3.1.10 - Equilbrium Constant Kp Flashcards
What is the definition of partial pressure?
The partial pressure of a gas in a mixture is the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture.
Calculate the partial pressure of N₂ in a mixture containing 0.2 moles N₂, 0.5 moles O₂, and 1.2 moles CO₂ with a total pressure of 3 kPa.
Mole fraction of N₂ = 0.2 / (0.2 + 0.5 + 1.2) = 0.105. Partial pressure of N₂ = 0.105 × 3 kPa = 0.315 kPa.
What is Kp and how is it calculated for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g)?
Kp is the equilibrium constant using partial pressures. For the reaction: Kp = (p(NH₃)²) / (p(N₂) × p(H₂)³).
How do changes in temperature affect the value of Kp and the position of equilibrium for an exothermic reaction?
For an exothermic reaction, increasing the temperature decreases Kp as the system shifts towards reactants (left shift). The position of equilibrium shifts towards the endothermic direction to oppose the increase in temperature.
Does increasing pressure affect the value of Kp?
No, increasing pressure does not change the value of Kp. While pressure changes can shift the position of equilibrium by affecting mole fractions, the value of Kp remains constant because it only varies with temperature.
Calculate the value of Kp for the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), given initial moles and that 20% of N₂ reacted at a total pressure of 2 kPa.
At equilibrium: N₂ = 0.8 moles, H₂ = 2.4 moles, NH₃ = 0.4 moles. Mole fractions (N₂ = 0.222, H₂ = 0.667, NH₃ = 0.111). Partial pressures: N₂ = 0.444 kPa, H₂ = 1.334 kPa, NH₃ = 0.222 kPa. Kp = (0.222²) / (0.444 × 1.334³) = 0.0469 kPa⁻².
How is Kp affected by temperature?
Kp only changes with temperature. They do not change if pressure or concentration are altered. A catalyst also has no effect on Kc or Kp.
What is partial pressure?
Pressure of each gas in a mixture of gases.
Why do catalysts not have an effect on the position of equilibrium?
Catalyst increases the rate of forward and backward reactions by the same amount. Equilibrium position is the same without a catalyst yet just reached quicker.
How does the system react to an increase in pressure in terms of shifting equilibrium for a reaction where the product side has fewer moles of gas?
If pressure is increased in a reaction where the product side has fewer moles of gas, the system shifts towards the side with fewer moles to oppose the change. This shift reduces volume and increases pressure, maintaining the value of Kp.
Provide a scenario where changing conditions affect Kc or Kp.
Changing the temperature affects both Kc and Kp because it alters the reaction’s kinetics and equilibrium position. For exothermic reactions, increasing temperature decreases Kp as the reaction favors reactants to absorb excess heat.
What is the effect of decreasing pressure on the equilibrium yield of SO₃ in the reaction 2SO₂(g) + O₂(g) ↔ 2SO₃(g)?
Decreasing the pressure reduces the yield of SO₃. The equilibrium shifts to the side with more moles of gas to oppose the change in pressure. In this case, the left side (reactants) has 3 moles versus 2 moles on the right (products).
What is the effect on the value of Kp if the pressure of the equilibrium mixture is increased at constant temperature?
The value of Kp remains unchanged. Kp is dependent only on temperature; changes in pressure do not affect Kp.
Does adding a catalyst affect the value of Kp?
No, adding a catalyst does not affect Kp. A catalyst speeds up the attainment of equilibrium by increasing the rate of the forward and backward reactions equally but does not influence the position of equilibrium or Kp.
