Question

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

Experimental	$[A]/mol{L}^{-1}$	$[B]/mol{L}^{-1}$	Initial rate/mol$L^{-1}mi{n}^{-1}$
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

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

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

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

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

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

$(3)÷(1)\Rightarrow x=1$

$(2)÷(3)\Rightarrow x=2$

So, order with respect to $A=1$

Order with respect to $B=2$

$(4)÷(3)$

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

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

$x=0.3M$

$(5)÷(4)$

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

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

$y=0.4M$