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 Le Chatelier’s Principle
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

5. Equilibrium positions Combination of concentrations that allow Q = K
Le Chatelier
Definition
Equilibrium positions
Combination of concentrations that allow Q = K
Infinite number of possible equilibrium positions
Le Châtelier’s principle
System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress
System said to “shift to right” when forward reaction is dominant (Q < K)
System said to “shift to left” when reverse direction is dominant (Q > K)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

6. Q = K reaction at equilibrium Q < K reactants go to products
Le Chatelier
Q & K Relationships
Q = K reaction at equilibrium
Q < K reactants go to products
Too many reactants
Must convert some reactant to product to move reaction toward equilibrium
Q > K products go to reactants
Too many products
Must convert some product to reactant to move reaction toward equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

7. Change in Concentration
Le Chatelier
Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow
Equilibrium mixture is blue-green
Add excess Cl– (conc. HCl)
Equilibrium shifts to products
Makes more yellow CuCl42–
Solution becomes green
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

8. Change in Concentration
Le Chatelier
Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow
Add Ag+
Removes Cl–: Ag+(aq) + Cl–(aq)  AgCl(s)
Equilibrium shifts to reactants
Makes more blue Cu(H2O)42+
Solution becomes increasingly more blue
Add H2O?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

9. Change in Concentration: Example
Le Chatelier
Change in Concentration: Example
For the reaction 2SO2(g) + O2(g) SO3(g)
Kc = 2.4 × 10–3 at 700 °C
Which direction will the reaction move if moles of O2 is added to an equilibrium mixture?
Towards the products
Towards the reactants
No change will occur
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

10. Change in Concentration
Le Chatelier
Change in Concentration
When changing concentrations of reactants or products
Equilibrium shifts to remove reactants or products that have been added
Equilibrium shifts to replace reactants or products that have been removed
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

11. Change in Pressure or Volume
Le Chatelier
Change in Pressure or Volume
Consider gaseous system at constant T and n
3H2(g) + N2(g) NH3(g)
If volume is reduced
Expect pressure to increase
To reduce pressure, look at each side of reaction
Which has less moles of gas
Reactants = 3 mol + 1 mol = 4 mol gas
Products = 2 mol gas
Reaction favors products (shifts to right)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

12. Change in Pressure or Volume
Le Chatelier
Change in Pressure or Volume
Consider gaseous system at constant T and n
H2(g) + I2(g) HI(g)
If pressure is increased, what is the effect on equilibrium?
nreactant = = 2
nproduct = 2
Predict no change or shift in equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

13. Change in Pressure or Volume
Le Chatelier
Change in Pressure or Volume
2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g)
If you decrease volume of reaction, what is the effect on equilibrium?
Reactants: All solids, no moles gas
Products: 2 moles gas
Decrease in V, causes an increase in P
Reaction shifts to left (reactants), as this has fewer moles of gas
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

14. Change in Pressure or Volume
Le Chatelier
Change in Pressure or Volume
Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can
Increasing pressure
Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids
Substances are already almost incompressible
Changes in V, P and [X ] effect position of equilibrium (Q), but not K
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

15. Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow
Le Chatelier
Change in Temperature
Ice water
Boiling water
Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow
Reaction endothermic
Adding heat shifts equilibrium toward products
Cooling shifts equilibrium toward reactants
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

16. Add heat energy, shift reaction right
Le Chatelier
Change in Temperature
Hf°=+6 kJ (at 0 °C)
Energy + H2O(s) H2O(l )
Energy is reactant
Add heat energy, shift reaction right
3H2(g) + N2(g) NH3(g) Hrxn= –47.19 kJ
3 H2(g) + N2(g) NH3(g) + energy
Energy is product
Add heat, shift reaction left
H2O(s) H2O(l)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

17. Le Chatelier Change in Temperature
Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change
Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

18. Increase in temperature of exothermic reaction makes K smaller
Le Chatelier
Change in Temperature
Changes in T change value of mass action expression at equilibrium, so K changed
K depends on T
Increase in temperature of exothermic reaction makes K smaller
More heat (product) forces equilibrium to reactants
Increase in temperature of endothermic reaction makes K larger
More heat (reactant) forces equilibrium to products
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

19. Catalyst lowers Ea for both forward and reverse reaction
Le Chatelier
Change with Catalyst
Catalyst lowers Ea for both forward and reverse reaction
Change in Ea affects rates k r and k f equally
Catalysts have no effect on equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

20. Addition of Inert Gas at Constant Volume
Le Chatelier
Addition of Inert Gas at Constant Volume
Inert gas
One that does not react with components of reaction
e.g. argon, helium, neon, usually N2
Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products
Since it doesn’t react with anything
No change in concentrations of reactants or products
No net effect on reaction
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

21. How To Use Le Chatelier’s Principle
Write mass action expression for reaction
Examine relationship between affected concentration and Q (direct or indirect)
Compare Q to K
If change makes Q > K, shifts left
If change makes Q < K, shifts right
If change has no effect on Q, no shift expected
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

22. H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq)
Group
Problem
Consider:
H3PO4(aq) + 3OH–(aq) H2O(l) + PO43–(aq)
What will happen if PO43– is removed?
Q is proportional to [PO43–]
Decrease [PO43–], decrease in Q
Q < K equilibrium shifts to right
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

23. H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq) is exothermic.
Group
Problem
The reaction
H3PO4(aq) + 3OH–(aq) H2O(aq) + PO43–(aq)
is exothermic.
What will happen if system is cooled?
heat
Since reaction is exothermic, heat is product
Heat is directly proportional to Q
Decrease in T, decrease in Q
Q < K equilibrium shifts to right
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

24. [Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O pink blue
Group
Problem
The equilibrium between aqueous cobalt ion and the chlorine ion is shown:
[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O
pink blue
It is noted that heating a pink sample causes it to turn violet.
The reaction is:
endothermic
exothermic
cannot tell from the given information
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

25. Group
Problem
The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products?
Ka = 2.2 × 10–3
Ka = 1.8 × 10–5
Ka = 4.0 × 10–10
Ka = 6.3 × 10–3
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E