Ammonia reacts with oxygen giving nitrogen and water. If the rate of formation of $N_{2}$ is $0.70 mol L = s^{- 1}$ the rate at which $O_{2}$ is consumed is

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$0.70 mol L^{- 1} s^{- 1}$

$1.05 mol L^{- 1} s^{- 1}$

$2.10 mol L - 1 s^{- 1}$

$0.35 mol L^{- 1} s^{- 1}$

answer is A.

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Detailed Solution

The reaction is $4 {NH}_{3} + 3 O_{2} \to 2 N_{2} + 6 H_{2} O$ $\frac{1}{2} \frac{d [N_{2}]}{d t} = - \frac{1}{3} \frac{d [O_{2}]}{d t}$

Hence, $- \frac{d [O_{2}]}{d t} = \frac{3}{2} \frac{d [N_{2}]}{d t} = (\frac{3}{2}) (0.7 mol L^{- 1} s^{- 1}) = 1.05 mol L^{- 1} s^{- 1}$

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