The rate of the reaction :

2N_2O_5 rightar | vedclass

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Question

Rate of disappearance of reactants $=$ Rate of appearance of products

$=\frac{1}{2} \frac{d\left(N_{2} O_{5}\right)}{d t}=\frac{1}{4} \frac{d\left(

Vedclass · Accepted Answer

$k' = 2k$ ;  $k'' = k/2$

Vedclass · Answer

Rate of disappearance of reactants $=$ Rate of appearance of products

$=\frac{1}{2} \frac{d\left(N_{2} O_{5}\right)}{d t}=\frac{1}{4} \frac{d\left(N O_{2}\right)}{d t}=\frac{d\left(O_{2}\right)}{d t}$

$=\frac{1}{2} k\left(N_{2} O_{5}\right)=\frac{1}{4} k'\left(N_{2} O_{5}\right)=k''\left(N_{2} O_{5}\right)$

$\Rightarrow \frac{k}{2}=\frac{k'}{4}=k''$

$\Rightarrow k'=2 k$,   $k''=\frac{k}{2}$

$A$ $(mol/l)$	$B$ $(mol/l)$	Rate
$0.05$	$0.05$	$1.2\times 10^{-3}$
$0.10$	$0.05$	$2.4\times 10^{-3}$
$0.05$	$0.10$	$1.2\times 10^{-3}$

The rate of the reaction :

$2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.

$\frac{-d[N_2O_5 ]}{dt} = k[N_2O_5]$

$\frac{d[NO_2 ]}{dt} = k'[N_2O_5]\,;$ $\frac{d[O_2 ]}{dt} = k"[N_2O_5]$

The relationship between $k$ and $k'$ and betweenk and $k''$ are

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$2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.

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