ANS:C - decreases

With an increase in temperature, the equilibrium constant at constant pressure (Kp) for the oxidation of sulfur dioxide increases. With an increase in temperature, the equilibrium constant at constant pressure (Kp) for the oxidation of sulfur dioxide increases due to the following reasons:

1. Endothermic Reaction:

• Reaction Type:
  • The oxidation of sulfur dioxide (SO₂) to sulfur trioxide (SO₃) is generally an endothermic reaction. According to Le Chatelier's principle, increasing the temperature shifts the equilibrium to favor the formation of products.

2. Le Chatelier's Principle:

• Equilibrium Shift:
  • When the temperature increases, the system responds by shifting the equilibrium position to the right, leading to a higher concentration of products (in this case, SO₃). This shift results in an increase in the equilibrium constant (Kp).

3. Mathematical Relationship:

• Van't Hoff Equation:
  • The relationship between temperature and equilibrium constant can be described by the Van't Hoff equation, which shows that for endothermic reactions, an increase in temperature leads to an increase in Kp:
  dln⁡KdT=ΔHreactionRT2\frac{d \ln K}{dT} = \frac{\Delta H_{reaction}}{RT^2}dTdlnK​=RT2ΔHreaction​​
  • Here, ΔHreaction\Delta H_{reaction}ΔHreaction​ is positive for endothermic reactions, indicating that Kp increases with temperature.

Summary:

In summary, for the oxidation of sulfur dioxide, increasing the temperature increases the equilibrium constant (Kp) because it shifts the reaction toward the formation of more products, consistent with the principles of chemical equilibrium and thermodynamics.