Find the rate of formation of NO_{2(g)} in th | vedclass

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(D) The rate of reaction is given by the expression:$	ext{Rate} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt} = +\frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]

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$0.04 \ mol \ dm^{-3} \ s^{-1}$

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(D) The rate of reaction is given by the expression:$	ext{Rate} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt} = +\frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$We are given that $-\frac{d[N_2 O_5]}{dt} = 0.02 \ mol \ dm^{-3} \ s^{-1}$.To find the rate of formation of $NO_2$,we use the relation:$\frac{1}{4} \frac{d[NO_2]}{dt} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt}$$\frac{d[NO_2]}{dt} = 2 	imes \left( -\frac{d[N_2 O_5]}{dt} ight)$$\frac{d[NO_2]}{dt} = 2 	imes 0.02 \ mol \ dm^{-3} \ s^{-1} = 0.04 \ mol \ dm^{-3} \ s^{-1}$

Find the rate of formation of $NO_{2(g)}$ in the following reaction.
$2 N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$
$\left[ \frac{-d[N_2 O_5]}{dt} = 0.02 \ mol \ dm^{-3} \ s^{-1} \right]$

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Find the rate of formation of $NO_{2(g)}$ in the following reaction.$2 N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$$\left[ \frac{-d[N_2 O_5]}{dt} = 0.02 \ mol \ dm^{-3} \ s^{-1} \right]$