Question

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

$L^{-1}$

(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)

$6.93\times 10^{-3}=k\times (0.1{)}^x(0.2{)}^y\dots \dots \dots \dots \dots \dots ..(i)$
$6.93\times 10^{-3}=k\times (0.1{)}^x(0.25{)}^y$
From the above equation, $y=0$
and $1.386\times 10^{-2}=k\times (0.2{)}^x(0.30{)}^y\dots \dots \dots \dots .$ (iii)
Divide equation (i) by (iii), we get
$<br/>\frac{1}{2}={\left(\frac{1}{2}\right)}^x\Rightarrow x=1<br/>$
So $r=k\times (0.1)\times (0.2{)}^0$
$6.93\times 10^{-3}=k\times 0.1\times (0.2{)}^0$
$k=6.93\times 10^{-2}$
$<br/>t_{1/2}=\frac{0.693}{2k}=\frac{0.693}{0.693\times 10^{-1}\times 2}=\frac{10}{2}=5<br/>$