Sciencemadness Discussion Board

Equilibrium constant for esterfication, urgent.

thunderfvck - 8-3-2004 at 22:10

Well it's that time again! I have another lab to do and I was wondering about something.

Okay, suppose acetic acid and 1-propanol were added to an RB flask. A 1 ml sample of this is titrated with, let's say 12 ml's of NaOH. About 8 drops of concentrated H2SO4 are added to the original solution and is put on reflux for 90 minutes. Titrated afterwards to find concentration of acetic acid.

A "blank" is created, by taking approxiametly the same volume of water as the original solution (with the acetic acid and 1-propanol) and adding 8 drops of H2SO4. Titrated. Is the reason for this because, when determining how much acetic acid there is (after reflux) via titration, we can now subtract from this the moles of H+ that was supplied by the H2SO4? This is what I originally thought but my teacher went on about volumes, and how we did this to find the volume of H2SO4 added, which is pretty damn insignificant if you ask me (there's about 25 ml's of reacting solution initially). We aren't even given the concentration of the acid, so what the hell? What's the deal, brotherins?

There's also a question about how one could experimentally prove that equilibrium has been attained. Could you just add some base and then wait and see if the concentration of acetic acid retains its original concentration? This kind of confuses me because of the past tense used in the question. It HAS been attained, so will the solution still work as though it is under equilibrium, after it has been refluxed and is sitting there, touching its genitals? The whole point of this lab is to determine the value for K at the reflux temperature, if the reaction is no longer being refluxed will it still strive to attain K?

Thanks a lot!

gritty_cryst - 9-3-2004 at 00:55

I think you have the right idea for the reason for titrating the blank. The volume wouldn't be important (besides affecting the volume part of the molarity) unless it is used to caclulate the moles, but by titrating you would be obtaining the moles directly.

I'm not exactly sure how it would be experimentally proven that equilibrium was obtained. I guess that one would take samples at time intervals and graph the concentrations of reactions until they stop changing, but whether this constitutes proof or not I don't know.

Quote:

The whole point of this lab is to determine the value for K at the reflux temperature, if the reaction is no longer being refluxed will it still strive to attain K?


For reactions whose rates are controlled by kinetics, it will reach equilibrium faster at higher temperatures because more molecules have enough energy to overcome the transition state barrier. Mixtures of compounds will always strive to attain equilibrium, its Le Chatelier's principle. The value of K for a reaction can vary with temperature however, and this is due both to entropy (thermodynamic) factors and kinetic reasons. At least, that is my limited understanding of it. Not thorough by any means.

thunderfvck - 9-3-2004 at 15:34

In order to determine whether or not equilibrium has been reached, after doing the reflux you take your 1 ml sample, titrate and then reflux the mixture again, then titrate another sample and compare the two. If they are the same, or relatively close anyway, you have just proved equilibrium has been achieved.

thunderfvck - 12-3-2004 at 23:39

I was just wondering about something else. I just want to verify whether my math on calculating the constant is correct.

Firstly, the reaction is:

1-butanol + acetic acid --> butyl acetate + water

Okay, so I found out how many moles of NaOH are required to titrate a 1 ml sample of A, let's say it was 10 ml's of 1 M NaOH.

So, 0.01L * 1 mol/L = 0.01 mol of OH- required to titrate sample, which implies that 0.01 mol of H+ were present. This is for acetic acid BTW. Okay, so to find the concentration of acetic acid...

(0.01 mol CH3COOH)/(0.001L) = 10 M. I divided by 0.001 L because the sample size was 1 ml, if this wasn't already obvious. So there's that number!

After refluxing, titration of sample B requires 5 ml's of NaOH (1M)...So that's 5 M CH3COOH...

So...K = ([butyl acetate][H2O])/([CH3COOH][1-butanol])

[CH3COOH] = 5 M = [1-butanol]
[butyl acetate] = 10 M (inital CH3COOH concentration) - 5 M = 5 M = [H2O]

Which gives K as 1. These are not my actual values but I'm just simplifiying. I was wondering if my subtracting concentrations as I had done just above was valid, along with everything else.

Well, that's about it. Butyl acetate smells lovely, bananas!

Okay, thank you duders!

gritty_cryst - 13-3-2004 at 12:29

Quote:

If they are the same, or relatively close anyway, you have just proved equilibrium has been achieved.


That seems reasonable. Do you have the math to demonstrate why, though? I'd be interested in seeing it. Or maybe just elaborate a bit if you can.

All your math seems good to me, too.

thunderfvck - 13-3-2004 at 12:51

Okay,

well, I won't give you my experimental results because I found that equilibrium actually WASN'T reached in my case (but a few changes can make the teacher happy ;)).

So, using the above example with my K = 1...okay, molarity of acetic acid after reflux is 5 M. Initially it was 10 M. So now we do the same thing again, let it sit on reflux for awhile (the whole bit of the solution). Then, after 30 minutes, let it cool, and take another 1 ml sample. It is titrated with 4.9 ml's of NaOH...

0.0049 L * 1 M = 0.0049 mol

(0.0049 mol)/(0.001L) = 4.9 M

THerefore...
[CH3COOH] = 4.9 M
[1-butanol] = 4.9 M
[butyl acetate] = 10 M - 4.9 M = 5.1 M
[H2O] = 5.1 M

K = 1.1 which is very close to the original value of 1. So we can conclude that equilibrium has been reached since after 30 minutes there was no MAJOR change in the concentration of one reactant, equilirbium must have been attained.