_____N2 + 3 H2 <=> 2 NH3______

____1-x____3-3x____2x___

moli totali = 1-x + 3-3x +2x = 4-2x

fraz_mol (NH3) = 2x/(4-2x)

2x/(4-2x-) = 0.676
2x = 0.676*(2-2x)
x = 0.806683

fraz_mol(N2) = (1-x)/(4-2x) = 0.081 (valore trovato con x = 0.806683)

fraz_mol(H2) = (3-3x)/(4-2x) = 0.243 ( idem)

Presssione parziale = fraz_mol *Pt

P(NH3) = 0.676 * 30 = 20.28 atm

P(N2) = 0.081 *30 = 2.43 atm

P(H2) = 0.242 *30= 7.26 atm

Calcolo Kp = P(NH3)^2/(P(N2)*P(H2)^3)
Kp = 20.28^2/(2.43*7.26^3) = 0.442 atm^-2