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Pressure & volume affecting equilibrium (1 Viewer)
- Thread starter nesstar
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A decrease in volume will cause an increase in pressure, and this will shift an equilibrium system towards whichever side has least GASEOUS particles.
So, a decrease in volume would cause the system:
CO(g) + Cl<sub>2</sub>(g) <---> COCl<sub>2</sub>(g) to move to the right
C(s) + H2O(g) <---> CO(g) + H<sub>2</sub>(g) to move to the left
whilst the system H<sub>2</sub>(g) + Cl<sub>2</sub>(g) <---> 2HCl(g) would not move.
An increase in volume will cause a decrease in pressure, and this will shift an equilibrium system towards whichever side has most GASEOUS particles.
So, an increase in volume would cause the system:
CO(g) + Cl<sub>2</sub>(g) <---> COCl<sub>2</sub>(g) to move to the left
C(s) + H2O(g) <---> CO(g) + H<sub>2</sub>(g) to move to the right
whilst the system H<sub>2</sub>(g) + Cl<sub>2</sub>(g) <---> 2HCl(g) would not move.
A change in pressure caused by some means without a change in volume, such as by adding some inert argon gas, will have no effect on the position of equilibrium.
So, a decrease in volume would cause the system:
CO(g) + Cl<sub>2</sub>(g) <---> COCl<sub>2</sub>(g) to move to the right
C(s) + H2O(g) <---> CO(g) + H<sub>2</sub>(g) to move to the left
whilst the system H<sub>2</sub>(g) + Cl<sub>2</sub>(g) <---> 2HCl(g) would not move.
An increase in volume will cause a decrease in pressure, and this will shift an equilibrium system towards whichever side has most GASEOUS particles.
So, an increase in volume would cause the system:
CO(g) + Cl<sub>2</sub>(g) <---> COCl<sub>2</sub>(g) to move to the left
C(s) + H2O(g) <---> CO(g) + H<sub>2</sub>(g) to move to the right
whilst the system H<sub>2</sub>(g) + Cl<sub>2</sub>(g) <---> 2HCl(g) would not move.
A change in pressure caused by some means without a change in volume, such as by adding some inert argon gas, will have no effect on the position of equilibrium.
Thank you so much! I had an assessment task today and it was so difficult (well, I found it difficult anyway). The equilibrium question was about Fe^3+ + SCN^- <--->FeSCN^2+ , which I had looked at, but the question required strange calculations that I couldn't understand. I understand K, which formed part of the question, but they were talking about a strange machine. I actually left most of it out. I know it's difficult without a question, and I know I'm blabbering on here, but how do you answer questions regarding K when you're not given all the concentrations? <I really must get hold of that question asap!>
well, an easier way... (for me anyway) is to imagine your reaction vessel as a jelly box.
If the volume decreases, then there are more jellys per unit area, that means that the pressure on the jellys have increased(they are unhappy, so they move to sides with less moles of jelly)
Likewise, if the volume increases, there is more space for jellys, and so they won't mind sharing and playing with their friends, (they congregate with each other, shift to more moles of gas)
edit: typo as pointed out by nesstar
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Thanks CM_Tutor for your help...i'm going to have to try and get hold of an actual question from somewhere...yeah, the stuff i can't do is probably prelim stuff, so i feel very stupid....
and thanx eagles...that actually makes sense (except im pretty sure u meant volume increases in the second part)...hmm...i like jelly!
and thanx eagles...that actually makes sense (except im pretty sure u meant volume increases in the second part)...hmm...i like jelly!