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?
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
(a) The number of moles of SO3 will decrease. As oxygen gas is removed, the reaction will shift to the left in order to produce more oxygen gas, resulting in a decrease in the number of moles of SO3.
(b) The number of moles of SO3 will increase. By increasing the pressure by decreasing the volume, the equilibrium will shift towards the side with fewer moles of gas. Since there are more moles of gas on the product side (2 moles of SO2 + 1 mole of O2) compared to the reactant side (2 moles of SO3), the equilibrium will shift towards the reactant side, resulting in an increase in the number of moles of SO3.
(c) The number of moles of SO3 will stay the same. Adding inert gas (argon) does not affect the equilibrium position, as it does not participate in the chemical reaction. Therefore, the number of moles of SO3 in equilibrium will remain unchanged.
(d) The number of moles of SO3 will stay the same. Decreasing the temperature will favor the exothermic reaction, which is the reverse reaction in this case. However, this will not directly affect the number of moles of SO3 in equilibrium.
(e) The number of moles of SO3 will decrease. Adding gaseous sulfur dioxide will increase the concentration of the reactant, causing the equilibrium to shift to the left in order to consume the excess sulfur dioxide. This shift will result in a decrease in the number of moles of SO3.
(a) If oxygen gas is removed, the equilibrium position will shift to the left to replace the lost oxygen gas. This means that the amount of SO3 will decrease, so the number of moles of SO3 will decrease.
(b) If the pressure is increased by decreasing the volume of the reaction container, the equilibrium position will shift to the side with fewer moles of gas. In this case, the products side has one mole of gas (SO2) and the reactant side has two moles of gas (SO3). Therefore, the equilibrium position will shift to the right, leading to an increase in the number of moles of SO3.
(c) If the pressure is increased by adding argon gas, it does not have any effect on the equilibrium position or the number of moles of SO3. Argon is an inert gas and does not participate in the reaction.
(d) If the temperature is decreased, the equilibrium position will shift to the side with the lower energy, which is the forward reaction in this case. This means that the amount of SO3 will decrease, so the number of moles of SO3 will decrease.
(e) If gaseous sulfur dioxide is added, it will disturb the equilibrium by increasing the concentration of one of the reactants. As a result, the equilibrium position will shift to the left to reduce the excess SO2. This means that the amount of SO3 will decrease, so the number of moles of SO3 will decrease.
To determine what will happen to the number of moles of SO3 in equilibrium with SO2 and O2 in each of the given cases, we can consider Le Chatelier's principle. This principle states that if a system in equilibrium is subjected to a change, it will adjust in a way that tends to counteract the effect of that change.
(a) If oxygen gas is removed:
According to Le Chatelier's principle, the system will shift in the direction that replenishes the "lost" oxygen, which means the equilibrium will shift to the right. As a result, the number of moles of SO3 will decrease.
(b) If the pressure is increased by decreasing the volume of the reaction container:
When the volume of the reaction container is decreased, the pressure increases. According to Le Chatelier's principle, the system will shift in the direction that reduces the number of gas molecules. In this case, the equilibrium will shift to the left, meaning the number of moles of SO3 will decrease.
(c) If the pressure is increased by adding argon gas:
Adding an inert gas, like argon, does not affect the equilibrium position as long as the volume remains constant. Therefore, the number of moles of SO3 will stay the same.
(d) If the temperature is decreased:
Lowering the temperature will decrease the kinetic energy of the molecules. According to Le Chatelier's principle, the system will shift in the direction that compensates for the decrease in temperature. In this case, since the forward reaction is exothermic (ΔH° = 197 kJ), the equilibrium will shift to the right, meaning the number of moles of SO3 will increase.
(e) If gaseous sulfur dioxide is added:
Adding sulfur dioxide will increase the concentration of the reactant. According to Le Chatelier's principle, the system will shift in the direction that reduces the excess sulfur dioxide. In this case, the equilibrium will shift to the left, so the number of moles of SO3 will decrease.
In summary:
(a) The number of moles of SO3 will decrease.
(b) The number of moles of SO3 will decrease.
(c) The number of moles of SO3 will stay the same.
(d) The number of moles of SO3 will increase.
(e) The number of moles of SO3 will decrease.
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.