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Equilibrium Constant and Standard Free Energy Change

In the esterf...

Question

In the esterfication $C_{2} H_{5} O H (l) + C H_{3} C O O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$ an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction = 0.333. Calculate the equilibrium constant.

K = 4

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K = 5

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K = 7

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K = 10

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Solution

The correct option is A $K = 4$
$C_{2} H_{5} O H (l) + C H_{3} C O O H (l) \to C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$
1 1 - -
1-x 1-x x x
$x / 2 = 0.33; x = 0.66$
So $K_{c} = 0.66 \times 0.66 / 0.33 \times 0.33 = 4.$

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Similar questions

C_{2} H_{5} O H (l) + C H_{3} C O O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)

= 0.333.

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{C H}_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ {C H}_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)

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$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$

In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of

little conc. $H_{2} S O_{4}$ . On equilibrium being attained

(C_{2} H_{5} O H)

27^{\circ} C, 33

K_{c}

C_{2} H_{5} O H (l) + C H_{2} C O O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)

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