In the following reaction $: x \mathrm{~A} \rightarrow y \mathrm{~B}$
$\log _{10}\left[-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right]=\log _{10}\left[\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right]+0.3010$
'A' and 'B' respectively can be:
Correct Option: , 3
$x A \rightarrow y B$
$\therefore \quad \frac{-\mathrm{d} \mathrm{A}}{\mathrm{xdt}}=\frac{1}{\mathrm{y}} \frac{\mathrm{dB}}{\mathrm{dt}}$
$\frac{-\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}} \times \frac{\mathrm{x}}{\mathrm{y}}$
$\log \left[\frac{-\mathrm{d} \mathrm{A}}{\mathrm{dt}}\right]=\log \left[\frac{\mathrm{dB}}{\mathrm{dt}}\right]+\log \left(\frac{\mathrm{x}}{\mathrm{y}}\right)$
Comparing this equation with the equation given in question. We get,
$\log \frac{x}{y}=0.3010$ or $\log \frac{x}{y}=\log 2$
$\therefore \quad \frac{x}{y}=2$
$\therefore$ The reaction is of type $2 \mathrm{~A} \rightarrow \mathrm{B}$.
Hence, option (3) is correct.