1. Chapter 13
Chemical Equilibrium

2. Chemical Equilibrium
When a reaction is not totally converted from reactants to products, a condition can be set up known as chemical equilibrium.
In this state, the concentrations of all reactants and products remain constant with time, as both forward and reverse reactions are occurring at the same rate.
On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

3. Ratef = Rater Chemical Equilibrium:
occurs when opposing reactions are proceeding at equal rates and the concentrations of products and reactants no longer change with time.
Ratef = Rater

4. Chemical Equilibrium
For this to occur, the system must be closed, no products or reactants may escape the system.

5. For Reaction: AB
Forward Reaction:
AB Rate = kf[A]
kf= forward rate constant
Reverse reaction:
BA Rate = kr[B]
kr= reverse rate constant

6. As forward reaction occurs,
[A] decreases and forward rate slows.
[B] increases and the rate of reverse reaction increases.
Eventually the forward and reverse reactions reach the same rate. (Chemical Equilibrium)

7. kf[A]= kr[B]
Once equilibrium is established, the concentrations of A and B do not change.

8. NOTE: [at equilibrium]
Writing equilibrium expressions
Keq = [products]
[reactants]
NOTE: [at equilibrium]
Note: eq is the general subscript for an equilibrium constant. Your textbook uses Keq throughout. Please look at the equilibrium constants given on the AP exam. These are the same as Keq, but are more specific to a type of reaction (examples: gaseous reaction or acid base reaction).

9. *Kp is used when the reactants and products are in the gaseous state.
Two common ways of describing equilibrium:
Kc = Concentration of substances in the
reaction are known.
Kp = Units of partial pressure are used
instead of concentration.
*Kp is used when the reactants and products are in the gaseous state.
*Solids and liquids are left out of equilibrium expressions because their concentrations do not change during chemical reactions.

10. Equilibrium Expressions Summary
The K value for a reaction that is reversed is the reciprocal of that K for the forward reaction
There are no units for K because of corrections to the nonideal behavior of substances in the reaction.

11. aA + bB  cC + dD Law of Mass Action: Kc = [C]c[D]d [A]a[B]b
Expresses the relationship between concentrations of reactants and products at equilibrium.
Example:
aA + bB  cC + dD
Kc = [C]c[D]d
[A]a[B]b

12. The expression above is known as the equilibrium-constant expression.
Kc = [C]c[D]d
[A]a[B]b
The expression above is known as the equilibrium-constant expression.
Kc is the equilibrium constant.
Subscript c indicates concentration in molarity.
In general, the numerator of an equilibrium constant expression is the product of all concentrations on the product side, each raised to the power of its coefficient.
Denominator is the product of concentrations on the reactant side.

13. Write the equilibrium constant expressions for the following reaction in the forward direction.
2SO2 (g) O2 (g)  2SO3(g)

14. Homework #22
K =
= = 3.2 × 1011

15. Example
Calculations can be made if the equilibrium concentrations of reactants and products are known.
If [NH3] = 3.1 x 10-2 mol/L, [N2]=8.5 x 10-1 mol/L, and
[H2] = 3.1 x 10-3 mol/L,
N2 (g) + 3H2 (g) NH3 (g)
a. What is the value of K?
b. What is the value of K’ (for the reverse reaction)?
c. What would be the K for this reaction?
½ N2 (g) + 3/2 H2 (g) NH3 (g)
(3.1 x 10-2 )2
= (8.5 x 10-1 )(3.1 x 10-3 )3 = 3.8 x 104
K’ = 1/K = 1/ 3.8 x 104 =2.6 x 10-5
For the reaction in (c), Kc = K ½ = (3.8 x 104 L2/mol2) ½ = 1.9 x 102

16. When complete all 3 components are present in the closed tank.
The Haber Process
N2(g) + 3H2(g)  2NH3(g)
This process for making ammonia is done at high temperature and pressure.
When complete all 3 components are present in the closed tank.

