પ્રક્રિયા 2A + B rightarrow C + D ના ગતિ | vedclass

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Question

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

Vedclass · Accepted Answer

દર $= k[A][B]^2$

Vedclass · Answer

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 $=\mathrm{k}(0.1)^{x}(0.1)^{y}=6.0 	imes 10^{-3}$

$II$. rate $=\mathrm{k}(0.3)^{x}(0.2)^{y}=7.2 	imes 10^{-2}$

$III$. rate $=\mathrm{k}(0.3)^{x}(0.40)^{y}=2.88 	imes 10^{-1}$

$IV$. rate $=\mathrm{k}(0.34)^{x}(0.1)^{y}=2.40 	imes 10^{-2}$

Dividing Eq. $(I)$ by Eq. $(IV)$, we get

$\left(\frac{0.1}{0.4}ight)^{x}\left(\frac{0.1}{0.1}ight)^{y}$

$\frac{6.0 	imes 10^{-3}}{2.4 	imes 10^{-2}}$

$\left(\frac{1}{4}ight)^{x} =\left(\frac{1}{4}ight)^{1}$

$	herefore x=1$

On dividing Eq. $(II)$ by Eq. $(III)$, we get

$\left(\frac{0.3}{0.3}ight)^{x}\left(\frac{0.2}{0.4}ight)^{y}=\frac{7.2 	imes 10^{-2}}{2.88 	imes 10^{-1}}$

$\left(\frac{1}{2}ight)^{y}=\frac{1}{4}$

$\left(\frac{1}{2}ight)^{y}=\left(\frac{1}{2}ight)^{2}$

$	herefore y=2$

Thus, rate law is,

$	ext {rate} =k[A]^{1}[B]^{2}$

$=k[A][B]^{2}$

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

ક્રમ	$[ NO ]_{0}$	$\left[ Cl _{2}\right]_{0}$	$r _{0}$
$1$	$0.10$	$0.10$	$0.18$
$2$	$0.10$	$0.20$	$0.35$
$3$	$0.20$	$0.20$	$1.40$

Similar Questions

પ્રક્રિયા $A+B \rightarrow$ નીપજ માટે વેગ નિયમ $r=k[ A ]^{1 / 2}[ B ]^{2}$ તરીકે દર્શાવાય છે. પ્રક્રિયાનો ક્રમ કેટલો હશે ?