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

Question

Chemistry

Riya Mcguire

5 May, 18:19

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

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Answers (1)

Know the Answer?

Jakob · Answer 1

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.