States of Matter - Result Question 80
81. At room temperature, the following reaction proceed nearly to completion. $2 NO+O _2 \longrightarrow 2 NO _2 \longrightarrow N _2 O _4$
The dimer, $N _2 O _4$, solidifies at $262 K$. A $250 mL$ flask and a $100 mL$ flask are separated by a stopcock. At $300 K$, the nitric oxide in the larger flask exerts a pressure of $1.053 \hspace {1mm} atm$ and the smaller one contains oxygen at $0.789 \hspace {1mm} atm.$
The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to $220 K$. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at $220 K$. (Assume the gases to behave ideally).
$(1992,4 M)$
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Answer:
Correct Answer: 81. $(0.221 { \hspace {1mm} atm})$
Solution:
First we calculate partial pressure of $NO$ and $O _2$ in the combined system when no reaction taken place.
$p V =\text { constant } \Rightarrow \quad p _1 V _1=p _2 V _2 $
$\Rightarrow \quad p _2(NO)=\dfrac{1.053 \times 250}{350}=0.752 \hspace {1mm} atm $
$p _2\left(O _2\right) =\dfrac{0.789 \times 100}{350}=0.225 \hspace {1mm} atm$
Now the reaction stoichiometry can be worked out using partial pressure because in a mixture.
$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad p _i \propto n _i $
$ \hspace{20mm} 2 NO \hspace{10mm} + \hspace{10mm} O _2 \hspace{10mm} \longrightarrow \hspace{10mm} 2 NO_2 \hspace{10mm} \longrightarrow \hspace{10mm} N_2O_4 $
$\text { Initial } \hspace{7mm} 0.752 {~\hspace {1mm} atm} \hspace{9mm} {0.225} {\hspace {1mm} atm} \hspace{29mm} 0 \hspace{35mm} 0 $
$\text { Final } \hspace{12mm} 0.302 \hspace{17mm} 0 \hspace{39mm} 0 \hspace{30mm} 0.225 {~\hspace {1mm} atm}$
Now, on cooling to $220 K, N _2 O _4$ will solidify and only unreacted $NO$ will be remaining in the flask.
$ \begin{aligned} & \because \quad p \propto T \\ & \therefore \quad \dfrac{p _1}{p _2}=\dfrac{T _1}{T _2} \\ & \Rightarrow \quad \dfrac{0.302}{p _2}=\dfrac{300}{220} \\ & \Rightarrow \quad p _2(NO)=0.221 \hspace {1mm} atm \end{aligned} $
BETA
