The half life period of first order decomposition of NH2NO2 is 2.1 hour at 15°C. The time required (in hour) for 5.2 g of NH2NO2 to decompose 99% will be ;
NH2NO2 (aq) → N2O(g) + H2O(l)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 13.95
For a first order reaction, the rate constant follows from the half-life, and the time for any fraction to react follows from the integrated rate equation:
\( \displaystyle k=\frac{0.693}{t_{1/2}} \)
\( \displaystyle t=\frac{2.303}{k}\log\frac{[R]_0}{[R]} \)
The time depends only on the fraction that reacts, not on the amount taken, so the mass 5.2 g is not needed.
When 99% decomposes, 1% is left, so [R]0/[R] = 100/1 = 100.
\( \displaystyle k=\frac{0.693}{2.1}=0.33\ \mathrm{h^{-1}} \)
\( \displaystyle t=\frac{2.303}{0.33}\log100=6.979\times2 \)
\( \displaystyle t=13.96\approx13.95\ \mathrm{h} \)
So 99% of \( \ce{NH2NO2} \) decomposes in about 13.95 hours.
Shortcut: For any first order reaction, t99% = (2.303 × 2/0.693) t1/2 ≈ 6.65 t1/2 = 6.65 × 2.1 ≈ 13.95 h.