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Nitrogen

The mixed anh...

Question

The mixed anhydride of nitrogen is:

N_{2} O_{2} (2 N O)

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N_{2} O_{4} (2 {N O}_{2})

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N_{2} O_{5}

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N_{2} O_{2}

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Solution

The correct option is B $N_{2} O_{2} (2 N O)$
Mixed anhydride or double acid anhydride is the compound that reacts with water to form a mixture of two acidic anhydride.
Two oxides of nitrogen are acid anhydrides; that is, they react with water to form two nitrogen-containing oxyacids.

$2 N O_{2} + H_{2} O ⟶ H N O_{3} + H N O_{2}$
Hence, the correct option is B.

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Similar questions

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2},

\frac{- d}{d t} [N_{2} O_{5}] = k_{1} [N_{2} O_{5}], \frac{d}{d t} [N O_{2}] = k_{2} [N_{2} O_{5}] a n d

\frac{d}{d t} [O_{2}] = k_{3} [N_{2} O_{3}],

N_{2} O_{5} ⟶ 2 N O_{2} + 1 / 2 O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

k_{1}, k_{2}

k_{3}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2} .

G i v e n : - \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}]

A n d \frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}]

The relation between $K_{1}, K_{2}$ and $K_{3}$ is

N_{2} O_{3}, N_{2} O_{4}

N_{2} O_{5}

N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{1}{2} O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

- \frac{d [{N O}_{2}]}{d t} = K_{2} [{N O}_{2}]

- \frac{d [O_{2}]}{d t} = K_{3} [O_{2}]

K_{1}, K_{2}, K_{3}

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