Would cgg(H2O) = x if “c0(H2O)= 0.1 mol/l” were not given, but the initial concentration c0(H2O) = 0 mol/l?
Assuming the initial concentration of water was exactly the same as that of the esters = 0 mol/l, it would then be cgg(ester) = cgg(water) = x
and in the mass action coefficient Kc then x^2 in the numerator?
I am a little confused by the cgg(water)= c0(water) + x
Is this just because we already specified an initial concentration for water on the product side? In all other calculations in class, we simply had 0 mol/l as the initial concentration for all products.
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I don’t see how you could have expected, but your result is accurate. The equilibrium concentrations are 1-0.65=0.35 mol/l for both starting materials, 0.65 mol/l for the ester and 0.75 mol/l for water. By entering into the Mass Act
c(ester)⋅c(H2O) / c(alcohol) / c(acid) = 0.65⋅0.75/0.352 ≅4
the predetermined equilibrium constant is obtained, so it is true.
I know that the result is true. After all, I only wrote this off the board 😂 Now I’m trying to trace the way. Now I understand. Recognition itself is not difficult. But at best, this “follows from the equation of reaction” still has a “how easy to see” there is a bit more
If c1=1 mol/l is the initial concentration of alcohol and acid and c2 =0.1 mol/l is the initial concentration of H2 O, then you name the ester concentration in equilibrium simply x and starts:
K = x⋅(c2+x)/(c1−x)2
and only need to dissolve after positive x.
All right, thanks
Yes, of course. I did so in the denominator because the initial contracts are randomly the same.
Could the equilibrium concentration of water and ester be combined with x^2 if the initial concentration of water was not 0.1 but 0 mol/liter?