Question
Two moles of an equimolar mixture of two alcohols R1–OH and R2–OH are esterified with one mole of acetic acid. If only 80% of the acid is consumed till equilibrium and the quantities of ester formed under equilibrium are in the ratio 3: 2. What is the value of equilibrium constant for the esterification of R1–OH?
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Solution
R1OH + CH3COOH $\rightleftharpoons$ CH3COOR1 + H2O
Initial moles 1 1 0 0
Equ. moles 1 – x 1 – (x + y) x x + y
R2OH + CH3COOH $\rightleftharpoons$ CH3COOR2 + H2O
Initial moles 1 1 0 0
Equ. moles 1 – y 1 – (x + y) y x + y
From question, x + y = 0.8 and x/y = 3/2
∴ x = 0.48 and y = 0.32
Now, $K_{1} = \frac{x.(x + y)}{(1 - x)\left\lbrack 1 - (x + y) \right\rbrack} = \frac{0.48 \times 0.8}{0.52 \times 0.2} = 3.69$
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