1. Chemical Equilibrium in Nature: (The formation of stalagmites and Stalactites)

2. Chemical Equilibrium Consider the following reactions:
CaCO3(s) + CO2(aq) + H2O(l)  Ca2+(aq) + 2HCO3-(aq) ..(1)
and
Ca2+(aq) + 2HCO3-(aq)  CaCO3(s) + CO2(aq) + H2O(l) ..(2)
Reaction (2) is the reverse of reaction (1).
At equilibrium the two opposing reactions occur at the same rate.
Concentrations of chemical species do not change once equilibrium is established.

3. Expression for Equilibrium Constant
Consider the following equilibrium system:
wA + xB ⇄ yC + zD
Kc =
The numerical value of Kc is calculated using the concentrations of reactants and products that exist at equilibrium.

4. Expressions for Equilibrium Constants
Examples:
N2(g) + 3H2(g) ⇄ 2NH3(g); Kc =
PCl5(g) ⇄ PCl3(g) + Cl2(g); Kc =
CH4(g) + H2(g) ⇄ CO(g) + 3H2(g);
Kc =

5. Different kinds of “K”
No matter what the subscript, K IS K IS K IS K IS K! They are all just equilibrium constants and they all get written and used the same way.

6. Kc vs Kp
When a reaction occurs in the gas phase, you can use the partial pressure of the gas instead of the concentration.
To separate the 2 different expressions, they are written differently:
Kc = equilibrium constant with concentrations of species
Kp = equilibrium constant with partial pressures of the species

7. Equilibrium Problems There are two main types of problems:
You know the concentrations and you are calculating the equilibrium constant
You know the equilibrium constant and you are calculating the concentrations
Of course, there are nuances to these problems.

8. Calculating the Equilibrium Constant
The only “trick” to calculating an equilibrium constant is that you must know the concentrations AT EQUILIBRIUM.
You cannot simply take any old concentrations at any old time. The reaction must have reached equilibrium

9. A simple example
Hydrogen and oxygen gas will react to form steam (gaseous water). Hydrogen and oxygen gases were mixed in a 2 L flask at 450 C. After equilibrium was established, it was determined that there were 3 moles of water, 1.2 moles of hydrogen and 0.8 moles of oxygen in the flask. What is the equilibrium constant for this reaction at 450 C?

10. A simple solution 1st you need a balanced equation:
2 H2 (g) + O2 (g)  2 H2O (g)
This allows us to immediately write the equilibrium constant expression:
Keq = Kc =[H2O]2
[H2]2[O2]
Since we know the equilibrium concentrations:
hydrogen: 1.2 moles/2 L = 0.6 M
oxygen: 0.8 moles/2 L = 0.4 M
water: 3 moles/2 L = 1.5 M
Kc =(1.5 M)2
(0.6 M)2(0.4 M)
Kc =15.63
Note: You could also calculate Kp by using the ideal gas law - more on that later.

11. A Note about Units Kc =(1.5 M)2 /(0.6 M)2(0.4 M) Kc =15.63
Notice that I wrote the equilibrium constant without units even though the concentrations have units:
Kc =(1.5 M)2 /(0.6 M)2(0.4 M)
Kc =15.63
Since the units of K will depend on stoichiometry, it is generally considered a UNITLESS QUANTITY!

12. A more complicated problem
Hydrogen and oxygen gas will react to form steam (gaseous water) g of hydrogen and 28.6 g of oxygen were mixed in a 2 L flask at 250 C. After equilibrium was established, it was determined that there was 6.6 g of water What is the equilibrium constant for this reaction at 250 C?

13. A series of simple calculations
1st you need a balanced equation:
2 H2 (g) + O2 (g)  2 H2O (g)
This allows us to immediately write the equilibrium constant expression:
Keq = Kc =[H2O]2
[H2]2[O2]
The question is: what are the equilibrium concentrations of all of the species?

14. Determining the concentrations
ICE - ICE - BABY - ICE – ICE
The easiest way to solve this problem is by using an I-C-E chart (“ice chart”) where I = initial concentration, C= change in concentration, and E = the equilibrium concentration.

