Equilibrium Constant, Kp
19.1 Equilibrium Constant Kp for Homogeneous Systems:
19.1 Equilibrium constant
- Many reversible reactions take place in the gas phase.
- Gaseous equilibria also obey the equilibrium law, but it is usual to express their Kp for homogeneous
concentrations in a different way using the idea of partial pressure. systems
!"#$% "' ()% *
- The mole fraction of gas A = +"+)# !"#$% "' ()%
- The partial pressure of a gas in a mixture is found by multiplying the mole fraction of that gas by the
total pressure.
- Partial pressure of gas A = mole fraction of A x total pressure
- It is worked out the same way as Kc but using the partial pressures rather than concentrations.
e.g., 1.00 moles of PCl5 vapour are heated to 500K in a sealed vessel. The equilibrium
mixture, at a pressure of 625kPa, contains 0.600 moles of chlorine. Calculate Kp
PCl5 ⇌ PCl3 + Cl2
Initial mol 1.00 0.00 0.00
Change mol -0.600 +0.600 +0.600
Equilibrium mol 0.400 0.600 0.600
Mol fraction 0.25 0.375 0.375
Partial pressure 156.25 234.375 234.375
(𝑃𝐶𝑙- )(𝐶𝑙. ) 234.375 × 234.375
𝐾, = =
(𝑃𝐶𝑙/ ) 156.25
Kc = 352 kPa
- It is affected by Le Chatelier’s principle in the same way as Kc. Increasing the temperature of an
exothermic reaction pushes the equilibrium to the left and decreases Kp
19.1 Equilibrium Constant Kp for Homogeneous Systems:
19.1 Equilibrium constant
- Many reversible reactions take place in the gas phase.
- Gaseous equilibria also obey the equilibrium law, but it is usual to express their Kp for homogeneous
concentrations in a different way using the idea of partial pressure. systems
!"#$% "' ()% *
- The mole fraction of gas A = +"+)# !"#$% "' ()%
- The partial pressure of a gas in a mixture is found by multiplying the mole fraction of that gas by the
total pressure.
- Partial pressure of gas A = mole fraction of A x total pressure
- It is worked out the same way as Kc but using the partial pressures rather than concentrations.
e.g., 1.00 moles of PCl5 vapour are heated to 500K in a sealed vessel. The equilibrium
mixture, at a pressure of 625kPa, contains 0.600 moles of chlorine. Calculate Kp
PCl5 ⇌ PCl3 + Cl2
Initial mol 1.00 0.00 0.00
Change mol -0.600 +0.600 +0.600
Equilibrium mol 0.400 0.600 0.600
Mol fraction 0.25 0.375 0.375
Partial pressure 156.25 234.375 234.375
(𝑃𝐶𝑙- )(𝐶𝑙. ) 234.375 × 234.375
𝐾, = =
(𝑃𝐶𝑙/ ) 156.25
Kc = 352 kPa
- It is affected by Le Chatelier’s principle in the same way as Kc. Increasing the temperature of an
exothermic reaction pushes the equilibrium to the left and decreases Kp
