Question

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

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

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

• A. $10$
• B. $5$
• C. $100$
• D. $1$

Answer 1

Answer: (B)

$5$

$6.93\times {10}^{-3}=k\times \left(0.1{)}^x\right(0.2{)}^y$...................(i)

$6.93\times {10}^{-3}=k\times \left(0.1{)}^x\right(0.25{)}^y$..................(ii)

From the above equation, $y=0$

and $1.386\times {10}^{-2}=k\times \left(0.2{)}^x\right(0.30{)}^y$...........(iii)

Divide equation (i) by (iii), we get

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

So $r=k\times \left(0.1\right)\times (0.2{)}^0$

$6.93\times {10}^{-3}=k\times 0.1\times (0.2{)}^0$

$k=6.93\times {10}^{-2}$

$t_{1/2}=\frac{0.693}{2k}=\frac{0.693}{0.693\times {10}^{-1}\times 2}=\frac{10}{2}=5$

Hence, the correct option is B