$37.8$ g $ ext{N}_2 ext{O}_5$ was taken in a $1$ L reaction vessel and allowed to undergo the following reaction at $500$ K: $2 ext{N}_2 ext{O}_{5(g)} ightleftharpoons2 ext{N}_2 ext{O}_{4(g)}+ ext{O}_{2(g)}$. The total pressure at equilibrium was found to be $18.65$ bar. Then, $K_p=$ ______ $ imes10^{-2}$ [nearest integer] Assume $ ext{N}_2 ext{O}_5$ to behave ideally under these conditions. Given: $R=0.082\ ext{bar L mol}^{-1} ext{K}^{-1}$

Question

$37.8$ g $	ext{N}_2	ext{O}_5$ was taken in a $1$ L reaction vessel and allowed to undergo the following reaction at $500$ K: $2	ext{N}_2	ext{O}_{5(g)}ightleftharpoons2	ext{N}_2	ext{O}_{4(g)}+	ext{O}_{2(g)}$. The total pressure at equilibrium was found to be $18.65$ bar. Then, $K_p=$ ______ $	imes10^{-2}$ [nearest integer] Assume $	ext{N}_2	ext{O}_5$ to behave ideally under these conditions. Given: $R=0.082\ 	ext{bar L mol}^{-1}	ext{K}^{-1}$

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $n(	ext{N}_2	ext{O}_5)=\frac{37.8}{108}=0.35$ mol; initial $P=nRT/V=0.35	imes0.082	imes500=14.35$ bar. If $2x$ bar of $	ext{N}_2	ext{O}_5$ reacts: $P_{N_2O_5}=14.35-2x$, $P_{N_2O_4}=2x$, $P_{O_2}=x$, total $=14.35+x=18.65\Rightarrow x=4.3$. $P_{N_2O_5}=5.75$, $P_{N_2O_4}=8.6$, $P_{O_2}=4.3$. $K_p=\frac{8.6^2	imes4.3}{5.75^2}=9.62=962	imes10^{-2}$. Answer: 962.

Explanation

Reviewed by Dr. Abubacker

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