15. An ICE Chart
2 H2 (g) + O2 (g)  2 H2O (g)
Initial
Change
Equilibrium

16. What do you know? 4.36 g hydrogen * 1 mol H2 = 2.16 mol H2 2.016 g H2
(this is the INITIAL amount)
28.6 g oxygen * 1 mol O2 = mol O2
32.0 g O2
6.6 g H2O * 1 mol H2O = mol H2O
18.02 g H2O
(this is the EQUILIBRIUM AMOUNT)

17. UNITS! UNITS! UNITS!
An ICE chart can use EITHER moles or concentration (molarity) or even pressure (atm), but you must use only one of these in any single ICE chart.
Kc uses molarity, so it is usually easiest to use concentration
I will do the problem both ways!

18. An ICE Chart 2 H2 (g) + O2 (g)  2 H2O (g) Initial Change Equilibrium
What is the change in quantities?
2.16 mol
0.894 mol
0 mol
?????
??????
????
0.366 mol

19. The “change” is all about stoichiometry!
2 H2 (g) + O2 (g)  2 H2O (g)
Initial
Change
Equilibrium
Now it is easy to finish filling in the ICE chart!
2.16 mol
0.894 mol
0 mol
- 2 x x
+ 2 x
????
?????
0.366 mol

20. An ICE chart is really just “accounting for moles”
2 H2 (g) + O2 (g)  2 H2O (g)
Initial
Change
Equilibrium
It is often helpful to use an ICE chart for other types of problems, it is a great way to keep track of what is going on.
2.16 mol
0.894 mol
0 mol
- 2 x x
+ 2 x
2.16 – 2 x
0.894 – x
2 x = mol

21. Determining x allows me to fill in the rest of the chart
2 H2 (g) + O2 (g)  2 H2O (g)
Initial
Change
Equilibrium
2 x = mol
x = mol
2.16 mol
0.894 mol
0 mol
- 2 x x
+ 2 x
2.16 – 2 x
0.894 – x
2 x = mol

22. Determining x allows me to fill in the rest of the chart
2 H2 (g) + O2 (g)  2 H2O (g)
Initial
Change
Equilibrium
2.16 mol
0.894 mol
0 mol
- 2 (0.183 mol)
0.183 mol
+ 2 (0.183 mol)
2.16 – 2 (0.183)
1.794 mol
0.894 – 0.183
0.711 mol
0.366 mol

23. Now we need to calculate the concentrations and put them into Kc
[H2] = mol/2L = M
[O2]= mol/2L =0.356 M
[H2O] = mol/2L = M
Keq = Kc = [H2O]2
[H2]2[O2]
Keq = Kc = [0.183]2
[0.897]2[0.356]
Kc = 0.117

24. We could do the same calculation using concentrations directly in the ICE chart
2.16 mol H2/2L = 1.08 M
(this is the INITIAL concentration)
0.894 mol O2/2L = M
0.366 mol H2O/2L = M
(this is the EQUILIBRIUM concentration)

25. An ICE Chart 2 H2 (g) + O2 (g)  2 H2O (g) Initial Change Equilibrium
2x = M
x = M
1.08 M
0.447 M
0 M
- 2 x -x
+ 2 x
1.08 – 2x x
2x = M

26. An ICE Chart 2 H2 (g) + O2 (g)  2 H2O (g) Initial Change Equilibrium
- 2 x -x
+ 2 x
1.08 – 2(0.0915)
0.897 M
0.447 –
0.356 M
0.183 M

27. Now we simply put the concentrations into Kc
Keq = Kc = [H2O]2
[H2]2[O2]
Keq = Kc = [0.183]2
[0.897]2[0.356]
Kc = 0.117
Same answer as before!

28. How to use Kc
What if you already know the equilibrium constant? How do you use it?
If you know the equilibrium constant, you can use it to determine the equilibrium concentrations of all of the reacting species!

29. How to use Kc
What if you already know the equilibrium constant? How do you use it?
If you know the equilibrium constant, you can use it to determine the equilibrium concentrations of all of the reacting species!

