1. Chemical Equilibrium CHAPTER 15
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop

2. CHAPTER 15 Chemical Equilibrium
Learning Objectives:
Reversible Reactions and Equilibrium
Writing Equilibrium Expressions and the Equilibrium Constant (K)
Reaction Quotient (Q)
Kc vs Kp
ICE Tables
Quadratic Formula vs Simplifying Assumptions
LeChatelier’s Principle
van’t Hoff Equation
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

3. CHAPTER 15 Chemical Equilibrium
Lecture Road Map:
Dynamic Equilibrium
Equilibrium Laws
Equilibrium Constant
Le Chatelier’s Principle
Calculating Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

4. CHAPTER 15 Chemical Equilibrium
Dynamic Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

5. Chemical equilibrium exists when
Dynamic Eq
Equilibrium
Chemical equilibrium exists when
Rates of forward and reverse reactions are equal
Reaction appears to stop
Concentration of reactants and products do not change over time
Remain constant
Both forward and reverse reaction never cease
Equilibrium signified by double arrows ( )
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

6. N2O4 2 NO2 Dynamic Eq Equilibrium Initially have only N2O4
Only forward reaction
As N2O4 reacts NO2 forms
As NO2 forms
Reverse reaction begins to occur
NO2 collide more frequently as concentration of NO2 increases
Eventually, equilibrium is reached
Concentration of N2O4 does not change
Concentration of NO2 does not change
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

7. Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

8. N2O4 2NO2 Dynamic Eq Equilibrium Closed system
Equilibrium can be reached from either direction
Independent of whether it starts with “reactants” or “products”
Always have the same composition at equilibrium under same conditions
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

9. Reactants Equilibrium Products N2O4 2NO2 Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

10. Mass Action Expression
Dynamic Eq
Mass Action Expression
Simple relationship among [reactants] and [products] for any chemical system at equilibrium
Called the mass action expression
Derived from thermodynamics
Forward reaction: A  B
Reverse reaction: A  B
At equilibrium: A B
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

11. Uses stoichiometric coefficients as exponent for each reactant
Dynamic Eq
Reaction Quotient
Uses stoichiometric coefficients as exponent for each reactant
For reaction: aA + bB cC + dD
Reaction quotient
Numerical value of mass action expression
Equals “Q ” at any time, and
Equals “K ” only when reaction is known to be at equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

12. Ex. 1 H2(g) + I2(g) 2HI(g) 440˚C Exp’t Initial Amts Equil’m Amts
Equil’m [M] I
1.00 mol H2
0.222 mol H2
M H2
10 L
1.00 mol I2
0.222 mol I2
M I2
0.00 mol HI
1.56 mol HI
0.156 M HI II
0.00 mol H2
0.350 mol H2
M H2
10 L
0.100 mol I2
0.450 mol I2
M I2
3.50 mol HI
2.80 mol HI
0.280 M HI

13. Ex. 1 H2(g) + I2(g) 2HI(g) 440 ˚C Exp’t Initial Amts Equil’m Amts
Equil’m [M]
III
mol H2
0.150 mol H2
M H2
10 L
0.00 mol I2
0.135 mol I2
M I2
1.27 mol HI
1.00 mol HI
0.100 M HI IV
0.00 mol H2
0.442 mol H2
M H2
10 L
0.00 mol I2
0.442 mol I2
M I2
4.00 mol HI
3.11 mol HI
0.311 M HI

14. Mass Action Expression
= same for all data sets at equilibrium
Equilibrium Concentrations (M )
Exp’t
[H2]
[I2]
[HI] I
0.0222
0.156 II
0.0350
0.0450
0.280
III
0.0150
0.0135
0.100 IV
0.0442
0.311
Average = 49.5

15. Write mass action expressions for the following: 2NO2(g) N2O4(g)
Group
Problem
Write mass action expressions for the following:
2NO2(g) N2O4(g)
2CO(g) + O2(g) CO2(g)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

16. Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?
Group
Problem
Which of the following is the correct mass action expression for the reaction:
Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

17. CHAPTER 15 Chemical Equilibrium
Equilibrium Laws
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

18. at equilibrium write the following equilibrium law
Equilibrium Laws
For reaction
H2(g) + I2(g) HI(g) at 440 ˚C
at equilibrium write the following equilibrium law
Equilibrium constant = Kc = constant at given T
Use Kc since usually working with concentrations in mol/L
For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

19. Predicting Equilibrium Laws
For general chemical reaction:
dD eE fF gG
Where D, E, F, and G represent chemical formulas
d, e, f, and g are coefficients
Mass action expression is
Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation.
Equilibrium law is:
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

