# Question a83c1

Aug 16, 2016

The reaction was initiated by volimetric ratio of Hydrigen and Nitrogen as $3 : 1$ and at equilibrium the ratio remained more or less same 36%:13%~~3:1  (Within experimental error.)
So the volume of $N {H}_{3}$ in the equilibrium mixture may be taken as  (100-36-13)%=51%

If the pressure of the reaction mixture at equilibrium be P atm then the partial pressures of the constituents of the reaction mixture will be as follows.

Partial presssure of Hydrogen
p_(H_2)=36%ofP=0.36P" atm"

Partial presssure of Nitrogen
p_(N_2)=13%ofP=0.13P" atm"

Partial presssure of Ammonia
p_(NH_3)=51%ofP=0.51P" atm"#

The equation of the gaseous reaction:

$3 {H}_{2} \left(g\right) + {N}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 N {H}_{3} \left(g\right)$

Now the equlibrium constant of the gaseous reaction in respect of pressure is given by

${K}_{p} = {\left({p}_{N {H}_{3}}\right)}^{2} / \left({p}_{{N}_{2}} \times {\left({p}_{{H}_{2}}\right)}^{3}\right)$

$\implies {K}_{p} = {\left(0.51 P\right)}^{2} / \left(0.13 P \times {\left(0.36 P\right)}^{3}\right) {\text{ atm}}^{-} 2$

$\approx 42.88 {P}^{-} 2 {\text{ atm}}^{-} 2$

If the value of P is known then ${K}_{p}$ can be calculated out.