30. Another Simple Problem
The Kc value for the formation of water from hydrogen and oxygen at 850 C is 4x If I mix 5.0 grams of hydrogen and 5.0 grams of oxygen in a 3 L flask at 850 C, what is the equilibrium concentration of the water?

31. Another simple solution
1st you need a balanced equation:
2 H2 (g) + O2 (g)  2 H2O (g)
This allows us to immediately write the equilibrium constant expression:
Kc =[H2O]2 = 4x10-6
[H2]2[O2]

32. Again, the Power of ICE 2 H2 (g) + O2 (g)  2 H2O (g) Initial Change
Equilibrium
The “Change” line is always just stoichiometry
-2x -x
+ 2 x

33. We already know a couple of things
5.0 g hydrogen * 1 mol H2 = 2.48 mol H2
2.016 g H2
2.48 mol H2 = M
3 L
5.0 g oxygen * 1 mol O2 = mol O2
32.0 g O2
0.156 mol O2 = M

34. Again, the Power of ICE 2 H2 (g) + O2 (g)  2 H2O (g) Initial Change
Equilibrium
Now, we know everything – well, sort of.
0.827 M M
0 M
-2x -x
+ 2 x
0.827 – 2 x
– x
2 x

35. We have all of the equilibrium concentrations in terms of x…
…we can use Kc to solve for x
Kc = [H2O]2 = 4x10-6
[H2]2[O2]
= (2x)2
( x)2( x)

36. It looks like a mess…
…and it sort of is (although your calculator can probably solve it)
BUT you can simplify it with a helpful assumption:
ASSUME x<<0.0533

37. If we assume x is small Kc =[H2O]2 / [H2]2[O2] = 4x10-6 4x10-6 = (2x)2

38. A very simple problem remains
4x10-6 = (2x)2
( x)2( x)
(0.827)2(0.0533)
4x10-6 = 4x2
1.458x10-7 = 4x2
3.645x10-8 = x2
X = 1.91x10-4

39. Was the assumption good?
We started by assuming x<<0.0533
We now “know” that, with this assumption, x is 1.91x10-4
Is 1.91x10-4 << ?

40. Critical Judgment
How small is small depends on how accurate an answer you need: If you need 1 sig fig, than any number that is a factor of smaller is insignificant. If you need 2 sig figs, then it must be about 100 times smaller. If you need 3 sig figs it must be about 1000 times smaller.
A good general rule for our purposes is that if a number is <5% of another number, it is significantly smaller. This maintains 2 sig figs in all the concentrations – usually enough.

41. We put x back into the ICE chart
2 H2 (g) + O2 (g)  2 H2O (g)
Initial
Change
Equilibrium
And we have our answer.
0.827 M M
0 M
-2 (1.91x10-4)
- 1.91x10-4
+ 2 (1.91x10-4)
0.827 – 2 (1.91x10-4)
= 0.827
– 1.91x10-4 =
3.8x10-4

42. Clickers!
The Kc value for the formation of H2O from H2 and O2 at 500 C is 3.2x I put 5.0 mol of H2 and 5.0 mol of O2 in a 2 L flask at 500 C, what is the [H2O] at equilibrium?
A M D M
B M E M
C M

43. Calculating Equilibrium Constant
Example-1:
1.000 mole of H2 gas and mole of I2 vapor are introduced into a 5.00-liter sealed flask. The mixture is heated to a certain temperature and the following reaction occurs until equilibrium is established.
H2(g) + I2(g) ⇄ 2HI(g)
At equilibrium, the mixture is found to contain mole of HI. (a) What are the concentrations of H2, I2 and HI at equilibrium? (b) Calculate the equilibrium constant Kc.

