Question

Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1 ΔG∘2 for second step=−35.8 kJ⋅mol−1 ΔG∘3 for third step=−24.3 kJ⋅mol−1 Calculate the overall formation constant (Kf) for the complex [Ni(en)3]2+.

Answer 1

kf = 1.16 x 10¹⁸

Step-by-step explanation:

Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹

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Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