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 $1.053 atm$ and the smaller one contains oxygen at $0.789 atm$.
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. $(2.20 atm)$
Solution:
- First we calculate partial pressure of $NO$ and $O _2$ in the combined system when no reaction taken place.
$$ \begin{aligned} p V & =\text { constant } \Rightarrow \quad p _1 V _1=p _2 V _2 \ \Rightarrow \quad p _2(NO) & =\frac{1.053 \times 250}{350}=0.752 atm \ p _2\left(O _2\right) & =\frac{0.789 \times 100}{350}=0.225 atm \end{aligned} $$
Now the reaction stoichiometry can be worked out using partial pressure because in a mixture.
$$ \begin{aligned} & p _i \propto n _i \ & \begin{array}{lrllc} & 2 NO & +O _2 & \longrightarrow & 2 NO _2 \ \text { Initial } & 0.752 atm & 0.225 atm \end{array} \longrightarrow \underset{0}{N _2 O _4} \ & \begin{array}{lllll} \text { Final } & 0.302 & 0 & 0 & 0.225 atm \end{array} \end{aligned} $$
Now, on cooling to $220 K, N _2 O _4$ will solidify and only unreacted $NO$ will be remaining in the flask.
$$ \begin{aligned} & \because \quad p \propto T \ & \therefore \quad \frac{p _1}{p _2}=\frac{T _1}{T _2} \ & \Rightarrow \quad \frac{0.302}{p _2}=\frac{300}{220} \ & \Rightarrow \quad p _2(NO)=0.221 atm \end{aligned} $$
