The half life of first order decomposition of nitramide is $2.1$ hours at $15^{o} C$ .
$N H_{2} N O_{2} (a q .) ⟶ N_{2} O (g) + H_{2} O (l)$
If $6.2$ $g$ of $N H_{2} N O_{2}$ is allowed to decompose calculate (i) time taken for $N H_{2} N O_{2}$ to decompose $99$ % and (ii) the volume of dry $N_{2} O$ produced at this point, measured at STP.

Open in App

Solution

(i) $l = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{2.1} = 0.33 {h r}^{- 1}$
Applying kinetic equation of first order reaction,
$k = \frac{2.303}{t} {log}_{10} \frac{a}{(a - x)}$
or $t = \frac{2.303}{0.33} {log}_{10} \frac{100}{(100 - 99)} = 13.96$ hrs
(ii) Number of moles of $N H_{2} N O_{2}$ decomposed
$= 0.99 \times \frac{6.2}{62} = 0.099$
Number of moles of $N_{2} O$ formed $= 0.099$
Volume of $N_{2} O$ at STP $= 0.099 \times 22400$ $m L$ $= 2217.6$ $m L$

Suggest Corrections

Similar questions

2.1 hr

15^{\circ} C

N H_{2} N O_{2} (s) \to N_{2} O (g) + H_{2} O (l)

6.2 g

N H_{2} N O_{2}

N H_{2} N O_{2}

99 %

N_{2} O

15^{\circ} C .

N H_{2} N O_{2} (a q .) \to N_{2} O (g) + H_{2} O (l)

N H_{2} N O_{2}

N H_{2} N O_{2}

N H_{2} N O_{2}

N_{2} O

Join BYJU'S Learning Program