17. Regardless of the starting concentrations, at equilibrium the relative concentrations of the 3 gases are the same.
Even if you start with all ammonia, you end up with all 3 at the same relative concentrations at equilibrium.
Equilibrium can be achieved from either direction

18. Equilibrium constant for Haber Process
The equilibrium-constant depends only on the stoichiometry (3,2,1) of the reaction, not on its mechanism.

19. The following equilibrium process has been studied at 230ºC:
2NO(g)+ O2(g) 2NO2(g)
In one experiment the concentration of the reacting species at equilibrium are found to be
[NO]=0.0542M [O2]=0.127M [NO2]=15.5M.
Calculate Kc of the reaction at this temperature.

20. 6.44 x 105 Kc has no units

21. Kc in terms of pressure aA + bB  cC + dD
Example:
aA + bB  cC + dD
PA is partial pressure of A in atm, PB is the partial pressure of B, etc.

22. Example The reaction forming nitrosyl chloride is
2 NO (g) + Cl2 (g) NOCl (g)
If the pressures at equilibrium are PNOCl = 1.2 atm,
PNO=5.0 x 10-2 atm, and PCl2 = 3.0 x 10-1 atm, what is the value of Kp?
Kp = PNOCl = (1.2)2 = x 103 atm-1
(PNO )2(PCl2) (5.0 x 10-2)2(3.0 x 10-1)

23. Δn = nproducts - nreactants
Values for Kc and Kp are usually different.
It is possible to calculate one from the other.
• n is the change in the number of moles of gas.
Δn = nproducts - nreactants
N/v has units of moles per liter (M)
n equals the sum of coefficients of the gas products minus sum of the coefficients of the gas reactants.

24. N2O4(g) 2NO2(g)
n = 2-1 = 1
For the above reaction, Kp=Kc(RT)

25. Calculate the value for Kp.
For the equilibrium 2SO3(g)  2SO2(g) + O2(g) at temperature 1000K, Kc has a value 4.08 x 10-3.
Calculate the value for Kp.
Kp = 4.08 x 10-3 (.0821 x 1000)1
0.335

26. A 3:1 starting mixture of hydrogen, H2, and Nitrogen, N2 comes to equilibrium at 500 °C. The mixture at equilibrium is 3.506% NH3, % N2, and % H2 by volume. The total pressure in the reaction vessel was 50.0 atm. What is the value of Kp and Kc for the reaction?
N2 + 3H NH3
Hint: Use Dalton’s Law of partial pressures to determine the partial pressure of each component in the reaction.
PA = XA PT Since we know we have 100%,
Partial Pressure = (% of substance /100 X total pressure)

27. ( ) ( ) = 4.79 x 104 Δn 1 Kc = Kp RT -2 1 Kc = Kp 0.0821 x 773
P NH3 = x 50.0 atm = atm
P N2 = x 50.0 atm = atm
P H2 = x 50.0 atm = atm
(P NH3)2
(1.75)2
Kp = =
= 11.9
(P N2)
(P H2)3
(48.1)
(0.175)3 Δn ( 1 RT )
Kc = Kp
The (1 / x 773) is just an easier way to write this problem in order to consider the -2 power for only RT. -2 ( 1 x )
= 4.79 x 104
Kc = Kp

28. Homework #28 KP = K(RT)Δn, K = Δn = 2 - 3 = -1;

29. Equilibrium constants
Magnitude of
Equilibrium constants
When Kc is very large the numerator is much larger than the denominator.
The equilibrium lies to the right, or to the product side of the equation (mostly products formed).
K = [products]
[reactants]

30. When the equilibrium constant is very small the equilibrium lies to the left, or the reactant side (mostly reactants).
K>>1 Products favored.
K<<1 Reactants favored.

31. Heterogenous Equilibria
Substances that are in equilibria and are in different phases.
CaCO3(s) CaO(s)+ CO2(gas)
Equilibrium constant
This presents a problem with concentration units.
Concentration equals density divided by its molar mass(M).
Density = g mol = mol
M cm g cm3

32. The concentration of a pure solid or liquid is a constant.
If a pure solid or pure liquid are involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression for the Rx.

