1. 3 Nov 97Entropy & Free Energy (Ch 20)1 CHEMICAL EQUILIBRIUM Chapter 16 equilibrium vs. completed reactions equilibrium constant expressions Reaction quotient computing positions of equilibria: examples Le Chatelier’s principle - effect on equilibria of: addition of reactant or product pressure temperature YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)

2. 3 Nov 97Entropy & Free Energy (Ch 20)2 For any type of chemical equilibrium of the type THE EQUILIBRIUM CONSTANT If K is known, then we can predict concentrations of products or reactants. a A + b B c C + d D the following is a CONSTANT (at a given T) :

3. 3 Nov 97Entropy & Free Energy (Ch 20)3 All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q. Q = = 0.35 0.25 = 1.40 Q - the reaction quotient If Q = K, then system is at equilibrium. To reach EQUILIBRIUM [Iso] must INCREASE and [n] must DECREASE. Since K =2.5, system NOT AT EQUIL. [iso] [n] Q has the same form as K,... but uses existing concentrations n-Butane iso-Butane 0.25 0.35

4. 3 Nov 97Entropy & Free Energy (Ch 20)4 Q/K Q  Typical EQUILIBRIUM Calculations 2 general types: a. Given set of concentrations, is system at equilibrium ? Calculate Q compare to K 1 Q = K IF: Q > K or Q/K > 1  REACTANTS Q < K or Q/K < 1  PRODUCTS Q=K at EQUILIBRIUM

5. 3 Nov 97Entropy & Free Energy (Ch 20)5 Step 2 Put equilibrium [ ] into K c. Step 1 Define equilibrium condition in terms of initial condition and a change variable [H 2 ][I 2 ][HI] At equilibrium1.00-x1.00-x2x H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Step 3. Solve for x. 55.3 = (2x) 2 /(1-x) 2 Square root of both sides & solve gives: x = 0.79 [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M At equilibrium

6. 3 Nov 97Entropy & Free Energy (Ch 20)6 “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” EQUILIBRIUM AND EXTERNAL EFFECTS The position of equilibrium is changed when there is a change in: –pressure –changes in concentration –temperature The outcome is governed by LE CHATELIER’S PRINCIPLE Henri Le Chatelier 1850-1936 - Studied mining engineering - specialized in glass and ceramics.

7. 3 Nov 97Entropy & Free Energy (Ch 20)7 If concentration of one species changes, concentrations of other species CHANGES to keep the value of K the same (at constant T) no change in K - only position of equilibrium changes. Shifts in EQUILIBRIUM : Concentration ADDING REACTANTS - equilibrium shifts to PRODUCTS ADDING PRODUCTS - equilibrium shifts to REACTANTS REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION - GAS-FORMING; PRECIPITATION

8. 3 Nov 97Entropy & Free Energy (Ch 20)8 Solution A. Calculate Q with extra 1.50 M n-butane. INITIALLY: [n] = 0.50 M [iso] = 1.25 M What happens ? CHANGE: ADD +1.50 M n-butane What happens ? Q < K. Therefore, reaction shifts to PRODUCT Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63 n-Butane Isobutane Effect of changed [ ] on an equilibrium 16_butane.mov (16m13an1.mov) K = [iso] [n] = 2.5

9. 3 Nov 97Entropy & Free Energy (Ch 20)9 Solution B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane][isobutane] Initial0.50 + 1.501.25 Change- x + x Equilibrium2.00 - x1.25 + x Butane/Isobutane x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane. A B

10. 3 Nov 97Entropy & Free Energy (Ch 20)10 Effect of Pressure (gas equilibrium) Increase P in the system by reducing the volume. N 2 O 4 (g) 2 NO 2 (g) 16_NO2.mov (16m14an1.mov) Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT P N 2 O 4 increases See Ass#2 - question #6 P NO 2 decreases

11. 3 Nov 97Entropy & Free Energy (Ch 20)11 EQUILIBRIUM AND EXTERNAL EFFECTS Temperature change  change in K Consider the fizz in a soft drink CO 2 (g) + H 2 O(liq) CO 2 (aq) + heat Increase T Equilibrium shifts left: [CO 2 (g)]  [CO 2 (aq)]  K decreases as T goes up. Decrease T [CO 2 (aq)] increases and [CO 2 (g)] decreases. K increases as T goes down K c = [CO 2 (aq)]/[CO 2 (g)] HIGHER T LOWER T Change T: New equilib. position? New value of K?

12. 3 Nov 97Entropy & Free Energy (Ch 20)12 Temperature Effects on Chemical Equilibrium K c = 0.00077 at 273 K K c = 0.00590 at 298 K N 2 O 4 + heat 2 NO 2 (colorless)(brown)  H o rxn = + 57.2 kJ Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction 16_NO2RX.mov (16m14an1.mov)

13. 3 Nov 97Entropy & Free Energy (Ch 20)13 EQUILIBRIUM AND EXTERNAL EFFECTS Add catalyst ---> no change in K A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

14. 3 Nov 97Entropy & Free Energy (Ch 20)14 CHEMICAL EQUILIBRIUM Chapter 16 equilibrium vs. completed reactions equilibrium constant expressions Reaction quotient computing positions of equilibria: examples Le Chatelier’s principle - effect on equilibria of: addition of reactant or product pressure temperature YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)

