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# Equlibria Question watch

1. Basically, when you are working out Kc concentration for products, you do initial moles + moles reacted when given an equilibrium concentration for a reactant, for example.

So if a reactant was 1 mole intially, but 0.3 at equilbrium and you were finding out the equilibrium moles for product, would it be (0+0.3) or (0+0.7) ?
Im getting confused.

Thanks
2. LPresuming that you are saying the reactant started at 1 mol and was found to have 0.3 mols of it left at equilibrium:

We can calculate a number of moles used up, x:

1 - 0.3 = 0.7

To calculate the number of moles of product at equilibrium we do:

number of moles of product x mol of reactant used up

so:

NaCl ==> Na + 1/2Cl

NaCl:
Initial: 1.0
Equilibrium moles: 0.4

Moles reacted = 0.6

so at equilibrium

Cl: 0.3 moles

Does that make sense?
3. (Original post by S.H.Rahman)
LPresuming that you are saying the reactant started at 1 mol and was found to have 0.3 mols of it left at equilibrium:

We can calculate a number of moles used up, x:

1 - 0.3 = 0.7

To calculate the number of moles of product at equilibrium we do:

number of moles of product x mol of reactant used up

so:

NaCl ==> Na + 1/2Cl

NaCl:
Initial: 1.0
Equilibrium moles: 0.4

Moles reacted = 0.6

so at equilibrium

Cl: 0.3 moles

Does that make sense?
So should I always find out moles reacted? or only equilibrium moles to then find out conc to put into Kc.

I thought Cl would be (0+0.4) then divide by two which is 0.2. So I should always use moles reacted value.
So should I always find out moles reacted? or only equilibrium moles to then find out conc to put into Kc.

I thought Cl would be (0+0.4) then divide by two which is 0.2. So I should always use moles reacted value.
Yes, you should use your 'moles reacted x moles of product' to find your product moles at equilibrium. I would suggest using the ICEC method, which is what I, and everyone else in my school, use to do Kc questions. It really helps!

Where:
x - change in moles (moles reacted)

I - Initial Moles
C - Change
E - Equilibrium moles
C - Conc (divide by volume)

e.g.

NaCl ==> Na + 1/2Cl

Q: 1 mole of NaCl decomposes in a container of volume 200cm^3. It is found that 0.4 moles of NaCl remain at equilibrium. Calculate Kc for this reaction.

Write down what you know:
NaCl | Na | Cl |
I: 1 | 0 | 0 |
C: 1-x | 0+x | 0 + 1/2x | <=== we only add half the change for Cl (1/2Cl)
E: 0.4| | |

NaCl | Na | Cl |
I: 1 | 0 | 0 |
C: 1-x | 0+x | 0 + 1/2x | <=== we only add half the change for Cl (1/2Cl)
E: 0.4| 0.6 | 0.3 | <=== x must be 0.6 (1 - x = 0.4)
C: [divide all above by 0.2]

Kc = ([Na][Cl]^1/2) / ([NaCl])
5. (Original post by S.H.Rahman)
Yes, you should use your 'moles reacted x moles of product' to find your product moles at equilibrium. I would suggest using the ICEC method, which is what I, and everyone else in my school, use to do Kc questions. It really helps!

Where:
x - change in moles (moles reacted)

I - Initial Moles
C - Change
E - Equilibrium moles
C - Conc (divide by volume)

e.g.

NaCl ==> Na + 1/2Cl

Q: 1 mole of NaCl decomposes in a container of volume 200cm^3. It is found that 0.4 moles of NaCl remain at equilibrium. Calculate Kc for this reaction.

Write down what you know:
NaCl | Na | Cl |
I: 1 | 0 | 0 |
C: 1-x | 0+x | 0 + 1/2x | <=== we only add half the change for Cl (1/2Cl)
E: 0.4| | |

NaCl | Na | Cl |
I: 1 | 0 | 0 |
C: 1-x | 0+x | 0 + 1/2x | <=== we only add half the change for Cl (1/2Cl)
E: 0.4| 0.6 | 0.3 | <=== x must be 0.6 (1 - x = 0.4)
C: [divide all above by 0.2]

Kc = ([Na][Cl]^1/2) / ([NaCl])
Ok, i understand. Thanks

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