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Ideal Gas Equation

11 moles N 2 ...

Question

$11 m o l e s$ $N_{2}$ and $12 m o l e s$ of $H_{2}$ mixture reacted in $20 L$ vessel at $800 K$ . After equilibrium was reached, $6 m o l e$ of $H_{2}$ was present. $3.58 L$ of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to $300 K$ . What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all $N H_{3}$ dissolved in water (ii) no change in volume of liquid (iii) no reaction of $N_{2}$ and $H_{2}$ at $300 K$

(Given $R = 0.821 L a t m m o l^{- 1} K^{- 1}$ )

A

18.47 a t m

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B

60 a t m

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C

22.5 a t m

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D

45 a t m

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Solution

The correct option is C $22.5 a t m$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$I n i t i a l m o l e s 11120$
$a t e q u i l i b r i u m 964$
$Moles of N_{2} a n d H_{2} present at equilibrium = 9 + 6 = 15$
After addition of water
$N H_{3} (g) + H_{2} O (l) \to N H_{4} O H (l)$
$Volume of vessel available for gaseous mixture of N_{2} a n d H_{2} = 20 - 3.58 \Rightarrow 16.42 L$
$Pressure exerted by gaseous mixture at 300 K By using the relation, P V = n R T \Rightarrow P = \frac{n R T}{V} = \frac{15 \times 0.0821 \times 300}{16.42} \Rightarrow 22.5 a t m$

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