graph explanation

1. Chemical Equilibrium
Focusing on Acid-Base
Systems
Unit 4

2. Chapter 15:

3. Chapter 15.1:
Reaction Graphs & Equilibrium
is a homogeneous system ( equilibrium)
system has same phase throughout
Rxn on graph:
NOTE: #s of molecules don't matter for equilibrium! Equilibrium is reached when:
Rates of rxn are equal
Conc. Remains constant
There are no visible changes (eg. colour)

4. Equilibrium: the extent of a reaction
In stoichiometry we talked about theoretical
yields, and the many reasons actual yields
may be lower.
Another critical reason actual yields may be
lower is the reversibility of chemical
reactions: some reactions may produce only
70% of the product you may calculate they
ought to produce.
Equilibrium looks at the extent of a chemical
reaction.

5. A chemical system that is separate
from its surroundings where no
matter can enter or leave is called
a closed system.
When a bottle of carbonated
beverage is opened, the pressure on
the system changes and dissolved
gas is allowed to leave the system.
The equilibrium has been disturbed.

6. Closed Systems at Equilibrium
Evidence from many chemical reactions occurring in a
closed system has shown us that after some reactions appear
to have stopped, there is a mixture of reactants and
products present.
Na2SO4(aq) + CaCl2(aq) CaSO4(s) + 2 NaCl(aq)
forward
reverse
We assume that any closed system with no observable
changes occurring is in a state of dynamic equilibrium.
The forward reaction (collisions between reactants to form
products) and the reverse reaction (collisions between
products to form reactants) are occurring simultaneously
and at the same rate.

7. Equilibrium is a state in which there are no observable
changes as time goes by.
Chemical equilibrium is achieved when:
• the rates of the forward and reverse reactions are equal
and
• the concentrations of the reactants and products remain
constant
Physical equilibrium
H2O (l)
Chemical equilibrium
N2O4 (g)
H2O (g)
2NO2 (g)

8. 2 2
H O(l) H O(g)
2+ 2
4 4
CuSO (s) Cu (aq) + SO (aq)

phase equilibrium
solubility equilibrium

9. Rate of sale of
cookies
=
Rate of replacing
cookies

10. Chemical Reaction Equilibrium
2 2
H (g) + I (g) 2 HI(g), 448 C
t  

11. 2 2
H (g) + I (g) 2 HI(g), 448 C
t  
1 – start with reactants only
2 – start with a mixture of reactants and product
3 – start with products only
No matter the starting conditions, the system reaches
a state of dynamic equilibrium each time.

12. 2 2
H (g) + I (g) 2 HI(g), 448 C
t  
The rate of the forward
reaction decreases as the
number of reactant molecules
decreases (fewer collisions).
The rate of the reverse
reaction increases as the
number of product molecules
increases (more collisions).
Dynamic equilibrium is reached when the rate of the forward
reaction is equal to the rate of the reverse reaction.

13. Percent Yield
Percent yield provides a way to refer to the chemicals
present in equilibrium systems.
2 2
H (g) + I (g) 2 HI(g), 448 C
t  
78%
The maximum possible yield is calculated using
stoichiometry, assuming all of the reactant molecules are
used up to form products.

14. Based on percent yield, equilibrium systems fall under one
of the above classifications.

15. The Concept of Equilibrium
• Consider colorless frozen N2O4. At room temperature, it
decomposes to brown NO2:
N2O4(g)  2NO2(g).
• At some time, the color stops changing and we have a
mixture of N2O4 and NO2.
• Chemical equilibrium is the point at which the rate of the
forward reaction is equal to the rate of the reverse reaction.
At that point, the concentrations of all species are constant.
• Using the collision model:
– as the amount of NO2 builds up, there is a chance that
two NO2 molecules will collide to form N2O4.
– At the beginning of the reaction, there is no NO2 so the
reverse reaction (2NO2(g)  N2O4(g)) does not occur.

16. The Concept of Equilibrium
• As the substance warms it begins to decompose:
N2O4(g)  2NO2(g)
• When enough NO2 is formed, it can react to form
N2O4:
2NO2(g)  N2O4(g).
• At equilibrium, as much N2O4 reacts to form NO2
as NO2 reacts to re-form N2O4
• The double arrow implies the process is dynamic.
N2O4(g) 2NO2(g)

17. N2O4 (g) 2NO2 (g)
Start with NO2 Start with N2O4 Start with NO2 & N2O4
equilibrium
equilibrium
equilibrium

18. constant

19. ICE Tables
I – initial
C – change
E – equilibrium
ICE tables are used for quantitative calculations
involving chemical equilibrium systems that
are not quantitative (i.e. < 99.9% yield).

20. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00
Change
Equilibrium
H2 (g) + I2 (g) 2HI (g)

21. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00
Change
Equilibrium
H2 (g) + I2 (g) 2HI (g)

22. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change
Equilibrium
H2 (g) + I2 (g) 2HI (g)

23. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change
Equilibrium 0.22
H2 (g) + I2 (g) 2HI (g)

24. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change
Equilibrium 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)

25. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78
Equilibrium 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)

26. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78
Equilibrium 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)
0.78 x 1 = 0.78
1

27. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78 - 0.78
Equilibrium 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)
0.78 x 1 = 0.78
1

28. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78 - 0.78
Equilibrium 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)
0.78 x 1 = 0.78
1
0.78 x 2 = 0.78
1

29. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78 - 0.78 + 1.6
Equilibrium 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)
0.78 x 1 = 0.78
1
0.78 x 2 = 1.6
1

30. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78 - 0.78 + 1.6
Equilibrium 0.22 0.22
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)
0.78 x 1 = 0.78
1
0.78 x 2 = 1.6
1

31. Set up ICE Table
Conc. [H2 (g)]
(mmol/L)
[I2 (g)]
(mmol/L)
[HI (g)]
(mmol/L)
Initial 1.00 1.00 0.00
Change - 0.78 - 0.78 + 1.6
Equilibrium 0.22 0.22 1.6
1.00 – x = 0.22
0.78
H2 (g) + I2 (g) 2HI (g)
0.78 x 1 = 0.78
1
0.78 x 2 = 1.6
1

32.  Read pgs. 676 – 687
 pg. 682 Practice #’s 3 – 7

34. The Equilibrium Constant, Kc
A + B C + D
a b c d
   
   
C D
A B
c d
c a b
K 
Consider the following generic reaction equation for a system at
equilibrium:
a, b, c, d – coefficients
A, B, C, D – chemical formulas
The equilibrium law expression allows us to calculate the value of
the equilibrium constant, Kc.
If Kc > 1, then the products are favoured at equilibrium.
If Kc < 1, then the reactants are favoured at equilibrium.
products
reactants
The greater the value of the equilibrium constant, the more the products
are favoured at equilibrium.

35.  
   
2
2
2
2
NO (g)
NO(g) O (g)
c
K 
2 2
2 NO(g) + O (g) 2 NO (g)
First, write the balanced equation with whole-number coefficients.

36. Using Equilibrium Constant (Law) more examples
2 important rules:
1. [H2O(l) ] is constant!
Does not change EVER!
DO NOT INCLUDE!
2. also solids (s) are considered constant
ALSO DO NOT INCLUDE!
Eg) 6HCl (aq) + 1Fe2O 3 (s) ↔ 2FeCl 3 (aq) + 3H2O (l)
K = [C]c [D]d = [FeCl 3] 2 [H2O] 3
[A]a [B]b [HCl ] 6 [Fe2O3] 3
Actually is = [FeCl 3] 2
[HCl ] 6

37. Calculating "K"
E.g.2) 1CO(g) + 2H2 (g) ↔ 1CH3OH(aq)
At 773°C the value for the equilibrium constant is 0.398 mol2/L2.
The equilibrium concentration for CO(g) & H2 (g) are 0.105 mol/L &
0.250 mol/L respectively. What is the equilibrium conc. of
methanol?
K = [C]c [D]d = [CH3OH] = [CH3OH] = 0.398 mol2
[A]a [B]b [H2 ] 2 [CO] [0.250] 2 [0.105] L2
[CH3OH] = 2.61 x 10 -3 mol
L

38. 2 2
H (g) + I (g) 2 HI(g) 2 2
2 HI(g) H (g) + I (g)
 
  
2
2 2
HI(g)
H (g) I (g)
c
K 
  
 
2 2
2
H (g) I (g)
HI(g)
c
K 
formation decomposition
reciprocal
Kc = 40 (given)
1
40
c
K 
Kc = 0.025
reciprocal

39. Equilibrium law expressions do NOT include solids or
liquids because their concentrations are fixed – the
chemical amount (number of moles) per unit volume is a
constant value.
  
3
NH (g) HCl(g)
c
K 
4 3
NH Cl(s) NH (g) + HCl(g)

40. 2+
2+
Zn (aq)
Cu (aq)
c
K
 
 

 
 
2+ 2+
Zn(s) + Cu (aq) Cu(s) + Zn (aq)
Ions in solution must be represented as single entities.
Equilibrium constant expressions are always written from the
net ionic form of reaction equations. Spectator ions are not
included.

41. Predicting Final Equilibrium Concentrations
RICE Tables
Are used to help determine changes in
equilibrium after a reaction.
R = Ratio
I = Initial
C = Change
E = Equilibrium

42. Example 1:
In a 1.00 L reaction vessel, hydrogen and
iodine react to form hydrogen iodide. If
0.100 mol of each gas are present initially
and at equilibrium is 0.020 mol of iodine
are present calculate the K for the reaction?

43. H2 (g) + I2 (g)  2HI (g)
R
I
C
E
H2 I2 HI
1 1 2
0.100 0.100 0
-0.08 -0.08 +0.16
0.02 0.020 0.16
Ratio, Initial, Change, Equilibrium
[H2] [I2]
K =
[HI]2
=
[0.020] [0.020]
[0.160]2
= 64

44. Example 2:
Carbon monoxide and water react in an
equilibrium to form carbon dioxide and
hydrogen gas. Each of the reactants is
present initially with a concentration of
0.10 mol/L and the value for K for the
reaction is 4.06. What are the final
concentrations of reactants and products?

45. CO (g) + H2O (g)  CO2 (g) + H2 (g)
CO H2O CO2
1 1 1
0.10 0.10 0
-x -x +x
0.10 - x 0.10 - x x
R
I
C
E
H2
1
0
+x
x
= 4.06
=
[0.10 - x]2
[x]2
K = [CO2][H2]
[CO][H2O]
K value tells us that reaction favors the products.

46. ___x___ = 2.01
[0.10 - x]
x = 0.201 - 2.01x
3.01x = 0.201
x = 0.0668
CO & H2O: 0.10 - 0.0668 = 0.033 M
CO2 & H2: x = 0.0668 = 0.067 M

47. Predicting Final Equilibrium Concentrations
Go over the example above on p. 686-7.

48.  Read pgs. 676 – 689
 pg. 688 Practice #’s 1,3 – 11