# In an experiment starting with 1 mole C2H5OH, 1 mole CH3COOH and 1 mole of water, the equilibrium mixture on analysis shows that 54.3% of the acid is esterified. Calculated Kc.

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## 1 Answers

ankitesh gupta
63 Points
9 years ago

C2H5OH + CH3COOH ---------------> C2H5CH3COO + H2O

INITIAL MOLES:                                 1              1                                      0              1

EQUILIBRIUM MOLES :           1-(54.3*1)/100    1-(54.3*1)/100        0+(54.3*1)/100    1+(54.3*1)/100

* SIGN DENOTES MULTIPLICATION

SINCE THEY ARE TELLING THAT ONLY 54.3% OF ACID IS CONVERTED THIS MEANS THAT 54.3% OF INITIAL MOLES IS CONVERTED INTO ESTER MEANS  (54.3*1)/100 MOLES IS REMOVED . AND THEN ACCORDING TO STICHIOMETERY SINCE ALL THE COEEFICIENTS ARE 1 THEREFORE THE NO. OF MOLES OF ACID WHICH IS CONVERTED INTO ESTER THAN MANY MOLES WILL ALSO BE REMOVED FROM  C2H5OH .

THEREFORE KC= [H2O]1 [C2H5CH3COO]1

----------------------

[CH3COOH ]1 [C2H5OH]1      [     ]  SIGN DENOTES CONCENTRATION MEANS MOLES / VOLUME

KC = [{1+(54.3*1)/100}/V] [{0+(54.3*1)/100}/V]

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[{1-(54.3*1)/100}/V]  [{1-(54.3*1)/100}/V]

SOLVE THIS EQUATION AND GET YOUR ANSWER

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