How will the following system at equilibrium shift in each of the following cases?
2 SO3(g) <--> 2 SO2(g) + O2(g) H° = 197 kJ
(a) SO2(g) is added
(b) the pressure is decreased by increasing the volume of the container
(c) the pressure is increased by adding Ne(g)
(d) the temperature is decreased
(e) O2(g) is removed
To determine how the system at equilibrium will shift in each of the given cases, we need to apply Le Chatelier's principle. Le Chatelier's principle states that when a system at equilibrium is disturbed, it will shift in a way that counteracts the disturbance and re-establishes equilibrium.
(a) When SO2(g) is added: According to Le Chatelier's principle, the equilibrium will shift to the left in order to consume some of the added SO2(g). This means that more SO3(g) will be formed, resulting in a decrease in the concentration of SO2(g) and O2(g), while the concentration of SO3(g) will increase.
(b) When the pressure is decreased by increasing the volume of the container: In this case, increasing the volume will decrease the pressure. According to Le Chatelier's principle, the equilibrium will shift to the side with more moles of gas to counteract the decrease in pressure. In this reaction, two moles of SO3(g) form two moles of SO2(g) and one mole of O2(g). Since the forward reaction involves a decrease in the number of moles of gas, the equilibrium will shift to the left, meaning that more SO3(g) will be formed.
(c) When the pressure is increased by adding Ne(g): Adding an inert gas like Ne(g) will not affect the equilibrium because the concentration of reactants and products will remain the same. Therefore, there will be no shift in the equilibrium.
(d) When the temperature is decreased: Decreasing the temperature will shift the equilibrium in the direction that produces heat. In this case, the reaction is exothermic (H° = 197 kJ). Therefore, a decrease in temperature will cause the equilibrium to shift to the right, favoring the exothermic reaction. This means more SO3(g) will be formed.
(e) When O2(g) is removed: According to Le Chatelier's principle, removing one of the products will shift the equilibrium to favor the formation of more of that product. In this case, if O2(g) is removed, the equilibrium will shift to the right, meaning that more SO2(g) and SO3(g) will be formed.
Please note that these predictions are based on the assumption that all other factors, such as concentration, remain constant. Also, it's important to note that the system may not fully shift to the predicted direction, but it will definitely move in that direction to counteract the given disturbance.
To determine how the given system at equilibrium will shift in each case, we need to apply Le Chatelier's principle. Le Chatelier's principle states that when a system at equilibrium is subjected to a stress, the system will shift in a way that reduces the effect of that stress and restores equilibrium.
(a) When SO2(g) is added: Adding more reactant (SO2) will cause the system to shift in the forward direction to consume the excess reactant. The equilibrium will shift to the right to form more products (SO3 and O2).
(b) When the pressure is decreased by increasing the volume of the container: Increasing the volume decreases the pressure. According to Le Chatelier's principle, the system will shift in the direction that produces more gas molecules. In this case, the forward reaction produces 3 moles of gas (2 SO2 + O2), while the reverse reaction produces 2 moles (2 SO3). Therefore, the equilibrium will shift to the left, favoring the side with fewer moles of gas.
(c) When the pressure is increased by adding Ne(g): Adding an inert gas like Ne does not affect the equilibrium position, as it does not participate in the reaction. Therefore, adding Ne will not cause a shift in the equilibrium.
(d) When the temperature is decreased: Decreasing the temperature is equivalent to removing heat from the system. According to Le Chatelier's principle, the system will shift in the direction that produces more heat. Since the reaction is exothermic (H° = -197 kJ), the forward reaction is favored at higher temperatures. Therefore, the equilibrium will shift to the right to increase the heat production and restore equilibrium.
(e) When O2(g) is removed: Removing product (O2) will cause the system to shift in the forward direction to compensate for the loss. The equilibrium will shift to the right to produce more O2 and restore equilibrium.
It is important to note that Le Chatelier's principle provides predictions about the direction of shift, but it does not quantify the extent of the shift. The actual shift in equilibrium depends on the relative magnitudes of the forwards and reverse rate constants and the concentrations of the reactants and products.
Don't you have some ideas about what happens. Remember, Le Chatelier's Principle says that when a system at equilibrium is subjected to a stress, the reaction will shift in such a way so as to relieve the stress.
I'll do the first one.
when SO2 is added (a product) what must the reaction do? It must shift so as to use up the SO2 that was added. How can it do that? It can shift to the left, using up some of the O2 and some of the SO2 and forming more SO3. That way the SO2 concentration (the material which stressed the equilibrium) is reduced. Now you try the others and give reasons.