What is δg° at 298 k for the following equilibrium? ag+(aq) + 2nh3(aq) ag(nh3)2+(aq); kf = 1.7 × 107 at 298 k a. 41 kj b. 0 c. –41 kj d. 18 kj e. –18 kj

c. –41 kJ

Explanation:

Let's consider the following equation for the formation of a complex.

Ag⁺(aq) + 2NH₃(aq) ⇄ Ag(NH₃)₂⁺(aq)

The equilibrium constant (Kf) is 1.7 × 10⁷.

We can find the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = - R × T × ln Kf

where,

R: ideal gas constant

T: absolute temperature

ΔG° = - (8.314 J/K.mol) × 298 K × ln 1.7 × 10⁷ = - 4.1 × 10⁴ J/mol = -41 kJ/mol

This is the energy per mole of the balanced reaction.

if you're asking what comes out, it's steam.

b. is the answer hoped i