1. If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress This shift occurs because the stress will cause a new set of initial concentrations ( a new Q). The equilibrium will have to shift in order to bring Q back to K. 3 Types of stress Change in concentration Change in temperature Change in pressure Topic B: Le Chatelier’s Principle and Optimum Conditions

2. Adding product makes Q>K Removing reactant makes Q>K Adding reactant makes Q<K Removing product makes Q<K Determine the effect on Q, will tell you the direction of shift Change amounts of reactants and/or products

3. Increase in concentration The shift is away from what increases (This reduces the increase) N 2 (g) + 3H 2 (g) 2NH 3 (g) Increase in [ H 2 ]: Equilibrium shifts to the right Change in concentration of reactant or product

4. Decrease in concentration The shift is toward what is decreased (This replaces the decrease) N 2 (g) + 3H 2 (g) 2NH 3 (g) Decrease in [ H 2 ]: Equilibrium shifts to the left Change in concentration of reactant or product

5. If volume is decreased, increasing pressure: System will shift in the direction that has the least moles of gas to reduce the pressure. If volume is increased, decreasing pressure: System will shift in the direction that has the most moles of gas to increase pressure. Do not count reactants or products that are liquids or solids (they are not part of the equilibrium) Change Pressure

6. By decreasing volume System will move in the direction that has the least moles of gas. Because partial pressures (and concentrations) change a new equilibrium must be reached. System tries to minimize the moles of gas. Increase Pressure

7. By adding an inert gas Partial pressures of reactants and product are not changed No effect on equilibrium position Change in Pressure

8. Affects the rates of both the forward and reverse reactions. Doesn’t just change the equilibrium position, changes the equilibrium constant. The direction of the shift depends on whether it is exo- or endothermic Change in Temperature

9. We can explain a change in K with increased temperature (decrease for exothermic and increase for endothermic) Consider the energy profiles in each type of reaction. For an exothermic forward reaction: The activation energy of the forward reaction is smaller than the activation energy of the reverse (endothermic) reaction For an endothermic forward reaction: The activation energy of the forward reaction is larger than the activation energy of the reverse (exothermic) reaction In short, the endothermic direction always has the larger activation energy in an equilibrium system.

11. an increase in temperature causes a larger increase in the rate for reactions with larger activation energies we conclude that an increase in temperature causes: the rate of the endothermic reaction to increase more than the rate of the exothermic reaction. This means that increasing temperature will always shift an equilibrium system in the endothermic direction For exothermic reaction Backward - more reactants - smaller K’s For endothermic reactions Forward - more products -- larger K’s.

12. DH<0 Releases heat Think of heat as a product Raising temperature is like increasing a concentration and the equilibrium shifts away from that point. N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat Shifts to left. Exothermic

13. DH>0 Absorbs heat Think of heat as a reactant Raising temperature shifts equilibrium away from that point. Ba(OH) 2 (s) + 2NH 4 SCN(s) + heat Ba(SCN) 2 (s) + 2 NH 3 (g) Shifts to right. Endothermic

14. As 4 O 6 (s) + 6C(s) As 4 (g) + 6CO(g) Predict the direction of the shift a. addition of carbon monoxide b. addition or removal of carbon or As 4 O 6 c. removal of gaseous arsenic (As 4 ) Sample Ex. 13.13

15. Predict the shift in equilibrium when volume is reduced. a. P 4 (s) + 6Cl 2 (g) 4PCl 3 (l) b. PCl 3 (g) + Cl 2 (g) PCl 5 (g) c. PCl 3 (g) + 3NH 3 (g) P(NH 2 ) 3 (g) + 3HCl(g) Sample Ex. 13.14

16. Predict how the value of K changes as the temperature is increased. a. N 2 (g) + O 2 (g) 2NO(g) ∆H o = 181 kJ b. 2SO 2 (g) + O 2 (g) 2SO 3 (g) ∆H o = -198 kJ Sample Ex. 13.15

17. Le Chatelier and Q versus K (non-standard electrochemical cells) We saw in unit 5 how to make predictions for voltage change in an electrochemical cell where the conditions were not standard. That explanation used a pseudo Le Chatelier argument. We can also explain the change in voltage under non- standard conditions by using ‘Q versus K’ (in the new curriculum this is probably the preferred argument)

18. Consider the reaction below, which has a standard voltage of 1.10 V, that eventually drops to 0.00 V once equilibrium has been established. Cu 2+( aq) + Zn(s)   Zn 2+( aq) + Cu(s) If the [Zn 2+( aq)] is increased above the standard 1.0 M (all other conditions are kept constant) then the value of Q increases To restore Q = K and to re-establish equilibrium, the reaction must shift back (to the left) and the voltage will decrease.

19. Cu 2+( aq) + Zn(s)   Zn 2+( aq) + Cu(s) If the [Cu 2+( aq)] is increased above the standard 1.0 M (all other conditions are keptconstant) Then the value of Q decrease To restore Q = K and to re-establish equilibrium, the reaction must shift forward (to the right) and the voltage will increase.