If the surface area of the reactants increases, then order of the reaction

The rate constant for the reaction, $2{N_2}{O_5} \to 4N{O_2}$ $ + {O_2}$ is $3 \times {10^{ - 5}}{\sec ^{ - 1}}$. If the rate is $2.40 \times {10^{ - 5}}\,mol\,\,litr{e^{{\rm{ - 1}}}}{\sec ^{ - 1}}$. Then the concentration of ${N_2}{O_5}$ (in mol litre $^{-1}$) is

Select the incorrect option :

By “the overall order of a reaction”, we mean

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

For a chemical reaction $A \rightarrow B$, it was found that concentration of $B$ is increased by $0.2\, mol\,L^{-1}$ in $30\, \mathrm{~min}$. The average rate of the reaction is $......\times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}$. (Nearest integer)