The rate equation provided in your question describes the rates of the forward and reverse reactions for this chemical system. The equilibrium constant, $K_c$, is the ratio of the forward rate constant ($k_f$) to the reverse rate constant ($k_r$). At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so the rate equation simplifies to:



Given the rate equation:



At equilibrium, $-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 0$, so:



Rearranging terms, we find:



Now, the equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reaction in question, we have:



Dividing both sides of our rate equation by $[\text{{PtCl}}_4]^{2-}$, we find that:

= 0.02

So, the equilibrium constant $K_c$ for this reaction is approximately 0, when rounded to the nearest integer.