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When looking at a reaction at equilibrium and trying to decide which way the reaction will proceed with a decrease in pressure, I know we are supposed to chose the side with less moles. But if there is a solid on one side of the reaction should we count the moles it contributes or do all of the elements need to be in the gaseous phase when looking at the pressure aspect of Le Chatelier's Principle?
In compliance with previous responses, you only have to take into account the moles of gas on either side of the reaction. If there is only a solid on one side, then that side would have 0 moles of gas and thus, fewer moles of gas than the other side.
We are always trying to see how Q compares to K when trying to see what will happen with non-equilibrium conditions, and solids and liquids are not included in the equation for K, so they should be disregarded. Also, pressure really only relates to gases anyways.
Only gases are affected by an increase or decrease in pressure through a change in volume specifically. It is important to note that a change in pressure due to an addition of an inert gas will have no effect on the overall reaction.
When changing the pressure of a system make sure to only focus on the moles of gas to see the effects on the reaction. Do not account for the moles of solid, liquid, or aqueous compounds when examining the effects of altering the pressure/ volume.
If there is a solid on one side of the reaction, don't count the moles it contributes. All the elements must be in the gaseous phase because gas is related to pressure. Only take into account gas molecules, not even aqueous, solids, or liquids when dealing with pressure.
You only look at molecules in the gas phase since those are the one's whose concentration is affected by a change in pressure. However, this change in pressure must cause a change in the volume of the container. If the pressure is changed by adding another gas, it won't affect the concentrations of the gases so K would not change and the reaction would still be at equilibrium.
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