00:02
Let's begin by writing in equilibrium expression here.
00:07
I'm going to write this as a word equation, products over reactants, molarity of ethyl acetate, water over cetic acid, and ethanol.
00:32
And we're told that the equilibrium constant here is equal to 3 .4.
00:37
So for part a, we're given equilibrium mixture in moles of each.
00:45
Since we're dealing with malarity in the equilibrium expression, we can assume a constant volume and therefore we can substitute moles in.
00:54
So 3 .4, and let's put our known values in.
00:57
The unknown value is ethyl acetate.
01:00
So this would be ethyl acetate.
01:08
And the value that we get will be in moles because we're plugging in moles.
01:13
This will be 12 moles of water, so 12 .0 over acetic acid is 4 .0 and ethanol will be 6 .0.
01:26
So let's solve for the moles of acetic acid, which will be 6 .8 moles.
01:42
So that would be the moles of acetic acid.
01:44
That would be present at equilibrium.
01:49
For part b, we're going to have to set up an ice table.
01:54
So the ice table look like this.
02:00
Acetic acid, ethanol, equilibrium with ethylacetate, and water...