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In a $25 \mathrm{~L}$ container, 15 moles of $\mathrm{N}_2$ reacted with 12 moles of $\mathrm{H}_2$ at 800 K. It was observed that 6 moles of $\mathrm{H}_2$ was present after equilibrium. The gaseous mixture was suddenly cooled to 300 K after addition of $4.35 \mathrm{~L} of \mathrm{H}_2 \mathrm{O} (l)$ in the mixture. What will be the final pressure of the mixture keeping in mind the following points.
- All $\mathrm{NH}_3$ dissolved in water.
- There is no change in volume of the liquid.
- No reaction took place between $\mathrm{N}_2$ and $\mathrm{H}_2$ at 300 K

Also, neglect the V.P. of the liquid solution.

Option: 1
$18.47 \text { atm }$

Option: 2
$60 \text { atm }$

Option: 3
$45 \text { atm }$

Option: 4
$22.6 \text { atm }$

Answers (1)

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Solution : $\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$

Initial Moles : 15 12 0

At Equilibrium : 13 6 4

Moles of $\mathrm{N}_2$ and $\mathrm{H}_2$ present at equilibrium=19

After addition of $\mathrm{H}_2 \mathrm{O}$

$\mathrm{\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}}$

Volume of vessel available for gaseous mixture of $\mathrm{N}_2$ and $\mathrm{H}_2$

$\mathrm{\begin{aligned} & =25-4.35 \\ & =20.65 \mathrm{~L} \end{aligned}}$

$\mathrm{\text { Pressure at } 300 \mathrm{~K}=\frac{19 \times 0.0821 \times 300}{20.65}=22.6 \mathrm{~atm}}$

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Pankaj

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