2SO3(g) -> 2SO2(g) + O2(g) Delta H = 197kj
For this reaction shown above, describe the changes that will occur to the number of moles of SO3 under the following conditions:
(a) Additional oxygen is added to the reaction vessel.
(b) The pressure in the reaction vessel is decreased by increasing its volume.
(c) The pressure is increased by the addition of helium gas.
(d)The temperature is increased.
(e) Oxygen gas is removed.
What do you not understand about Le Chatlier's principle?
a)Additional oxygen makes the reaction shift so as to remove oxygen. That means the reactions shifts to the left and SO3 will increase.
(a) Additional oxygen is added to the reaction vessel:
Adding additional oxygen to the reaction vessel will increase the concentration of the reactant, O2(g), on the right side of the equation. According to Le Chatelier's principle, this shift will cause the reaction to shift to the left, favoring the formation of more SO3(g). Consequently, the number of moles of SO3 will increase.
(b) The pressure in the reaction vessel is decreased by increasing its volume:
Increasing the volume of the reaction vessel will lower the pressure inside. According to Le Chatelier's principle, if the pressure is decreased, the reaction will shift to the side with higher total moles of gas to counterbalance the change. In this case, since there are less total moles of gas on the product side (2SO2(g) + O2(g)) compared to the reactant side (2SO3(g)), the reaction will shift to the right and decrease the number of moles of SO3.
(c) The pressure is increased by the addition of helium gas:
Increasing the pressure in the reaction vessel by adding helium gas will not directly affect the number of moles of SO3. Since helium is considered an inert gas and does not participate in the reaction, it will not cause a shift in the equilibrium position.
(d) The temperature is increased:
Increasing the temperature will favor the endothermic reaction as indicated by the positive value of ΔH. According to Le Chatelier's principle, increasing the temperature will cause the reaction to shift in the direction that absorbs heat, which, in this case, is the forward reaction. Hence, increasing the temperature will favor the formation of more SO3, and the number of moles of SO3 will increase.
(e) Oxygen gas is removed:
Removing oxygen gas from the reaction vessel will decrease the concentration of the reactant, O2(g), on the right side of the equation. According to Le Chatelier's principle, this shift will cause the reaction to shift to the right, favoring the formation of more SO3(g). Consequently, the number of moles of SO3 will increase.
To determine the changes that will occur to the number of moles of SO3 under different conditions, we need to consider the stoichiometry of the balanced equation and the Le Chatelier's principle.
(a) Additional oxygen is added to the reaction vessel:
According to the balanced equation, 2 moles of SO3 reacts to form 2 moles of SO2 and 1 mole of O2. If additional oxygen is added, it will shift the equilibrium to the right side to compensate for the increase in the amount of oxygen. This means that more SO3 will react to form additional SO2 and O2. As a result, the number of moles of SO3 will decrease.
(b) The pressure in the reaction vessel is decreased by increasing its volume:
Decreasing the pressure by increasing the volume of the reaction vessel will shift the equilibrium in the direction that produces fewer moles of gas. In the given reaction, only the reactant side has a larger number of moles of gas (2 moles of SO3). Therefore, the equilibrium will shift to the left to produce more SO3. This means that the number of moles of SO3 will increase.
(c) The pressure is increased by the addition of helium gas:
When pressure is increased, the equilibrium shifts in the direction that reduces the total moles of gas. In the given reaction, both the reactant side (2 moles of SO3) and the product side (2 moles of SO2 and 1 mole of O2) have the same total number of moles of gas. Therefore, increasing the pressure by adding helium gas will not affect the number of moles of SO3.
(d) The temperature is increased:
Raising the temperature will favor the endothermic reaction (the reaction with the positive delta H value), as it tends to absorb heat. In this case, the forward reaction is the endothermic reaction since the delta H value is positive. Therefore, increasing the temperature will shift the equilibrium to the right, resulting in more SO3 decomposing to form SO2 and O2. Consequently, the number of moles of SO3 will decrease.
(e) Oxygen gas is removed:
If oxygen gas is removed from the reaction vessel, it will shift the equilibrium to the left to replace the lost oxygen. In order to maintain the stoichiometric balance, more SO3 will decompose to form SO2 and O2. Hence, the number of moles of SO3 will decrease.
It is important to note that changes in moles of substances in a reaction depend on various factors and may also be influenced by other factors not mentioned above. The explanation provided above is based on the given conditions and the general principles of chemical equilibrium.