1. Unit II - Equilibrium - Chapter 13

2. 2.1. Characterize chemical reactions in terms of reversibility and relative concentrations of reactants and products. [Readings 13.1 Problems 18]

3. Demonstration:
Fe+3 + SCN- --> FeSCN+2
Fe+3
SCN-
FeSCN+2

4. Demonstration:
Fe+3 + SCN- --> FeSCN+2
Fe+3
SCN-
FeSCN+2

5. Fe+3 + SCN- --> FeSCN+2
Add Fe+3

6. Fe+3 + SCN- --> FeSCN+2
Add Fe+3

7. Fe+3 + SCN- --> FeSCN+2
Add Fe+3
Darker
More FeSCN+2
Excess SCN-
The relative amounts of reactants and products in a chemical reaction are effected by concentration changes.

8. Fe+3 + SCN- --> FeSCN+2
Add Fe+3
Darker
More FeSCN+2
Excess SCN-
Add SCN-

9. Fe+3 + SCN- --> FeSCN+2
Add Fe+3
Darker
More FeSCN+2
Excess SCN-
Add SCN-

10. Fe+3 + SCN- --> FeSCN+2
Add Fe+3
Darker
More FeSCN+2
Excess SCN-
Expt. Shows that Fe3+ and SCN- must be present after the rxn. All three species must exist at the same time.
Add SCN-
Darker
More FeSCN+2
Excess Fe+3
All reactants and products are present during the reaction.

11. Observations on the chemical reaction: Fe+3 + SCN- --> FeSCN+2
All reactants and products are present during the reaction.
Reactions are reversible.
The relative amounts of reactants and products in a chemical reaction are effected by concentration changes.

12. AgNO3(aq) + NaC2H3O2(aq) --> AgC2H3O2(s)+ NaNO3(aq)
25 mL .08 M = .002 moles
AgNO3
25 mL .08 M = .002 moles
NaC2H3O2
AgNO3(aq) + NaC2H3O2(aq) --> AgC2H3O2(s)+ NaNO3(aq)
Ag+ + NO3- + Na+ + C2H3O2- --> AgC2H3O2(s) + Na+ + NO3-
Net Ionic Equation Ag+(aq) + C2H3O2-(aq) --> AgC2H3O2(s)

13. Ag+ + C2H3O2- --> AgC2H3O2(s)
Initial moles moles moles
Final (expected) moles moles moles
Final (experimental) g = moles
Where did the rest go?
moles moles moles
= ==> M reactants are left.

14. AgC2H3O2(s) -->Ag+ + C2H3O2-
Initial moles moles moles
Final moles moles moles
Initial moles moles moles
Final moles moles moles
Turn the reaction around to show silver acetate dissolving.
The moles Ag+ and Ac- are the max soluble
If start with .02 moles of Silver acetate ==> ??
same

15. AgC2H3O2(s) <==> Ag+(aq) + C2H3O2-(aq)
[Ag+]i
[C2H3O2-]i
[Ag+]e
[C2H3O2-]e
0.1 M
0.1 M
0.062 M
0.062 M
0.1 M
0.2 M
0.1 M
0.3 M
Initial concentrations | final concentrations after rxn.
0.038 M Ag+ form the silver acetate solid.
0.1 M
0.4 M

16. AgC2H3O2(s) <==> Ag+(aq) + C2H3O2-(aq)
[Ag+]i
[C2H3O2-]i
[Ag+]e
[C2H3O2-]e
0.1 M
0.1 M
0.062 M
0.062 M
0.1 M
0.2 M
0.029 M
0.124 M
0.1 M
0.3 M
0.016 M
0.209 M
Final data table
0.1 M
0.4 M
0.012 M
0.302 M

17. [Ag+] vs. [C2H3O2-] Final concentrations 0.07  0.06 0.05 0.04 [Ag+]
0.03 
0.02  
0.01
Final concentrations
0.05
0.1
0.15
0.2
0.25
0.3
0.35
[C2H3O2-]

18. [Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -]
0.07
0.07  
0.06
0.06
0.05
0.05
0.04
0.04
[Ag+]
[Ag+]
0.03 
0.03 
0.02
0.02    
0.01
0.01
Silver vs. 1/Ac gives a linear relationship
[Ag+] = K 1/[Ac-] or
K = [Ag+][Ac-]
0.05
0.1
0.15
0.2
0.25
0.3
0.35 2 4 6 8 10 12 14 16 18
[C2H3O2-]
1/[C2H3O2-]

