1. Disturbing Equilibrium and Non-equilibrium conditions
Le Châtelier’s Principle, the van’t Hoff equation, the Reaction Quotient (Q)

2. Lesson Objectives Know
-Le Châtelier’s principle outlines how equilibrium is affected by changes in a system
-Equilibrium can be affected by changes in reactant/product amounts, changes in volume or pressure, or changes in temperature
-temperature is the only factor that affects the equilibrium constant
Understand
-when disturbed from equilibrium, a reaction will shift whichever way (to products or to reactants) to relieve “stress” in the system
-the reaction quotient is used to determine which way a reaction needs to shift to reach equilibrium Do
-analyze changing conditions to determine which way a reaction needs to shift to regain equilibrium
-solve the van’t Hoff equation for a missing variable, whether the variable is a K value, DH, or a temperature.
-solve the Q (reaction quotient) equation for non-equilibrium conditions
-use Q to determine which way a reaction will shift to regain equilibrium

3. Le châtelier’s principle
Disturbing Equilibrium
Le châtelier’s principle

4. Le Chatelier’s Principle
If a system at equilibrium is disturbed, it will shift to partially counteract the change
Factors that can disturb equilibrium:
Adding or removing reactant or product (physically or chemically)
Increasing or decreasing volume (which results in a pressure change for gases)
Increasing or decreasing temperature (adding or removing heat)

5. Adding or removing reactant or product
Add reactant or product: reaction shifts to use up the added substance (stress)
Remove reactant or product: reaction shifts to add it back in
Substances can be removed by reacting them out – making precipitate, water, gas, etc.
Only affects equilibrium if the added or removed substance is:
In the equation (inert or irrelevant substances have no affect on equilibrium)
Not a solid or liquid (solids and liquids don’t affect equilibrium because their concentrations don’t change)

6. Example 1: changing amounts
Iodine chloride decomposes at high temperatures to iodine and chlorine:
2 ICl(g)  I2(g) + Cl2(g)
If some iodine condenses, which direction would this reaction shift from equilibrium?

7. Example 2: changing amounts
Sulfur dioxide and oxygen can combine to make sulfur trioxide
2 SO2(g) + O2(g)  2SO3(g)
If more oxygen is introduced into this system while it is at equilibrium, which way will the reaction shift?

8. 2H2(g) + CO(g)  CH3OH(g) H = -90.2 kJ
If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after more H2 is added.
[CO] will increase
[CO] will decrease
[CO] will stay the same

9. 2H2(g) + CO(g)  CH3OH(g) H = -90.2 kJ
If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after CH3OH is added.
[CO] will increase
[CO] will decrease
[CO] will stay the same

10. Changing the volume Increase volume (decrease pressure of gases)
The system will shift to side with more moles of gas to try to raise pressure
Decrease volume (increase pressure of gases)
The system will shift to the side with fewer moles of gas to try to lower pressure

11. Example 3: changing volume
Sulfur dioxide and oxygen can combine to make sulfur trioxide
2 SO2(g) + O2(g)  2SO3(g)
If the container holding this reaction is expanded (volume increases), which way will the reaction shift from equilibrium?

12. 2H2(g) + CO(g)  CH3OH(g) H = -90.2 kJ
If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after the volume of the system is decreased.
[CO] will increase
[CO] will decrease
[CO] will stay the same

13. Changing the temperature
Increasing temp. is like adding (stressing) heat
System will shift in the endothermic direction
Decreasing temp. is like removing heat
System will shift in the exothermic direction
If heat was shown in the equation, the system shifts to counteract the addition or removal of heat (just like reactants or products)
Endothermic: ΔH = positive, written in reactants
Exothermic: ΔH = negative, written in products

14. Example 4: changing temperatures
Sulfur dioxide and oxygen can combine to make sulfur trioxide 2 SO2(g) + O2(g)  2SO3(g) This reaction is highly exothermic, releasing nearly 200 kJ of energy per mole of product created. If the temperature of this reaction was increased, which way would the reaction shift from equilibrium?

15. 2H2(g) + CO(g)  CH3OH(g) H = -90.2 kJ
If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after the temperature of the system is increased.
[CO] will increase
[CO] will decrease
[CO] will stay the same

16. 2H2(g) + CO(g)  CH3OH(g) H = -90.2 kJ
If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after more catalyst is added.
[CO] will increase
[CO] will decrease
[CO] will stay the same

17. Which of the following is NOT true according to LeChâtelier’s Principle?
A system at equilibrium that has been disturbed will shift away from the stressor
Removing a product from a system will cause the system to shift back to reactants
Changing amounts of solids or liquids in a reaction does not affect a system’s equilibrium
An increase in a gaseous system’s volume will shift the reaction toward the side with more moles of gas
A decrease in the temperature of a system at equilibrium will cause the reaction to shift in the exothermic direction.

18. Example 5: Calculating new values
Iodine chloride decomposes at high temperatures:
2 ICl(g)  I2(g) + Cl2(g) At equilibrium the partial pressures are (in atm)
ICl: 0.43, I2: 0.16 and Cl2: 0.27
Calculate K
If enough iodine condenses to drop its partial pressure to 0.10 atm, what is the partial pressure of iodine gas when the equilibrium is reestablished?

19. CO(g) + Cl2(g) ⇄ COCl2 (g) + heat

20. The reaction quotient (q)
Non-equilibrium conditions
The reaction quotient (q)

21. Q: the reaction quotient
Q is just like K, except it is for ANY point in time, not just at equilibrium. You will find Q (like finding K) and then compare Q to K
Q = [products]
[reactants]

22. COMPARING Q TO K Q<K Q = [products] [REACTANTS]
reaction must shift to right to produce more products
Q>K
Q = [PRODUCTS]
[reactants]
reaction must shift to left to produce more reactants
Q=K
reaction is AT EQUILIBRIUM

23. Example 6: Calculating and Using Q
For the reaction: 2NO2 (g)  2NO (g) + O2 (g) K at a certain temperature is Predict the direction the system will move if one starts with:
PO2 = PNO = PNO2 = 0.10 atm
PNO2 = atm, PO2 = atm
PNO2 = 0.20atm, PO2 = 0.010atm, PNO = 0.040atm

24. The van’t hoff equation
Effect of Temperature on K
The van’t hoff equation

25. Effect of Temperature on K
Of all types of ways to disturb equilibrium, (adding/removing a substance, changing volume/pressure, changing temperature), temperature is the only factor to affect the equilibrium constant.
Use van’t Hoff equation to find new K , or if given new K, to find ΔH
ln K2 = ΔH 1 – 1
K1 R T1 T2
R = 8.31 J/mol·K
Look familiar?
ln k2 = Ea
k R T T2

26. Example 7: van’t Hoff equation
For the system H2(g) + I2(g)  2HI(g) ΔH = -9.4 kJ and K= 62.5 at 800 K. What is the equilibrium constant at 333°C?