Chemistry - Chemical Kinetics Question with Solution | TestHub

Assume rate law expression

$\text{Rate =}\text{k}{\left[\text{A}\right]}^x{\left[\text{B}\right]}^y$

$0.05=\text{k}{\left[0.1\right]}^x{\left[0.1\right]}^y⟶\left(\text{I}\right)$

$0.10=\text{k}{\left[0.2\right]}^x{\left[0.1\right]}^y⟶\left(\text{II}\right)$

$0.05=\text{k}{\left[0.1\right]}^x{\left[0.2\right]}^y⟶\left(\text{III}\right)$

$\frac{\text{II}}{\text{I}}\Rightarrow 2={\left[2\right]}^x\therefore x=1$

$\frac{\text{I}\text{II}}{\text{I}}\Rightarrow 1={\left[2\right]}^y=2{}^0\therefore y=0$

Order w.r.t. A is one

Order w.r.t. B is Zero

Rate law expression is

$\text{Rate =}\text{k}\left[\text{A}\right]$

$0.05=\text{k}{\left[0.1\right]}^1$

$\text{k}=\frac{0.05}{0·1}=0·5{\text{sec}}^{-1}$