1. A + B C + D Chemical Equilibrium and Equilibrium Law
p.01
Chemical Equilibrium and Equilibrium Law
It is the study of Reversible Rxns!
A + B C + D
The rxn makes products which themselves react to give back the reactants.
The rxn never stops.
C. Y. Yeung (CHW, 2009)

2. How does the Concentrations change with time in a reversible rxn?
p.02
How does the Concentrations change with time in a reversible rxn?
conc.
time
[reactant]
eqm reached: no further change
[reactant]  0
i.e. Not all reactants converted to products.
[product] t’
At time = t’, an equilibrium is established No further change in the concentrations.
At eqm, [reactant] and [product] may not be the same.

3. How does the Rate change with time in a reversible rxn?
p.03
How does the Rate change with time in a reversible rxn?
rate
time
forward rate t’
rate  0
i.e. The reaction is Dynamic!
backward rate
At time = t’, an equilibrium is established Rate of FWR = Rate of BWR  0
FWR and BWR do not stop, but have the same rate.

4. A Set of [Reactant] and [Product] at eqm.
“Chemical Eqm” can exist only if:
A chemical eqm occurs when we have a reversible reaction in a closed system and the conditions do not change.
(temperature / concentration / pressure)
When a condition changes, the eqm will shift to a new “eqm position”.
A Set of [Reactant] and [Product] at eqm.

5. p.05
Dynamic Equilibrium: Observe the “Eqm Shift” from the Colour Change in rxn.
e.g. Br2 + H2O H+ + Br- + HOBr
Adding conc. OH- turns the colour lighter (paler).
[If no eqm is involved, no colour change would be observed.]
(OH- decreases [H+], eqm shifts forward to increase [H+], and decreases [Br2] at the same time).
Adding conc. H+ turns the colour darker (deeper).
[If no eqm is involved, no colour change would be observed.]
([H+] increases, eqm shifts backward to decrease [H+], and increases [Br2] at the same time).
(ref. p. 92 for another example)

6. Calculation about Equilibrium: “Eqm Law” …. (1)
p.06
Calculation about Equilibrium: “Eqm Law” …. (1)
When a condition changes, the eqm would shift to a new eqm positions.
e.g. N2 + 3H NH3
Consider the following 3 eqm positions at temp = T:
[H2]/mol dm-3
[N2]/mol dm-3
[NH3]/mol dm-3
0.50
1.00
0.087
1.35
1.15
0.412
2.44
1.85
1.27
[NH3]2/[N2][H2]3
0.060
eqm constant (Kc)

7. Kc = [NH3(g)]2 [N2(g)][H2(g)]3 Kc = [FeSCN2-(aq)] [Fe3+(aq)][SCN-(aq)]
p.07
Calculation about Equilibrium: “Eqm Law” …. (2)
e.g.1 N2(g) + 3H2(g) NH3(g)
Kc =
[NH3(g)]2
[N2(g)][H2(g)]3
e.g.2 Fe3+(aq) + SCN-(aq) FeSCN2-(aq)
Kc =
[FeSCN2-(aq)]
[Fe3+(aq)][SCN-(aq)]
e.g.3 CaCO3(s) CaO(s) + CO2(g)
Kc = [CO2(g)]
p. 95 Check Point 16-4

8. Position lies to the LEFT Position lies to the RIGHT
Implication of “Kc” ?
p.08
e.g.1 CH3COOH + H2O CH3COO- + H3O+
Kc = 1.0010-5
Position lies to the LEFT
i.e. reactants dominate. ( CH3COOH is a weak acid)
e.g.2 HCl + H2O H3O+ + Cl-
Kc = 5.56108
Position lies to the RIGHT
i.e. products dominate. ( HCl is a strong acid)

9. p. 98 Check Point 16-5A Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) Kc =
at start
at eqm
0.05M 0M
0.0122M
(6.10/1000)(0.05)
25/1000
0.0122M
0.0378M
Kc =
(0.0378)
(0.0122)(0.0122)
= 254 M-1
Note that Kc is quite large, the FWR is more favorable.

10. p. 100 Check Point 16-5B PCl5(g) PCl3(g) + Cl2(g) Kc =
at start (mol dm-3)
at eqm (mol dm-3)
0.009
0.25
0.007
0.252
0.002
Kc =
(0.252)(0.002)
(0.007)
= mol dm-3
Note that Kc is quite small, the BWR is more favorable.

11. Calculation about Eqm involving Gases: “Kp” …. (partial pressures)
e.g. N2O4(g) NO2(g)
At 350K, the eqm pressure is 7104 Pa in a closed container. The mixture is pale brown in colour, and mole fraction of N2O4(g) = 0.46 at eqm. Find Kp. P
N2O4
= 0.46(7104) = 3.22 104 Pa
= 7104 – 3.22 104 = 3.78 104 Pa P
NO2
(3.22 104)
Kp =
(3.78 104)2
= 4.44104 Pa

12. p. 102 Check Point 16-6 H2(g) + I2(g) 2HI(g) at start (mol)
at eqm (mol) 1
0.2
0.2
1.6
total no. of mol = = 2.0 mol P H2
= PT 
0.2
2.0
= P I2
= PT P HI
= PT 
1.6
2.0
= PT
Kp =
= 64
(0.80)2
(0.10)(0.10)

13. p. 230 Q.6(a)(i),(ii) (1996 --- Kp) 2SO2(g) + O2(g) 2SO3(g)
at start (mol)
at eqm (mol) 3 1
1.5
0.25
1.5
Kp =
(P )2
SO3
(P ) O2
SO2
(a) Expression for Kp:
(b)
total no. of mol = = 3.25 mol P
SO2
= 373 
1.5
3.25
= P
SO3
= 172 kPa
Kp =
(172)2
(28.7)
= kPa-1 P O2
= 373 
0.25
3.25
= kPa

14. p. 231 Q.10 (1998 --- Kc) N2(g) + O2(g) 2NO(g) Kc = 1.210-2
at start (mol)
at eqm (mol) 2 1
2-x
1-x 2x
Kc = 1.210-2 =
(2x/2)2
[(2-x)/2][(1-x)/2]
4x2
(2-x)(1-x) =
x = 0.077
 [N2]eqm = (2 – 0.077)/2 = 0.96 mol dm-3

15. Assignment Study the examples in p. 98-104, p.104 Check Point 16-7
p.127 Q.1-5, 7
p. 231 Q.9(a) [due date: 9/3(Wed)]
Quiz on Chemical Kinetics (Ch )
[9/3(Mon)]

16. p.16
Next ….
Significances of Eqm Constant and Le Chatelier’s Principle (p )
[Tuesday]