1. Chapter 15 Chemistry the Central Science 12th Ed.
Chemical Equilibrium
Chapter 15
Chemistry the Central Science
12th Ed.

2. Chemical Equilibrium
The state where the concentrations of all reactants and products remain constant with time.
On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

3. The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
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4. 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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5. A System at Equilibrium
Once equilibrium is achieved, the amount of each reactant and product remains constant.
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6. 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)
2NO2(g)
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7. The Equilibrium Constant
Forward reaction:
N2O4(g)  2NO2(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. Problem
Suppose that the gas-phase reactions A  B and B  A are both elementary processes with rate constants of 3.8 × 10−2s−1 and 3.1 × 10−1s−1, respectively.
What is the value of the equilibrium constant for the equilibrium A  B.
Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? Explain.

13. Figure: 15-08

14. Equilibrium Expression
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)

16. Notes on Equilibrium Expressions (EE)
The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse.
When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)n
The units for K depend on the reaction being considered.

17. 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)
2NO2(g)
Kc = = 4.72 at 100 C
[N2O4]
[NO2]2
N2O4(g)
2NO2(g)
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18. 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)
2NO2(g)
Kc = = (0.212)2 at 100 C
[NO2]4
[N2O4]2
4NO2(g)
2N2O4(g)
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19. 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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20. K v. Kp For jA + kB  lC + mD Kp = K(RT)n
n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

21. 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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22. 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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23. 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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24. Heterogeneous Equilibria
. . . are equilibria that involve more than one phase.
CaCO3(s)  CaO(s) + CO2(g)
K = [CO2]
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

25. 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) + 2Cl(aq)
Kc = [Pb2+] [Cl]2
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26. CaCO3(s)  CaO(s) + CO2(g)

27. Reaction Quotient
. . . helps to determine the direction of the move toward equilibrium.
The law of mass action is applied with initial concentrations.

28. 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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29. Reaction Quotient (continued)
H2(g) + F2(g)  2HF(g)

30. Reaction Quotient

31. Solving Equilibrium Problems
1. Balance the equation.
2. Write the equilibrium expression.
3. List the initial concentrations.
4. Calculate Q and determine the shift to equilibrium.

32. Solving Equilibrium Problems (continued)
5. Define equilibrium concentrations.
6. Substitute equilibrium concentrations into equilibrium expression and solve.
7. Check calculated concentrations by calculating K.

33. 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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34. N2 + 3H2  2NH3

35. N2 + 3H2  2NH3

36. Effects of Changes on the System
1. Concentration: The system will shift away from the added component.
2. Temperature: K will change depending upon the temperature (treat the energy change as a reactant).
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37. 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.
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38. END