A2 equilibrium question

5 moles of ethanol, 6 moles of ethanoic acid, 6 moles of ethyl ethanoate and 4 moles of water were mixed together in a stoppered bottle at 15 degrees celcius.

After equilibrium had been attained the bottle was found to contain only 4 moles of ethanoic acid.

a) write an equation for the reaction between ethanol and ethanoic acid to form ethyl ethanoate and water. - I got this.

b) Write an expression for the equilibrium constat Kc for the reaction. Got this as well.

c) How many moles of ethanl, ethyl ethanoate and water are present in the equilbrium mixture. Don't understand this one, is it 5 + 6 + 6 + 4? Or does it want the number of moles after equilbrium has been achieved?

d) what is the value of Kc for this reaction?

Please Help :tsr:

Reply 1

mc_watson87

a) CH3CH2OH + CH3COOH <==> CH3COOCH2CH3 + H2O

b) Kc = [H2O]e[CH3COOCH2CH3]e/[CH3CH2OH]e[CH3COOH]e

c) The number of moles after equilibrium has been reached.

Initial Moles are 5, 6, 6, 4
At equilibrium there are 4 moles of ethanoic acid. 2 moles of ethanoic acid have been used. These 2 moles would have reacted with 2 moles of ethanol as they are 1:1 ratio as shown in the equation. So there would be 3 moles of ethanol at equilibrium. There is a 1:1 ration between reactants and products, so 2 additional moles of water and ethylethanoate would be formed. Giving 8 and 6 moles respectively. Total moles is 21

d) The concentrations of each substance is the equilibrium amount/ over total moles at equilibrium. Shove these concentrations into your Kc equation.