For the reaction,$2 \ NO_{(g)} + O_{2_{(g)}} \rightarrow 2 \ NO_{2_{(g)}}$. If $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$,calculate the rate of consumption of $NO_{(g)}$.

Consider the elementary reaction $A_{(g)} + B_{(g)} \rightarrow C_{(g)} + D_{(g)}$. If the volume of the reaction mixture is suddenly reduced to $\frac{1}{3}$ of its initial volume,the reaction rate will become '$x$' times the original reaction rate. The value of $x$ is:

$NO_2$ required for a reaction is produced by the decomposition of $N_2O_5$ in $CCl_4$ as per the equation $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$. The initial concentration of $N_2O_5$ is $3.00 \, mol \, L^{-1}$ and it is $2.75 \, mol \, L^{-1}$ after $30 \, minutes$. The rate of formation of $NO_2$ is ............... $mol \, L^{-1} \, min^{-1}$

In the reaction $xA \to yB,$
$0.48 + \log \left\{ -\frac{d[A]}{dt} \right\} = \log \left\{ +\frac{d[B]}{dt} \right\} + 0.7$
Then,$x : y$ is

Consider the reaction: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$. The equality relationship between $\frac{d[NH_3]}{dt}$ and $\frac{d[H_2]}{dt}$ is

The rate of a gaseous reaction is given by $r = K[x][y]$. If the volume of the reaction vessel is suddenly reduced to $1/4$ of its initial volume,the rate of the reaction will ............