44. Calculating Equilibrium Constant for reaction: H2(g) + I2(g) ⇄ 2HI(g)
————————————————————————————
H2(g) I2(g) ⇄ HI(g)
Initial [ ], M:
Change in [ ], M:
Equilibrium [ ], M
Kc = = = 57

45. Calculating Equilibrium Constant
Example-2:
0.500 mole of HI is introduced into a 1.00 liter sealed flask and heated to a certain temperature. Under this condition HI decomposes to produce H2 and I2 until an equilibrium is established. An analysis of the equilibrium mixture shows that mole of HI has decomposed. Calculate the equilibrium concentrations of H2, I2 and HI, and the equilibrium constant Kc for the following reaction:
H2(g) + I2(g) ⇄ 2HI(g),

46. Calculating Equilibrium Constant
The reaction: H2(g) + I2(g) ⇄ 2HI(g), proceeds from right to left.
————————————————————————————
H2(g) I2(g) ⇄ 2HI(g)
Initial [ ], M:
Change in [ ], M:
Equil’m [ ], M
Kc = = 56.6

47. CALCULATIONS INVOLVING Kp
Example When nitrogen (1 mole) and hydrogen (3 moles) react at constant
temperature at a pressure of 8 x 106 Pa, the equilibrium mixture
was found to contain 0.7 moles of ammonia. Calculate Kp .
N2(g) H2(g) NH3(g)
moles (initially)
moles (equilibrium) 1 - x x x (x moles of N2 reacted)
mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) x/(4-2x) (total moles = 4 - 2x)
partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P )
at equilibrium there are 0.7 moles of ammonia, so 2x = and x = 0.35
the total pressure (P) = 8 x 106 Pa.
applying the equilibrium law Kp = (PNH3) with units of Pa-2
(PN2) . (PH2)3
Substituting in the expression gives Kp = x Pa-2

48. Expression and Value of Equilibrium Constant for a Reaction
The expression for K depends on the equation;
The value of K applies to that equation; it does not depend on how the reaction occurs;
Concentrations used to calculate the value of K are those measured at equilibrium.

49. Relationships between chemical equations and the expressions of equilibrium constants
The expression of equilibrium constant depends on how the equilibrium equation is written. For example, for the following equilibrium:
H2(g) + I2(g) ⇄ 2 HI(g);
For the reverse reaction:
2HI(g) ⇄ H2(g) + I2(g);
And for the reaction: HI(g) ⇄ ½H2(g) + ½I2(g);

50. Expression and Values of Equilibrium Constant Using Partial Pressures
Consider the following reaction involving gases:
2SO2(g) + O2(g) ⇄ 2SO3(g)
Kp =

51. The Relationship between Kc and Kp
Consider the reaction: 2SO2(g) + O2(g) ⇄ 2SO3(g)
Kc = and Kp =
Assuming ideal behavior,
where PV = nRT and P = (n/V)RT = [M]RT
and PSO3 = [SO3]RT; PSO2 = [SO2]RT; PO2 = [O2]RT

52. Relationship between Kc and Kp
For reaction: PCl5(g)  PCl3(g) + Cl2(g);

53. Relationship between Kc and Kp
In general, for reactions involving gases such that,
aA + bB ⇄ cC + dD
where A, B, C, and D are all gases, and a, b, c, and d are their respective coefficients,
Kp = Kc(RT)Dn
and Dn = (c + d) – (a + b)
(In heterogeneous systems, only the coefficients of the gaseous species are counted.)

54. Relationship between Kc and Kp
For other reactions:
1. 2NO2(g) ⇄ N2O4(g); Kp = Kc(RT)-1
2. H2(g) + I2(g) ⇄ 2 HI(g); Kp = Kc
3. N2(g) + 3H2(g) ⇄ 2 NH3(g); Kp = Kc(RT)-2

55. Homogeneous & Heterogeneous Equilibria
Homogeneous equilibria:
CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g);
CO(g) + H2O(g) ⇄ CO2(g) + H2(g);
Heterogeneous equilibria:
CaCO3(s) ⇄ CaO(s) + CO2(g);
HF(aq) + H2O(l) ⇄ H3O+(aq) + F-(aq);
PbCl2(s) ⇄ Pb2+(aq) + 2 Cl-(aq);

