QJEEM21Ch9f2d: JEE Main 2021 Chemistry - Class 11 JEEM21Ch9f2d Solution: The reaction rate for the reaction [PtCl 4 ] 2 + H 2 O [Pt(H 2…

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:

$k_f[\text{{PtCl}}_4]^{2-} = k_r[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Given the rate equation:

$-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

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

$0 = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Rearranging terms, we find:

$4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} = 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

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:

$K_c = \frac{[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]}{[\text{{PtCl}}_4]^{2-}}$

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

$K_c = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = \frac{1}{50}$ = 0.02

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