Question #a83c1

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Question #a83c1

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Answer 1 · 2016-08-16T05:44:15

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 $NH_3$ 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:

$3H_2(g)+N_2(g)rightleftharpoons2NH_3(g)$

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

$K_p=(p_(NH_3))^2/(p_(N_2)xx(p_(H_2))^3)$

$=>K_p=(0.51P)^2/(0.13Pxx(0.36P)^3)" atm"^-2$

$~~42.88P^-2" atm"^-2$

If the value of P is known then $K_p$ can be calculated out.

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