Transcript Slide 1

Homogeneous Gaseous Equilibria
• Many industrial processes involve gaseous systems eg.
Haber Process
3H2 (g) + N2 (g)
2NH3 (g)
• 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
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
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
Total pressure
Equilibrium mixture
is at 3000 kPa
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
Partial pressure, p = mole fraction x total pressure
pA = 8/20 x 3000
pB = 7/20 x 3000
pC = 5/20 x 3000
= 1200 kPa
= 1050 kPa
= 750 kPa
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
Total pressure
Equilibrium mixture
is at 3000 kPa
A(g) + 3B(g)
P = pA + pb + pc
8
2C(g)
7
Total = 20
5
pA = 8/20 x 3000
pB = 7/20 x 3000
pC = 5/20 x 3000
= 1200 kPa
= 1050 kPa
= 750 kPa
Kp =
pC2
pA x pB3
=
7502
= 4.25 x 10-4 kPa-2
1200 x 10503
Kp Expressions & Units
N2(g) + 3H2(g)
2NH3(g)
Kp = (pNH3)2
(pN2) x (pH2)3
Kp =
Kp =
kPa2
kPa x kPa3
1
kPa2
= kPa-2
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
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)
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
Kp =
(pZ)2
(pX) x (pY)
=
(10)2
20 x 70
(kPa)2
(kPa) x (kPa)
= 0.071
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.
Partial pressure, p
PCl5
pPCL5 = 80kPa
PCl3
pPCl3 =
Cl2
pCl2 =
Total
120kPa
Ptot = pPCl5+ pPCl3+ pCl2
120 = 80+ pPCl3+ pCl2
pPCl3= pCl2(both 1 mole)
pPCl3 = 20kPa
pCl2 = 20kPa
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 =
Partial pressure, p
PCl5
pPCL5 = 80kPa
PCl3
pPCl3 = 20kPa
Cl2
pCl2 = 20kPa
Total
120kPa
Kp = 5
(pPCl3)(pCl2)
(pPCl5)
(kPa)(kPa)
(kPa)
Kp = 5 kPa
=
20 x 20
80
Calculating Kp values
2SO2 (g) + O2 (g)
2SO3 (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
SO2
12
O2
6
SO3
0
Total
Mole fraction
Partial pressure
200 kPa
Calculating Kp values
2SO2 (g) + O2 (g)
2SO3 (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
SO2
12
O2
6
SO3
0
Total
90% SO2 reacted to form SO3
90% x 12 = 10.8 moles SO3
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
Calculating Kp values
2SO2 (g) + O2 (g)
2SO3 (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
At equilibrium
O2
6
0.6
0.6/12.6
SO3
0
10.8
10.8/12.6
Mole fraction = moles
total moles
Total
12.6
Calculating Kp values
2SO2 (g) + O2 (g)
2SO3 (g)
Partial pressure, p = mole fraction x total pressure
Initial moles Moles at equlib
Mole fraction
SO2
12
1.2
1.2/12.6
O2
6
0.6
0.6/12.6
SO3
0
10.8
10.8/12.6
Total
pSO2 = 1.2/12.6 x 200 kPa
pO2 = 0.6/12.6 x 200 kPa
pSO2 = 10.8/12.6 x 200 kPa
12.6
= 19.05 kPa
= 9.52 kPa
= 171.43 kPa
Partial pressure
200 kPa
Calculating Kp values
2SO3 (g)
2SO2 (g) + O2 (g)
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
0
10.8
10.8/12.6
171.43
Total
12.6
Kp =
(pSO3)2 .
(pSO2)2(pO2)
Kp =
kPa2 .
(kPa)2(kPa)
200 kPa
= 171.43 2
(19.05)2 x 9.52
Kp =
= 8.51
8.51 kPa-1.