Sulfur trioxide gas, one of the causes of acid rain, is produced in the upper atmosphere when oxygen reacts with sulfur dioxide gas in the reaction shown below:

2SO2(g) + O2(g) <--> 2SO3(g) deltaH0 = -197kJ

The gases are placed in a reaction vessel and allowed to come to equilibrium at temperature T, pressure P, and volume V. Predict and explain the effects that each of the following will have on the equilibrium composition of the reaction:

1. The partial pressure of SO3(g) is increased by the addition of SO3(g).
2. The pressure in the vessel is increased by the addition of He.
3. The total pressure in the vessel is increased by decreasing the volume of the vessel.
4. The temperature of the system is decreased.
5. The partial pressure of O2(g) is decreased.

This is all about Le Chatelier's Principle. In a nutshell, that says that a system will shift to undo what we've done to it.

Here is the same equation rewritten to include the heat.
2SO2(g) + O2(g) <--> 2SO3(g) + heat

1. If we add SO3 the reaction will try to get rid of the added SO3. How can it do that? By reacting to the left.
2. Addition of He has no effect since the partial pressures of the components remain the same.
3. Total pressure is increased. The system will try to decrease the pressure. How can it do that? By decreasing the number of moles (it's the molecules impinging on the walls that contributes to the pressure). You have 3 mol on the left; two on the right. Therefore, it will move to the right (to the side with fewer moles of gas)
I'll be glad to critique your thoughts on the others.

To predict the effects of the given changes on the equilibrium composition of the reaction, we need to consider Le Chatelier's principle. This principle states that when a system at equilibrium is subjected to a change, it will adjust to minimize the effect of that change and restore equilibrium.

1. The partial pressure of SO3(g) is increased by the addition of SO3(g):
- According to Le Chatelier's principle, increasing the concentration of a reactant will shift the equilibrium in the direction that consumes the added substance. In this case, increasing the partial pressure of SO3(g) will shift the equilibrium to the left, causing the reverse reaction to occur, and forming more SO2(g) and O2(g). The concentration of SO3(g) will decrease.

2. The pressure in the vessel is increased by the addition of He:
- Adding an inert gas, like helium (He), does not affect the concentrations of the reactants or products. It only increases the total pressure of the system. Hence, the equilibrium composition remains unchanged.

3. The total pressure in the vessel is increased by decreasing the volume of the vessel:
- In this case, decreasing the volume increases the pressure, causing a shift in the equilibrium to the side with fewer gas molecules. Since the forward reaction has two moles of gas, while the reverse reaction has three moles of gas, decreasing the volume will favor the forward reaction, resulting in an increase in the concentration of SO3(g).

4. The temperature of the system is decreased:
- The reaction is exothermic, as indicated by the negative deltaH0 value. According to Le Chatelier's principle, decreasing the temperature of an exothermic reaction favors the reaction that produces heat. Thus, the equilibrium will shift to the right, favoring the formation of more SO3(g) to replace the heat lost. As a result, the concentration of SO3(g) will increase.

5. The partial pressure of O2(g) is decreased:
- Decreasing the partial pressure of O2(g) will create an imbalance in the reaction. To restore equilibrium, the reaction will shift to the left, resulting in a decrease in the concentration of SO3(g). More SO3(g) will react to produce more SO2(g) and O2(g) to compensate for the decreased partial pressure of O2(g).

It's important to note that these predictions are based on the understanding of Le Chatelier's principle and the stoichiometry of the chemical reaction.