NO _ 2 required for a reaction is produced by decomposition of N _...

The given reaction is,

$2{N}_{2}{O}_{5 ( g )} \to 4{NO}_{2 ( g )} + {O}_{2 ( g )}$

The initial concentration of ${N}_{2}{O}_{5}$ is $3mol{L}^{- 1}$ and it is $2 . 75mol{L}^{- 1}$ after $30$ minutes.

Now, 

$- \dfrac{\Delta\left[{N}_{2}{O}_{5}\right]}{\Delta t} = - \dfrac{( 2 . 75 - 3 )}{30 - 0}mol{L}^{- 1}\min^{- 1} = \dfrac{0 . 25}{30}$

$\dfrac{1}{2} \times \dfrac{\Delta\left[{N}_{2}{O}_{5}\right]}{\Delta t} = \dfrac{1}{4}\dfrac{- \Delta\left[{NO}_{2}\right]}{\Delta t}\dfrac{- \Delta\left[{NO}_{2}\right]}{\Delta t} = \dfrac{0 . 25}{30} \times 2 = 1 . 66 \times {10}^{- 2}M / Min$

$≃ 17 \times {10}^{- 3}M / Min$