33. Write the equilibrium expression Kc and Kp for the following reaction:
3Fe(s)+ 4H2O(g) Fe3O4(s) + 4H2(g)

34. Equilibrium Constants
Applications of
Equilibrium Constants
Magnitude of K determines whether reaction will proceed.
Using K we can predict:
Direction of reaction
Calculate concentrations of reactants and products at equilibrium.

35. Reaction Quotient (Q) N2(g) + 3H2(g)  2NH3(g)
This is used when you are given information about reactants and products when they have not yet reached equilibrium.
Allows us to determine direction reaction WILL go.
--may be given concentrations before reaction begins.
--may be given concentrations determined during a reaction.
Calculate Q as you do K.
N2(g) + 3H2(g)  2NH3(g)
• Compare Q and K as follows.

36. Reaction Quotient (Q) and reaction direction
Q=K when system is at equilibrium
Q>K reaction goes to left(reactants)
Q<K reaction goes to right(products)

37. Direction of a Reaction
Step 1: Calculate the reaction quotient, Qc
Step 2: Compare Qc to Kc

38. case 1: Qc > Kc At equilibrium case 2: Qc = Kc Too much reactants!
Too much product!
Large numerator.
At equilibrium
case 2: Qc = Kc
Too much reactants!
case 3: Qc < Kc
Large denominator.

39. N2(g)+ 3H2(g)  2 NH3(g)
At the start of a reaction, there are mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50 L reaction vessel at 200º C. If the equilibrium constant (Kc) for the reaction is 0.65 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.

40. Initial Concentrations
Example 14.8 page 580 chang

41. N2(g)+ 3H2(g)  2 NH2(g)
Kc= 0.65
Qc is smaller than Kc. Not at equilibrium; too much reactants.
Net Rx goes to right.

42. Problem
N2(g) + 3H2(g) NH3(g)
In the reaction above, the value of Kc at 500°C is 6.0 x At some point during the reaction, the concentrations of each material were measured. At this point, the concentrations of each substance were [N2] = 1.0 x 10-5 M, [H2] = 1.5 x 10-3 M, and [NH3] = 1.5 x 10-3 M. Determine the direction that the reaction was most likely to proceed when the measurements were taken.
(.0015)3
= (1.0 x 10-5) (1.5 x 10-3 )3
Qc=1.0 x Most likely will go to the left (reactant side)

43. Extent of reaction Kc >> 1 (any # > 10)
Forward rxn goes to near completion
Kc << 1 (any # < 0.1)
Reverse rxn goes to near completion

44. Homework #34
Determine Q for each reaction, and compare this value to Kp (2.4 × 103) to determine which direction the reaction shifts to reach equilibrium. Note that, for this reaction, K = Kp since Δn = 0.
Q = = 2.2 × 103
Q < Kp so the reaction shifts right to reach equilibrium.

45. Homework #34 Cont. b. Q = = 4.0 × 103 > Kp
 Reaction shifts left to reach equilibrium.
c. Q = = 2.4 × 103 = Kp; at equilbrium

46. Calculating Equilibrium Constant Concentrations are not known.”
“When all Equilibrium
Concentrations are not known.”

47. Calculating Equilibrium Constants
Equilibrium concentrations are often unknown.
If one equilibrium concentration is known, we can use stoichiometry to find the equilibrium concentration of the other species.

48. Equilibrium Concentrations
Initial concentration
Change in concentration
Equilibrium concentration
Called an ICE diagram

49. H2 I2 2HI Start Δ Finish H2(g) + I2(g) 2HI(g)
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium constant, Kc? Procedure for calculating equilibrium concentrations.
H2(g) + I2(g) HI(g) H2 I2
2HI
Start Δ
Finish
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00
L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium
contant, Kc?
1) First, we design our table so there is a column for each component in the equation, and rows for start, Δ, and finish.

