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In an experiment starting with 1 mole C 2 H 5 OH, 1 mole CH 3 COOH and 1 mole of water, the equilibrium mixture on analysis shows that 54.3% of the acid is esterified. Calculated K c . 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.
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.
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] ------------------------------------------------ [{1-(54.3*1)/100}/V] [{1-(54.3*1)/100}/V] SOLVE THIS EQUATION AND GET YOUR ANSWER IF YOU ARE SATISFIED PLZ DO APPROVE THE ANSWER
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]
------------------------------------------------
[{1-(54.3*1)/100}/V] [{1-(54.3*1)/100}/V]
SOLVE THIS EQUATION AND GET YOUR ANSWER
IF YOU ARE SATISFIED PLZ DO APPROVE THE ANSWER
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