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What is the equilibrium constant of #CH_3COOH(aq) + C_2H_5OH(aq) -> CH_3COOC_2H_5 (aq) + H_2O(l)#?

Answer 1

#K_"eq"=(CH_3COOC_2H_5)/((CH_3COOH)(C_2H_5OH))#

#K_"eq"="Products"/"Reactants"#

Reactions always find equilibrium at the least restrictive phase which in this problem, is aq. #H_2O# is then represented as a one because it is in a liquid state. Technically you could put a one in the equation but anything times one is itself so there's no need.

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Answer 2

Ellie Andrews

The equilibrium constant, (K_c), for the reaction (CH_3COOH(aq) + C_2H_5OH(aq) \rightarrow CH_3COOC_2H_5 (aq) + H_2O(l)) is given by:

[K_c = \frac{{[CH_3COOC_2H_5][H_2O]}}{{[CH_3COOH][C_2H_5OH]}}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the equilibrium constant of #CH_3COOH(aq) + C_2H_5OH(aq) -&gt; CH_3COOC_2H_5 (aq) + H_2O(l)#?

What is the equilibrium constant of #CH_3COOH(aq) + C_2H_5OH(aq) -> CH_3COOC_2H_5 (aq) + H_2O(l)#?