50. H2 I2 2HI Initital 2.0 Change +3.50 Equilibrium 3.50
H2(g) + I2(g) HI(g) H2 I2
2HI
Initital
2.0
Change
+3.50
Equilibrium
3.50
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00
L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium
contant, Kc?
Next, we fill in what we know.
*Since HI is a product, it is 0 at the start of the reaction.
*We also know that at equilibrium, 3.50 mol of HI exist.

51. H2 I2 2HI Start 2.0 Δ +3.50 Finish 3.50 H2(g) + I2(g) 2HI(g)Δ
+3.50
Finish
3.50
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00
L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium
contant, Kc?
3) We now use stoichiometry to fill in the rest of the table.

52. H2 I2 2HI Start 2.0 Δ -1.75 +3.50 Finish 3.50 H2(g) + I2(g) 2HI(g)Δ
-1.75
+3.50
Finish
3.50
3.50 mol HI mol H2
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00
L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium
contant, Kc?
= 1.75 mol H2
2 mol HI
*It is negative because it is being used up in the reaction.

53. H2(g) + I2(g) HI(g) H2 I2
2HI
Start
2.0 Δ
-1.75
+3.50
Finish
0.25
3.50
3.50 mol HI mol I2
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00
L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium
contant, Kc?
= 1.75 mol I2
2 mol HI
*Now we can fill in the “finish” column.

54. (HI)2 (3.50)2 = Kc = = 196 (H2) (I2) (0.25) (0.25) H2 I2 2HI Start 2.0
H2(g) + I2(g) HI(g) H2 I2
2HI
Start
2.0 Δ
-1.75
+3.50
Finish
0.25
3.50
The quantities in the finished row are used to calculate Kc.
When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00
L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium
contant, Kc?
(HI)2
(3.50)2 =
Kc =
= 196
(H2)
(I2)
(0.25)
(0.25)

55. You try this one!
Nitric oxide gas, NO, and oxygen gas, O2, react to form the poisonous gas nitrogen dioxide, NO2, in the reaction shown below:
2NO(g) + O2(g) NO2(g)
10.0 moles of NO and 6.00 moles of O2 are placed into an evacuated 1.00 L vessel, where they begin to react. At equilibrium, there are 8.80 moles of NO2 present. Calculate the value of Kc, assuming that the temperature remains constant throughout the reaction.

56. 2NO O2 2NO2 Initial 10 6 Change 8.8 4.4 1.2 1.6 2NO(g) + O2(g) 2NO2(g)
10.0 moles of NO and 6.00 moles of O2 are placed into an evacuated 1.00 L vessel, where they begin to react. At equilibrium, there are 8.80 moles of NO2 present. Calculate the value of Kc, assuming that the temperature remains constant throughout the reaction.
2NO(g) + O2(g) NO2(g)
2NO O2
2NO2
Initial 10 6
Change
8.8
4.4
Equilibrium
1.2
1.6
Kc = 33.6
Kc= (NO2)2
(O2) (NO) = 33.6

57. When 3. 0 mol of I2 and 4. 0 mol of Br2 are placed in a 2
When 3.0 mol of I2 and 4.0 mol of Br2 are placed in a 2.0 L reactor at 150oC, the following reaction occurs until equilibrium is reached:
I2(g) + Br2(g)  2IBr(g)
Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the value of the equilibrium constant Kc for the reaction?
Kc=3.0
Consider M since we have a 2.0 L vessel.

