posted by <3 on .
H2 (g) + I2 (g) -> 2HI (g)
What would happen to the amount of iodide present if hydrogen iodide is added? Removed? The volumeof the system is decreased?
I got increase, decrease & same as my answer, but I don't completely understand it. Please help.
Your answers are correct but you don't understand why? Le Chatelier's Principle says, in somewhat unsophisticated terms, that a system in equilibrium will try to undo what we do to it. So if we add HI, the reaction will shift to use up the added HI. The ONLY way it can use HI is for the reaction to go to the left. If HI is removed the system will try to ADD HI. The ONLY way it can do that is to react more H2 with I2 to form HI. Got it?
The same kind of reasoning applies for changes in volume (that changes the concn) BUT I remember another way to do it. When P increases, the reaction shifts to the side with fewer mols (gas reactions only since liquids and solids are not affected by volume/pressure changes). In this case, volume goes up or down the pressure stays the same so the reaction doesn't shift either way since the number of mols is the same on each side (2 vs 2).
For this reaction,
3H2 + N2 ==> 2NH3
Add H2 shifts to the right to form more NH3.
Add N2 shifts to the right.
Increase pressure it shifts to the right because there are 4 mols gas on the left and 2 on the right. It shifts to the side with the fewer mols and that increases NH3.
Increase volume? Increase volume means decrease P so it will shift to the left meaning more H2 and more N2 and less NH3.