15. 3 Nov 97Entropy & Free Energy (Ch 20)15 Entropy and Free Energy (Kotz Ch 20) Spontaneous vs. non-spontaneous thermodynamics vs. kinetics entropy = randomness (S o ) Gibbs free energy (  G o )  G o for reactions - predicting spontaneous direction  G rxn versus  G o rxn predicting equilibrium constants from  G o rxn

16. 3 Nov 97Entropy & Free Energy (Ch 20)16 Entropy and Free Energy ( Kotz Ch 20 ) How can we predict if a reaction can occur, given enough time? Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur. To predict if a reaction can occur at a reasonable rate, one needs to consider: some processes are spontaneous; others never occur. WHY ? THERMODYNAMICS KINETICS 9-paper.mov 20m02vd1.mov

17. 3 Nov 97Entropy & Free Energy (Ch 20)17 Thermodynamics state of a chemical system (P, T, composition) From a given state, would a chemical reaction decrease the energy of the system? If yes, system is favored to react — a product-favored system which will have a spontaneous reaction. Most product-favored reactions are exothermic. Spontaneous does not imply anything about time for reaction to occur. (kinetics)

18. 3 Nov 97Entropy & Free Energy (Ch 20)18 Thermodynamics versus Kinetics Diamond to Graphite –spontaneous from thermodynamics – but not kinetically favored. Paper burns. - product - favored reaction. - Also kinetically favored once reaction is begun.

19. 3 Nov 97Entropy & Free Energy (Ch 20)19 Product-Favored Reactions E.g. thermite reaction Fe 2 O 3 (s) + 2 Al(s)  2 Fe(s) + Al 2 O 3 (s)  H = - 848 kJ In general, product- favored reactions are exothermic.

20. 3 Nov 97Entropy & Free Energy (Ch 20)20 Non-exothermic spontaneous reactions But many spontaneous reactions or processes are endothermic... NH 4 NO 3 (s) + heat  NH 4 + (aq) + NO 3 - (aq)  H sol = +25.7 kJ/mol or have  H = 0...

21. 3 Nov 97Entropy & Free Energy (Ch 20)21 Entropy, S One property common to product-favored processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S. Reaction of K with water

22. 3 Nov 97Entropy & Free Energy (Ch 20)22 PROBABILITY - predictor of most stable state WHY DO PROCESSES with  H = 0 occur ? Consider expansion of gases to equal pressure: This is spontaneous because the final state, with equal # molecules in each flask, is much more probable than the initial state, with all molecules in flask 1, none in flask 2 SYSTEM CHANGES to state of HIGHER PROBABILITY THIS IS USUALLY the more RANDOM state.

23. 3 Nov 97Entropy & Free Energy (Ch 20)23 Gas expansion - spontaneity from greater probability Consider distribution of 4 molecules in 2 flasks P 1 < P 2 P 1 > P 2 P 1 = P 2 With more molecules (>10 20 ) P 1 =P 2 is most probable by far

24. 3 Nov 97Entropy & Free Energy (Ch 20)24 WHAT about EXOTHERMIC REACTIONS ? Consider 2 H 2 (g) + O 2 (g)  2 H 2 O (l) HIGHLY EXOTHERMIC final state (liquid) is MUCH MORE ORDERED (less random arrangement) than initial state (2 gases) BUT PROBABILITY of final state is higher when one considers change in the surroundings. WHY ? Heat evolved increases motion of molecules in the surroundings > Increased disorder of surroundings decreased disorder of system MUST consider change in disorder in BOTH SYSTEM and SURROUNDINGS to predict DIRECTION of SPONTANEITY

25. 3 Nov 97Entropy & Free Energy (Ch 20)25 Directionality of Reactions How probable is it that reactant molecules will react? PROBABILITY suggests that a product-favored reaction will result in the dispersal of energy or dispersal of matter or both.

26. 3 Nov 97Entropy & Free Energy (Ch 20)26 Probability suggests that a product-favored reaction will result in the dispersal of energy or of matter or both. Directionality of Reactions Energy DispersalMatter Dispersal 9_gasmix.mov 20m03an1.mov 9_exorxn.mov 20m03an2.mov

27. 3 Nov 97Entropy & Free Energy (Ch 20)27 Directionality of Reactions Energy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings. The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. The final state—with energy dispersed—is more probable and makes a reaction product-favored.

28. 3 Nov 97Entropy & Free Energy (Ch 20)28 Standard Entropies, S o Every substance at a given temperature and in a specific phase has a well-defined Entropy At 298 o the entropy of a substance is called S o - with UNITS of J.K -1.mol -1 The larger the value of S o, the greater the degree of disorder or randomness e.g. S o (in J.K -1 mol -1 ) : Br 2 (liq) = 152.2 Br 2 (gas) = 245.5 For any process:  S o =  S o (final) -  S o (initial)  S o (vap., Br 2 ) = (245.5-152.2) = 93.3 J.K -1 mol -1

29. 3 Nov 97Entropy & Free Energy (Ch 20)29 S (gases) > S (liquids) > S (solids) S o (J/Kmol) H 2 O(gas)188.8 H 2 O(liq) 69.9 H 2 O (s) 47.9 Ice Water Vapour Entropy, S