In the following reaction: $xA \to yB$,$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$,where the negative sign indicates the rate of disappearance of the reactant. Thus,$x : y$ is:

In the reaction of formation of sulphur trioxide by contact process $2SO_2 + O_2 \rightleftharpoons 2SO_3$,the rate of reaction was measured as $\frac{d[O_2]}{dt} = -2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. The rate of reaction in terms of $[SO_2]$ in $mol \ L^{-1} \ s^{-1}$ will be:

The rate of disappearance of $SO_2$ in the reaction $2SO_2 + O_2 \to 2SO_3$ is $1.28 \times 10^{-3} \ g/sec$. What is the rate of formation of $SO_3$?

Thermal decomposition of $N_2O_5$ occurs as per the equation below:
$2N_2O_5 \longrightarrow 4NO_2 + O_2$
The correct statement is:

For the reaction $3 \,A \rightarrow 2 \,B$, the rate of reaction $+\frac{d[B]}{d t}$ is equal to:

For the reaction $2 N_2O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,the initial concentration of $N_2O_5$ is $2.0 \ mol \ L^{-1}$ and after $300 \ min$,it is reduced to $1.4 \ mol \ L^{-1}$. The rate of production of $NO_2$ (in $mol \ L^{-1} \ min^{-1}$) is