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# Desperately need help! :( :( :( Watch

1. 0.10 moles of propan-1-ol and 0.20 moles of ethanoic acid was roasted together at 25.C

C3H7OH(l) + CH3COOH(l) <--> CH3COOC3H7 (l) + H2O (l)

At equilibrium the mixture contained 0.05 moles of propyl ethanote. The volume of the mixture was 1dm^3

Calculate the equilibrium constant, Kc at 25C.

2. (Original post by ILoveUSA)
0.10 moles of propan-1-ol and 0.20 moles of ethanoic acid was roasted together at 25.C

C3H7OH(l) + CH3COOH(l) <--> CH3COOC3H7 (l) + H2O (l)

At equilibrium the mixture contained 0.05 moles of propyl ethanote. The volume of the mixture was 1dm^3

Calculate the equilibrium constant, Kc at 25C.

Calculate the moles of each component at equilibrium using the fact that originally there were 0.00 mol of propyl ethanoate and at equilibrium the mixture contains 0.05 mol.

All of the components react in a 1:1 ratio, so you should be able to use stoichiometry to get the equilibrium moles of each.

The total volume = 1 litre, so the concentration of each component is the same as the moles.

Now write out the equilibrium law expression with the equilibrium values ....
3. (Original post by charco)
Calculate the moles of each component at equilibrium using the fact that originally there were 0.00 mol of propyl ethanoate and at equilibrium the mixture contains 0.05 mol.

All of the components react in a 1:1 ratio, so you should be able to use stoichiometry to get the equilibrium moles of each.

The total volume = 1 litre, so the concentration of each component is the same as the moles.

Now write out the equilibrium law expression with the equilibrium values ....
Is this right?

C3H7OH
initial-0.10
reacted - 0.050
equilbirum - 0.05

CH3COOH
initial-0.20
reacted - 0.050
equilbirum - 0.15

CH3COOC3H7
initial - 0.00
equiblirum - 0.050

H2O

I'm stuck on what to put for H2O
4. 0.10 moles of propan-1-ol and 0.20 moles of ethanoic acid was roasted together at 25.C

C3H7OH(l) + CH3COOH(l) <--> CH3COOC3H7 (l) + H2O (l)

At equilibrium the mixture contained 0.05 moles of propyl ethanote. The volume of the mixture was 1dm^3

Calculate the equilibrium constant, Kc at 25C.

(0.05)*(0.05)
/ (0.05) * (0.15)

= 0.33

Is that right?
Someone suggested that it is 0.15

Who is right?
5. (Original post by ILoveUSA)
Is this right?

C3H7OH
initial-0.10
reacted - 0.050
equilbirum - 0.05

CH3COOH
initial-0.20
reacted - 0.050
equilbirum - 0.15

CH3COOC3H7
initial - 0.00
equiblirum - 0.050

H2O

I'm stuck on what to put for H2O
Yes, you've got the idea.

Just apply the same logic to water ... 1 mole of water is produced for every 1 mol of reactant reacting ...
6. (Original post by ILoveUSA)
0.10 moles of propan-1-ol and 0.20 moles of ethanoic acid was roasted together at 25.C

C3H7OH(l) + CH3COOH(l) <--> CH3COOC3H7 (l) + H2O (l)

At equilibrium the mixture contained 0.05 moles of propyl ethanote. The volume of the mixture was 1dm^3

Calculate the equilibrium constant, Kc at 25C.

(0.05)*(0.05)
/ (0.05) * (0.15)

= 0.33

Is that right?
Someone suggested that it is 0.15

Who is right?
If 0.05 mol of ester are produced then the same number of moles of water must be made.

Equilibrium law is [products]/[reactants]

7. (Original post by charco)
If 0.05 mol of ester are produced then the same number of moles of water must be made.

Equilibrium law is [products]/[reactants]

Sorry! So my answer is right then? It should be (0.05*0.05)?
8. 0.10 moles of propan-1-ol and 0.20 moles of ethanoic acid was roasted together at 25.C

C3H7OH(l) + CH3COOH(l) <--> CH3COOC3H7 (l) + H2O (l)

At equilibrium the mixture contained 0.05 moles of propyl ethanote. The volume of the mixture was 1dm^3

Calculate the equilibrium constant, Kc at 25C.

(0.05)*(0.05)
/ (0.05) * (0.15)

= 0.33

Is that right?
Someone suggested that it is 0.15

Who is right?

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Updated: April 16, 2013
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