Ethanolic acid and ethane-1,2-diol react together to form the diester (C4H10O4) as shown - AQA - A-Level Chemistry - Question 5 - 2017 - Paper 2
Question 5
Ethanolic acid and ethane-1,2-diol react together to form the diester (C4H10O4) as shown.
2CH3COOH(l) + HOCH2CH2OH(l) ⇌ C4H10O4(l) + 2H2O(l)
1. Draw a structural f... show full transcript
Worked Solution & Example Answer:Ethanolic acid and ethane-1,2-diol react together to form the diester (C4H10O4) as shown - AQA - A-Level Chemistry - Question 5 - 2017 - Paper 2
Step 1
Draw a structural formula for the diester C4H10O4
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Answer
The structural formula for the diester C4H10O4 can be represented as:
O O
|| ||
CH3-C-O-CH2-CH2-O-C-CH3
This shows the ester bonds connecting the acid and alcohol components.
Step 2
Complete Table 1
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Answer
To complete Table 1, we need to find the equilibrium amounts:
At the start:
CH3COOH: 0.470 mol
HOCH2CH2OH: 0.205 mol
C4H10O4: 0 mol
H2O: 0 mol
At equilibrium:
CH3COOH: 0.180 mol (given)
HOCH2CH2OH: 0.205 - x mol
C4H10O4: x mol
H2O: 2x mol
Setting up the equilibrium expression:
Kc=[CH3COOH]2[HOCH2CH2OH][C4H10O4][H2O]2
Substituting the equilibrium values will give the complete table.
Step 3
Write an expression for the equilibrium constant, Kc, for the reaction
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Answer
The equilibrium constant expression for the reaction is:
Kc=[CH3COOH]2[HOCH2CH2OH][C4H10O4][H2O]2
This expression shows the concentrations of the products over the reactants.
Step 4
Justify this statement
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Answer
The justification for the statement is that the total volume of the mixture does not affect the ratio of concentrations in the equilibrium constant expression. Since the volume cancels out when calculating Kc, it is sufficient to know the molar amounts of each component at equilibrium without measuring the total volume.
Step 5
Calculate the amount of ethanolic acid present
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Answer
Using the data from Table 2 and the known value of Kc:
Kc=6.45=(x)2(0.264)(0.802)(1.15)2
Rearranging gives:
x2=6.45(0.264)(0.802)(1.15)2
Calculating:
x2=1.69080.802∗1.3225=0.5071
Thus, x=0.5071≈0.711
The amount of ethanolic acid present at equilibrium is approximately 0.711 mol (rounded to three significant figures).
