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# calculations using Kc watch

1. I'm really stuck with this passage in the book 'if you have a question in which Kc has no units (with the same total moles on both sides of the equilibrium), then the volume cancels. If you have a problem like this, then use V to represent the volume. So don't panic if you haven't been given the volume in such questions.'

I tried writing the equation to figure it out, but I'm even more confused now:

Kc = (mol/V) (mol/V) / (mol/V) (mol/V)

So if the moles are the same then why don't they cancel out? Why does the volume cancel?
2. The amounts of each species can vary (mol), the volume of the solution/container, however, is a constant throughout.
3. (Original post by EierVonSatan)
The amounts of each species can vary (mol), the volume of the solution/container, however, is a constant throughout.
But don't they use the same volume in all experiments? So why doesn't it cancel in those cases but cancels here?
4. (Original post by tazmaniac97)
But don't they use the same volume in all experiments? So why doesn't it cancel in those cases but cancels here?
It's not the initial volumes. When they are all mixed together (by logic) they must all be in the same volume ...
5. (Original post by tazmaniac97)
But don't they use the same volume in all experiments? So why doesn't it cancel in those cases but cancels here?
It depends upon the exact equilibrium under investigation as to whether the volumes will or will not cancel

The example you gave has a form which can be cancelled. Two volumes on the top of the equation and two on the bottom.

Each equilibrium will have a single volume, regardless. If they didn't then they would be in separate solutions/containers and couldn't be reacting together
6. (Original post by EierVonSatan)
It depends upon the exact equilibrium under investigation as to whether the volumes will or will not cancel

The example you gave has a form which can be cancelled. Two volumes on the top of the equation and two on the bottom.

Each equilibrium will have a single volume, regardless. If they didn't then they would be in separate solutions/containers and couldn't be reacting together
Oh I get it! I can't believe I didn't think of this before

Thank you

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