In Haber's process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of the gaseous mixture under the aforesaid condition in the end?

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20 litres ${NH}_{3}$ , $25 litres$ $N_{2}$ , 20 litres $H_{2}$

10 litres ${NH}_{3}$ , 25 litres $N_{2}$ , 15 litres $H_{2}$

20 litres ${NH}_{3}$ , 10 litres $N_{2}$ , 30 litres $H_{2}$

20 litres ${NH}_{3}$ , 25 litres $N_{2}$ , 15 litres $H_{2}$

answer is B.

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Detailed Solution

$N_{2} + 3 H_{2} \overset{50 %}{⟶} 2 {NH}_{3}$

$30 30 -$

$25 15 10$

Here,

$∴$ $Vol. of N_{2} at eq = 30 - 5 = 25 lit$

$∴ Vol. of H_{2} at eq = 30 - 3 \times 5 = 15 lit$

$∴ Volume of {NH}_{3} at equilibrium = 2 \times 5 = 10 lit$

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