Q.

1.0 mole of ethyl alcohol and 1.0 mole of acetic acid are mixed. At equilibrium, 0.666 mole of ester is formed. The value of equilibrium constant is

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a

14

b

12

c

4

d

3

answer is C.

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Detailed Solution

large {C_2}{H_5}OH + C{H_3}COOH rightleftharpoons C{H_3}COO{C_2}{H_5} + {H_2}O

large {left( {{n_{{C_2}{H_5}OH}}} right)_i} = 1

large {left( {{n_{{CH_3}COOH}}} right)_i} = 1

1 mole of C2H5OH react with 1 mole of CH3COOH gives 1 mole of CH3COOC2H5 and 1 mole of H2O

'X1' mole of C2H5OH react with 'X2' mole of CH3COOH gives 0.666 mole of COOC2H5 and X3 moles of H2O

X1 = X2 = X3 = 0.666

Stoichiometry large mathop {{C_2}{H_5}OH}limits^{1,mole} + large mathop {C{H_3}COOH}limits^{1,mole} large rightleftharpoons large mathop {C{H_3}COO{C_2}{H_5}}limits^{1,mole} + large mathop {{H_2}O}limits^{1,mole}
Initial moles 1 1   0 0
Moles at equilibrium (1 - 0.666) (1 - 0.666)   0.666 0.666

large {K_C} = frac{{left[ {Ester} right]left[ {{H_2}O} right]}}{{left[ {{C_2}{H_5}OH} right]left[ {C{H_3}COOH} right]}}

large {K_C} = frac{0.666times0.666}{0.333times0.333}

large boxed{{K_C} = 4}

 

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