58. 3. 0 mol of I2 and 4. 0 mol of Br2 are placed in a 2
3.0 mol of I2 and 4.0 mol of Br2 are placed in a 2.0 L reactor at 150oC. Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the value of the equilibrium constant Kc for the reaction?
I2(g) + Br2(g)  2IBr(g) I2
Br2
2IBr
Initial 3 4
Change
3.2
Equilibrium
Kc=3.0
Kc=3.0

59. If you are dealing with the pressures of gases at equilibrium, you will be given: moles, temperature, and volume.
is the key to calculating the pressure of each gas at equilibrium. Stoichiometric equivalents may also be used.
PV = nRT

60. Example: 2SO3(g) 2SO2(g) + O2(g)
Initially, moles of SO3 gas in a 2.0 L vessel are heated to
a temperature of 1000K where it reaches equilibrium. At equilibrium
the vessel is found to contain moles of SO3 gas.
A) Calculate the equilibrium partial pressures of SO2(g) and O2(g).
B) Calculate Keq.

61. 2SO3 2SO2 O2 initial .500 Δ finish PV = nRT P = nRT/V
1) We must first calculate the pressures from the information given in the problem.
PV = nRT P = nRT/V
P = (0.8206)(1000) / (2.0) = atm
This is the initial pressure of SO3 gas.
2SO3(g) SO2(g) O2(g)
2SO3
2SO2 O2
initial
.500 Δ
finish

62. 2SO3 2SO2 O2 initial .500 Δ finish .200 PV = nRT P = nRT/V
P = (0.8206)(1000) / (2.0) = atm
This is the pressure of SO3 gas at equilibrium.
2SO3(g) SO2(g) O2(g)
2SO3
2SO2 O2
initial
.500 Δ
finish
.200

63. 2SO3 2SO2 O2 initial .500 Δ -.300 finish .200
2) Next, we must place the information we know into the table.
2SO3(g) SO2(g) O2(g)
2SO3
2SO2 O2
initial
.500 Δ
-.300
finish
.200

64. 2SO3 2SO2 O2 initial .500 Δ -.300 +.300 +.150 finish .200
Now we use stoichiometric relationships to fill in the rest.
Notice the 2:2 ratio of SO3 and SO2.
Notice the 2:1 ratio of SO3 and O2.
2SO3(g) SO2(g) O2(g)
2SO3
2SO2 O2
initial
.500 Δ
-.300
+.300
+.150
finish
.200

65. 2SO3 2SO2 O2 initial .500 Δ -.300 +.300 +.150 finish .200 .300 .150
3) Now we add the columns. The finish row gives us the partial pressures of each gas at equilibrium.
2SO3(g) SO2(g) O2(g)
2SO3
2SO2 O2
initial
.500 Δ
-.300
+.300
+.150
finish
.200
.300
.150

66. 2SO3 2SO2 O2 initial .500 Δ -.300 +.300 +.150 finish .200 .300 .150
4) Write equilibrium constant expression and solve for Keq.
2SO3(g) SO2(g) O2(g)
2SO3
2SO2 O2
initial
.500 Δ
-.300
+.300
+.150
finish
.200
.300
.150

67. Homework #42 K = 2 NH3(g) ⇌ N2(g) + 3 H2(g) Initial 2M 0 0
Change M → M M
Equil M M M

68. Homework #42 Cont.
[N2] = 0.50 M; [H2] = 1.5 M [NH3]= 1M
K = = 1.7

69. Calculation of Equilibrium
Concentration
Often only Kc and beginning concentrations are known.
Must be able to solve for equilibrium concentrations.

70. Problem
H2(g) + I2(g)  2HI(g)
A mixture of mol H2 and mol I2 was placed in a 1.00L stainless steel flask at 430ºC. Calculate the concentrations of H2, I2, and HI at equilibrium. The equilibrium constant Kc for the reaction is 54.3 at this temperature.

71. H2(g) + I2(g) 2HI(g) Initial Molarity(M) 0.500 0.00 Change(M) -x +2x
Equlibrium (M)
0.500-x 2x
Start by putting in what we know.

72. H2(g) + I2(g) 2HI(g) H2 I2 2HI Start (M) 0.500 M Δ (M) -x 2x
Δ (M) -x 2x
Finish (M)
0.500-x

73. Must now plug x back into “finish” row to calculate actual concentrations.

74. At equilibrium the concentrations are:
[H2]= ( ) = 0.107M
[I2] = ( ) = 0.107M
[HI] = 2(0.393) = 0.786M
Answers can be checked by calculating Kc using the equilibrium concentrations.

