Chemistry - Chemical Kinetics Question with Solution | TestHub

For the reaction
$\mathrm{x}\mathrm{A}\to \mathrm{y}\mathrm{B}$
$-\frac{1}{\mathrm{x}}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{d}\mathrm{t}}=+\frac{1}{\mathrm{y}}\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{d}\mathrm{t}}$
$-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{d}\mathrm{t}}=+\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{d}\mathrm{t}}\times \frac{\mathrm{x}}{\mathrm{y}}$
$\log \left\{-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{d}\mathrm{t}}\right\}=\log \left\{\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{d}\mathrm{t}}\right\}+\log \left(\frac{\mathrm{x}}{\mathrm{y}}\right)$…(i)
Given$\log \left\{-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{d}\mathrm{t}}\right\}=\log \left\{\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{d}\mathrm{t}}\right\}+0.3$…(ii)
Comparing (i) and (ii), we get
$\log \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=0.3 \Rightarrow \log \frac{\mathrm{x}}{\mathrm{y}}= \log 2$
So$\frac{\mathrm{x}}{\mathrm{y}}=\frac{2}{1}$.