6.8 CO + H2O ⇄ CO2 + H2
initial : a b c
au reactionat: x x
la echilibru: (a - x) (b - x) (c + x) x
" " " " """ " ' 0,2 0,5 0,5 0,25
x = 0,25 moli a-0,25 = 0,2 a = 0,45 b - 0,25 = 0,5 b = 0,75 c = 0,25
a:b:c = 0,45:0,75:0,25 = 9:15:5
b) Kc = x(c +x)/(a-x)(b-x) = 0,25·0,5 /0,2·0,5 = 1,25
c) initial : 1,5 1,5
react.: x x
la ech.: (1,5 - x) 1,5-x) x x
x² /(1,5-x)² = 1,25 x/(1,5-x) = 1,118 2,118 x = 1,677 x = 0,79
% CO = %H2O = (1,5 - 0,79)·100/3 = 23,66%
%CO2 = %H2 = 79/3 = 26,33%
6.9 PCl5 ⇄ PCl3 + Cl2
initial: a b c
au reactionat: x x
la echilibru : (a-x) (b+x) (c+x)
(a-x)/3 = (b+x)/7 = (c+x)/2 c+x = 56,8 /71 = 0,8moli Cl2 a-x = 1,2 moliPCl5
b+x = 2,8 moli PCl3
b) c = 0 x = 0,8 a = 2 b = 2 a : b = 2 : 2 = 1 : 1
c) Pr. = x·100/a = 80/2 = 40%
6.10 N2 + 3H2 ⇄ 2NH3
initial : 70 moli 160moli 130 moli
au reactionat: x 3x
la echilibru : 70-x 160-3x 130+2x
PV = nRT P = 360·0,082·500 /50 = 295,2atm ⇒ la presiune marita, procesul se desfasoara in sensul reactiei inverse
b) (130+2x)/(360 - 2x) = 0,7145 130 + 2x = 257,22 - 1,429x
3,429x = 127,22 x = 37 70-37 = 33moliN2 160 - 111 = 49 moli H2
130 + 74 = 204 moli NH3
V = 50l T = 500K n = 286 moli P = 286·0,082·500/50 = 234,52atm
