Find the equilibrium moles for the following reaction with initially 0.051 mol for Ethyl acetate, 0.260 mol for water and 0.03045 mol of acetic acid at equilibrium.
I'm not sure how to get the ones at equilibrium for the left side or even how much you have to take away for the reaction.
CH3CH2COOCH3 + H2O ---> CH3CH2OH + CH3COOH
initial:
EthylAc: 0.051 mol
Water: 0.260 mol
EthylOH: --------
Acetic: --------
rxn:
EthylAc:
Water:
EthylOH:
Acetic:
eq:
EthylAc: ?
Water: ?
EthylOH: ?
Acetic: 0.03045 mol
Thank you
does it mean that because acetic acid has 0.03045 mol at equilibrium that x = 0.03045 mol? So then Ethanol would also have 0.03045 mol at eq and EtAc and H20 would both be their initial [conc] - x?
because it's a 1:1 ratio for all, then at rxn, we would have -x for both EtAc and H2O and + x on the right, so x = 0.03045.
Is this correct?
You have run your sentence together and I can't be sure about the concentrations. From the table, however, it appears initially you have EtOAc of 0.051 and H2O of 0.260 (so initially EtOH = 0 and acetic acid = 0).
Yes, if you have 0.03045 mols acetic acid at equilibrium you will have 0.03045 mols EtOH at equilibrium.
.......EtOAc + H2O ==> EtOH + HAc
I......0.051..0.260.....0......0
C.......-x.....-x.......x.......x
E.....0.051-x.0.260-x...x......x
So = 0.03045 which allows you to place numbers for all and calculate Keq.(at least I suppose that is what you are to do).
okay thanks Dr.Bob, that's exactly what I did too.
To find the equilibrium moles for the given reaction, we need to use the concept of stoichiometry and the equilibrium constant.
First, let's write the balanced chemical equation for the reaction:
CH3CH2COOCH3 + H2O ---> CH3CH2OH + CH3COOH
The stoichiometry of the reaction tells us that for every one mole of ethyl acetate (CH3CH2COOCH3) that reacts, one mole of ethanol (CH3CH2OH) and one mole of acetic acid (CH3COOH) are formed.
Now, let's examine the initial amounts given:
Ethyl Acetate: 0.051 mol
Water: 0.260 mol
Acetic Acid: 0.03045 mol
From the balanced chemical equation, we see that for every one mole of ethyl acetate that reacts, one mole of acetic acid is formed. Therefore, the amount of acetic acid at equilibrium will be the same as the amount of ethyl acetate initially, which is 0.051 mol.
Now, let's determine the change in moles for water. Since there is a 1:1 stoichiometric ratio between water and ethanol, for every 1 mole of water that reacts, 1 mole of ethanol is formed. Therefore, the decrease in moles of water is equal to the number of moles of ethanol formed.
To determine the equilibrium moles of ethanol, we need to subtract the moles of water that reacted from the initial moles of water. The change in moles of water is calculated as follows:
Change in moles of water = Initial moles of water - Moles of ethanol formed
Change in moles of water = 0.260 mol - 0.051 mol = 0.209 mol (decrease)
Since the decrease in moles of water is equal to the number of moles of ethanol formed, we can conclude that the equilibrium moles of ethanol are 0.209 mol.
Now, let's summarize the results:
Initial:
Ethyl Acetate: 0.051 mol
Water: 0.260 mol
Ethanol: -
Acetic Acid: 0.03045 mol
At Equilibrium:
Ethyl Acetate: ?
Water: ?
Ethanol: 0.209 mol
Acetic Acid: 0.03045 mol
To find the equilibrium moles of ethyl acetate and water, we need to apply the principle of conservation of mass. Since the reaction reaches equilibrium, the amount of ethyl acetate that has reacted will be equal to the amount of ethanol formed, and the amount of water that has reacted will be equal to the amount of ethanol formed.
Therefore, the equilibrium moles of ethyl acetate can be calculated as follows:
Equilibrium moles of ethyl acetate = Initial moles of ethyl acetate - Moles of ethanol formed
Equilibrium moles of ethyl acetate = 0.051 mol - 0.209 mol
Equilibrium moles of ethyl acetate = -0.158 mol
However, it is not physically meaningful to have a negative amount of substance. Therefore, we conclude that the equilibrium moles of ethyl acetate are 0 mol.
Similarly, since the decrease in moles of water is equal to the number of moles of ethanol formed, we can conclude that the equilibrium moles of water are also 0.209 mol.
Final results:
At Equilibrium:
Ethyl Acetate: 0 mol
Water: 0.209 mol
Ethanol: 0.209 mol
Acetic Acid: 0.03045 mol
Note: It is important to mention that the equilibrium concentrations and moles can only be determined accurately if the value of the equilibrium constant (K) is known or if additional information about the reaction conditions is provided. In this case, the information about the equilibrium constant is not given, so we cannot determine the concentrations accurately without that information.