56. Equilibrium Constant Expressions for Heterogeneous System
Examples:
CaCO3(s) ⇄ CaO(s) + CO2(g);
Kc = [CO2] Kp = PCO2; Kp = Kc(RT)
HF(aq) + H2O(l) ⇄ H3O+(aq) + F-(aq);

57. Solubility Eqilibrium
PbCl2(s) ⇄ Pb2+(aq) + 2Cl-(aq);
Ksp = [Pb2+][Cl-]2
(Ksp is called solubility product)

58. Combining Equations and Equilibrium Constants
when two or more equations are added to yield a net equation, the equilibrium constant for the net equation, Knet, is equal to the product of equilibrium constants of individual equations.
For example,
Eqn(1): A + B ⇄ C + D;
Eqn(2): C + E ⇄ B + F;

59. Combining Equations and Equilibrium Constants
Net equation: A + E ⇄ D + F;
= K1 x K2
If Eqn(1) + Eqn(2) = Net equation,
then K1 x K2 = Knet

60. Equilibrium Exercise #1
A flask is charged with 2.00 atm of nitrogen dioxide and 1.00 atm of dinitrogen tetroxide at 25 oC and allowed to reach equilibrium. When equilibrium is established, the partial pressure of NO2 has decreased by 1.24 atm. (a) What are the partial pressures of NO2 and N2O4 at equilibrium?
(b) Calculate Kp and Kc for following reaction at 25 oC.
2 NO2(g) ⇄ N2O4(g)
(Answer: Kp = 2.80; Kc = 68.6)

61. Equilibrium Exercise #2a
Methanol is produced according to the following equation:
CO(g) + 2H2(g) ⇄ CH3OH(g)
In an experiment, mol each of CO and H2 were allowed to react in a sealed 10.0-L reaction vessel at 500 K. When the equilibrium was established, the mixture was found to contain mole of CH3OH. What are the equilibrium concentrations of CO, H2 and CH3OH? Calculate the equilibrium constants Kc and Kp for this reaction at 500 K?
(R = L.atm/Mol.K)
(Answer: [CO] = M; [H2] = M; [CH3OH] = M;
(b) Kc = 14.5; Kp = 8.60 x 10-3)

62. Equilibrium Exercise #2b
For the reaction: CO(g) + 2H2(g) ⇄ CH3OH(g)
Kc = 15.0 at a certain temperature. Is a reaction mixture that contains 0.40 M CO, 0.80 M H2, and 0.10 M CH3OH at equilibrium? If not, in which direction will the net reaction occur? What will be their concentrations when equilibrium is established?

63. Equilibrium Exercise #3
1. The reaction: N2(g) + 3H2(g) ⇄ 2NH3(g),
has equilibrium constant, Kc = at 500oC.
What is the equilibrium constant for the following reaction?
NH3(g) ⇄ ½N2(g) + 3/2H2(g)
For the reaction: 2SO2(g) + O2(g) ⇄ 2 SO3(g),
Kc = 280 at 1000 K.
What is the equilibrium constant for the decomposition of SO3 at 1000 K according to the following equation?
SO3(g) ⇄ SO2(g) + ½ O2(g).

64. Equilibrium Exercise #4
If N2(g) + ½ O2(g) ⇄ N2O(g); Kc(1) = 2.4 x 10-18
and N2(g) + O2(g) ⇄ 2 NO(g); Kc(2) = 4.1 x 10-31
What is the equilibrium constant for the reaction?
N2O(g) + ½ O2(g) ⇄ 2NO(g)
(Answer: Knet = 1.7 x 10-13)

65. Applications of Equilibrium Constant
For any system or reaction:
Knowing the equilibrium constant, we can predict whether or not a reaction mixture is at equilibrium, and we can predict the direction of net reaction.
Qc = Kc  equilibrium (no net reaction)
Qc < Kc  a net forward reaction;
Qc > Kc  a net reverse reaction
The value of K tells us whether a reaction favors the products or the reactants.

66. Equilibrium constant is used to predict the direction of net reaction
For a reaction of known Kc value, the direction of net reaction can be predicted by calculating the reaction quotient, Qc.
Qc is called the reaction quotient, where for a reaction such as:
aA + bB ⇄ cC + dD;
Qc has the same expression as Kc , but
Qc is calculated using concentrations that are not necessarily at equilibrium.

