Calculate ΔG_rxn at 298 K under the conditions shown below for the following reaction. 2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ
P(Hg) = 0.025 atm,P(O₂) = 0.037 atm

Question

Calculate ΔG_rxn at 298 K under the conditions shown below for the following reaction. 2 Hg(g)+ O₂(g)→ 2 HgO(s)ΔG° = -180.8 kJ P(Hg)= 0.025 atm,P(O₂)= 0.037 atm

Quizplus · Accepted Answer

First, we need to calculate the ΔG for the given reaction using the formula: ΔG = ΔG° + RT ln(Q) where ΔG° is the standard free energy change, R is the gas constant (8.314 J/K·mol), T is the temperature in Kelvin (298 K), and Q is the reaction quotient. Q can be calculated using the given partial pressures of Hg and O2: Q = (P(Hg))^2/ P(O₂) Q = (0.025 atm)^2/0.037 atm Q = 0.0169 atm Substituting values into the formula, we get: ΔG = -180.8 kJ + (8.314 J/K·mol × 298 K × ln(0.0169)) ΔG = -283.9 kJ/mol Finally, we can calculate the ΔG for the desired conditions using the formula: ΔG = ΔG° + RT ln(Q/P°) where P° is the standard state pressure (1 atm). For Hg and O2, we get: ΔG = -180.8 kJ + (8.314 J/K·mol × 298 K × ln(0.0169/1)^2) ΔG = -154.4 kJ/mol Therefore, the answer is B, which is the only option with a negative value. A negative ΔG indicates that the reaction is spontaneous under the given conditions.

Calculate ΔGrxn at 298 K Under the Conditions Shown Below

Calculate ΔG_rxn at 298 K Under the Conditions Shown Below