States of Matter 1 Question 80
81. At room temperature, the following reaction proceed nearly to completion. $2 \mathrm{NO}+\mathrm{O}{2} \longrightarrow 2 \mathrm{NO}{2} \longrightarrow \mathrm{N}{2} \mathrm{O}{4}$
The dimer, $\mathrm{N}{2} \mathrm{O}{4}$, solidifies at $262 \mathrm{~K}$. A $250 \mathrm{~mL}$ flask and a $100 \mathrm{~mL}$ flask are separated by a stopcock. At $300 \mathrm{~K}$, the nitric oxide in the larger flask exerts a pressure of $1.053 \mathrm{~atm}$ and the smaller one contains oxygen at $0.789 \mathrm{~atm}$.
The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to $220 \mathrm{~K}$. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at $220 \mathrm{~K}$. (Assume the gases to behave ideally).
$(1992,4 \mathrm{M})$
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Answer:
Correct Answer: 81. $(2.20 \mathrm{~atm})$
Solution:
- First we calculate partial pressure of $\mathrm{NO}$ and $\mathrm{O}_{2}$ in the combined system when no reaction taken place.
$$ \begin{aligned} p V & =\text { constant } \Rightarrow \quad p_{1} V_{1}=p_{2} V_{2} \ \Rightarrow \quad p_{2}(\mathrm{NO}) & =\frac{1.053 \times 250}{350}=0.752 \mathrm{~atm} \ p_{2}\left(\mathrm{O}_{2}\right) & =\frac{0.789 \times 100}{350}=0.225 \mathrm{~atm} \end{aligned} $$
Now the reaction stoichiometry can be worked out using partial pressure because in a mixture.
$$ \begin{aligned} & p_{i} \propto n_{i} \ & \begin{array}{lrllc} & 2 \mathrm{NO} & +\mathrm{O}{2} & \longrightarrow & 2 \mathrm{NO}{2} \ \text { Initial } & 0.752 \mathrm{~atm} & 0.225 \mathrm{~atm} \end{array} \longrightarrow \underset{0}{\mathrm{~N}{2} \mathrm{O}{4}} \ & \begin{array}{lllll} \text { Final } & 0.302 & 0 & 0 & 0.225 \mathrm{~atm} \end{array} \end{aligned} $$
Now, on cooling to $220 \mathrm{~K}, \mathrm{~N}{2} \mathrm{O}{4}$ will solidify and only unreacted $\mathrm{NO}$ will be remaining in the flask.
$$ \begin{aligned} & \because \quad p \propto T \ & \therefore \quad \frac{p_{1}}{p_{2}}=\frac{T_{1}}{T_{2}} \ & \Rightarrow \quad \frac{0.302}{p_{2}}=\frac{300}{220} \ & \Rightarrow \quad p_{2}(\mathrm{NO})=0.221 \mathrm{~atm} \end{aligned} $$
