Question

If equilibrium constant for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 298 K is 2.54, the value of equilibrium constant for the reaction \(\frac{1}{2}\)N₂ + \(\frac{3}{2}\)H₂ ⇌ NH₃ will be ………

Answer 1

 1.59

Since the reaction is halved \(K_P^"=\sqrt{K'_P}=\sqrt{(2.54)}=1.59\)