For the equilibrium 2SO2(g) + O2(g) = 2SO3(g), what will change the value of Kp?
A adding a catalyst
B adding more O2
C increasing the pressure
D increasing the temperature
what is the oxidizing agent?
C
The value of Kp, the equilibrium constant, is determined by the stoichiometry of the balanced equation and the temperature.
Option A: Adding a catalyst does not change the value of Kp. A catalyst only increases the rate at which the reaction reaches equilibrium, but it does not affect the position of the equilibrium or the value of Kp.
Option B: Adding more O2 will increase the concentration of O2 and shift the equilibrium towards the products (SO3). This will increase the value of Kp.
Option C: Increasing the pressure will also shift the equilibrium towards the side with fewer moles of gas. In this case, increasing the pressure will favor the formation of SO3 because it has fewer moles of gas. Thus, increasing pressure will increase the value of Kp.
Option D: Increasing the temperature affects the equilibrium constant according to Le Chatelier's principle. In an exothermic reaction, like this one, increasing the temperature will shift the equilibrium in the reverse direction to absorb the added heat. This will decrease the concentration of SO3 and decrease the value of Kp.
In conclusion, options B and C will increase the value of Kp, while option D will decrease the value of Kp. Thus, the correct answer is either B or C.
To understand what will change the value of Kp, we need to analyze the given equilibrium reaction, which is 2SO2(g) + O2(g) ⇌ 2SO3(g).
Kp is the equilibrium constant expressed in terms of partial pressures (or concentrations when dealing with aqueous solutions). It describes the relationship between the concentrations (or partial pressures) of the reactants and products at equilibrium.
Now, let's consider the options and analyze how each one would potentially affect the value of Kp:
A) Adding a catalyst: Adding a catalyst does not change the position of equilibrium or the concentrations (or partial pressures) of the reactants and products at equilibrium. Therefore, adding a catalyst would not change the value of Kp.
B) Adding more O2: According to Le Chatelier's principle, if we add more O2 to the reaction, the equilibrium will shift to the right in an attempt to counteract the increase in O2 concentration. This means that more SO3 will be produced. As a result, both the partial pressures of SO3 and O2 will increase, while the partial pressure of SO2 will decrease. Consequently, the value of Kp will increase due to the increase in the partial pressures of the products.
C) Increasing the pressure: By increasing the pressure, the equilibrium will shift in the direction that reduces the total number of moles of gas particles. In this case, for every 2 moles of SO2 and 1 mole of O2 reacting, 2 moles of SO3 are formed. Since the total number of moles of gas particles decreases from the left to the right side of the reaction, increasing the pressure will shift the equilibrium to the right, resulting in an increase in the concentration (or partial pressure) of SO3. As a result, the value of Kp will increase.
D) Increasing the temperature: When the temperature is increased, the equilibrium will shift in the endothermic direction to absorb the added heat. In this case, the forward reaction (formation of SO3) is exothermic. Therefore, increasing the temperature will shift the equilibrium to the left, resulting in a decrease in the concentration (or partial pressure) of SO3. As a result, the value of Kp will decrease.
Based on the analysis above, option B (adding more O2) and option C (increasing the pressure) will both change the value of Kp. Option D (increasing the temperature) will also change the value of Kp, but in the opposite direction. Therefore, the correct answer is either B or C, depending on the specific conditions and how the equilibrium system is manipulated.