Question

4.6 g of ethyl alcohol mixed with 6g of acetic acid.Find out the value of Kc if 2g of acetic acid remains unused in chemical reaction.

Answer 1

Heya mate
The answer is here



Question
4.6 g of ethyl alcohol mixed with 6g of acetic acid.Find out the value of Kc if 2g of acetic acid remains unused in chemical reaction.





Answer






According to the definition Equilibrium constant is the ratio of products.


C2H5OH+CH3COOH+-->CH3COOC2H5+H2O





So , VALUE OF KC IS


KC = fraction of [H2O]×[CH3COOC2H5][CH3COOH]×[C2H5OH]

KC=2/3 ×2/3
= 4 / 9





hope it helps

Answer 2

Heya mate
The answer of ur question is

QUESTION

4.6 g of ethyl alcohol mixed with 6g of acetic acid.Find out the value of Kc if 2g of acetic acid remains unused in chemical reaction.

As we know that the , Equilibrium constant is the ratio of the product.

Given Equation is here

C2H5OH + CH3COOH->> CH3COOC2H5 + H2O

KC = frac of [H2O]×[CH3COOC2H5][CH3COOH]×[C2H5OH]

KC=2/3×2/3

=4/9

KC =0.44

hope you got it