Question

The reaction
$2N_2{O}_5(g)\to 4N{O}_2(g)+O_2(g)$
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 $30min$. The pressure exerted by the gases after $60min$. will be (Assume temperature remains constant) :

• A. 106.25 mm Hg
• B. 116.25 mm Hg
• C. 125 mm Hg
• D. 150 mm Hg

Answer 1

Answer: (A)

$2N_2{O}_5(g)\to 4N{O}_2(g)+O_2(g)$

$50+3x=87.5$

$\Rightarrow 3x=37.5$

$\Rightarrow x=12.5$

$\Rightarrow 2x=25$

(Half $r$ acetate) $\Rightarrow T_{1/2}=30min$

Hence after $60min,P_{N_2{O}_5}$ remaining $=\frac{50}{4}=12.5$ torr

Hence decrease in $P_{N_2{O}_s}=50-12.5=37.5$ torr

$P_{N{O}_2}=2\times 37.5=75$ torr;

$P_{O_2}=\frac{37.5}{2}=18.75$ torr;

$P_{\text{total }}=12.5+75+18.75=106.25$ torr