1. Chapter 15 Chemical Equilibrium
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 15 Chemical Equilibrium
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
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3. The Concept of Equilibrium
As a system approaches equilibrium, both the forward and reverse reactions are occurring.
At equilibrium, the forward and reverse reactions are proceeding at the same rate.
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4. A System at Equilibrium
Once equilibrium is achieved, the amount of each reactant and product remains constant.
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5. Depicting Equilibrium
Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow.
N2O4 (g)
2 NO2 (g)
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6. The Equilibrium Constant
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7. The Equilibrium Constant
Forward reaction:
N2O4 (g)  2 NO2 (g)
Rate Law:
Rate = kf [N2O4]
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8. The Equilibrium Constant
Reverse reaction:
2 NO2 (g)  N2O4 (g)
Rate Law:
Rate = kr [NO2]2
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9. The Equilibrium Constant
Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
Rewriting this, it becomes kf kr
[NO2]2
[N2O4] =
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10. The Equilibrium Constant
The ratio of the rate constants is a constant at that temperature, and the expression becomes
Keq = kf kr
[NO2]2
[N2O4] =
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11. The Equilibrium Constant
Consider the generalized reaction
aA + bB
cC + dD
The equilibrium expression for this reaction would be
Kc =
[C]c[D]d
[A]a[B]b
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12. The Equilibrium Constant
AHL Only!!
The Equilibrium Constant
Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp =
(PCc) (PDd)
(PAa) (PBb)
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13. Relationship Between Kc and Kp
AHL Only!!
Relationship Between Kc and Kp
From the Ideal Gas Law we know that
PV = nRT
Rearranging it, we get
P = RT n V
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14. Relationship Between Kc and Kp
AHL Only!!
Relationship Between Kc and Kp
Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes
Kp = Kc (RT)n
where
n = (moles of gaseous product) - (moles of gaseous reactant)
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15. Temperature vs. Vapour Pressure
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16. Equilibrium Can Be Reached from Either Direction
As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.
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17. Equilibrium Can Be Reached from Either Direction
This is the data from the last two trials from the table on the previous slide.
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18. Equilibrium Can Be Reached from Either Direction
It doesn’t matter whether we start with N2 and H2 or whether we start with NH3: we will have the same proportions of all three substances at equilibrium.
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19. What Does the Value of K Mean?
If K>>1, the reaction is product-favored; product predominates at equilibrium.
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20. What Does the Value of K Mean?
If K>>1, the reaction is product-favored; product predominates at equilibrium.
If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.
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21. Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.
Kc = = at 100 C
[NO2]2
[N2O4]
N2O4 (g)
2 NO2 (g)
Kc = = 4.72 at 100 C
[N2O4]
[NO2]2
N2O4 (g)
2 NO2 (g)
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22. Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.
Kc = = at 100 C
[NO2]2
[N2O4]
N2O4(g)
2 NO2(g)
Kc = = (0.212)2 at 100 C
[NO2]4
[N2O4]2
2 N2O4(g)
4 NO2(g)
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23. Manipulating Equilibrium Constants
The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.
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24. Heterogeneous Equilibrium
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25. The Concentrations of Solids and Liquids Are Essentially Constant
Both can be obtained by multiplying the density of the substance by its molar mass — and both of these are constants at constant temperature.
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26. The Concentrations of Solids and Liquids Are Essentially Constant
Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression.
PbCl2 (s)
Pb2+ (aq) + 2 Cl-(aq)
Kc = [Pb2+] [Cl-]2
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27. CaCO3 (s)
CO2 (g) + CaO(s)
As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.
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28. Equilibrium Calculations
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29. An Equilibrium Problem
A closed system initially containing x 10-3 M H2 and x 10-3 M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for the reaction taking place, which is
H2 (g) + I2 (s)
2 HI (g)
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30. What Do We Know? [H2], M [I2], M [HI], M Initially Change
At equilibrium
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31. What Do We Know? [H2], M [I2], M [HI], M Initially 1.000 x 10-3 Change
At equilibrium
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32. What Do We Know? [H2], M [I2], M [HI], M Initially 1.000 x 10-3
Change
At equilibrium
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33. [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 Change
Change
At equilibrium
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34. [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 Change
Change
At equilibrium
1.87 x 10-3
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35. [HI] Increases by 1.87 x 10-3 M [H2], M [I2], M [HI], M Initially
Change
+1.87 x 10-3
At equilibrium
1.87 x 10-3
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36. Stoichiometry tells us [H2] and [I2] decrease by half as much.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
At equilibrium
1.87 x 10-3
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37. We can now calculate the equilibrium concentrations of all three compounds…
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
At equilibrium
6.5 x 10-5
1.065 x 10-3
1.87 x 10-3
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38. …and, therefore, the equilibrium constant.
Kc =
[HI]2
[H2] [I2]
= 51 =
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
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39. The Reaction Quotient (Q)
Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
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40. the system is at equilibrium.
If Q = K,
the system is at equilibrium.
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41. there is too much product, and the equilibrium shifts to the left.
If Q > K,
there is too much product, and the equilibrium shifts to the left.
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42. there is too much reactant, and the equilibrium shifts to the right.
If Q < K,
there is too much reactant, and the equilibrium shifts to the right.
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43. Le Châtelier’s Principle
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44. Le Châtelier’s Principle
“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”
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45. Industrial Processes Many industrial processes involve equilibria.
The aim in industry is produce your chosen substance as efficiently as possible - that is, quickly but with the least input of energy.
This requires studying the kinetics and equilibrium of the reaction.

