Question

Help!
How can one determine the equilibrium constant?
(Experiment procedure along with calculations are needed).
For this, I chose the reaction between ethanol and ethanoic acid (esterification rxn).

Answer 1

okay, you need to find a way to monitor either the concentration of product or one of the reactants.

I can think of a few ways that are relatively simple.

For both, you'll have to run the reaction several times (repeats)

1. using UV-VIS spectrophotometry if one of the reactants or the product have different absorption spectrum.

You'll have to use the beer-lambert law.

2. Do a titration to measure how much ethanoic acid you have left.

From here you can find how much product formed and the constant.

Answer 2

a problem with 2 is that you might hydrolyze the ester, so you would have to separate them...

Answer 3

I'll still confused. I have a procedure but I'm not quite certain if it'll be effective. Here it is:
1) Burettes are to be set up in the lab containing the following: ethyl ethanoate, ethanoic acid, 3 mol/dm^3, ethanol and distilled water.

2) Before starting, make sure the reagent bottle has a well-fitting stopper. Run the liquids into the reagent bottle from the burettes and stopper immediately to prevent evaporation. Shake well, then allow the bottle to stand at room temperature for a week, to allow the mixture to reach equilibrium.

3) After one week- Titrate the whole of the mixture in the bottle with 1.0 mol/dm^3 NaOH using phenolphthalein as the indicator.

4) In order to find the exact conc. of the HCl catalyst, titrate 5cm^3 of the 3 mol/dm^3 HCl with 1.0 mol/dm^3 NaOH.
5) Finally, weigh 5cm^3 of each of the liquids used to make up your mixture (including the HCl).

As for the calculations, I have no idea how to begin. Every time I research, I keep getting different things that leaves me confused :/

Answer 4

Overall it's good. but there are a few problems to address

"3) After one week- Titrate the whole of the mixture in the bottle with 1.0 mol/dm^3 NaOH using phenolphthalein as the indicator."

If you add NaOH to the solution, you'll hydrolyze some of the ester, leaving you with more acid. This will make the concentration of acid appear larger.

"4) In order to find the exact conc. of the HCl catalyst, titrate 5cm^3 of the 3 mol/dm^3 HCl with 1.0 mol/dm^3 NaOH."

You already know the amount of catalyst (you added it in the beginning); it's concentration does not change, but you do have to take this into account when titrating.

The calculations are easy. you could find the molarity of a sample from the titration with:

\(M_{acid}V_{acid}=M_{base}V_{base}\)

then subtract it from the initial amount (that you added). Then you'd also know the concentrations of the ester and the alcohol.

You could then just plug these into the equilibrium expresion.

Answer 5

Can I have an example, please? If I have an example, I'll understand it completely.

Answer 6

An example of what exactly?

Answer 7

The calculation.

Answer 8

Also, during titration, when a slight purple colour is seen , do I stopper immediately and measure the volume? Is that the equilibrium point?

Answer 9

oh okay, sure.

Lets say you titrate the acid with NaOH (note that acid is monoprotic).

You end up using 15 mL of 0.1 M NaOH on 20 mL of a sample of the solution (remember that you need to separate the acid from the ester because you'll have some hydrolysis).

The concentration of the acid is: \(0.1M*0.015~L=M_{acid}*0.02~L\rightarrow M_{acid}=0.075~M\)

Now let's say you initially put (any volume) of 0.1 M ethanoic acid - and 0.1 M ethanol.

Using an ICE table we can find their values at equilibrium
acid alcohol ester
I 0.1 M 0.1 M 0
C -x -x +x (this is apparent from the balanced equation)
E 0.1 -x 0.1-x x

we know that at equilibrium, the acid concentration is 0.075 M

so 0.075 M = 0.1 -x; x=0.025 M

so we plug everything into the equation:

\(K=\dfrac{[ester]}{[acid]*[alcohol]}=\dfrac{0.025~M}{0.075~M*0.075~M}=4.44\)

Answer 10

Also, during titration, when a slight purple colour is seen , do I stopper immediately and measure the volume? Is that the equilibrium point?


that would be the equivalence point, where you have added the same amount of base as there was of acid.

Answer 11

I see.

I understand it completely now. My questions have been answered and for that, thank you so much. Also, how do I tabulate my results? That's actually the last thing I need in order to complete this planning and designing lab.

Answer 12

So the results are the equilibrium constants. Since you'll repeat this a number of times, you'll have to take an average (add the K's up and divide the by the number of trials); you could take this a step further and calculate the standard deviation.

Answer 13

Got it. okay, I'll do it since my teacher loves calculating the standard deviation for test scores.

A massive thank you for helping me!

One more thing, do I have to find the number of moles for anything?

Answer 14

no problem at all!

not directly, it's hidden in \(M_1V_1=M_2V_2\);

its a shortcut you can use because the coefficients in the balanced reaction are 1 for both.

You could do this "the long way", but it works out the same.

Answer 15

Got it. Thanks again!