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This question is about equilibrium - AQA - A-Level Chemistry - Question 5 - 2021 - Paper 2

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This question is about equilibrium. 0 5.1 1 mol of a diester with molecular formula C2H4O2 is added to 1 mol of water in the presence of a small amount of catalyst.... show full transcript

Worked Solution & Example Answer:This question is about equilibrium - AQA - A-Level Chemistry - Question 5 - 2021 - Paper 2

Step 1

0 5.1 Amount in the mixture

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Answer

At the start, we have 1 mol of diester and 1 mol of water. The equilibrium equation is given as:

C2H4O2(l) + 2 H2O(l) ⇌ 2 CH3COOH(l) + HO(CH2)2OH(l)

Since 2 mol of ethanoic acid are formed at equilibrium, we denote the changes in terms of x. The amounts at equilibrium can be defined as:

  • Diester: 1x1 - x
  • Water: 1x1 - x
  • Acid: 22
  • Diol: yy

Step 2

0 5.2 Deducing the structure

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Answer

The structure of the diester C2H4O2 can be represented as:

  O
  ||
HO-C-(CH2)-C-OH
  ||
  O

This structure shows the diester with the hydroxyl groups that participate in the reaction.

Step 3

0 5.3 Calculate the amount of water

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Answer

Using the equilibrium concentrations from Table 3:

  • Amount of diester: 0.9710.971 mol
  • Amount of acid: 0.4520.452 mol
  • Amount of diol: 0.2730.273 mol

The equilibrium expression is given by:

Kc=[Acid]2[Diol][Diester][Water]2K_c = \frac{[\text{Acid}]^2 [\text{Diol}]}{[\text{Diester}][\text{Water}]^2} Substituting the known values:

0.161=(0.452)2(0.273)(0.971)([Water]2)0.161 = \frac{(0.452)^2(0.273)}{(0.971)([\text{Water}]^2)} Solving for [Water], we calculate:

  1. Calculate the numerator: (0.4522)(0.273){(0.452^2)(0.273)}
  2. Rearranging the equation: [Water]2=0.45220.2730.1610.971[\text{Water}]^2 = \frac{0.452^2 * 0.273}{0.161 * 0.971}
  3. Extracting the square root gives the final amount of water in mol.

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