Question

The half time of first order decomposition of nitramide is $2.1\text{hr}$ at ${15}^{\circ }\text{C}$.
$N{H}_{2}N{O}_2\left(s\right)\to N_{2}O\left(g\right)+H_{2}O\left(l\right)$
If $6.2\text{g}$ of $N{H}_{2}N{O}_2$ is allowed to decompose, then time taken for $N{H}_{2}N{O}_2$ to decompose $99\%$ and volume of dry $N_{2}O$ produced at this point measured respectively at STP will be

• A. $13.95\text{hr},22.4\text{L}$
• B. $13.95\text{hr},2.217\text{L}$
• C. $2.1\text{hr},22.4\text{L}$
• D. $2.1\text{hr},2.217\text{L}$

Answer 1

The correct option is B $13.95\text{hr},2.217\text{L}$

For a first-order reaction, $k=\frac{2.303}{t}log\frac{[A{]}_0}{[A{]}_t}$
$t_{1/2}=\frac{0.693}{k}$
$\Rightarrow k=\frac{0.693}{t_{1/2}}$
$\Rightarrow k=\frac{0.693}{2.1}$
$\text{Initial moles of nitramide}=\frac{6.2}{62}=0.1$
$\therefore t=\frac{2.303\times 2.1}{0.693}log\frac{0.1}{0.001}=13.95\text{hr}$
Since, the decomposition is 99%, so 99% of the initial moles of $N{H}_{2}N{O}_2$ would be converted to $N_{2}O$.
$\text{Moles of}{N}_{2}O=\frac{0.1\times 99}{100}$
$\therefore \text{Volume of}{N}_{2}OatSTP=\frac{0.1\times 99\times 22.4}{100}=2.217\text{litre}.$