$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

At $3000 \ K$ the equilibrium pressures of $CO_2$,$CO$ and $O_2$ are $0.6 \ atm$,$0.4 \ atm$ and $0.2 \ atm$ respectively. $K_p$ for the reaction,$2CO_2 \rightleftharpoons 2CO + O_2$ is

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 $K_c = ...$. (Nearest integer)

Match List-$I$ (Hypothetical reaction) with List-$II$ (Ratio of $K_p/K_c$ for the given reaction) and select the correct answer using the options given below.

$(1)$ $A_{2(g)} + 3B_{2(g)} \rightleftharpoons 2AB_{3(g)}$	$(i)$ $(RT)^{-2}$
$(2)$ $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$	$(ii)$ $(RT)^0$
$(3)$ $A_{(s)} + 1.5B_{2(g)} \rightleftharpoons AB_{3(g)}$	$(iii)$ $(RT)^{1/2}$
$(4)$ $AB_{2(g)} \rightleftharpoons AB_{(g)} + 0.5B_{2(g)}$	$(iv)$ $(RT)^{-1/2}$

The relation between equilibrium constant $K_p$ and $K_c$ is

$A_{(g)} + 3B_{(g)} \rightleftharpoons 4C_{(g)}$. The starting concentrations of $A$ and $B$ are equal. At equilibrium,the concentrations of $A$ and $C$ are the same. The value of $K_c$ is: