N2O5→2NO2+21O2 At t=0 At t=50 min5050−p102p102p1 Total pressure at 50 minutes $\begin{aligned}
&=50-\mathrm{p}_1+2 \mathrm{p}_1+\frac{\mathrm{p}_1}{2}=87.5 \
&50+\frac{3 \mathrm{p}_1}{2}=87.5 \
&\frac{3 \mathrm{p}_1}{2}=37.5 \
&\therefore \quad \mathrm{p}_1=\frac{37.5 \times 2}{3}=25
\end{aligned}At\mathrm{t}=100 \mathrm{min} \quad 50-\mathrm{p}_2 \quad 2 \mathrm{p}_2 \quad \frac{\mathrm{p}_2}{2}50minutesishalflifeperiodFor100minutesi.e.for2halflives50-p_2=12.5\therefore \mathrm{p}_2=37.5 \mathrm{mm} \text { of } \mathrm{Hg}Totalpressureat100minutes=50-\mathrm{p}_2+2 \mathrm{p}_2+\frac{\mathrm{p}_2}{2}=50+\frac{3 \mathrm{p}_2}{2}=50+\frac{3}{2} \times 37.5=50+56.25=106.25 \mathrm{~mm}of\mathrm{Hg}$
N_2 O_5 decomposes to NO_2 and O_2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 ~mm ~Hg to…
JEE Main 2018 — Chemistry Physical Chemistry
2018mcqhard
N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 minutes at constant temperature will be ______.
Official previous-year question
Held on 15 Apr 2018 · Verified 6 Jul 2026.
Options
- A
136.25 mm Hg
- B
106.25 mm Hg
- C
175.0 mm Hg
- D
116.25 mm Hg
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