In Haber process 3 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded…

In Haber process 3 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 gaseous mixture under the above said condition in the end?
  1. 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
  2. 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
  3. 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
  4. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

Solution

$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$ $1 \mathrm{~mol} \quad 3 \mathrm{~mol} 2 \mathrm{~mol}$ 2 vol $3 \mathrm{~mol} 2 \mathrm{vol}$ $(\operatorname{mol} \%=\operatorname{vol} \%)$ $\begin{aligned} & \begin{array}{l} 10 \mathrm{vol} \quad 30 \mathrm{vol} \quad 20 \mathrm{vol} \\ \text { (expected = product) } \end{array} \\ & (30-x)(30-x) 2 x \\ & \text { Given that } 2 x=20 \times \frac{50}{100}=10 \\ & \therefore \quad x=5 \end{aligned}$ Composition of gaseous mixture $\begin{aligned} \mathrm{N}_2 & =(30-x)=25 \text { litre } \\ \mathrm{H}_2 & =(30-3 x)=15 \text { litre } \\ \mathrm{NH}_3 & =2 x=10 \text { litre } \end{aligned}$

Asked in: NEET 2003

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