What is ΔG° for the following reaction? (F = 96,500 C • mol^-1) I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l)
Given:

Question

What is ΔG° for the following reaction? (F = 96,500 C • mol^-1) I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l) Given:

Quizplus · Accepted Answer

We can use the equation ΔG° = -RT ln(K), where R is the gas constant, T is the temperature, and K is the equilibrium constant. At standard conditions (25°C, 1 atm), ΔG° can be calculated as ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. We can use standard thermodynamic data to calculate ΔH° and ΔS° for this reaction: I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l) ΔH° = (-73.2 kJ/mol) - 2(-79.3 kJ/mol) + (0 kJ/mol) + (0 kJ/mol) = +25.8 kJ/mol ΔS° = (34.9 J/mol•K) + 2(198.7 J/mol•K) - (95.7 J/mol•K) - (99.9 J/mol•K) = +137.1 J/mol•K Plugging these values into the equation for ΔG°, we get: ΔG° = (+25.8 kJ/mol) - (298 K)(+137.1 J/mol•K)(1/1000 J/kJ) = +22.2 kJ/mol Finally, we convert from kJ/mol to kJ/C by dividing by the Faraday constant: ΔG° = (+22.2 kJ/mol) / (96,500 C/mol) = +0.23 V This is a positive value, which tells us that the reaction is not spontaneous under standard conditions. Therefore, we choose answer A) +104 kJ/mol, which is the closest to the calculated value of +22.2 kJ/mol when converted to kJ/mol.

What Is ΔG° for the Following Reaction? (F = 96,500