Chemical EquilibriumHard

Question

The approach to the following equilibrium was observed kinetically from both directions.

PtCl42– + H2O$\rightleftharpoons$Pt(H2O)Cl3– + Cl–

At 25°C, it was found that $- \frac{d\left\lbrack PtCl_{4}^{2 -} \right\rbrack}{dt}$ = (3.9 × 10–5 s–1) [PtCl42–]–(2.1×10–3 L mol–1 s–1) [Pt (H2O) Cl3–] [Cl–]

The value of Keq (equilibrium constant) for the complexation of the fourth Cl− by Pt (II) is

Options

A.53.8 mol L–1
B.0.018 mol L–1
C.53.8 L mol–1
D.0.018 L mol–1

Solution

Keq for the reaction in backward direction $= \frac{K_{b}}{K_{f}} = \frac{2.1 \times 10^{- 3}\text{ L mo}\text{l}^{- 1}\text{ }\text{s}^{- 1}}{3.9 \times 10^{- 5}\text{ }\text{s}^{- 1}} = 53.846\text{ L mo}\text{l}^{- 1}$

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