19. [Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -]
0.07
0.07  
0.06
0.06
0.05
0.05
0.04
0.04
[Ag+]
[Ag+]
0.03 
0.03 
0.02
0.02    
0.01
0.01
Silver vs. 1/Ac gives a linear relationship
[Ag+] = K 1/[Ac-] or
K = [Ag+][Ac-]
0.05
0.1
0.15
0.2
0.25
0.3
0.35 2 4 6 8 10 12 14 16 18
[C2H3O2-]
1/[C2H3O2-]
[Ag+] = 4.0 x 10-3 / [C2H3O2-] + 0
[Ag+]
[C2H3O2-] =
4.0 x 10-3

20. [Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -]
0.07
0.07  
0.06
0.06
0.05
0.05
0.04
0.04
[Ag+]
[Ag+]
0.03 
0.03 
0.02
0.02    
0.01
0.01
Silver vs. 1/Ac gives a linear relationship
[Ag+] = K 1/[Ac-] or
K = [Ag+][Ac-]
0.05
0.1
0.15
0.2
0.25
0.3
0.35 2 4 6 8 10 12 14 16 18
[C2H3O2-]
1/[C2H3O2-]
[Ag+] = 4.0 x 10-3 / [C2H3O2-] + 0
K = [Ag+] [C2H3O2-]
[Ag+]
[C2H3O2-] =
4.0 x 10-3

21. 2.2. Determine equilibrium expressions for homogeneous and heterogeneous chemical reactions from stoichiometry. [Readings 13.2, 13.3, & 13.4 Problems 1, 2, 26, 28, 34, 36, 38, & 46]

22. aA + bB <==> cC + dD[ C ] [ D ] K = a b [ A ] [ B ]
In heterogeneous equilibria, pure liquids or solids are not included.
In aqueous equilibria, pure water is not included.

23. Superscripts? A + B <==> 2C A + B <==> C + C
K = [C] [C] / [A] [B]
K = [C]2 / [A] [B]

24. Example
Reaction
N2 (g) + 3 H2 (g) –––> 2 NH3 (g)

25. Example
Reaction
N2 (g) + 3 H2 (g) –––> 2 NH3 (g)
K = ?

26. Example
Reaction
N2 (g) + 3 H2 (g) –––> 2 NH3 (g)

27. Example Reaction N2 (g) + 3 H2 (g) –––> 2 NH3 (g) For rxn:
2 NH3 (g) –––> N2 (g) + 3 H2 (g)
K’ = ?

28. Example Reaction N2 (g) + 3 H2 (g) –––> 2 NH3 (g) For rxn:
2 NH3 (g) –––> N2 (g) + 3 H2 (g)
K’ = 1/K

29. Pure solids or liquids? What is the [AgC2H3O2] ?
= moles / Liter {moles = weight / MW}
= weight / Lx MW
= density / MW = constant for pure substance

30. Water not included? What is the concentration of water in pure water?
1000g / L (1 mole / 18 g) = 55.6 M
What is the concentration of water in 0.1 M NaCl solution? (5.9 g)
994g / L (1 mole / 18 g) = 55.2 M
Therefore [H2O] in solutions is a constant.

31. aA + bB <==> cC + dD[ C ] [ D ] K = a b [ A ] [ B ]
In heterogeneous equilibria, pure liquids or solids are not included.
In aqueous equilibria, pure water is not included.

32. Example
Write equilibruim expressions for the following reactions: (in a 1 liter container)
A. 1/8 S8 (s) + O2 (g) –––> SO2 (g)
B. NH3 (aq) + H2O (l) –––> NH4+ (aq) + OH- (aq)
A. K = [SO2]/[O2] B.
H2O is not included

33. Gas Phase Reactions
Often partial pressures are used instead of concentrations for gas phase reactions.
Rxn:
N2O4 (g) –––2 NO2 (g)
What is the eauilibrium constant?
Solve for Kp

34. Gas Phase Reactions
Often partial pressures are used instead of concentrations for gas phase reactions.
Rxn:
N2O4 (g) –––2 NO2 (g)
What is the eauilibrium constant?
Solve for Kp

35. K is temperature dependent:
Units: Kc M
Kp atm
Kp = Kc (RT) ∆n
K > 1
K < 1
K = 1
N2 + 3H2 ––> 2 NH3 {>,<, OR = 1}
M=mol/L = n/v
Thus: P = N/V * RT = M * RT

