Question

The results given in the below table were obtained during kinetic studies of the following reaction:
$2 A + B \longrightarrow C + D$

Experimental	$[A]/molL^{-1}$	$[B]/molL^{-1}$	Initial rate/mol$L^{-1} \,min^{-1}$
I	0.1	0.1	$6.00 \times 10^{-3}$
II	0.1	0.2	$2.40 \times 10^{-2}$
III	0.2	0.1	$1.20 \times 10^{-2}$
IV	X	0.2	$7.20 \times 10^{-2}$
V	0.3	Y	$2.88 \times 10^{-1}$

$X$ and $Y$ in the given table are respectively :

• A. $0.3, 0.4$
• B. $0.4, 0.3$
• C. $0.4, 0.4$
• D. $0.3, 0.3$

Answer 1

Answer: (A)

From rate law

$r =-\frac{1}{2} \frac{ d [ A ]}{ dt }=\frac{- d [ B ]}{ dt }$

$= K [ A ]^{ x }[ B ]^{ y }$

$6 \times 10^{-3}= K (0.1)^{ x }(0.1)^{ y } \quad \ldots \ldots(1)$

$2.4 \times 10^{-2}= K (0.1)^{ x }(0.2)^{ y } \quad \ldots \ldots(2)$

$1.2 \times 10^{-2}= K (0.2)^{ x }(0.1)^{ y } \quad \ldots \ldots(3)$

$(3) \div(1) \Rightarrow x =1$

$(2) \div(3) \Rightarrow x =2$

So, order with respect to $A =1$

Order with respect to $B =2$

$(4) \div(3)$

$\left(\frac{x}{0.2}\right) \times\left(\frac{0.2}{0.1}\right)^{2}=\frac{7.2 \times 10^{-2}}{1.2 \times 10^{-2}}$

$x=\frac{6 \times 0.2}{4}$

$x=0.3 M$

$(5) \div(4)$

$\left(\frac{y}{0.2}\right)^{2}=\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}$

$y ^{2}=4 \times 0.2^{2}$

$y=0.4 M$