46. The Haber Process
The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.
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47. The Haber Process
If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.
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48. Haber Process
This process combines hydrogen and nitrogen to form ammonia.
A mixture of N2(g) and H2(g) in a 1:3 ratio by volume is compressed and passed over a heated, finely divided, iron catalyst where the equilibrium is established.
N2(g) + 3H2(g)  2NH3(g)
∆H = -92 kJ mol-1

49. Haber Process
From the equation it can be seen that high pressure will favor formation of the product as 4 mol of gas react to form 2 mol of gas.
It is expensive to create a high pressure environment so there has to be a compromise between pressure and price!

50. Haber Process
The forward reaction is exothermic so a low temperature would favor products.
However low temperatures result in a low rate so it would take a long time to make the NH3.
A moderate (compromise) temperature is found to produce the most NH3 per hour.
A catalyst is also used to increase the rate of reaction – this does not change the amount of product formed (yield).

51. Haber Process
Typical conditions for the Haber process is a pressure of 200 atm (20 MPa) and temperatures around 450°C (723K).
The reaction is not left long enough for equilibrium to be reached as the rate would decrease at eqm.

52. Haber Process
At any time only about 20% of the reactants are converted to ammonia.
The leftover reactants are not wasted though.
The reaction system is cooled and NH3 condenses to a liquid as it forms hydrogen bonds so can be separated and the N2 and H2 gases are recycled.
Separating the product in this way also favors the forward reaction to keep occurring.

53. The Haber Process
This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.
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54. Haber Process
About 80% of ammonia (NH3) is used to make fertilizers which are important for growing crops especially for food.
Other uses of ammonia and its compounds are explosives, plastics and refrigerants.
About 120 million tonnes of ammonia are produced worldwide in a year so it is to significant to the economy.

55. The Contact Process
This is the production of sulfuric acid via a series of reactions. It has the highest production of any chemical in the world with 150 tonnes made annually.
Sulfuric acid is used to make fertilizers, detergents, dyes, explosives, drugs, plastics plus other products.

56. The Contact Process Step 1. The combustion of sulfur:
S(s) + O2(g)  SO2(g)
Step 2. The oxidation of SO2:
2SO2(g) + O2(g)  2SO3(g) H= -196kJ mol-1
Step 3. Combination of SO3 with water:
SO3(g) + H2O(l)  2H2SO4(aq)

57. The Contact Process
The 3rd step is actually shown below. The SO3 is added to a flowing solution of concentrated sulfuric acid and then water is added. This avoids the violent direct reaction between H2SO4 and H2O:
SO3(g) + H2SO4(l)  H2S2O7(l) + H2O(l) 2H2SO4(aq)
A vanadium (V) oxide (V2O5) is used as a catalyst in this reaction.

58. Contact Process
The Contact process is the production of sulfuric acid by the oxidation of sulfur.
Initially sulfur is burnt in the air to make sulfur dioxide:
S(s) + O2(g)  SO2(g)
Sulfur dioxide is mixed with air and passed over a vanadium (V) oxide catalyst to produce sulfur trioxide:
2SO2(g) + O2(g)  2SO3(g) H= -196kJ mol-1

59. Contact Process
As with the Haber process a high pressure would favor the formation of the product.
In this case a high conversion rate occurs at only 2 atm so it is not as expensive as the Haber process.
The chosen compromise temperature is K (450°C) and the use of a catalyst (V2O5) help to increase the forward reaction rate.

60. The Effect of Changes in Temperature
Co(H2O)62+(aq) + 4 Cl(aq)
CoCl4 (aq) + 6 H2O (l)
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61. Catalysts
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62. Catalysts
Catalysts increase the rate of both the forward and reverse reactions.
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63. Catalysts
When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.
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