67. What does the reaction quotient tell us?
If Qc = Kc,  the reaction is at equilibrium;
If Qc < Kc,  the reaction is not at equilibrium and there’s a net forward reaction;
If Qc > Kc,  the reaction is not at equilibrium and there’s a net reaction in the opposite direction.

68. Why is Equilibrium Constant Important?
Knowing Kc and the initial concentrations, we can determine the concentrations of components at equilibrium.

69. Equilibrium Exercise #5
For the reaction:
CO(g) + 2 H2(g) ⇄ CH3OH(g), Kc = 14.5 at 500 K.
Predict whether a mixture that contains 1.50 mol of H2, 1.00 mol of CO, and 0.50 mol of CH3OH in a 10.0-L vessel at 500 Kc is at equilibrium.
If not, indicate the direction in which the net reaction will occur to reach equilibrium.
(Answer: Qc = 22.2 > Kc; net reaction is to the left)

70. Consider the reaction: H2(g) + I2(g) ⇄ 2 HI(g),
Calculating equilibrium concentrations using initial concentrations and value of Kc
Consider the reaction:
H2(g) + I2(g) ⇄ 2 HI(g),
where Kc = 55.6 at 425oC.
If [H2]0 = [I2]0 = M, and [HI]0 = 0.0 M, what are their concentrations at equilibrium?

71. Using the ICE table to calculate equilibrium concentrations
Equation: H2(g) I2(g) ⇄ 2 HI(g),

Initial [ ], M
Change [ ], M x x x
Equilibrium [ ], M ( x) ( x) x
 

72. Calculation of equilibrium concentrations

73. Equilibrium Exercise #6
For the reaction:
2 NO2(g) ⇄ N2O4(g); Kp = 1.27 at 353 K.
If the initial pressure of NO2 was 3.92 atm, and initially there was no N2O4, what are the partial pressures of the gases at equilibrium at 353 K? What is the total gas pressure at equilibrium?
(Answer: PNO2 = 1.06 atm; PN2O4 = 1.43 atm; Ptotal = 2.49 atm)

74. Equilibrium Exercise #7
The reaction:
PCl5(g) ⇄ PCl3(g) + Cl2(g) has Kc =
A mol sample of PCl5 is placed in an empty 1.00-L flask and the above reaction is allowed to come to equilibrium at a certain temperature. How many moles of PCl5, PCl3, and Cl2, respectively, are present at equilibrium?
(Answer: PCl5 = mol; PCl3 = Cl2 = mol)

75. Le Châtelier’s Principle
The Le Châtelier's principle states that:
when factors that influence an equilibrium are altered, the equilibrium will shift to a new position that tends to minimize those changes.
Factors that influence equilibrium:
Concentration, temperature, and partial pressure (for gaseous)

76. The Effect of Changes in Concentration
Consider the reaction: N2(g) + 3H2(g) ⇄ 2 NH3(g);
If [N2] and/or [H2] is increased, Qc < Kc
 a net forward reaction will occur to reach new equilibrium position.
If [NH3] is increased, Qc > Kc, and a net reverse reaction will occur to come to new equilibrium position.

77. Effects of Pressure Change on Equilibrium
If the volume of a gas mixture is compressed, the overall gas pressure will increase. In which direction the equilibrium will shift in either direction depends on the reaction stoichiometry.
However, there will be no effect to equilibrium if the total gas pressure is increased by adding an inert gas that is not part of the equilibrium system.

78. Consider the reaction: 2SO2(g) + O2(g) ⇄ 2SO3(g),
Reactions that shift right when pressure increases and shift left when pressure decreases
Consider the reaction:
2SO2(g) + O2(g) ⇄ 2SO3(g),
The total moles of gas decreases as reaction proceeds in the forward direction.
If pressure is increased by decreasing the volume (compression), a forward reaction occurs to reduce the stress.
Reactions that result in fewer moles of gas favor high pressure conditions.