75. Homework #46 H2(g) + I2(g) ⇌ 2 HI(g) K = 100.
Initial 1.00 M 1.00M M
Change -x -x → x
Equil – x 1.00 – x x
K = 100. = ;
Taking the square root of both sides: 10.0 =

76. 46 Cont.
10.0 =
x = x,
12.0 x = 9.0,
x = 0.75 M
[H2] = [I2] = = 0.25 M;
[HI] = (0.75) = 2.50 M

77. Shouldn’t need for AP exam but you never know!!
H2(g) + I2(g) 2HI(g)
For the same reaction and temperature, suppose the initial concentrations of HI, H2, and I2 are M, M, & M respectively. Calculate the concentrations at equilibrium.

78. H2(g) + I2(g)  2HI(g) Initial Molarity(M) 0.00623 0.00414 0.0224
Change(M) -x
+2x
Equlibrium (M) x x x

80. ax2 + bx + c = 0 The solution is:
This is a quadratic equation of the form:
ax2 + bx + c = 0
The solution is:
a= 50.3; b = ; c = 8.98x10-4
X = M or x = M
The first solution in impossible since it says more I2 reacted
X = M or x = M The first solution in impossible since it says more I2 reacted

81. At equilibrium the concentrations are:
[H2]= ( ) = M
[I2] = ( ) = M
[HI] = ( ( )= M

82. 5% Rule
Some problems have an x that is really really small in comparison to the equilibrium number.
Usually happens when K is very small also
Assume to ignore the x in the equilibrium concentration to avoid using quadratic equation.
Is a valid assumption when the x value compared to the equilibrium concentration is less than 5%

83. Example of 5% rule #52 N2O4(g) ⇌ 2 NO2(g) K = = 4.0 × 10-7
Initial mol/10.0 L
Change -x +2x
Equil – x 2x

84. #52 Continued
K = = = 4.0 × 10-7;
Because K has a small value, assume that x is small
compared to 0.10 so that x ≈ 0.10.
Solving: 4.0 × ≈ , 4x2 = 4.0 × 10-7 ,
x = 1.0 × 10-4 M
Checking the assumption by the 5% rule:
× 100 = 0.10%
Because this number is less than 5%, we will say that the assumption is valid.

85. #52 Final Answer [N2O4] = 0.10 - 1.0 × 10-4 = 0.10 M;
[NO2] = 2x = 2(1.0 × 10-4) = 2.0 × 10-4 M

86. Le Chatelier's Principle
If a stress is applied to a system in a state of equilibrium, the equilibrium will shift so to relieve that stress.
Stress could be a change in:
temperature
pressure
concentration of one component.

87. Effects of Changes on the System
Concentration: The system will shift away from the added component or toward a removed component.
Temperature: K will change depending upon the type of reaction. Since exothermic produces heat, it will shift away from added, endo the opposite.

88.  In Reactant or Product Concentration
Adding a substance to a system at equilibrium will shift the reaction by consuming some of the added item.
Removing a substance from a system at equilibrium will shift the reaction to form more of the removed substance.

89. Effects of Changes on the System (continued)
3. Pressure:
a. Addition of inert gas does not affect the equilibrium position.
b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.

90. Effects of Volume & Pressure
Reducing the volume of a gas equilibrium system (at constant T) will cause a shift in the direction that reduces the number of moles of gas in the system.
N2O4  2NO2
Reducing volume favors which direction?
Shift to left.
Reduced volume of a gas mixture means that pressure is increased.
Pressure and volume changes do not change the value of K as long as the Temperature remains constant.

91. H2(g) + I2(g)  2HI(g)
Changing pressure or volume of this gaseous state of equilibrium causes what change?
No Change.

92. Effect of Temperature Change
Equilibrium constant values (Keq) always change with temperature change, but stays the same when pressure or concentration is changed.
Enthalpy of the reaction plays a major roll in how heat effects the equilibrium.
Pressure changes do not cause a change in the Kc.

