The following results were obtained during kinetic studies of the...

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

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

${\left(\dfrac{0 .1}{0 .1}\right)}^{x}{\left(\dfrac{0 .2}{0 .25}\right)}^{y} = \dfrac{6 .93 \times {10}^{- 3}}{6 .93 \times {10}^{- 3}} ⇒ y = 0$

From equation \left(1\right)$and$\left(3\right)$and put the value$y = 0$,<br /><br />${\left(\dfrac{0 .1}{0 .2}\right)}^{x}{\left(\dfrac{0 .2}{0 .3}\right)}^{0} = \dfrac{6 .93 \times {10}^{- 3}}{1 .386 \times {10}^{- 2}} ∴ y = 0$<br /><br />${\left(\dfrac{1}{2}\right)}^{x} = \left(\dfrac{1}{2}\right) ⇒ x = 1$<br /><br />The rate law will be$R = K{[ A ]}^{1}{[ B ]}^{0}$<br /><br />Hence, the reaction is of the first order.<br /><br />$K = \dfrac{R}{\left[A\right]} = \dfrac{6 .93 \times {10}^{- 3}}{0 .1} = 6 .93 \times {10}^{- 2}$<br /><br />Here$t\dfrac{1}{2}$for the first-order reaction is given by,$t\dfrac{1}{2} = \dfrac{0 .693}{K} = \dfrac{0 .693}{( 6 .93 \times {10}^{- 2})} =10