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 if 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

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