The reaction rate for the reaction [PtCl_4]^{ | vedclass

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Question

(A) For the reaction $[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-$,the rate of the forward reaction is $r_f = k_f [[PtCl_4]^{2-}]$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) For the reaction $[PtCl_4]^{2-} + H_2O ightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-$,the rate of the forward reaction is $r_f = k_f [[PtCl_4]^{2-}]$ and the rate of the backward reaction is $r_b = k_b [[Pt(H_2O)Cl_3]^-] [Cl^-]$.At equilibrium,the rate of the forward reaction equals the rate of the backward reaction $(r_f = r_b)$.Therefore,$k_f [[PtCl_4]^{2-}] = k_b [[Pt(H_2O)Cl_3]^-] [Cl^-]$.The equilibrium constant $K_c$ is given by $K_c = \frac{k_f}{k_b} = \frac{[[Pt(H_2O)Cl_3]^-] [Cl^-]}{[[PtCl_4]^{2-}]}$.From the given rate expression,$k_f = 4.8 	imes 10^{-5} \ s^{-1}$ and $k_b = 2.4 	imes 10^{-3} \ M^{-1} s^{-1}$.$K_c = \frac{4.8 	imes 10^{-5}}{2.4 	imes 10^{-3}} = 0.02$.The nearest integer to $0.02$ is $0$.

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