93. Le Châtelier’s Principle
Effect of Temperature Changes
The equilibrium constant is temperature dependent.
For an endothermic reaction, H > 0 and heat can be considered as a reactant.
For an exothermic reaction, H < 0 and heat can be considered as a product.
Adding heat (i.e. heating the vessel) favors away from the increase:
if H > 0, adding heat favors the forward reaction,
if H < 0, adding heat favors the reverse reaction.

94. When temperature is increased the reaction is shifted in the direction that absorbs heat.
Endothermic: + (shift to rt)
Reactants + heat  Products
Exothermic: - (shift to left)
Reactants  Products + heat

95. 2PbS(s) + 3 O2(g) 2PbO(s) + 2SO2(g)
Consider the following equilibrium systems, and predict the direction of the net reaction in each case as a result of  pressure (volume) on the system at constant temperature.
2PbS(s) + 3 O2(g) 2PbO(s) + 2SO2(g)
Shift to right.
PCl5(g) PCl3(g)+Cl2(g)
First: consider only the gaseous molecules. 3moles reactants and 2 moles of products.
Net rx shifts to the right (products)
2nd reaction: 2 moles products/ 1mole reactants Reaction favors left when pressure is increased, products are favored.
Shift to left.
H2(g) + I2(g)  2HI(g)
No influence on position of equilibrium.

96. Consider the following equilibrium process:
N2F4(g) 2 NF2(g) º= 38.5kJ
Predict the changes in the equilibrium if :
The reacting mixture is heated at constant pressure.
Shift to right.
NF2 gas is removed at constant T and Pressure.
a,.) Endothermic, therefore the forward reaction is favored.
Kc is increased with increasing temperature.
b.) Kc remains unchanged, but reaction shifts left, favoring replacing the NF2 that was removed.

97. Example & 16
15—Predict the shift in equilibrium when volume is reduced in each of these(causing increased pressure):
P4 (s) +6 Cl2 (g) PCl3 (l)
Since there are no moles of gas on the right, volume reduction will favor that side.
b. PCl3 (g) + Cl2 (g) PCl5 (g)
Since there are fewer moles of gas on the right, the equilibrium will shift that direction.
c. PCl3 (g) + 3 NH3 (g) P(NH2)3 (g) + 3 HCl (g)
Since there are 4 moles on either side of the equation, there is no shift due to pressure change.
16—For each of these reactions, predict the shift of equilibrium with an increase in temperature
N2 (g) + O2 (g) NO (g) DHº= 181 kJ
This reaction is endothermic, so heat acts as an added reactant, favoring the products to the right
2 SO2 (g) + O2 (g) SO3 (g) DHº= -198 kJ
This reaction is exothermic, heat a product, so it shifts to the left.

98. Summary of LeChatelier
Concentration: Equilibruim shifts away from what is added, toward what is removed.
Temperature: Equilibrium shifts away from added temperature, depending on type of reaction,
Exo-toward reactants Endo-toward products
Pressure: Equilibrium shifts toward less molecules
when pressure is added

99. Effects of Catalysts
Catalysts lower the energy barrier between reactant and products.
Ea for both forward and reverse are lowered to the same extent.
Catalysts increase the rate at which equilibrium is reached.
Does NOT change the composition of the equilibrium mixture.
See diagram on page 582// figure 15.16

100. 3O2(g) 2O3(g) º=284kJ When the above reaction is at equilibrium:
What is the effect of  pressure
by  volume?
 O2?
 Temperature?
Adding catalyst?
Shift to right.
Shift to left.
No change, catalysts only help increase the rate at which equilibrium is reached.
1.) Shift is left to right
favors lowering volume (less moles to the right)
2.) Favors reaction left to right, must decrease the added O2
3.) Favors right to left due to being endothermic.
4.) No effect on the equilibrium
Catalyst only get to equilibrium faster.