chem eqm

LetV dm³ bethe volume of the reaction mixture at eqm.
Let y mol be the number of moles of ester at eqm.

C3H7COOH + C2H5OH ⇌ester + H2O

At eqm :
[C3H7COOH] = (2 - y)/V mol
[C2H5OH] = (2 - y)/V mol
[ester] = y/V mol
[H2O] = (2.4 + y)/V mol

Kc = [ester] [ H2O] / [C3H7COOH] [C2H5OH]
(y/V) [(2.4 + y)/V] / [(2 - y)/V] [(2 - y)/V] = 2.45
y(2.4 + y) / (2 - y)² = 2.45
(2.4y + y²) /(4 - 4y + y²) =2.45
2.4y + y² = 9.8- 9.8y + 2.45y²
1.45y² - 12.2y + 9.8 = 0
y = [12.2 ± √(12.2² - 4x1.45x9.8)] / (2x1.45)
y = 7.51 (rejected) or y = 0.899

The number of moles of the ester at eqm = 0.899mol ≈ 0.9 mol

The equilibrium concentration of the ester = 0.899/V mol dm⁻³ ≈ 0.9/Vmol dm⁻³
where V dm³ isthe volume of the equilibrium mixture.