Chemistry - Chemical Kinetics Question with Solution | TestHub

Let's assume$\mathrm{x}$and$\mathrm{y}$are the order of the reaction with respect to$\mathrm{A}$and$\mathrm{B}$, respectively.

From equation$\left(1\right)$and$\left(2\right)$,

${\left(\frac{0.1}{0.1}\right)}^{\mathrm{x}}{\left(\frac{0.2}{0.25}\right)}^{\mathrm{y}}=\frac{6.93\times {10}^{-3}}{6.93\times {10}^{-3}}\Rightarrow \mathrm{y}=0$

From equation$\left(1\right)$and$\left(3\right)$and put the value$\mathrm{y}=0$,

${\left(\frac{0.1}{0.2}\right)}^{\mathrm{x}}{\left(\frac{0.2}{0.3}\right)}^0=\frac{6.93\times {10}^{-3}}{1.386\times {10}^{-2}}$$\therefore \mathrm{y}=0$

${\left(\frac{1}{2}\right)}^{\mathrm{x}}=\left(\frac{1}{2}\right)\Rightarrow \mathrm{x}=1$

The rate law will be$\mathrm{R}=\mathrm{K}{\left[\mathrm{A}\right]}^1{\left[\mathrm{B}\right]}^0$

Hence, the reaction is of the first order.

$\mathrm{K}=\frac{\mathrm{R}}{\left[\mathrm{A}\right]}=\frac{6.93\times {10}^{-3}}{0.1}=6.93\times {10}^{-2}$

Here${\mathrm{t}}_{\frac{1}{2}}$for the first-order reaction is given by,${\mathrm{t}}_{\frac{1}{2}}=\frac{0.693}{\mathrm{K}}=\frac{0.693}{(6.93\times {10}^{-2}\mathrm{)}}=10$