Question

$2A+B\to$ Product

$\left[A\right]$	$\left[B\right]$	Rate ($\text{M}{\text{min}}^{-1}$)
$0.1$	$0.20$	$6.93\times {10}^{-3}$
$0.1$	$0.25$	$6.93\times {10}^{-3}$
$0.2$	$0.3$	$1.386\times {10}^{-2}$

Time in minutes, when concentration of $A$ becomes half

Answer 1

Rate $=k\left[A{]}^x\right[B{]}^y$
$6.93\times {10}^{-3}=k\left(0.1{)}^x\right(0.20{)}^y...\left(1\right)$
$6.93\times {10}^{-3}=k\left(0.1{)}^x\right(0.25{)}^y...\left(2\right)$
$1.386\times {10}^{-2}=k\left(0.2{)}^x\right(0.30{)}^y...\left(3\right)$
Divide equation (1) by (2)
$\Rightarrow 1=(0.20/0.25{)}^y$
$y=0$
Divide equation (1) by (3)
$\Rightarrow \left(1/2\right)=(1/2{)}^x\times 1$
$x=1$
From equation (1) and (3) we get x = 1, so it is first order with respect to A.
$6.93\times {10}^{-3}=k\left(0.1\right)$
$\Rightarrow k=6.93\times {10}^{-2}{\text{min}}^{-1}$
We know for firts order reaction:
$t=\frac{2.303}{k}log\frac{A_o}{A_t}$
$\Rightarrow Ast\to t_{1/2}:A_t\to \frac{A_o}{2}$
$\Rightarrow t_{1/2}=\frac{2.303}{k}log\frac{A_o}{\frac{A_o}{2}}$
$\Rightarrow t_{1/2}=\frac{2.303}{k}log2$
$\Rightarrow t_{1/2}=0.693/k$
$\Rightarrow t_{1/2}=0.693/(6.93\times {10}^{-2})=10\text{min}$