The oxidation of ammonia produces nitrogen and water according to the equation :
$4 {NH}_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (g)$
If the rate of formation of $N_{2}$ at a certain temperature is $3.0 mol \cdot L^{- 1} \cdot s^{- 1}$ what is the rate of disappearance of $O_{2}$ ?

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$2.0 mol \cdot L^{- 1} \cdot s^{- 1}$

$9.0 mol \cdot L^{- 1} \cdot s^{- 1}$

$3.0 mol \cdot L^{- 1} \cdot s^{- 1}$

$4.5 mol \cdot L^{- 1} \cdot s^{- 1}$

answer is C.

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Detailed Solution

The given reaction is,

$4 {NH}_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (g)$

Here, rate of disappearance of $O_{2}$

$- \frac{d [O_{2}]}{d t} = \frac{3}{2} \frac{d [N_{2}]}{d t} - \frac{d [O_{2}]}{d t} = \frac{3}{2} (3) - \frac{d [O_{2}]}{d t} = 4.5 mol . L^{- 1} s^{- 1}$

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