What amount of acid is esterified if a mixture consist initially of 1.0mole of ethanoic acid,1.0mole of ethanol and 1.0mole of water. Given that the equilibrium constant for the reaction at 100degree Celsius is 4.0 CH3COOH+CH3CH2OH>CH3COOCH2CH3+H2 O

0.54 mole

Explanation:

                                     CH3COOH    CH3CH2OH   CH3COOCH2CH3    H2O

Initial concentration   1.0 mole         1.0 mole                0 mole                  1.0mol

Change                          - x                   - x                      + x                          + x  

Equilibrium                   (1.0 - x)          (1.0 - x)                   x                       (1.0 + x)

K = [CH3COOCH2CH3]*[H2O]/[CH3COOH]*[CH3CH2OH]

x*(1.0+x)/(1.0-x)(1.0-x) = 4.0

x+x²=4*(1-x)²

x+x² = 4(1² - 2x + x²)

x + x² = 4 - 8x + 4x²

4 - 8x + 4x²- x² - x= 0

3x² - 9x + 4 = 0

x=2.5 , x=0.54

2.5 mole of acid cannot be esterified, because there is only 1.0 mole of acid,

so answer is 0.54 mole.

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