પ્રક્રિયા $2A + B \rightarrow C + D$ ના ગતિમય અભ્યાસ દરમિયાન નીચેના પરિણામો મળે છે.

Run $[A]/mol\,L^{-1}$ $[B]/mol\,L^{-1}$ $D$ ઉત્પન્ન થવાનો શરૂઆતનો દર $mol\,L^{-1}\,min^{-1}$

$I.$ $0.1$ $0.1$ $6.0 \times 10^{-3}$

$II.$ $0.3$ $0.2$ $7.2 \times 10^{-2}$

$III.$ $0.3$ $0.4$ $2.88 \times 10^{-1}$

$IV.$ $0.4$ $0.1$ $2.40 \times 10^{-2}$

ઉપરની વિગત પરથી નીચેનામાંથી ક્યું સાચુ છે ?

Question

પ્રક્રિયા $2A + B \rightarrow C + D$ ના ગતિમય અભ્યાસ દરમિયાન નીચેના પરિણામો મળે છે.

Run	$[A]/mol\,L^{-1}$	$[B]/mol\,L^{-1}$	$D$ ઉત્પન્ન થવાનો શરૂઆતનો દર $mol\,L^{-1}\,min^{-1}$
$I.$	$0.1$	$0.1$	$6.0 \times 10^{-3}$
$II.$	$0.3$	$0.2$	$7.2 \times 10^{-2}$
$III.$	$0.3$	$0.4$	$2.88 \times 10^{-1}$
$IV.$	$0.4$	$0.1$	$2.40 \times 10^{-2}$

ઉપરની વિગત પરથી નીચેનામાંથી ક્યું સાચુ છે ?

AIPMT 2010, Advanced

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Solution

d
Let the order of reaction with respect to $\mathrm{A}$ is $\mathrm{x}$

and with respect to $\mathrm{B}$ is $y$.

Thus,

rate $=\mathrm{k}[A]^{x}[B]^{y}$

($x$ and $y$ are stoichiometric coefficient)

For the given cases,

$I$. rate $=\mathrm{k}(0.1)^{x}(0.1)^{y}=6.0 \times 10^{-3}$

$II$. rate $=\mathrm{k}(0.3)^{x}(0.2)^{y}=7.2 \times 10^{-2}$

$III$. rate $=\mathrm{k}(0.3)^{x}(0.40)^{y}=2.88 \times 10^{-1}$

$IV$. rate $=\mathrm{k}(0.34)^{x}(0.1)^{y}=2.40 \times 10^{-2}$

Dividing Eq. $(I)$ by Eq. $(IV)$, we get

$\left(\frac{0.1}{0.4}\right)^{x}\left(\frac{0.1}{0.1}\right)^{y}$

$\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$

$\left(\frac{1}{4}\right)^{x} =\left(\frac{1}{4}\right)^{1}$

$\therefore x=1$

On dividing Eq. $(II)$ by Eq. $(III)$, we get

$\left(\frac{0.3}{0.3}\right)^{x}\left(\frac{0.2}{0.4}\right)^{y}=\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}$

$\left(\frac{1}{2}\right)^{y}=\frac{1}{4}$

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

$\therefore y=2$

Thus, rate law is,

$\text {rate} =k[A]^{1}[B]^{2}$

$=k[A][B]^{2}$

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