Question

A mixture of gaseous nitrogen and gaseous hydrogen attains equilibrium with gaseous Ammonia according to the reaction $\frac{1}{2}{N}_{2(g)}+\frac{3}{2}{H}_{2(g)}\rightleftharpoons N{H}_{3(\epsilon )}$It is found that at equilibrium when temperature is 673 K, atmospheric pressure is 100, moles of gaseous nitrogen and gaseous hydrogen are in ratio 1:3 and mole percent of gaseous ammonia is 24. The equilibrium constant Kp for the reaction at the above condition is-

• (A) 0.0128 atm^-1
• (B) 0.1876 atm^-1
• (C) 0.0186 atm^-1
• (D) 0.558 atm^-1

Answer 1

The Correct Option is A

Explanation:
$\frac{1}{2}{\mathrm{N}}_{2(\mathrm{g})}+\frac{3}{2}{\mathrm{H}}_{2(\mathrm{g})}\rightleftharpoons {\mathrm{NH}}_{3(\mathrm{g})}$Partial pressure of ${\mathrm{NH}}_3=\frac{24}{100}\times 100$$=24.0\mathrm{atm}$Pressure of ${\mathrm{N}}_2+{\mathrm{H}}_2=100-24=76\mathrm{atm}$Partial pressure of${\mathrm{H}}_2=\frac{3}{4}\times 76=57\mathrm{atm}$Partial pressure of ${\mathrm{N}}_2$$=\frac{1}{4}\times 76=19\mathrm{atm}$${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{{\mathrm{NH}}_1}}{{\mathrm{p}}_{{\mathrm{N}}_2}^{12\mathrm{p}}+{\mathrm{p}}_{\mathrm{H}}^{3/2}}$$=\frac{24}{(19{)}^{12}\times (57{)}^{3/2}}$$=\frac{24}{4.358\times (7.549{)}^3}$$=\frac{24}{4.358\times 430}=0.0128{\mathrm{atm}}^{-1}$