Ethanolic acid and ethane-1,2-diol react together to form the diester (C₈H₁₈O₄) as shown - AQA - A-Level Chemistry - Question 5 - 2017 - Paper 2

From the initial amounts and considering the formation of the product C₈H₁₈O₄:

• At start:

  • CH₃COOH: 0.470
  • HOCH₂CH₂OH: 0.205
  • C₈H₁₈O₄: 0
  • H₂O: 0

• At equilibrium:

  • CH₃COOH: 0.180 (0.470 - 0.290)
  • HOCH₂CH₂OH: 0.205 - 0.290 = 0
  • C₈H₁₈O₄: 0.180
  • H₂O: 0 + 0.290 = 0.290

The expression for the equilibrium constant (Kc) for the reaction is:

$K_c = \frac{[C₈H₁₈O₄][H₂O]^2}{[CH₃COOH]^2 [HOCH₂CH₂OH]}$

where the brackets denote the equilibrium concentrations of the respective species.

The total volume of the mixture is irrelevant in this case because the equilibrium constant is defined in terms of the ratios of concentrations (or partial pressures) of the reactants and products. As such, any volume factor cancels out as it is present in all terms. Thus, the value of Kc remains consistent regardless of the overall volume used in the mixture.

Given Kc = 6.45, using the concentrations provided in Table 2:

Let the amount of CH₃COOH = x mol. The equilibrium compositions can be expressed as:

$K_c = \frac{(0.802)(1.15)^2}{(x)(0.264)}$

Substituting the values into the equation:

$6.45 = \frac{(0.802)(1.15)^2}{(x)(0.264)}$

Solving for x:

$x = \frac{(0.802)(1.15)^2}{6.45 \times 0.264} \approx 0.789$

Thus, the amount of ethanolic acid present in the new equilibrium mixture is approximately 0.789 mol.