Question

The following results were obtained during the kinetic studies of the reaction:$2A+B\to Products$

Experiment	[A] (in mol $L^{-1}$	[B] (in mol $L^{-1}$)	Initial rate of the reaction (in mol $L^{-1}mi{n}^{-1}$ )
(I)	0.10	0.20	6.93x${10}^{-3}$
(II)	0.10	0.25	6.93x${10}^{-3}$
(III)	0.20	0.30	1.386x${10}^{-2}$

The time (in minutes) required to consume half of A is:

• A. 10
• B. 5
• C. 100
• D. 1

Answer 1

The correct option is B 5

$6.93\times {10}^{-3}$ =$K\times \left(0.1{)}^x\right(0.2{)}^y$
$6.93\times {10}^{-3}$ =$K\times \left(0.1{)}^x\right(0.25{)}^y$
So, $y=0$
and $1.386\times {10}^{-2}=K\times \left(0.2{)}^x\right(0.30{)}^y$
$x=1$
so, $r=K\times \left(0.1\right)\times (0.2{)}^0$
$6.93\times {10}^{-3}=K\times 0.1\times (0.2{)}^0$
K=$6.93\times {10}^{-2}$
$t_{\frac{1}{2}}$=$\frac{0.693}{2K}$
=$\frac{10}{2}$=5