JEE Main 2021 — Chemical Equilibrium Question with Solution

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.