36. K Values for Related Chemical Equations
(1) A <=> B K1= x
(2) B <=> A K2= K1-1 = x-1
(3) 2A <=> 2B K3= K12 = x2

37. 2.3. Determine the stoichiometric relationship between initial and equilibrium concentrations of reactants and products. [Readings 13.5 Problems] Determine values for K from equilibrium concentrations of reactants and products in a chemical reaction. [Readings 13.5 Problems 80, ]

38. H2O + HC7H5O2 <=> H3O+ + C7H5O2-
Initial M M M
Equilibrium ? x ?
Change x x x 10-3
Equilibrium x x x 10-3
Initial amount based on experiment
Change amount based on stoichiometry
Equilibrium amount based on equilibrium

39. 2.5. Determine the equilibrium concentrations of reactants and products of a chemical reaction from initial concentrations and values of K. [Readings 13.5 Problems 10, 11, 25, 48, 52, 54, 56, & 58]

40. H2O + HC7H5O2 <=> H3O+ + C7H5O2-
If 61 g of HC7H5O2 is dissolved in 1 L of solution, what is [H3O+] at equilibrium? Kc = 6.3x10-5
61 g = 0.50 mol (122g/mol)
ICE Table!!
H2O + HC7H5O2 <=> H3O+ + C7H5O2-
I M
C x x x
E x x x
Because magnitude is -4, very little of the reactant breaks down, thus we can assume that 0.5-x ≈.5 (If the magnitude is -1 or -2, then use the quadratic eqn.

41. 2.6. Determine if equilibrium has been reached in a chemical reaction; determine the direction the reaction will shift if equilibrium has not been reached. [Readings 13.5 Problems ex 13.6 & prob. 13.7]

42. rR <==> pP
K = [P]ep / [R]er
Q = [P]ip / [R]ir

43. 2.7. Use Le Châtelier’s Principle to predict the direction a reaction at equilibrium will shift as a result of changes in concentration, pressure/volume, and temperature as it approaches a new equilibrium. [Readings 13.6, 13.7, 13.8, & 13.9 Problems 66, 68, 70, & 72]

44. Le Châtelier’s Principle
If a stress is put on a system in equilibrium, it will shift (adjust) to minimize (offset, reverse) the effect of the stress.

45. Le Châtelier’s Principle
If a stress is put on a system in equilibrium, it will shift (adjust) to minimize (offset, reverse) the effect of the stress.

46. Equilibrium: r a t e f o r w a r d R P r a t e r e v e r s e
A system is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.
This does not mean:
The amounts are equal
The reaction has stopped

47. . .
A <==> B

48. Stress -- changes that effect rates of forward and reverse reactions differently
concentration
temperature
catalyst?
pressure?

49. Shift -- An adjustment in the concentrations of reactants and products
Shift to right -- increase [P], decrease [R]
Shift to left -- increase [R], decrease [P]

50. Minimize A(g) + B(g) <=> 2C(g) K = [C]2 / [A] [B] I 2 5 0
E K = 1
add 19 A Q ≠ 1 C
new E K = 1

51. N2 + 3 H2 <==> 2 NH3 .

52. N2 + 3 H2 <==> 2 NH3 Concentration H 2 NH 3 N 2 Time ------->. .
N2 + 3 H2 <==> 2 NH3
Concentration H 2 NH 3 N 2
Time >

53. Demo: 2 NO(g) + O2(g) --> 2 NO2(g)
Initial mL mL
Predicted mL
Actual mL
2 NO2(g) <==> N2O4(g) + ∆
brown colorless

54. Change Temperature
2 NO2(g) <==> N2O4(g) + ∆ Increase T -- More Heat -- P constant Equilibrium Shifts to Reduce Heat Color Deepens
K = [N2O4] / [NO2]2 , K decreases
Therefore K is T dependent

55. Change Pressure (by decreasing V)
2 NO2(g) <==> N2O4(g) + ∆ PT Kp
Equilibrium atm atm atm
P x 2 (1/2 V) atm atm atm
Change atm atm
New atm atm atm Equilibrium
Increase pressure ==> a decrease in the number of moles of gas
decrease in pressure ==> an increase in the number of moles of gas, thus achieving equilibrium.

56. Self Test X(g) + H2O(l) <==> Q (s) + Y(g) ∆H = +150 kJ
Change Shift Change in K
increase Px
increase [Y]
remove 1/2 Q
increase PT
increase T
catalyst
Change Shift ∆K?
increase Px right no
increase [Y] left no
remove 1/2 Q right no
increase PT no chng no
increase T right yes
catalyst no chng no