$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3} 1 mole N_{2} and 4 moles H_{2} are taken$
in 15 L flask at 27°C. After complete conversion of N₂ into NH₃, 5 L of H₂O is added. Pressure set up in the flask is

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$\frac{1 \times 0 .0821 \times 300}{10} atm$

$\frac{2 \times 0 .0821 \times 300}{10} atm$

$\frac{1 \times 0 .0821 \times 300}{15} atm$

$\frac{3 \times 0 .0821 \times 300}{15} atm$

answer is D.

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Detailed Solution

$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}$

Taken 1 4 0

Reacts 1 3

H₂ Ieft = 1 mol NH₃ formed = 2 mol but NH₃(g) dissolves in water forming NH₄OH. Residual gas is only H₂ which exerts pressure.

Volume = 15 - 5 = 10 L (5 L is occupied by H₂O added)

$∴ p = \frac{n}{V} RT = \frac{1 \times 0 .0821 \times 300}{10} atm$

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