N _ 2 + 3 H _ 2 ⟶ 2N H _ 3 1 mole N _ 2 and 4 moles H _ 2 are taken in 15L flask at 27 ^ o C . After complete conversion of N _ 2 into N H _ 3 , 5L of H _ 2 O is added. Pressure set up in the flask is (assume water

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Correct answer: D. (1 × 0.0821 × 300)/(10)atm

cases & N _ 2 & + & 3 H _ 2 & ⟶ & 2 NH _ 3 Initial mole & 1 & & 4 & & 0 Reacted mole & 1 & & 3 & Left mole & 0 & & 4-3=1 & & 2 cases H _ 2 left = 1 mol N H _ 3 formed = 2 mol But N H _ 3 (g) dissolves in water forming N H _ 4 OH . Residual gas is only H _ 2 which exerts pressure. Volume = 15 - 5 = 10L (5L occupied by H _ 2 O added) ∴ p = (n)/(V) RT = (1 × 0.0821 × 300)/(10) atm

N _ 2 + 3 H _ 2 ⟶ 2N H _ 3 1 mole N _ 2 and 4 moles H _ 2 are taken...

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