The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25âC:

CH3CO2H(soln)+ C2H5OH(soln)â CH3CO2C2H5(soln)+H2O(soln)Kc=3.4

Acetic acid Ethanol
Ethyl acetate

Calculate the number of moles of all reactants and products in an equilibrium mixture prepared by mixing 1.00 mol of acetic acid and 10.00 mol of ethanol.

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Beverley SmithLv2

17 Dec 2019

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Ok this is a three part questions but I think i correctly answered#1 but i get lost after that..

1. An aqueous solution of ethyl acetate at room temperature has aninitial concentration of 0.15 M. Calculate the equilibriumconcentrations of ethyl acetate (CH3COOC2H5), ethanol (C2H5OH) andacetic acid if the equilibrium constant at this temperature is Kc =3.00.
CH3COOC2H5(aq) + H2O ? CH3COOH(aq) + C2H5OH(aq)

Here I set up an ice table an ice table and got a quadratic ofx^2+3.00x-.45=0
and i got x=.143.

Now here is what I get confused..

2) 2. A solution contains 0.15 M acetic acid and 0.15 ethanol. Ifthe equilibrium constant defined by the reaction in Problem 1 is Kc= 3.00, calculate the equilibrium concentrations of ethyl acetate,acetic acid and ethanol.

Im guessing you just do products/reactants?
so
[0.15 M] x [0.15 M] / [CH3COOC2H5] = 3.00
and CH3COO2H5 = .0075 M.

I think im right up to this point then im not sure what to donext?

then there's

3. A solution contains 0.01 M ethyl acetate, 0.10 M acetic acid and0.10 M ethanol. For the same equilibrium constant of 3.00,calculate the equilibrium concentrations of ethyl acetate, ethanol,and acetic acid in the solution.

orangemoose0

this is my chemistry lab discussion&conclusion
please help me revise it (mainly focus on conclusion ifpossible)
thank you!

Discussion
In the equilibrium experiment we were able to observe the directionof the equilibrium by looking at our chemical equation CH3COOH +C2H5OH <-> CH3COOC2H + H2O. The reaction in flasks b-eshifted left towards equilibrium, while flask f shifts to the righttowards equilibrium. Our Kc for flask b was 6.864, flask c was6.889, flask d was 5.695, and flask e was 98.73. Flask f was 6.833and our mean for our Kc was 25.003. The temperature for ourtitration was 22 degree Celsius.
For our titration for flask b we used 43.10mL of NaOH(equivalent to 0.043 moles of NaOH), flask c we used up 40.61 mL ofNaOH (equivalent to.04028 moles of NaOH), for flask d we used36.42mL of NaOH (equivalent to .03611 moles of NaOH), flask e weused 36.95mL of NaOH (equivalent to .03665 of NaOH), and flask f weused 22.10mL of NaOH (equivalent to .02192 moles of NaOH). We knowthat NaOH is one to one ration to HCl + acetic acid(CH3COOH). Since the moles HCl + moles acetic acid is equalto total moles of NaOH, then the moles HCl + acetic acid for flaskb-f is 0.043 moles, 0.04028 moles .03611 moles .03665 moles .02192moles. The moles of HCl in each flask were 0.01488. Therefore wewere able to subtract the total moles of NaOH with the moles HClleaving us with the final moles of acetic acid (CH3COOH). The molesacetic acid (CH3COOH) for flask b-f was 0.02787, 0.02540, 0.02124,0.02177, and 0.007042. The final moles of ethanol (C2H5OH) wereobtained by taking the initial moles of ethanol (C2H5OH) minus thechange of moles in acetic acid from initial to final. The finalmoles of ethanol (C2H5OH) in flasks b-f were 0.02787 moles, 0.02540moles, 0.03836 moles, 0.004309 moles, and 0.05808 moles. In orderto calculate the final moles of Ethyl acetate (CH3COOC2H) we had toadd the initial moles of ethyl acetate (CH3COOC2H) with the changeof moles of ethanol (C2H5OH) from initial to final. For flask b-fthe final moles of Ethyl acetate (CH3COOC2H) was 0.02319 moles,0.01544 moles, 0.01961 moles, 0.03654 moles, and 0.01042. In orderto obtain the final moles of H2O we had to add the initial moles ofwater (H2O) with the change of moles of ethanol (C2H5OH) frominitial to final. The final moles of water (H2O) are 0.2300 moles,0.2878 moles, 0.2366 moles, 0.2535 moles, and 0.2682 moles. Finallywe were able to obtain the equilibrium constant Kc. In thebeginning when we were calculating for the initial moles ofconcentrations we had to divide the mass percent by 100 to changeit back into its decimal form. So when we calculated Kc we had tomultiple each of the final moles by 100. We calculated Kc byeach of the final moles of products times 100 divided by each ofthe final moles of reactants times 100.
In conclusion our values of Kc were all between therange of 0.0001 to 1000. The data we recorded proves that they wereall at equilibrium. Our standard deviation for KcÂ¬ was 41.22percent. Our percent deviation was so high due to flask e which hada 98.73 for its Kc value. Knowing that our numbers are withinequilibrium range of 0.0001 to 1000, this proves that our Kc is valid. Some sources of errors were not accurately pipetting thecorrect of each solution into its proper flask. Other errorsinclude not accurately weighing out the KHP to its plus or minus0.0001 gram, contaminated beakers, flasks, pipet, burette, weightboat, and scoopula, also including not reading the exactmeasurement of NaOH titrated.

The equilibrium constant for the esterification of ethanol and aceticacid in an aqueous solution CH3COOH(aq) + C2H5OH(aq) -----> CH3COOC2H5(aq) + H2O(l) amounts to 0.566 at 298 K. The reaction mixture initially contains 0.040 mol dm^â3 of acetic acid and 0.025 mol dm^â3 of ethanol. Find themolar concentration of ethyl acetate at the equilibrium. Assume that the pH is low enough that CH3COOH is not dissociated at all and thatthe activity of water equals 1 just like for pure liquid water (after all,the solution is quite dilute). Hint:This problem requires you to keep an extra large number ofsignificant digits in all intermediate calculations: otherwise, your finalanswer will be very inaccurate. Do not forget to check your answer in the end.

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CH3CO2H(soln)+	C2H5OH(soln)â	CH3CO2C2H5(soln)+H2O(soln)Kc=3.4
Acetic acid	Ethanol	Ethyl acetate