1. Homogeneous Gaseous Equilibria
Many industrial processes involve gaseous systems eg. Haber Process
It is more convenient to use pressure measurements to calculate equilibrium constants
This generates expressions for equilibrium constants, Kp, in terms of the partial pressures of the reactant and products
3H2 (g) + N2 (g)
2NH3 (g)

2. Partial Pressures
In an equilibrium mixture of gases, each component will contribute to the pressure
The total pressure is the sum of the pressures of all the gases
Each gas contributes a partial pressure, p, to the total pressure, P.
However, only total pressure can be measured

3. Total pressure
P = pA + pb + pc 8 7
Total = 20 5
Partial pressure, pA
Partial pressure, pB
Partial pressure, pC
Mole fraction = 8/20
Mole fraction = 7/20
Mole fraction = 5/20
Mole fraction, x = moles of component
total moles of all components

4. Equilibrium mixture is at 3000 kPa
Total pressure
P = pA + pb + pc
Equilibrium mixture is at 3000 kPa 8 7
Total = 20 5
Partial pressure, pA
Partial pressure, pB
Partial pressure, pC
Mole fraction = 8/20
Mole fraction = 7/20
Mole fraction = 5/20
Partial pressure, p = mole fraction x total pressure
pB = 7/20 x 3000
= 1050 kPa
pC = 5/20 x 3000
= 750 kPa
pA = 8/20 x 3000
= 1200 kPa

5. The Equilibrium Constant Kp
The general equation for any homogeneous gaseous reaction at equilibrium is…
aA(g) + bB(g)
cC(g) + dD(g)
Product pressures
Kp = pCc pDd pAa pBb
Reactant pressures
pA represents the partial pressure in kPa or Pa
a,b,c & d are the numbers of moles of substances A, B, C & D

6. Equilibrium mixture is at 3000 kPa
Total pressure
P = pA + pb + pc
Equilibrium mixture is at 3000 kPa 8
A(g) + 3B(g)
2C(g) 7
Total = 20 5
pA = 8/20 x 3000
= 1200 kPa
pB = 7/20 x 3000
= 1050 kPa
pC = 5/20 x 3000
= 750 kPa
Kp = pC pA x pB3
= x
= 4.25 x 10-4 kPa-2

7. Kp Expressions & Units N2(g) + 3H2(g) 2NH3(g)
Kp = (pNH3) (pN2) x (pH2)3
Kp = kPa kPa x kPa3
Kp = kPa2
= kPa-2

8. Partial pressure, p = mole fraction x total pressure
Calculating Kp values
A gaseous mixture contains 7 moles of X, 2 moles of Y and 1 mole of Z at equilibrium at a total presssure of 100kPa. Calculate the value of Kp.
X (g) + Y(g)
2Z (g)
1. Calculate the mole fractions of each gas
Mole fraction, x = moles of component
total moles of all components
2. Calculate partial pressure of each gas
Partial pressure, p = mole fraction x total pressure
Moles
Mole fraction, x
Partial pressure, p X 7
7/10 = 0.7
px = 0.7 x 100 = 70kPa Y 2
2/10 = 0.2
py = 0.2 x 100 = 20kPa Z 1
1/10 = 0.1
pz = 0.1 x 100 = 10kPa
Total 10

9. Calculating Kp values = (10)2 20 x 70 (kPa)2 (kPa) x (kPa)
A gaseous mixture contains 7 moles of X, 2 moles of Y and 1 mole of Z at equilibrium at a total presssure of 100kPa. Calculate the value of Kp.
X (g) + Y(g)
2Z (g)
Moles
Mole fraction, x
Partial pressure, p X 7
7/10 = 0.7
px = 0.7 x 100 = 70kPa Y 2
2/10 = 0.2
py = 0.2 x 100 = 20kPa Z 1
1/10 = 0.1
pz = 0.1 x 100 = 10kPa
Total 10
3. Calculate equilibrium constant Kp, and work out units
= (10) x 70
(kPa)2 (kPa) x (kPa)
Kp = (pZ) (pX) x (pY)
= 0.071