79. Reaction that shifts left when pressure increases, but shifts right when pressure decreases
Consider the reaction: PCl5(g) ⇄ PCl3(g) + Cl2(g);
Forward reaction results in more gas molecules.
Pressure increases as reaction proceeds towards equilibrium.
If mixture is compressed, pressure increases, and reverse reaction occurs to reduce pressure;
If volume expands and pressure drops, forward reaction occurs to compensate.
This type of reactions favors low pressure condition

80. Reactions not affected by pressure changes
Consider the following reactions:
CO(g) + H2O(g) ⇄ CO2(g) + H2(g);
H2(g) + Cl2(g) ⇄ 2HCl(g);
Reactions have same number of gas molecules in reactants and products.
Reducing or increasing the volume will cause equal effect on both sides – no net reaction will occur.
Equilibrium is not affected by change in pressure.

81. The Effect Temperature on Equilibrium
Consider the following exothermic reaction:
N2(g) + 3H2(g) ⇄ 2NH3(g); DHo = -92 kJ,
The forward reaction produces heat => heat is a product.
When heat is added to increase temperature, reverse reaction will take place to absorb the heat;
If heat is removed to reduce temperature, a net forward reaction will occur to produce heat.
Exothermic reactions favor low temperature conditions.

82. The Effect Temperature on Equilibrium
Consider the following endothermic reaction:
CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g), DHo = 205 kJ
Endothermic reaction absorbs heat  heat is a reactant;
If heat is added to increasing the temperature, it will cause a net forward reaction.
If heat is removed to reduce the temperature, it will cause a net reverse reaction.
Endothermic reactions favor high temperature condition.

83. Equilibrium Exercise #8
Determine whether the following reactions favor high or low pressures?
2SO2(g) + O2(g) ⇄ 2 SO3(g);
PCl5(g) ⇄ PCl3(g) + Cl2(g);
CO(g) + 2H2(g) ⇄ CH3OH(g);
N2O4(g) ⇄ 2 NO2(g);
H2(g) + F2(g) ⇄ 2 HF(g);

84. Equilibrium Exercise #9
Determine whether the following reactions favors high or low temperature?
2SO2(g) + O2(g) ⇄ 2 SO3(g); DHo = -180 kJ
CO(g) + H2O(g) ⇄ CO2(g) + H2(g); DHo = -46 kJ
CO(g) + Cl2(g) ⇄ COCl2(g); DHo = -108 kJ
N2O4(g) ⇄ 2 NO2(g); DHo = +57 kJ
CO(g) + 2H2(g) ⇄ CH3OH(g); DHo = -270 kJ

85. Chemical Equilibria in Industrial Processes
Production of Sulfuric Acid, H2SO4;
S8(s) O2(g)  8SO2(g)
2SO2(g) + O2(g) ⇄ 2SO3(g); DH = -198 kJ
SO3(g) + H2SO4(l)  H2S2O7(l)
H2S2O7(l) + H2O(l)  2H2SO4(l)
The second reaction is exothermic and has high activation energy;
though thermodynamically favored the reaction is very slow at low temperature,.
At high temperature reaction goes faster, but the yield would be very low.
An optimum condition is achieved at moderate temperatures and using catalysts to speed up the reaction. Reaction also favors high pressure.

86. Chemical Equilibria in Industrial Processes
The production of ammonia by the Haber-Bosch process:
N2(g) + 3H2(g) ⇄ 2NH3(g); DH = -92 kJ
This reaction is exothermic and very slow at low temperature.
Increasing the temperature will increase reaction rate, but will lower the yield.
An optimum condition is achieved at moderate temperature of 250 to 300oC with catalyst added to increase the reaction rate.
Increasing the pressure will favor product formation.
Reaction favors low temperature and high pressure conditions.

87. Chemical Equilibria in Industrial Processes
The production of hydrogen gas:
Reaction: CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g);
This reaction is endothermic with DH = 206 kJ
Increasing the reaction temperature will increase both the rate and the yield.
This reaction favors high temperature and low pressure conditions.