The rate of certain reaction depends on concentration according to the equation $\frac{{ - dc}}{{dt}} = \frac{{{K_1}C}}{{1 + {K_2}C}},$ what is the order, when concentration $(c)$ is very-very high

Consider the reaction :

$Cl_2(aq) + H_2S(aq) \to  S(s) + 2H^+(aq) + 2Cl^-(aq)$

The rate equation for this reaction is rate $= k[Cl_2][H_2S]$ Which of these mechanisms is/are consistent with this rate equation ?

$A.\,C{l_2} + {H_2}S \to {H^ + } + C{l^ - } + C{l^ + } + H{S^- }$  (slow)

$C{l^ + } + H{S^ - } \to {H^ + } + C{l^ - } + {S}$   (fast)

$B.\, H_2S  \Leftrightarrow  H^+ + HS^-$   (fast equilibrium)

$Cl_2 + HS^-\to  2Cl^-+ H^+ + S$ (slow)

Rate of reaction is given by following rate law $ - \frac{{d\left[ c \right]}}{{dt}} = \frac{{{k_1}\,\left[ c \right]}}{{1 + {k_2}\,\left[ c \right]}}$ order of reaction when concentration is verh high

In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction ?

In a chemical reaction $A$ is converted into $B$ . The rates of reaction, starting with initial concentrations of $A$ as $2 \times {10^{ - 3}}\,M$ and $1 \times {10^{ - 3}}\,M$ , are equal to $2.40 \times {10^{ - 4}}\,M{s^{ - 1}}$ and $0.60 \times {10^{ - 4}}\,M{s^{ - 1}}$ respectively. The order of reaction with respect to reactant $A$ will be

The following data was obtained for chemical reaction given below at $975\, \mathrm{~K}$.

$2 \mathrm{NO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$

The order of the reaction with respect to $\mathrm{NO}$ is ..... .