10. Calculating Kp values PCl5 (g) PCl3 (g) + Cl2(g)
A sample of phosporus (V) chloride is introduced to the reaction vessel. The reaction is carried out at 120kPa. At equilibrium the partial pressure of PCl5 is 80kPa. Calculate Kp.
Ptot = pPCl5+ pPCl3+ pCl2
Partial pressure, p
PCl5
pPCL5 = 80kPa
PCl3
pPCl3 =
Cl2
pCl2 =
Total
120kPa
120 = 80+ pPCl3+ pCl2
pPCl3= pCl2(both 1 mole)
pPCl3 = 20kPa
pCl2 = 20kPa

11. Kp = (pPCl3)(pCl2) (pPCl5)
Calculating Kp values
PCl5 (g)
PCl3 (g) + Cl2(g)
A sample of phosporus (V) chloride is introduced to the reaction vessel. The reaction is carried out at 120kPa. At equilibrium the partial pressure of PCl5 is 80kPa. Calculate Kp and state its units.
Kp = (pPCl3)(pCl2) (pPCl5)
= x
Partial pressure, p
PCl5
pPCL5 = 80kPa
PCl3
pPCl3 = 20kPa
Cl2
pCl2 = 20kPa
Total
120kPa
Kp = (kPa)(kPa) (kPa)
Kp = 5 kPa

12. Calculating Kp values 2SO3 (g) 2SO2 (g) + O2 (g)
Initially a vessel contained 12 moles of SO2 and 6 moles of O2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO2 had reacted to form SO3. Calculate Kp for this reaction, and state its units.
Initial moles
Moles at equlib
Mole fraction
Partial pressure
SO2 12 O2 6
SO3
Total
200 kPa

13. Calculating Kp values 2SO3 (g) 2SO2 (g) + O2 (g)
Initially a vessel contained 12 moles of SO2 and 6 moles of O2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO2 had reacted to form SO3. Calculate Kp for this reaction, and state its units.
90% SO2 reacted to form SO3
90% x 12 = 10.8 moles SO3
Initial moles
Moles at equlib
SO2 12 O2 6
SO3
Total
10% SO2 remaining
10% x 12 = 1.2 moles SO2
SO2 reacts with O2 so.....
10% O2 remaining
10% x 6 = 0.6 moles O2

14. Calculating Kp values 2SO3 (g) 2SO2 (g) + O2 (g)
Initially a vessel contained 12 moles of SO2 and 6 moles of O2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO2 had reacted to form SO3. Calculate Kp for this reaction, and state its units.
Initial moles
Moles at equlib
Mole fraction
SO2 12
1.2
1.2/12.6 O2 6
0.6
0.6/12.6
SO3
10.8
10.8/12.6
Total
12.6
At equilibrium
Mole fraction = moles total moles

15. Partial pressure, p = mole fraction x total pressure
Calculating Kp values
2SO3 (g)
2SO2 (g) + O2 (g)
Partial pressure, p = mole fraction x total pressure
Initial moles
Moles at equlib
Mole fraction
Partial pressure
SO2 12
1.2
1.2/12.6 O2 6
0.6
0.6/12.6
SO3
10.8
10.8/12.6
Total
12.6
200 kPa
pSO2 = 1.2/12.6 x 200 kPa = kPa
pO2 = 0.6/12.6 x 200 kPa = 9.52 kPa
pSO2 = 10.8/12.6 x 200 kPa = kPa

16. Calculating Kp values 2SO3 (g) 2SO2 (g) + O2 (g) SO2 O2 SO3
Initial moles
Moles at equlib
Mole fraction
Partial pressure
SO2 12
1.2
1.2/12.6
19.05 O2 6
0.6
0.6/12.6
9.52
SO3
10.8
10.8/12.6
171.43
Total
12.6
200 kPa
= 8.51
Kp = (pSO3) (pSO2)2(pO2)
= (19.05)2 x 9.52
Kp = kPa (kPa)2(kPa)
Kp = kPa-1.