Question

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

Answer 1

c. –41 kJ

Step-by-step 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.

Answer 2

ΔG° = 41.248 KJ/mol (298 K); the correct answer is a) 41 KJ

Step-by-step explanation:

Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)

⇒ Kf = 1.7 E7; T =298K

⇒ ΔG° = - RT Ln Kf.....for aqueous solutions

∴ R = 8.314 J/mol.K

⇒ ΔG° = - ( 8.314 J/mol.K ) * ( 278 K ) ln ( 1.7 E7 )

⇒ ΔG° = 41248.41 J/mol * ( KJ / 1000J )

⇒ ΔG° = 41.248 KJ/mol