Question

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

(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)

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