Question

How do I find the `K_c`? (Question in description)

1.40mol of propanoic acid (`"CH"_3"CH"_2"COOH"`) and 5.40mol of ethanol are added together at `30^circ"C"`. After the system has come to equilibrium, a portion of the mixture is titrated against `0.4mol` `dm^-3` of `"NaOH"` solution. The titration shows that the whole of the equilibrium mixture requires `149cm^3` of standard alkali for neutralisation. Find the value of `K_c` for the esterification process.

Answer 1

`K_c~=0.20`

Explanation:

`"H"_3"CCH"_2"C(=O)OH" + "HOCH"_2"CH"_3 stackrel(Delta)rarr"H"_3"CCH"_2"C(=O)OCH"_2"CH"_3+"H"_2"O"`

We started with `0.140*mol` of the acid and excess ethanol....and we finished with ....

`0.40*mol*dm^-3xx149*cm^3xx10^-3*dm^3*cm^-3=0.0596*mol` with respect to the carboxylic acid....

And therefore....

`underbrace(0.140*mol)_"starting moles of acid"-0.0596*mol=underbrace(0.0804*mol)_"moles of acid that reacted"`

And these `"moles of acid that reacted"` are equal to the `"moles of ethanol that reacted"`, and the `"moles of ester that were produced"`

And since `K_c=({"moles of ester"}{"moles of water"}xx1/V^2)/({"moles of acid"}{"moles of alcohol"}xx1/V^2)`

.....we gots...`K_c-=(0.0804)^2/[(0.0596)(0.540-0.0804)]=0.24`

Do you follow?