What will happen to the number of moles of SO3 in equilibrium with SO2 and O2 in the following reaction in each of the following cases?

Question

2 SO3(g) equilibrium reaction arrow 2 SO2(g) + O2(g) ΔH° = 197 kJ
(a) Oxygen gas is removed.
increase
decrease
stay the same

(b) The pressure is increased by decreasing the volume of the reaction container.
increase
decrease
stay the same

(c) The pressure is increased by adding argon gas.
increase
decrease
stay the same

(d) The temperature is decreased.
increase
decrease
stay the same

(e) Gaseous sulfur dioxide is added.
increase
decrease
stay the same

Answer 1

Le Chatelier's Principle reworded in easier to understand language. When a system in equilibrium is subjected to a stress, it will shift so as to undo what we've done to it.
That means when we add a reagent it shift to get rid of the added stuff.
When we increase P it will shift to the side with fewer mols.

Now rewrite the equation to include the heat. Here is what you wrote.
2 SO3(g) equilibrium reaction arrow 2 SO2(g) + O2(g) ΔH° = 197 kJ

Here is my rewrite.
2SO3(g) + heat <==> 2SO2(g) + O2(g)

a. Remove O2 gas. Removing O2 gas means the reaction will shift to try to ADD O2 gas. Which way is that. It will shift to the right. What does that do to the SO3? It decreases it, of course. The problem doesn't ask for it but SO2 will increase.
Now you do the others.