- 154.2 kJ/mol
- B- 178.9 kJ/mol
- C- 129.5 kJ/mol
- DNone of these
Correct answer
C. - 129.5 kJ/mol
Step-by-step solution
Δ r G ∘ = 0 - 77.1 × 2 = - 154.2 kJ/mol Q = H + 2 P H 2 · Ag + 2 = 1 0 - 6 0.5 × 1 0 - 1 0 = 2 × 1 0 4 Δ G = Δ r G ∘ + RT ln Q Δ r G = - 154.2 + 8.314 × 300 ln 2 × 1 0 4 1 0 0 0 = - 129.5 kJ/mol