CH3CH2OH (l) + CH3COOH (l) --><--- CH3COOCH2CH3 (l) + H2O (l)
This is a reversible reaction.
a) the equilibrium constant Kc for the above reaction at 25 deg ce is 4.0. Calculate the mass of ethanoic acid that must be mixed with 2.0 mol of ethanol to produce 1.5 mol of ethyl ethnoate at equilibrium.
I solved this question w/o drawing the ICE table, after all when I tried doing the ICE method, there are 2 unknowns, making it difficult for me to find anything at all. So I used stoichiometry. 1 mol of ethanol reacts with 1 mol of ethanoic acid to give 1 mol of ethyl ethanoate and water. So, to produce 1.5 mol of ethyl ethanoate, 1.5 mol of ethanoic acid (limiting agent) is need, The 2.0 mol of ethanol stated in the question, I took it as the excess. From there, I take 1.5 X 60.0 = 90g of ethanoic acid.
b) Suppose 1 mole of ethanol, 1 mole of ethanoic acid,3 moles of ethyl ethanoate and 3 moles of water are mixed together ina separate stoppered flask. How many moles of ethyl ethanoate will be present at equilibrium at 25 deg cel?
LOL. no idea how to solve. Dont mind if someone guides me. Gonna try this question again. Anyway, the Kc value given in the question, when do I use it?
thanks
Ok, I'll post the final answer here, and let you keep trying.
I'm having work (ie. tuition class) now all the way till 10pm tonight.
If no one helps you out by tonight and you still can't get the correct answers, I'll post the working then.
Final Answers : 157.5g and 2.667mol
Both to 4 sig fig. In the exams, always give the 4 or 5 sig fig value first, *then* give the 3 sig fig value. Note that when changing values from 4 or 5 sig fig to 3 sig fig, it's not "approximately equals to", it's "exactly equals to" whatever value, to 3 sig fig.
a) you can't just assume that any reactants, like ethanoic acid as a limiting reagent. But we can assume that there are no products CH3COOCH2CH3 and H2O initially, since it's not stated in the qn.
as this is a homogenous reaction(all reactants and products have the same state), H20 is included in the Kc equation
Kc = [H2O][CH3COOC2H5] / [CH3COOH][C2H5OH]
for unknowns, we can use algebra, letting the initial no.of moles of CH3COOH be x, and volume of the solution be V (for calculating the concentration of the reactants and products). as you know, 1.5 moles of CH3COOH react with 1.5 moles, so the equilibirum no.of moels of CH3COOH is (x - 1.5). substitute the eqm concentration values in the Kc eqn, and you will get the answer!
b) i'm not sure, but it seems that there would be a backward reaction of the previous eqn since there are more no.of moles of original products than the original reactants. the Kc value of the backward reaction is actually the inverse of Kc of the forward reaction under the same temperature (values of Kc are also temperature dependent). you can see it from here:
foward reaction's eqn: Kc = [H2O][CH3COOC2H5] / [CH3COOH][C2H5OH]
backward reaction's Kc eqn: K'c = [CH3COOH][C2H5OH]/ [H2O][CH3COOC2H5]
Kc = 1/ K'c
letting the change in no.moles of CH3COOC2H5 be -x, and substuiting them into the eqn, you will get 2 different values of x. (x = -5 or x= 1/3). you should reject x= -5 because it meant that there will be increase in the no.of moles of CH3COOC2H5.
then u shld be able to get answer, eqm no. of moles of CH3COOC2H5 is 2.67mol and not 2.67g haha.. other situations of rejecting a value of x is that when it makes the no. of moles of a reactant or product negative.
