[MitchTwitchita]

C2H5CO2H(l) + C3H7OH(l) ---> C2H5COOC3H7(l) + H2O(l)

18.5 mL propanoic acid are mixed with 19.0 mL n-propanol to produce n-propyl propanoate when heated and distilled.

At room temperature, the equilibrium constant, Keq is approximately 3.  Calculate the number of moles of reactants and products expected in the equilibrium mixture of this reaction.  (Hint: Let X be the number of moles of each product formed at equilibrium.  Substitute into the equilibrium equation using the moles of each reactant and product at equilibrium each divided by the total reaction volume.  Leave the equilibrium concentrations as fractions.  You can then take the square root of both sides of the equation after canceling like terms, i.e. the volumes.)

Other Data:

Boiling Point of n-propyl propanoate = 122 degrees C
Volume of water collected = 3.31 mL
Weight of product (C2H5COOC3H7) = 7.4846 g

I have no idea of how to start this problem, can anybody give me a push in the right direction please?

[Borek]

Do you now what the equation for equilibrium constant looks like?

[MitchTwitchita]

Yeah, Keq = [C2H5COOC3H7][H2O]/[C2H5CO2H][C3H7OH]  I think.

[Borek]

Looks OK.

Obviosuly at this point you should convert volumes of reactants to moles, but I wonder how they want you to do so without giving densities... Try to look them up in CRC Handbook or Merck index or something like that.

[MitchTwitchita]

O.k. I found the densities and calculated the moles.  The density for propanoic acid is 0.992 g/mL and the density for n-propanol is 0.803 g/mL.  This correlates into 0.248 moles of the acid and 0.255 moles of the alcohol.  Therefore, the propanoic acid would be the limiting reagent and 0.248 of  n-propyl propanoate and water would be formed.  Is this correct?  If so, do I make an ICE table with the products equilibrium(M) being 0.248/0.185L = (13.4 - x) and 0.255/0.019 L = (13.4 -x) and the products being (x) and (x) respectively? ie.Keq = (x)^2/(13.4 - x)^2 ?

[Borek]

Therefore, the propanoic acid would be the limiting reagent and 0.248 of  n-propyl propanoate and water would be formed.  Is this correct?

No. I mean - you are right about the stoichiometry, but the reaction doesn't go to completion.

Taking into account accuracy of volume measurements you may safely assume you took 0.25 mole of both reactants.

If so, do I make an ICE table with the products equilibrium(M) being 0.248/0.185L = (13.4 - x) and 0.255/0.019 L = (13.4 -x) and the products being (x) and (x) respectively? ie.Keq = (x)^2/(13.4 - x)^2 ?

You are at least partially right, but I have no idea what is 0.248/0.185L = (13.4 - x) ? Or rather - I know what you did, but it is wrong

Remember - you have mixed them, so your final volume is close to the sum of volumes. And what is left is 0.25-x.

[MitchTwitchita]

Oh, I thought M = n/v would take the place of 0.25.  Swing and a miss, I guess.  So, then keq = (x)^2/(0.25 - x)^2 , and I would just solve for x to figure out the moles?

[Borek]

Oh, I thought M = n/v would take the place of 0.25.

That's a goof idea, just your n is not 0.25 but 0.25-x. It happens that in this case volumes cancel out, so you may try equation in the form

keq = (x)^2/(0.25 - x)^2

I would. Which doesn't necesarilly mean it is OK

[MitchTwitchita]

So, 3 = x^2/(0.25 - x)^2
ignoring the lower x, 3 = x^2/0.0625
x = 0.433
I don't think that this can be right because it would give a negative for the reactants (0.25 - 0.433 = -0.183)

Using the quadratic equation, I come up with 2x^2 - 1.5x + 0.1875 = 0.1585 or 0.5915

I take 0.1585 because it won't give me a negative.

Therefore, 0.25 - 0.1585 = 0.0915

Does this mean that at equilibrium there is 0.0915 mol of reactants and 0.1585 mol of products?

[Borek]

Does this mean that at equilibrium there is 0.0915 mol of reactants and 0.1585 mol of products?

Sounds reasonable. But honestly I am not 100% sure. You know, it is Sunday evening and chemicalforums is not the only thing I am involved in at the moment

[MitchTwitchita]

Thanks for your help Borek!