Question

Constanta de echilibru,Kc,a reactiei de esterificare a acidului acetic cu etanolul este 4.Stiind ca se supun esterificarii 48 g acid acetic si 46 g etanol,calculeaza compozitia,in procente molare,a amestecului obtinut la echilibru.

Answer 1

Trasformam in moli masele de acid si alcool
Macidacetic=60g/mol si Metanol=46g/mol
nacid=48/60=8/10=0,8moli
nalcool=46/46=1mol
TABELUL LA ECHILIBRU
................Acid.....Alcool......Ester......Apa
initial:.....0,8..........1..............0.............0
Tr/Rez:...x............x...............x.............x
Echilibru:0,8-x.....1-x............x.............x
Kc=x^2/(1-x)(0,8-x)=4
x^2=4(0,8-1,8x+x^2) |*5
5x^2=16-36x+20x^2
15x^2-36x+16=0
delta=1296-960=336
x1,2=(-b+√delta)/4a
√delta=18,33
x1=(36+18,33)/30=1,811 moli nu convine
x2=(36-18,33)/30=0,589moli convine
la echilibru avem: 0,211moli acid; 0,411moli alcool; 0,589 moli ester si 0,589moli apa
1,8moli amestec...........100%
0,589moli...........................z%
0,411moli...........................y%
0,211moli...........................z%
x=0,589*100/1,8=32,72% ester=32,72%apa
y=0,411*100/1,8=22,83% alcool z=0,211*100/1,8=11,72% acid