Question

Consider the reaction,

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

The equilibrium constant of the above reaction is K_p. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that \(\;{p_{N{H_3}}} < < {p_{total}}\) at equilibrium).

1. \(\frac{{{3^{3/2}}K_p^{1/2}{P^2}}}{4}\)
2. \(\frac{{{3^{3/2}}K_p^{1/2}{P^2}}}{{16}}\)
3. \(\frac{{K_p^{1/2}{P^2}}}{{16}}\)
4. \(\frac{{K_p^{1/2}{P^2}}}{4}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{{3^{3/2}}K_p^{1/2}{P^2}}}{{16}}\)

Concept:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

\({\rm{At\;}}:{\rm{\;}}{p_{{{\rm{N}}_2}}} = P,\;{P_{{{\rm{H}}_2}}} = 3P,\;{p_{{\rm{N}}{{\rm{H}}_3}}} = 2P\)

\( \Rightarrow {\rm{\;}}{p_{\left( {{\rm{total}}} \right){\rm{\;}})}} = {p_{{{\rm{N}}_2}}} + {p_{{{\rm{H}}_2}}} + {p_{{\rm{N}}{{\rm{M}}_3}}} \simeq {p_{{{\rm{N}}_2}}} + {p_{{{\rm{H}}_2}}}\;\left[\because {{P_{\left( {{\rm{total}}} \right){\rm{\;}}}} > > {p_{{\rm{N}}{{\rm{H}}_3}}}} \right]\)

= p + 3p = 4p

Now, \({K_p} = \frac{{p_{N{H_3}}^2}}{{{p_{{N_2}}} \times p_{{H_2}}^3}} = \frac{{p_{{N_{N{H_3}}}}^2}}{{p \times {{(3p)}^3}}}\)

\(= \frac{{p_{N{H_3}}^2}}{{27 \times {p^4}}} = \frac{{p{{_{N{\rm{H}}}^2}_3}}}{{27 \times {{\left( {\frac{P}{4}} \right)}^4}}}{\rm{\;}}\left[\because {P = 4p} \right]\)

\({K_\rho } = \frac{{p_{N{H_3}}^2 \times {4^4}}}{{{3^2} \times 3 \times {P^4}}}\)

\(\Rightarrow p_{N{H_3}}^2 = \frac{{{3^2} \times 3 \times {P^4} \times {K_p}}}{{{4^4}}}\)

\( \Rightarrow {p_{N{H_3}}} = \frac{{3 \times {3^{1/2}} \times {P^2} \times K_P^{1/2}}}{{{4^2}}}\)

\(= \frac{{{3^{3/2}} \times {P^2} \times K_P^{1/2}}}{{16}}\)