A2B4 ⇄ 2AB2
s-au luat in lucru: 3 0
au reactionat : x 2 x (acestia s-au obtinut)
la echilibru : 3-x 2x
1 4
2x = 4 ⇒ x = 2mol/l
Kc = [AB2]² / [A2B4] = 16/1 = 16 mol/l
PV = nRT P = n/V·RT = c·RT P = 5RT P1 = 1/5 ·RT P2 = 4/5 ·RT
Kp = P(AB2) ² / P(A2B4) = P2² / P1 = 1/25 ·(RT)² /4/5 ·RT = 1/20 ·RT = 0,082·1773/20=
= 7,27 atm