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Characteristics of Equilibrium Constant

30 litre H2...

Question

$30$ litre $H_{2} (g)$ and $30$ litre $N_{2} (g)$ were taken for the reaction in Haber's process which yields only
$50$ % of the expected ammonia due to the reversibility of the reaction. What will be the composition of the reaction mixture under the given condition?

A

{N H}_{3} : 20 L

;

N_{2} : 20 L

;

H_{2} : 20 L

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B

{N H}_{3} : 10 L

;

N_{2} : 25 L

;

H_{2} : 15 L

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C

{N H}_{3} : 20 L

;

N_{2} : 10 L

;

H_{2} : 30 L

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D

{N H}_{3} : 20 L

;

N_{2} : 25 L

;

H_{2} : 15 L

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Solution

The correct option is A ${N H}_{3} : 10 L$ ; $N_{2} : 25 L$ ; $H_{2} : 15 L$
$E x p a l n a t i o n :$

From given
Equation: $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
At the time $t = 0$ $30$ $30$ $0$
At the time $t = t$ $(30 - x)$ $(3 - 3 x)$ $2 x$

At $100 %$ yield, the reaction will consume $3 l i t e r s$ of $H_{2}$ gas and $1 l i t e r$ of $N_{2}$ gas and produce $2 l i t e r s$ of $N H_{3}$ .

But given In reaction, the yield produce is 50% of the expected product. 15 liters of hydrogen gas will be used since hydrogen gas is the limiting agent.

$∴$ Volume of $N_{2}$ left $= 30 - 5 = 25 l i t$

$∴$ Volume of $H_{2}$ left $= 30 - 3 \times 5 = 15 l i t$

$∴$ Volume of $N H_{3}$ at equilibrium $= 2 \times 5 = 10 l i t$

Hence the correct answer is option is $B$ .

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