The reaction rate for the reaction
$$[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-$$
was measured as a function of concentrations of different species. It was observed that
$$\frac{-d[PtCl_4]^{2-}}{dt} = 4.8 \times 10^{-5}[PtCl_4]^{2-} - 2.4 \times 10^{-3}[Pt(H_2O)Cl_3]^-][Cl^-]$$
where square brackets are used to denote molar concentrations.
The equilibrium constant K$$_c$$ = X (Nearest integer). Value of $$\frac{1}{X}$$ is _________ 
$$K_c = X$$ (Nearest integer)

We start with the elementary reversible reaction

$$[PtCl_4]^{2-} + H_2O \;\rightleftharpoons\; [Pt(H_2O)Cl_3]^- + Cl^-$$

The experimentally observed rate law for the disappearance of $$[PtCl_4]^{2-}$$ is

$$-\dfrac{d[PtCl_4]^{2-}}{dt}=4.8\times10^{-5}[PtCl_4]^{2-}-2.4\times10^{-3}[Pt(H_2O)Cl_3]^-\, [Cl^-]$$

We recognise the right-hand side as the difference between a forward rate and a backward rate, so we write

$$\text{rate}_{\text{net}}=\text{rate}_{\text{forward}}-\text{rate}_{\text{backward}}$$

and compare term by term:

$$\text{rate}_{\text{forward}}=k_f[PtCl_4]^{2-}$$ $$\text{rate}_{\text{backward}}=k_b[Pt(H_2O)Cl_3]^-\, [Cl^-]$$

Matching coefficients gives

$$k_f = 4.8\times10^{-5}\;\text{s}^{-1}$$ $$k_b = 2.4\times10^{-3}\;\text{M}^{-1}\text{s}^{-1}$$

At equilibrium the net rate is zero, so the forward and backward rates are equal:

$$k_f[PtCl_4]^{2-}=k_b[Pt(H_2O)Cl_3]^-\, [Cl^-]$$

Rearranging we obtain

$$\dfrac{[Pt(H_2O)Cl_3]^-\, [Cl^-]}{[PtCl_4]^{2-}}=\dfrac{k_f}{k_b}$$

By definition the equilibrium constant for the reaction is

$$K_c=\dfrac{[Pt(H_2O)Cl_3]^-\, [Cl^-]}{[PtCl_4]^{2-}}$$

Therefore

$$K_c=\dfrac{k_f}{k_b}$$

Substituting the numerical values, we have

$$K_c=\dfrac{4.8\times10^{-5}}{2.4\times10^{-3}}$$

Now we divide the mantissas first:

$$\dfrac{4.8}{2.4}=2$$

Next we divide the powers of ten:

$$10^{-5}\div10^{-3}=10^{(-5)-(-3)}=10^{-2}$$

Multiplying these two results together we find

$$K_c = 2\times10^{-2}=0.02$$

The problem asks us to denote this value by $$X$$ and then evaluate $$\dfrac{1}{X}$$. Hence

$$X=0.02\quad\Longrightarrow\quad\dfrac{1}{X}=\dfrac{1}{0.02}=50$$

So, the answer is $$50$$.