The following equilibrium was studied kinetically from both directions: PdCl 4 2- + H 2 O ⇌ Pd(H 2 O)Cl 3 - + Cl - . At 25°C, -(d[PdCl 4 2- ])/dt = (2.4×10 -4 s -1 )[PdCl 4 2- ] - (6.0×10 -3 L mol -1

Question

The following equilibrium was studied kinetically from both directions: PdCl 4 2- + H 2 O ⇌ Pd(H 2 O)Cl 3 - + Cl - . At 25°C, -(d[PdCl 4 2- ])/dt = (2.4×10 -4 s -1 )[PdCl 4 2- ] - (6.0×10 -3 L mol -1 s -1 )[Pd(H 2 O)Cl 3 - ][Cl - ]. The equilibrium constant for the complexation of the fourth Cl - by Pd(II) is

TestHub · Accepted Answer

At equilibrium, the forward reaction rate equals the backward reaction rate:

k_f[PdCl₄^2-] = k_b[Pd(H₂O)Cl₃^-][Cl^-]

The complexation of Cl⁻ is the reverse of the written step.

The equilibrium constant K is given by:

K = ([PdCl₄^2-])/([Pd(H₂O)Cl₃^-][Cl^-]) = (k_b)/(k_f)

Substituting the given rate constants:

K = (6.0 ×10^-3)/(2.4 ×10^-4)

Calculating the value of K:

K = 25 L mol^-1

This corresponds to option (c).

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