Chemistry - Chemical Kinetics Question with Solution | TestHub

From the experiment (1) and (2)
${\left(\frac{0.05}{0.10}\right)}^{\mathrm{x}}{\left(\frac{0.05}{0.05}\right)}^{\mathrm{y}}=\frac{0.045}{0.090}$
${\left(\frac{1}{2}\right)}^{\mathrm{x}}=\frac{1}{2}\Rightarrow \mathrm{x}=1$

From the experiment (2) and (3)
${\left(\frac{0.1}{0.2}\right)}^{\mathrm{x}}{\left(\frac{0.05}{0.1}\right)}^{\mathrm{y}}=\left(\frac{0.09}{0.72}\right)$
$\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^{\mathrm{y}}=\frac{0.09}{0.72}$
${\left(\frac{1}{2}\right)}^{\mathrm{y}}=\frac{0.09\times 2}{0.72}$
${\left(\frac{1}{2}\right)}^{\mathrm{y}}=\left(\frac{1}{4}\right)$
$\mathrm{y}=2$
Order of reaction with respect to A is $2$ and $\mathrm{B}=2$
$\therefore \mathrm{R}\propto \left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$
Or $\mathrm{R}=\mathrm{K}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$