Question

The reaction, $2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$, follows first order kinetics. The pressure of a vessel containing only $N_2{O}_5$ was found to increase from $50mmHg$ to $87.5mmHg$ in $30$ min. The pressure exerted by gases after $60$ min. will be:

[Assume temperature remains constant]

• A. $106.25mmHg$
• B. $150mmHg$
• C. $116.25mmHg$
• D. $125mmHg$

Answer 1

Answer: (A)

$106.25mmHg$

	$N_2{O}_5\to$	$N{O}_2+$	$O_2$
Initial pressure (mm Hg)	50	0	0
Pressure after 30 min (mm Hg)	50-2P	4P	P
Pressure after 60 min (mm Hg)	50-2P'	4P'	P'

After 30 min, the total pressure will be $50-2P+4P+P=87.5mmHg$.
$50+3P=87.5$
$3P=37.5$
$P=12.5$.
Thus after 30 min, the pressure of $N_2{O}_5$ decreases from 50 mm Hg to 25 mm Hg.
This corresponds to half life period.
In 60 min, the pressure of $N_2{O}_5$ will decrease to one fourth (two half life periods) i., 12.5 mm Hg.
Hence, $50-2P^{\prime }=12.5$
$2P^{\prime }=37.5$
$P^{\prime }=18.75$
Hence, the total pressure after 30 min is $50+3P^{\prime }=50+3\left(18.75\right)=106.25mmHg$.