What is ΔG° at 298 K for the following equilibrium?
Ag⁺(aq) + 2NH₃(aq) Ag(NH₃) ₂⁺(aq) ; K_f = 1.7 × 10⁷ at 298 K

Question

What is ΔG° at 298 K for the following equilibrium? Ag⁺(aq)+ 2NH₃(aq) Ag(NH₃)₂⁺(aq); K_f = 1.7 × 10⁷ at 298 K A)-41 kJ B)41 kJ C)-18 kJ D)0 E)18 kJ

Quizplus · Accepted Answer

The standard free energy change for the equilibrium can be calculated using the equation:

ΔG° = -RT ln(K)

where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.

Plugging in the values given:

ΔG° = - (8.314 J/mol*K) * (298 K) * ln(1.7 × 10^17)

ΔG° = -41 kJ/mol

Therefore, the answer is A, -41 kJ.

What Is ΔG° at 298 K for the Following Equilibrium