Chemical EquilibriumHard

Question

A quantity of 60 g of CH3COOH and 46 g of CH3CH2OH reacts in 5 L flask to form 44 g of CH3COOC2H5 at equilibrium. On taking 120 g of CH3COOH and 46 g of CH3CH2OH, CH3COOC2H5 formed at equilibrium is

Options

A.44 g
B.29.33 g
C.66 g
D.58.67 g

Solution

CH3COOH + C2H5OH $\rightleftharpoons$ CH3COOC2H5 + H2O

Case-I $\frac{60}{60} = 1\text{ mole}$ $\frac{46}{46} = 1\text{ mole}$ 0 0

Moles at Equ. $1 - x$ $1 - x$ $x = \frac{44}{88} = 0.5$ $x$

Case-II $\frac{120}{60} = 2\text{ mole}$ $\frac{46}{46} = 1\text{ mole}$ 0 0

Moles at Equ. 2 – y 1 – y y y

$K_{eq} = \frac{x.x}{(1 - x).(1 - x)} = \frac{y.y}{(2 - y).(1 - y)} \Rightarrow y = \frac{2}{3}$

∴ Mass of CH3COOC2H5 at equilibrium = $\frac{2}{3} \times 88$= 58.67 g

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