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

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


Equation.

C2H5OH+CH3COOH--->CH3COOC2H5+H2O

In the above equation.


☺️ Coefficients are 1

☺️ Exponent are disregard.

☺️1 mole of water is produced.

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

KC=2/3×2/3

=4/9

=0.44

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Answer 2

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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 (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.





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