A mixture of N₂ and H₂ in the molar ratio 1:3 attains equilibrium when 50% of mixture has reacted. If P is the  total pressure of the mixture, the partial pressure of NH₃ formed is P/y. The value of y is: 

$$\begin{gathered} {\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3 \\ \mathrm{Initial} \mathrm{mole} 1 3 0 \\ \mathrm{At} \mathrm{eq}. 1 - \mathrm{x} 3 - 3\mathrm{x} 2\mathrm{x} \end{gathered}$$

Out of 4 mole of mixture, 2 mole have reacted and hence 2 mole are left. 

$\therefore 1-\mathrm{x}+3-3\mathrm{x}=2$

$\text{ or }4\mathrm{x}=2\text{ or }\mathrm{x}=0.5$

$\therefore \text{ Total mole at equilibrium }=1-x+3-3x+2x$

= 4-2x = 4-2 × 0.5 = 3

No. of mole of NH₃ at equilibrium = 2x = 2 × 0.5 = 1

$\therefore {\mathrm{p}}_{{\mathrm{NH}}_3}=\frac{1}{3}\times \mathrm{P}=\frac{\mathrm{P}}{3}\therefore \mathrm{y}=3$