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

According to the definition, "Equilibrium constant (Kc) is the ratio of the products of the product concentration that is raised to the powers of their coefficient to the products of the reactant concentration that is raised to powers of their coefficient."

The reaction equation will be - 

C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O

In this equation, the coefficients are all 1 and so, the exponents are disregarded. Also, in this esterification reaction, 1 mole of water is produced for every mole of ethyl acetate produced.

Kc = frac of {[H₂O] x [CH₃COOC₂H₅]} {[CH₃COOH] x [C₂H₅OH]}

Kc = ((2/3) x (2/3)/1 = (4/9)

Therefore, Kc = 0.44.

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