Ammonia and oxygen react at high temperature as 4NH3+5O2→4NO+6H2O. The rate of appearance of NO is 3.6×10−3 M.sec−1. Then the rate of formation of water in mole.L−1sec−1

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Ammonia and oxygen react at high temperature as $4 {NH}_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$ . The rate of appearance of NO is $3.6 \times 10^{- 3} M . {sec}^{- 1}$ . Then the rate of formation of water in $mole . L^{- 1} \sec^{- 1}$

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$5.4 \times 10^{- 3}$

$3.6 \times 10^{- 3}$

$2.7 \times 10^{- 3}$

$8.1 \times 10^{- 3}$

answer is C.

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Detailed Solution

As per the reaction $4 {NH}_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$

$- \frac{1}{4} . \frac{d [{NH}_{3}]}{dt} = - \frac{1}{5} . \frac{d [O_{2}]}{dt} = \frac{1}{4} . \frac{d [NO]}{dt} = + \frac{1}{6} . \frac{d [H_{2} O]}{dt}$