# Equilibria Kc question

#1
Ethanoic acid reacts with ethanol in a reversible reaction represented by the equation below.
In an experiment 3.0 mol of ethanoic acid were mixed with 1.0 mol of ethanol and when the reaction had reached equilibrium 0.9 mol of water had been formed.

CH3COOH(l) + C2H5OH(l) <--> CH3COOC2H5(l) + H2O(l)

The percentage of ethanoic acid converted into the ester CH3COOC2H5 in this reaction is

A 22.5%

B 30%

C 43%

D 90%

Can someone send a worked solution to get the answer B? I'm getting C each time.
0
4 years ago
#2
0
4 years ago
#3
(0.9/3)*100 = 30%
0
#4
Won't let me upload to TSR
0
#5
Won't let me upload to TSR. I've posted the link to my solution, but it's getting approved first...
0
#6
(Original post by TutorsChemistry)
Put the link together all as one,
i.imgur
.com/
0
4 years ago
#7
(Original post by jacky33041)
Put the link together all as one,
i.imgur
.com/
BO1z0Dz.jpg

The problem is that you are dividing by 2.1, the equilibrium quantity of ethanoic acid.
The question asks how much has been converted, so you must divide by the initial quantity - divide by 3 rather than 2.1.
2
#8
Thank you.
1
9 months ago
#9
In order to solve this, you need to make an ICE table, im not sure if you've heard of that before. So basically you need to find the equilibrium moles for all of the molecules. since you know that 0.9 of water is formed, the change is 0.9 for all as its a 11:1 ratio. From this you can find that there has been 0.9 of the ester made hence you just do a percentage calculation of 0.9 / 3.0 after to get 43%
0
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