Ammonia and oxygen reacts at higher temperature as
$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)$
In an experiment, the concentration of NO increases by $1.08 * 10^{- 2} m o l {l i t r e}^{- 1}$ in 3 seconds.
Calculate,
(1) rate of reaction
(2) rate of disappearance of ammonia
(3) rate of formation of water

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Solution

$4 {N H}_{3} (g) + 5 O_{2} (g) ⟶ 4 N O (g) + 6 H_{2} O (g)$
rate $= \frac{1}{4} \frac{[△ {N H}_{3}]}{△ t} = \frac{1}{4} \frac{△ [N O]}{△ t} = \frac{1}{6} \frac{△ [H_{2} O]}{△ E}$
rate $= \frac{1}{4} \times \frac{1.08 \times 10^{- 2}}{3} = 0.09 \times 10^{- 2} = 9 \times 10^{- 4}$
rate of dissappearance of ${N H}_{3}$ = rate of formation of $N O = \frac{1.08 \times 10^{- 2}}{3}$
rate of formation of $H_{2} O \Rightarrow \frac{1}{6} \frac{[△ H_{2} O]}{△ t} = \frac{1}{4} \frac{△ [N O]}{△ t}$
$\frac{△ [H_{2} O]}{△ t} = \frac{6}{4} \frac{△ [N O]}{△ t} = \frac{6}{4} \times \frac{1.08 \times 10^{- 2}}{3} = 0.54 \times 10^{- 2}$

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