Originally posted by qdtimes2:a) you can't just assume that any reactants, like ethanoic acid as a limiting reagent. But we can assume that there are no products CH3COOCH2CH3 and H2O initially, since it's not stated in the qn.
as this is a homogenous reaction(all reactants and products have the same state), H20 is included in the Kc equation
Kc = [H2O][CH3COOC2H5] / [CH3COOH][C2H5OH]for unknowns, we can use algebra, letting the initial no.of moles of CH3COOH be x, and volume of the solution be V (for calculating the concentration of the reactants and products). as you know, 1.5 moles of CH3COOH react with 1.5 moles, so the equilibirum no.of moels of CH3COOH is (x - 1.5). substitute the eqm concentration values in the Kc eqn, and you will get the answer!
b) i'm not sure, but it seems that there would be a backward reaction of the previous eqn since there are more no.of moles of original products than the original reactants. the Kc value of the backward reaction is actually the inverse of Kc of the forward reaction under the same temperature (values of Kc are also temperature dependent). you can see it from here:
foward reaction's eqn: Kc = [H2O][CH3COOC2H5] / [CH3COOH][C2H5OH]
backward reaction's Kc eqn: K'c = [CH3COOH][C2H5OH]/ [H2O][CH3COOC2H5]
Kc = 1/ K'c
letting the change in no.moles of CH3COOC2H5 be -x, and substuiting them into the eqn, you will get 2 different values of x. (x = -5 or x= 1/3). you should reject x= -5 because it meant that there will be increase in the no.of moles of CH3COOC2H5.
then u shld be able to get answer, eqm no. of moles of CH3COOC2H5 is 2.67mol and not 2.67g haha.. other situations of rejecting a value of x is that when it makes the no. of moles of a reactant or product negative.
Yes, qdtimes2 is correct in his helpful explaination (including pointing out my typo error regarding the units. It should indeed be "moles", not "grams").
Audi, based on qdtimes2's guidance, are you able to solve the questions yet?
Whenever you're given non-zero quantities or molarities of both LHS reactants and RHS products, you should determine the Qc value, and compare it to Kc.
Qc, known as Equilibrium Quotient (in contrast to Kc, Equilibrium Constant), has the same formula as Kc (ie. product of molarities of RHS vs molarities of LHS, each raised to the power of their stoichiometric coefficients), except that it can be applied at any time, eg. initial. In contrast, Kc is a value derived from the molarities of reactants and products only at equilibrium point itself.
Notice in this question, Qc = 9.0 > Kc = 4.0. Imagine you are Qc, standing on a number line with Kc. Do you turn your head to the left or to the right, to look at Kc? Kc represents position of equilibrium. Notice that in this question, since Qc > Kc, you (ie. Qc), have to turn your head to the left to look at Kc (ie. position of equilibrium). Therefore we say "position of equilibrium lies to the left."
When we say "position of equilibrium lies to the left", it means the rate of the backward reaction will exceed the rate of the forward reaction, until equilibrium is reached, ie. Qc = Kc.
Hence in your ICE table, your Change will see "+x, +x, -x, -x" instead of the usual "-x, -x, +x, +x".
In most scenarios, when we add, or take away, some reactants or some products, or increase or decrease total pressure (by changing volume of container), and we say "position of equilibrium has shifted to the...", do you realize it is *not* actually Kc which has changed, but Qc?
So if Qc shifts to the right (eg. when you add some product), the Kc value or position of equilibrium has appeared to have "shifted to the left", *relative* to Qc (even though the Kc value has not changed at all).
The only time when Kc value actually changes, rather than the Qc value, is when temperature changes. If the forward reaction is endothermic (ie. heat may be regarded as a reactant), increasing the temperature will increase the Kc value (th Qc value remains unchanged). And therefore, it is indeed the Kc value or "position of equilibrium, has shifted to the right", relative to Qc.
H2 Chemistry students need not feel concerned if their JC did not teach them the above discussion, because there is much that JCs do not teach (because they simply do not have the time or opportunity to do so in a school setting; I know because I'm an ex-MOE teacher), which I teach my tuition students, in order to help them to understand, appreciate and apply Chemistry better.
If the above paragraphs confuse you, ignore them. If you understand them, then good for you. Use it to help you understand Chemistry better.
Audi, with qdtimes2's helpful post above, can you solve your questions yet? Or do yo need further help from us?