- Joined
- May 20, 2008
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- 159
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I will try to give only the relevant info.
CO2(g) + C(s) --><-- 2CO(g)
(where --><--- indicates the reaction is reversible)
Carbon dioxide at 298 K and 1 atm and an excess of powdered carbon dioxide were introduced into a furnace, which was then sealed so that the pressure would increase as the temperature rose. The furnace was heated to, and held constant at, a predetermined temperature.
Q. When the system stabilized at 1,200 K, a sample of helium was injected into the furnace. What should happen to the amount of carbon dioxide in the system?
Ok, so I thought this was just standard Le Chatlier. Amount of gas (and total pressure) goes up, so the equilibrium should shift to the left since that produces less moles of gas and would return the system to its prior condition.
The correct answer is that it should stay the same because the Helium is inert and thus does not react. The explanation also goes into the fact that the partial pressures of CO and CO2 gas will not change just because the Helium gas is introduced, which I understand.
I understand the question and answer now for the most part, my concern is that apparently my concept of Le Chatlier could use some fine tuning. (i.e. blindly saying "number of moles of gas goes up so shift to the side producing fewer moles of gas" can get you in trouble). I'm not sure what qualifications are needed though, and whether I could handle this situation if the volume of the container were NOT held constant or other twists on the question were introduced.
CO2(g) + C(s) --><-- 2CO(g)
(where --><--- indicates the reaction is reversible)
Carbon dioxide at 298 K and 1 atm and an excess of powdered carbon dioxide were introduced into a furnace, which was then sealed so that the pressure would increase as the temperature rose. The furnace was heated to, and held constant at, a predetermined temperature.
Q. When the system stabilized at 1,200 K, a sample of helium was injected into the furnace. What should happen to the amount of carbon dioxide in the system?
Ok, so I thought this was just standard Le Chatlier. Amount of gas (and total pressure) goes up, so the equilibrium should shift to the left since that produces less moles of gas and would return the system to its prior condition.
The correct answer is that it should stay the same because the Helium is inert and thus does not react. The explanation also goes into the fact that the partial pressures of CO and CO2 gas will not change just because the Helium gas is introduced, which I understand.
I understand the question and answer now for the most part, my concern is that apparently my concept of Le Chatlier could use some fine tuning. (i.e. blindly saying "number of moles of gas goes up so shift to the side producing fewer moles of gas" can get you in trouble). I'm not sure what qualifications are needed though, and whether I could handle this situation if the volume of the container were NOT held constant or other twists on the question were introduced.
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