Q. In the following reaction, we start with $2 \, mol$ of $N_{2}$ and $5 \, mol$ of $H_{2}$ exerting a total pressure of $7 \, atm$ at a given temperature in a closed vessel. When $50\%$ of $N_{2}$ is converted into $NH_{3}$
$N_{2}+3H_{2} \rightarrow 2NH_{3}$
Partial pressure of $NH_{3}$ is :

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Solution:

Given that 50% of nitrogen gas converted, i.e., the partial pressure of nitrogen gas decreased by $1atm$ . $ & \left(\text{N}\right)_{2 \left(\text{g}\right)} & + & 3\left(\text{H}\right)_{2 \left(\text{g}\right)} & \rightarrow & 2\left(\text{NH}\right)_{3 \left(\text{g}\right)} \\ \text{Initial Partial pressures} & \text{2 atm} & & \text{5 atm} & & 0 \\ & & & & & \\ \text{During Reaction} & -\text{1 atm} & & -\text{3 atm} & & +\text{2 atm} \\ & \text{--------------} & & \text{--------------} & & \text{--------------} \\ \text{Final Partial pressures} & \text{1 atm} & & \text{2 atm} & & \text{2 atm}$
Partial pressure of $\text{NH}_{3} = 2 \, \text{atm}$