A reaction involving two different reactants

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

For the reaction $A + B \rightarrow$ products, it is observed that

$(i)\,\,$on doubling the initial concentration of $A$ only, the rate of reaction is also doubled and

$(ii)$  on doubling the initial concentration of both $A$ and $B,$ there is a change by a factor of $8$ in the rate of the reaction.

The rate of this reaction is given by

In a reaction, $A + B \rightarrow$ product, rate is doubled when the concentration of $B$ is doubled, and rate increases by a factor of $8$ when the concentration of both the reactants $(A$ and $B)$ are doubled, rate law for the reaction can be written as

The decomposition of dimethyl ether leads to the formation of $CH _{4}, H _{2}$ and $CO$ and the reaction rate is given by

Rate $=k\left[ CH _{3} OCH _{3}\right]^{3 / 2}$

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.

Rate $=k\left(p_{ CH _{3} OCH _{3}}\right)^{3 / 2}$

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

The reaction of formation of phosgene from $CO$ and $Cl_2$ is $CO + Cl_2 \to COCl_2.$ The proposed mechanism is

$(i)$  $C{l_2}\,\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}\,2Cl$

$(ii)$  $Cl + CO\,\underset{{{k_4}}}{\overset{{{k_3}}}{\longleftrightarrow}}\,COCl$

$(iii)$  $COCl\, + \,C{l_2}\,\,\xrightarrow{{{k_5}}}\,COC{l_2}\, + \,Cl$ (slow)

Find the correct expression of rate law