Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !