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Autor: brrrm1, Sre, 02/10/2013 - 15:16.

Konstantna ravnoteže reakcije A+2B=2C iznosi 0,056.Kolika je ravnotežna koncentracija produkta C ako he
a) C(A)=0,2 mol/l

b) C(A)=C(B)=0,65 mol/L

c) C(A)=3*C (B)=0,27 mol/L

Average:

Autor: micaloknica, Sre, 02/10/2013 - 16:52.

A+2B---------------->2C

b) c(A)=c(B)=0.65 mol/l

K={C}rav.*2/({A}rav.*{B}rav.*2)

{C}=x

0.056=x*2/0.65*(0.65)*2

x*2=0.056*0.65*(0.65)*2=0.0154

x=(koren(0.0154))=0.124

{C}rav.=0.124 mol/l

c) c(A)=3*C c(B)=0.27 mol/l

c(C)=x

K=x*2/(3*x)*(0.27)*2

0.056*3*x*(0.27)*2=x*2

0.0123*x=x*2 skratis jedno x i dobijes

x=0.0123

{C]rav.=0.0123 mol/l

a) {A}=0.2 mol/l a {B}=2*{A}=0.2*2=0.4 mol/l

K=x*2/0.2*(0.4)*2

0.056=x*2/0.2*(0.4)*2

x*2=0.056*0.2*(0.4)*2

x*2=0.001792

x=(koren(0.001792))=0.0423

{C}rav.=0.0423 mol/l