A2 Equilibrium Calculations Watch

Jamie_
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The question is:

A mixture of 2.00 mole of A and 1.00 mole of B was allowed to reach equilibrium according to the equation.

A(l) + B(l) ----> C(l) + D(l)

The equilibrium mixture was found to contain 0.400 mole of B. Calculate the value for Kc at the temperature of the reaction.

I constructed this table.

Name:  Screen Shot 2013-11-16 at 20.38.36.png
Views: 256
Size:  12.6 KB

The problem I have is that I haven't been given a volume? At this point I would normally divide the Equilibrium amount (shown in table), by the volume of the container.

So for this question should I just do

1.4V-1 , 0.4V-1 , 0.6V-1 , 0.6V-1



Where V is the volume
?



Thanks for your help.
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propagation
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(Original post by Jamie_)
The question is:

A mixture of 2.00 mole of A and 1.00 mole of B was allowed to reach equilibrium according to the equation.

A(l) + B(l) ----> C(l) + D(l)

The equilibrium mixture was found to contain 0.400 mole of B. Calculate the value for Kc at the temperature of the reaction.

I constructed this table.

Name:  Screen Shot 2013-11-16 at 20.38.36.png
Views: 256
Size:  12.6 KB

The problem I have is that I haven't been given a volume? At this point I would normally divide the Equilibrium amount (shown in table), by the volume of the container.

So for this question should I just do

1.4V-1 , 0.4V-1 , 0.6V-1 , 0.6V-1



Where V is the volume
?



Thanks for your help.
You don't need the volume as the concentration units will cancel out.
top: Mol2Dm-6
bottom: Mol2Dm-6
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Jamie_
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(Original post by propagation)
You don't need the volume as the concentration units will cancel out.top: Mol2Dm-6bottom: Mol2Dm-6
I see that, but since I don't know the concentrations of any of the reactants or products, what do I actually plug into the Kc formula? Kc = [0.6V^-1][0.6V^-1] / [1.4V^-1][0.4V^-1] ?
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propagation
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(Original post by Jamie_)
I see that, but since I don't know the concentrations of any of the reactants or products, what do I actually plug into the Kc formula? Kc = [0.6V^-1][0.6V^-1] / [1.4V^-1][0.4V^-1] ?
Just the values, the Volume is only used to produce the concentration.
Because: Concentration= Moles/Volume
Yet, because we have not been given the volume we cannot work out a concentration so we must plug in the vaues we have into the equation and Kc thereby will not have any units just a value.
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Borek
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Note: in general such problems can't be solved, but this is a particular case, where units cancel out. Assume volume of V, calculate all concentrations, and enter them into the Kc formula. You'll have V2/V2, so the volume doesn't matter.
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riashat
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(Original post by Jamie_)
The question is:

A mixture of 2.00 mole of A and 1.00 mole of B was allowed to reach equilibrium according to the equation.

A(l) + B(l) ----> C(l) + D(l)

The equilibrium mixture was found to contain 0.400 mole of B. Calculate the value for Kc at the temperature of the reaction.

I constructed this table.

Name:  Screen Shot 2013-11-16 at 20.38.36.png
Views: 256
Size:  12.6 KB

The problem I have is that I haven't been given a volume? At this point I would normally divide the Equilibrium amount (shown in table), by the volume of the container.

So for this question should I just do

1.4V-1 , 0.4V-1 , 0.6V-1 , 0.6V-1



Where V is the volume
?



Thanks for your help.
jus sub in the moles at equil for each substance into Kc formula.
if there is no volume then you don't need to use it, in this case it cancels out.
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Borek
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(Original post by riashat)
if there is no volume then you don't need to use it
This is wrong as stated. No information about volume can mean question can't be solved.

Most questions are worded in such a way answer exists, but I have seen many questions that were worded wrong.
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