1.0 mol of ethanol and 1.0 mol of acetic acid are dissolved in water and kept at 100 °C. The volume of the solution is 250 mL. At equilibrium, 0.25 mol of acetic acid has been consumed in producing ethyl acetate. Calculate Kc at 100 °C for the reaction C2H5OH(aq) + CH3CO2H(aq) CH3CO2C2H5(aq) + H2O(l)

Kc = 0.333

Explanation:

First of all we state the equilibrium reaction:

C₂H₅OH(aq) + CH₃CO₂H(aq)  ⇄  CH₃CO₂C₂H₅(aq) + H₂O(l)

Initially we start with 1 mol of alcohol and 1 mol of acid. After the reaction goes complete, 0.25 moles of acid have been consumed. As ratio is 1:1, we also consumed 0.25 moles of alcohol. Then, in the product side we finally produced 0.25 moles of water and 0.25 moles of ethyl acetate.

How many moles are available in the reactant side, after the equilibrium?

1 mol - 0.25 moles (that have been reacted) = 0.75 moles.

To determine Kc, we need molar concentrations, so let's see M in products and reactants:

0.75 mol / 0.25 L = 3 M

0.25 mol / 0.25 L = 1 M

We needed to convert 0.250 mL to L → 0.250 mL . 1L/ 1000mL = 0.250 L

Let's determine expression for Kc:

Kc = [H₂O] . [CH₃CO₂C₂H₅]  / [C₂H₅OH] . [CH₃CO₂H]

1 . 1 / 3 . 3 = 1/9 → 0.333

20 drops hope this

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