Question

The reaction
$2 N _{2} O _{5}(g) \rightarrow 4 NO _{2}(g)+ O _{2}(g)$
follows first order kinetics. The pressure of a vessel containing only $N_2O_5$ was found to increase from $50\, mm\, Hg$ to $87.5\, mm\, Hg$ in $30\, min$. The pressure exerted by the gases after $60\, min$. 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)

$2 N _{2} O _{5}( g ) \rightarrow 4 NO _{2}( g )+ O _{2}( g )$

$50+3 x=87.5$

$\Rightarrow 3 x=37.5$

$\Rightarrow x=12.5$

$\Rightarrow 2 x=25$

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

Hence after $60\, min,\, P_{ N _{2} O _{5}}$ remaining $=\frac{50}{4}=12.5$ torr

Hence decrease in $P_{ N _{2} O _{ s }}=50-12.5=37.5$ torr

$P_{ NO _{2}}=2 \times 37.5=75$ torr;

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

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