After 24.9 g benzoic acid, 17.9 mL of methanol, and a trace of anhydrous sulfuri

Question

After 24.9 g benzoic acid, 17.9 mL of methanol, and a trace of anhydrous sulfuric acid are refluxed together, 3.4 g benzoic acid are recovered from the final mixture. What is the equilibrium constant for this reaction based on benzoic acid?

After 20.1 g benzoic acid, 18.6 mL of methanol, and a trace of anhydrous sulfuric acid are refluxed together, 11.5 g methyl benzoate are recovered from the final mixture. What is the equilibrium constant for this reaction based on methyl benzoate?

If Keq for the Fischer esterification of benzoic acid with methanol = 10, what is the percent conversion after an equimolar mixture of benzoic acid and methanol comes to equilbrium?

When determining Keq for the Fischer esterification of benzoic acid with methanol, how will your calculation be affected if you lose crystals of benzoic acid on the filter paper during vacuum filtration?

Explanation / Answer

1)moles of benzoic acid = 24.9/122.12 = 0.2038 Moles

weight of methanol = volume*density = 0.7918*17.9 = 14.173 gm

moles of methanol = 14.173/32.04 = 0.4424 moles.

CH3OH + C6H5COOH -----%u25BAC6H5COOCH3+H2O
Methanol + Benzoic acid ---------%u25BA Methyl-benzoate ester + Water

no.of moles of unreacted benzoic acid = 3.4 g = 3.4/122.12 = 0.0278 moles.

no.of moles of unreacted methanol = 0.4424-0.176 = 0.2664moles

no.of moles of reacted benzoic acid = 0.2038 -0.0278 = 0.176 moles.

no.of moles of ester formed = 0.176 moles

no.of moles of water formed = 0.176 moles

equilibrium constant for this reaction based on benzoic acid = [Ester][water]/[Methanol][Benzoic Acid]

Keq = 0.176*0.176/(0.0278*0.2664) = 4.1826

2)1)moles of benzoic acid = 20.1/122.12 = 0.1646 Moles

weight of methanol = volume*density = 0.7918*18.6 = 14.727gm

moles of methanol = 14.727/32.04 = 0.45966 moles.

CH3OH + C6H5COOH -----%u25BAC6H5COOCH3+H2O
Methanol + Benzoic acid ---------%u25BA Methyl-benzoate ester + Water

no.of moles o f Methyl Benzoate formed =11.5 /136.15 = 0.08446 moles.

no.of moles of unreacted benzoic acid = 0.1646-0.08446 = 0.08moles.

no.of moles of unreacted methanol = 0.45966-0.08446 = 0.3752moles

no.of moles of water formed = 0.08446 moles.

equilibrium constant for this reaction based on benzoic acid = [Ester][water]/[Methanol][Benzoic Acid]

Keq = 0.08446 *0.08446 /(0.08*0.3752) = 0.2376

3)Keq = 10 = [Ester][water]/[Methanol][Benzoic Acid]

Let say x moles of initial methanol and benzoic acid,

then

CH3OH + C6H5COOH -----%u25BAC6H5COOCH3+H2O

x x - -

x- y x-y y y

then Keq = (y*y)/(x-y)*(x-y)

then y^2/(x-y)^2 = 10

then y/(x-y) = sqrt(10) = 3.162

(x-y)/y = 1/3.162 = 0.3162

x/y = 1.3162

y = 0.7597 x

percentage conversion = y/x)*100 = (0.7597*100 ) = 75.97 %

4)the concentration of Benzoic acid keeps decreasing,

as the conc of benzoic acid is in the denimonator in the Keq, the value of Keq increases.

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