20. Predicting Equilibrium Laws
Only concentrations that satisfy this equation are equilibrium concentrations
Numerator
Multiply concentration of products raised to their stoichiometric coefficients
Denominator
Multiply concentration reactants raised to their stoichiometric coefficients
is scientists’ convention
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

21. What is equilibrium law?
Example
3H2(g) N2(g) NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

22. Various operations can be performed on equilibrium expressions
1. When direction of equation is reversed, new equilibrium constant is reciprocal of original
A + B C + D
C +D A + B
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

23. Equilibrium Operations
1. When direction of equation is reversed, new equilibrium constant is reciprocal of original
3H2(g) + N2(g) NH3(g) at 25˚C
2NH3(g) H2(g) + N2(g) at 25 ˚C
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

24. Equilibrium Operations
2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor.
A + B C + D
3A + 3B C + 3D
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

25. Equilibrium Operations
When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor
3H2(g) + N2(g) NH3(g) at 25 ˚C
Multiply by 3
9H2(g) + 3N2(g) NH3(g)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

26. Equilibrium Operations
3. When chemical equilibria are added, their equilibrium constants are multiplied
A + B C + D
C + E F + G
A + B + E D + F + G
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

27. NO3(g) + CO(g) NO2(g) + CO2(g) NO2(g) + CO(g) NO(g) + CO2(g)
Equilibrium
Operations
3. When chemical equilibria are added, their equilibrium constants are multiplied
2 NO2(g) NO3(g) + NO(g)
NO3(g) + CO(g) NO2(g) + CO2(g)
NO2(g) + CO(g) NO(g) + CO2(g)
Therefore
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

28. For: N2(g) + 3H2(g) 2NH3(g) Kc = 500 at a particular temperature.
Group
Problem
For: N2(g) + 3H2(g) NH3(g) Kc = 500 at a particular temperature.
What would be Kc for following?
2NH3(g) N2(g) + 3H2(g)
1/2N2(g) + 3/2H2(g) NH3(g)
0.002
22.4
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

29. CHAPTER 15 Chemical Equilibrium
Equilibrium Constant
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

30. Equilibrium Constant Kc
Most often Kc is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides
Sometimes partial pressures, in atmospheres, may be used in place of concentrations
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

31. Based on reactions in which all substances are gaseous
Equilibrium Kp
Based on reactions in which all substances are gaseous
Gas quantities are expressed in atmospheres in mass action expression
Use partial pressures for each gas in place of concentrations
e.g. N2(g) + 3H2(g) NH3(g)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

32. Relationship between Kp and Kc
Equilibrium
Relationship between Kp and Kc
Start with ideal gas law
PV = nRT
Rearranging gives
Substituting P/RT for molar concentration into Kc results in pressure-based formula
∆n = moles of gas in product – moles of gas in reactant
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

33. Kc = 11.7 Consider the reaction: 2NO2(g) N2O4(g)
Group
Problem
Consider the reaction: 2NO2(g) N2O4(g)
If Kp = for the reaction at 25 ˚C, what is value of Kc at same temperature?
n = nproducts – nreactants = 1 – 2 = –1
Kc = 11.7
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

34. Δn = (4 – 3) = 1 Kp = Kc(RT)Δn Kp= 0.99 × (0.082057 × 298.15)1 Kp = 24
Group
Problem
Consider the reaction A(g) + 2B(g) C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp?
0.99
2.0 24
2400
None of these
Δn = (4 – 3) = 1
Kp = Kc(RT)Δn
Kp= 0.99 × ( × )1
Kp = 24
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

35. Homogeneous and Hetergeneous
Equilibrium
Homogeneous and Hetergeneous
Homogeneous reaction/equilibrium
All reactants and products in same phase
Can mix freely
Heterogeneous reaction/equilibrium
Reactants and products in different phases
Can’t mix freely
Solutions are expressed in M
Gases are expressed in M
Governed by Kc
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

36. 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Equilibrium Law =
Heterogeneous
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Equilibrium Law =
Can write in simpler form
For any pure liquid or solid, ratio of moles to volume of substance (M ) is constant
e.g. 1 mol NaHCO3 occupies 38.9 cm mol NaHCO3 occupies 77.8 cm3
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

37. 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Equilibrium
Heterogeneous
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size
Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size
So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc
Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

38. Write equilibrium laws for the following: Ag+(aq) + Cl–(aq) AgCl(s)
Heterogeneous
Write equilibrium laws for the following:
Ag+(aq) + Cl–(aq) AgCl(s)
H3PO4(aq) + H2O H3O+(aq) + H2PO4–(aq)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

39. 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
Group
Problem
Given the reaction:
3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
What is the mass action expression?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

40. 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
Group
Problem
Given the reaction:
3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
What is mass action expression for the reverse reaction?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E