The correct option is A 4:5 N_2+ 3H_2⇌ 2NH_3 I #160; 0.2 #160; 0.6 #160; 0C #160; x #160; 3x #160; #160;+2xE #160; (0.2 x)(0.6 3x) ...

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Characteristics of Equilibrium Constant

Question

$40$ % mixture of $0.2 m o l e$ of $N_{2}$ and $0.6 m o l e$ of $H_{2}$ react to give ${N H}_{3}$ according to the equation $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ , at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:

4 : 5

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5 : 4

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7 : 10

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8 : 5

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Solution

The correct option is A $4 : 5$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
I 0.2 0.6 0
C -x -3x +2x
E (0.2-x)(0.6-3x) +2x
According to question $\div \frac{40}{100} [0.2 - x + 0.6 - 3 x] = 2 x$
$0.4 [0.8 - 4 x] = 2 x$
$0.32 - 1.6 x = 2 x$
$0.32 = 3.6 x$
$x = 0.0888$
Final volume $= (0.2 - x) + (0.6 - 3 x) + 2 x$
$= 0.2 - 0.0888 + 0.6 - 3 \times 0.0888 + 2 \times 0.0888$
$= 0.624$
Initial volume $= 0.2 + 0.6 = 0.8$
Ratio $= \frac{0.624}{0.8} ≃ 0.78 ≃ \frac{4}{5} \to (A)$

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40

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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