[narutodemonkill]

Consider C(s) + CO2(g) + heat <--> 2CO(g)

1. Carbon dioxide removed from mixture= system shifts left to produce more CO2(g)
2. The internal pressure of reaction vessel increased by addition of N2(g)( I have a feeling answer is no change but why)
3. Reaction vessel cooled so you want more heat system therefore shits left.
4. Volume of reaction vessel increased so pressure decreases not sure what would happen.
5. More graphite is added well graphite is solid does not affect equilibrium so no change.


This is extra homework I'm doing and I think I just don't get how pressure affects the system.
so number 2 and 4

[Mike_the_Tutor]

To understand how the changes in pressure affect the position of the equilibrium stick to this example and imagine you increase the pressure of the reaction vessel. It should be apparent that the right side of the equation will be more 'uncomfortable' with that change as there is more gas on that side. These two moles of carbon monoxide will become more 'squeezed' with the increased pressure and the equilibrium will move to the left.

Let's state clearly according to Le Chatelier:
If you increase the pressure, the equilibrium will shift to the side with least moles of gas.
If you decrease the pressure, the equilibrium will shift to the side with most moles of gas.

This should help with points 2 and 4.

1, 3, 5 you have absolutely correct! 

[samiam]

I get that now but number 2 does not affect the system, is this because.. its adding molecules and at is not really changing pressure of entire system.