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For the reaction, $2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2},$ , the rate equation can be expressed in two ways $-\frac{d\left [ N_{2}O_{5} \right ]}{dt}=k\left [ N_{2}O_{5} \right ]$ and $+\frac{d\left [ NO_{2} \right ]}{dt}=k'\left [ N_{2}O_{5} \right ]$

$k\, and \, k'$ are related as :
Option: 1
$k=k'$

Option: 2
$2k=k'$

Option: 3
$k=2k'$

Option: 4
$k=4k'$

Answers (1)

best_answer

As we discussed in the concept

Instantaneous Rate -

The rate of a reaction calculated at a particular instant of time is called Instantaneous Rate

- wherein

$r_{av}=\frac{-\Delta R}{\Delta t}=\frac{+\Delta P}{\Delta t}$

$\lim_{\Delta\rightarrow 0}r_{av} = r_{inst}$

Formula = $\frac{-d[R]}{dt}= \frac{-d[P]}{dt}$

$2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}$

given that

$\frac{-d\left [ N_{2}O_{5} \right ]}{dt}=k\left [ N_{2} O_{5}\right ]$

$+\frac{d\left [ NO_{2} \right ]}{dt}=k^{1}\left [ N_{2}O _{5}\right ]$

$\frac{d\left [ N_{2}O_{5} \right ]}{dt}=\frac{d\left [ NO_{2} \right ]}{2dt}$

$k\left [ N_{2} O_{5}\right ]=\frac{k^{1}\left [ N_{2}O_{5} \right ]}{2}$

$2k=k^{1}$

Therefore, option (2) is correct.

Posted by

vishal kumar

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