1. 1 Equilbrium Constant and EXTERNAL EFFECTS Formation of stalactites and stalagmites CaCO 3 (s) + H 2 O(liq) + CO 2 (g)  Ca 2+ (aq) + 2 HCO 3 - (aq)

2. 2 EQUILIBRIUM Temperature, catalysts, and changes in concentration/ pressure affect equilibria. The outcome is governed by LE CHATELIER’S PRINCIPLE “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

3. 3 Equilibrium constant and Concentration Concentration changes –no change in K –only the position of equilibrium changes.

4. 4 Butane- Isobutane Equilibrium butane isobutane

5. 5 Butane Isobutane butane isobutane At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]?

6. 6 Solution Calculate Q immediately after adding more butane and compare with K. Q is LESS THAN K. Therefore, the reaction will shift to the ____________. Butane  Isobutane

7. 7 Q is less than K, shifts right toward isobutane. Set up ICE table [butane][isobutane] Initial Change Equilibrium 0.50 + 1.501.25 - X + X 2.00 – x 1.25 + x Butane  Isobutane

8. 8 x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M.. Equilibrium has shifted toward isobutane. Butane  Isobutane

9. 9 Equilibrium Constant and Catalyst Add catalyst: NO change in K A catalyst only affects the RATE it approach equilibrium. Catalytic exhaust system

10. 10 Pressure and Equilibrium N 2 O 4 (g)  2 NO 2 (g) Increase P in the system by reducing the volume (at constant Temp). 

11. 11 N 2 O 4 (g)  2 NO 2 (g) Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO 2 decreases and P of N 2 O 4 increases.

12. 12 Temperature Effects on Equilibrium Figure 16.6

13. 13 Temperature Effects on Equilibrium N 2 O 4 (colorless) + heat  2 NO 2 (brown) ∆H o = + 57.2 kJ (endo) K c (273 K) = 0.00077 K c (298 K) = 0.0059

14. 14 Every T has a unique K Temperature change = change in K Consider the fizz in a soft drink CO 2 (aq) + HEAT  CO 2 (g) + H 2 O(l) K = P (CO 2 ) / [CO 2 ] Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because P(CO 2 ) increases and [CO 2 ] decreases. Decrease T. Now what? Equilibrium shifts left and K decreases.

15. 15 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat K = 3.5 x 10 8 at 298 K NH 3 Production

16. 16 Le Chatelier’s Principle Change T - changes K – causes change in P or concentrations at equilibrium Use a catalyst: K not changed. Reaction comes more quickly to equilibrium. Add or take away reactant or product: K does not change Reaction adjusts to new equilibrium “position”

17. 17 Examples of Chemical Equilibria Phase changes such as H 2 O(s) H 2 O(liq) 