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.

☺️ Coefficient are 1

☺️ Exponent are disregard

☺️1 mole of water is produced.

KC=frac of [H20]×[CH3COOC3H5][CH3COOH]×[C2H5OH]

KC=2/3×2/3

=4/9

=0.44

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Answer 2

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