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Topic: theoretical yield of equilib process  (Read 3678 times)

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Offline xc630

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theoretical yield of equilib process
« on: April 16, 2009, 02:09:56 AM »
Hi I would like some help with the following problem:

esterification is an equib process between an acid + alcohol and an ester + water

K = [ester]^2/ ([acid] [alcohol])

K=4.0
volume of mixture is 25.0 mL
5 g benzoic acid used
25 mL methanol used
assume cahnge in methanol conc is negligible as rxn proceeds and that the contribution of the acid catalyst, benzoic acid, and products to the volume is negligible.

Calculate the theoretical yield for this equilibrium

I did 5g/ 122.12 g = 0.0409 mol/0.025 L = 1.64 M acid
also from the density of  methanol I calculated the grams in 25 mL to be 19.795 g and the concentration of methanol to be 24.7M--- ((19.795/32.04)/0.035)
I plugged those into the equilib equation above and got 12.72 M for the ester conc
12.72 x 0.025 L =.318 mol x 136.15 g = 43.3g
so the theoretical yield would be 43.3/5 which is wrong. What did I do incorrectly? Thanks!

Offline Borek

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Re: theoretical yield of equilib process
« Reply #1 on: April 16, 2009, 03:26:54 AM »
I plugged those into the equilib equation above and got 12.72 M for the ester conc

Show how.
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Offline xc630

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Re: theoretical yield of equilib process
« Reply #2 on: April 16, 2009, 09:10:56 AM »
4= ([ester]^2)/ (1.64 x 24.7)
[ester]^2 = 162.03
[ester] = 12.72

??

Offline Borek

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Re: theoretical yield of equilib process
« Reply #3 on: April 16, 2009, 10:08:26 AM »
Concentration of acid at equilibrium is not 1.64.
« Last Edit: April 16, 2009, 10:38:15 AM by Borek »
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