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.

In the given reaction H2 (g) + I2 (g) -> 2HI (g), the reaction represents the forward direction where hydrogen gas and iodine gas react to form hydrogen iodide gas.

1. If hydrogen iodide gas is added to the system:
- According to Le Chatelier's principle, adding a product (hydrogen iodide) to the reaction mixture will shift the equilibrium towards the reactant side to counteract the change. This means that the reaction will tend to consume some of the added hydrogen iodide gas. Consequently, the amount of iodide present in the system will decrease.

2. If hydrogen iodide gas is removed from the system:
- Removing a product (hydrogen iodide) from the reaction mixture will drive the equilibrium in the forward direction to replenish the concentration of the removed species. As a result, the system will produce more hydrogen iodide gas to compensate for the loss. Consequently, the amount of iodide present in the system will increase.

3. If the volume of the system is decreased:
- When the volume is decreased, the pressure will increase. According to Le Chatelier's principle, if the pressure increases, the system will shift in the direction that reduces the pressure. In this case, the reaction will shift towards the side with fewer gas molecules, which is the reactants' side. Therefore, the system will produce more hydrogen iodide gas. As a result, the amount of iodide present in the system will increase.

Overall, the amount of iodide present in the system will increase when hydrogen iodide is removed or when the volume of the system is decreased. Conversely, adding hydrogen iodide to the system will reduce the amount of iodide present in the system.

In the given reaction, hydrogen gas (H2) and iodine gas (I2) react to form hydrogen iodide gas (HI). Let's analyze the effect of adding or removing hydrogen iodide, as well as the effect of decreasing the volume of the system, on the amount of iodide present.

1. Adding Hydrogen Iodide:
When hydrogen iodide is added to the system, it means that the concentration of the HI molecules increases. According to Le Chatelier's principle, adding a substance to a reaction that appears on the product side will shift the equilibrium towards the reactants. In this case, the equilibrium will shift towards the reactants, favoring the formation of more iodine (I2) and hydrogen gas (H2). Consequently, the concentration of iodide (I) would decrease.

2. Removing Hydrogen Iodide:
If hydrogen iodide is removed from the system, the concentration of HI will decrease. According to Le Chatelier's principle, removing a substance that appears on the product side will shift the equilibrium towards the products. In this case, the equilibrium will shift towards the products, leading to the formation of more iodide. Therefore, removing hydrogen iodide will increase the concentration of iodide (I).

3. Decreasing Volume of the System:
If the volume of the system is decreased, the pressure will increase. In this reaction, there is a change in the number of moles of gas particles. Two moles of HI gas are produced from one mole each of H2 and I2 gas. Therefore, the total number of moles of gas decreases when the reaction proceeds forward. According to Le Chatelier's principle, decreasing the volume will shift the equilibrium towards the side with fewer moles of gas. In this case, it will shift towards the products (HI) to reduce the pressure. As a result, the concentration of iodide (I) would increase.

In summary:
- Adding hydrogen iodide decreases the concentration of iodide (I).
- Removing hydrogen iodide increases the concentration of iodide (I).
- Decreasing the volume of the system increases the concentration of iodide (I).