Question

A $250\, ml$ flask containing $NO ( g )$ at $0.46\, atm$ is connected to a $100\, ml$ flask containing oxygen gas at $0.86$ atm by means of a stop cock at $350 \,K$. the gases are mixed by opening the stop cock where the following equilibrium is established.
$\underset{g}{2 NO} + O _{2} \longrightarrow \underset{g}{2 NO _{2}} \rightleftharpoons \underset{g}{N _{2} O _{4}}$
The first reaction is complete while the second is at equilibrium. If the total pressure is $0.37 \,atm , K _{ p }$ is

• A. $0.645 \,atm ^{-1}$
• B. $1.290 \,atm ^{-1}$
• C. $20 \,atm ^{-1}$
• D. $5.46\, atm ^{-1}$

Answer 1

Answer: (A)

Partial pressure of gases before reaction
$NO =0.46 \times \frac{250}{350}=0.3285 \,atm $
$O _{2}=0.86 \times \frac{100}{350}=0.2457\, atm$
Here, $NO$ is limiting reagent:
Partial pressure after reaction
Ist step: $NO =0, O _{2}=0.0814, $
$NO _{2}=0.3285\, atm$
Now, let $P$ be the decrease in partial pressure of $NO _{2}$ due to dimerization
$0.37=0.0814+0.3285- P + P / 2$
$P =0.0798$
$K _{ P }=\frac{ P _{ N _{2} O _{4}}}{ P _{ NO _{2}}^{2}}=\left(\frac{ P }{2}\right) \frac{1}{(0.3285- P )^{2}}$
$=\frac{0.0399}{(0.2487)^{2}}=0.645 \,atm ^{-1}$