yasmine_x
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Question 4 I need help on pleaseee
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kurtwhite23
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1) Get a balanced equation (but, this esterification reaction is already balanced)
2) Do the ICE moles method -- initial - change - equilibrium

-------E.acid ------- Ethanol ------- E.Ethanoate ------- Water

I-------0.6---------------0.5------------------0.6-----------------0.4

C-------0.6-x -------- 0.5-x ---------------0.6+x ------------ 0.4+x

E -----0.4

Use the acid's value to calculate x to then find out the rest of moles at equilibrium

Then, divide each value by the total system volume (0.1dm3) to get the concentration at equilibrium, (esterification is a special reaction that doesn't need you to divide by volume, but its good practise)

And sub into the Kc equation (Products / Reactants)

I got Kc = 4 - with no units
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yasmine_x
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(Original post by kurtwhite23)
1) Get a balanced equation (but, this esterification reaction is already balanced)
2) Do the ICE moles method -- initial - change - equilibrium

-------E.acid ------- Ethanol ------- E.Ethanoate ------- Water

I-------0.6---------------0.5------------------0.6-----------------0.4

C-------0.6-x -------- 0.5-x ---------------0.6+x ------------ 0.4+x

E -----0.4



Use the acid's value to calculate x to then find out the rest of moles at equilibrium

Then, divide each value by the total system volume (0.1dm3) to get the concentration at equilibrium, (esterification is a special reaction that doesn't need you to divide by volume, but its good practise)

And sub into the Kc equation (Products / Reactants)

I got Kc = 4 - with no units

What I don't understand is that why isn't the change 0.2? You are given 0.6 initial for ethanoic acid and eqiblirum 0.4 so change is 0.2? and the rest the change would be 0.2? because 1 . 1 ratio. Basically is x = 0.2 I'm so confused
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kurtwhite23
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Yes x = 0.2, sorry that's how I've learnt it with the Change being an algebra equation. So if C = 0.2, then ethanol would be 0.5-0.2 = 0.3 mol at equilibrium etc
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