1. 27 Oct 97Chemical Equilibrium1 Chemical Bonding and Molecular Structure (Chapter 9) Ionic vs. covalent bonding Molecular orbitals and the covalent bond (Ch. 10) Valence electron Lewis dot structures octet vs. non-octet resonance structures formal charges VSEPR - predicting shapes of molecules Bond properties bond order, bond strength polarity, electronegativity

2. 27 Oct 97Chemical Equilibrium2 Bond Polarity HCl is POLAR because it has a positive end and a negative end (partly ionic). Polarity arises because Cl has a greater share of the bonding electrons than H. Calculated charge by CAChe: H (red) is +ve (+0.20 e - ) Cl (yellow) is -ve (-0.20 e - ). (See PARTCHRG folder in MODELS.)

3. 27 Oct 97Chemical Equilibrium3 Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond. Bond Polarity (2) BONDENERGY “pure” bond339 kJ/mol calculated real bond432 kJ/mol measured ELECTRONEGATIVITY, . Difference 92 kJ/mol. This difference is the contribution of IONIC bonding It is proportional to the difference in

4. 27 Oct 97Chemical Equilibrium4 Electronegativity,   is a measure of the ability of an atom in a molecule to attract electrons to itself. Concept proposed by Linus Pauling (1901-94) Nobel prizes: Chemistry (54), Peace (63) See p. 425; 008vd3.mov (CD)

5. 27 Oct 97Chemical Equilibrium5 F has maximum . Atom with lowest  is the center atom in most molecules. Relative values of  determines BOND POLARITY (and point of attack on a molecule). Electronegativity,  Figure 9.7

6. 27 Oct 97Chemical Equilibrium6 Bond Polarity  (A) -  (B) 3.5 - 2.1  1.4  H 2.1 O F 3.5 4.0 Also note that polarity is “reversed.” Which bond is more polar ? (has larger bond DIPOLE) O—HO—F 3.5 - 4.0 0.5  (O-H) >  (O-F) Therefore OH is more polar than OF

7. 27 Oct 97Chemical Equilibrium7 Molecular Polarity Molecules such as HCl and H 2 O are POLAR They have a DIPOLE MOMENT. Polar molecules turn to align their dipole with an electric field. A molecule will be polar ONLY if a) it contains polar bonds AND b) the molecule is NOT “symmetric” Symmetric molecules

8. 27 Oct 97Chemical Equilibrium8 Molecular Polarity: H 2 O Water is polar because: a) O-H bond is polar b) water is non-symmetric The dipole associated with polar H 2 O is the basis for absorption of microwaves used in cooking with a microwave oven

9. 27 Oct 97Chemical Equilibrium9 B—F bonds are polar molecule is NOT symmetric B—F bonds are polar molecule is symmetric Molecular Polarity in NON-symmetric molecules B +ve F -ve Atom Chg.  B +ve 2.0 H +ve 2.1 F -ve 4.0 BF 3 is NOT polar HBF 2 is polar

10. 27 Oct 97Chemical Equilibrium10 Fluorine-substituted Ethylene: C 2 H 2 F 2 CIS isomer both C—F bonds on same side  molecule is POLAR. C—F bonds are MUCH more polar than C—H bonds. TRANS isomer both C—F bonds on opposite side  molecule is NOT POLAR.  (C-F) = 1.5,  (C-H) = 0.4

11. 27 Oct 97Chemical Equilibrium11 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)

12. 27 Oct 97Chemical Equilibrium12 16_CoCl2.mov (16z01vd1.mov) Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O ---> Co(H 2 O) 6 Cl 2 Co(H 2 O) 6 Cl 2 (aq) Co(H 2 O) 6 Cl 2 (aq) + 2 H 2 O

13. 27 Oct 97Chemical Equilibrium13 Chemical Equilibrium After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained. Fe 3+ + SCN - FeSCN 2+ 16_FeSCN.mov 16m03an1.mov FeCl 3 (aq) NaSCN(aq) FeSCN (aq)

14. 27 Oct 97Chemical Equilibrium14 CaCO 3 (s) + H 2 O(l) + CO 2 (g) Chemical Equilibria At a given T and pressure of CO 2, [Ca 2+ ] and [HCO 3 - ] can be found from the EQUILIBRIUM CONSTANT. Ca 2+ (aq) + 2 HCO 3 - (aq)

15. 27 Oct 97Chemical Equilibrium15 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) :

16. 27 Oct 97Chemical Equilibrium16 Determining K Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution 1. Set up a table of concentrations: [NOCl][NO][Cl 2 ] 2 NOCl(g) 2 NO(g) + Cl 2 (g) Before2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

17. 27 Oct 97Chemical Equilibrium17 Calculate K from equil. [ ] 2 NOCl(g) 2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] Before2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33 K = (0.66) 2 (0.33) (1.34) 2 = 0.080

18. 27 Oct 97Chemical Equilibrium18 Writing and Manipulating Equilibrium Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2 (g) SO 2 (g) NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq) S OO

19. 27 Oct 97Chemical Equilibrium19 K tot = K 1 x K 2 S(s) + 3/2 O 2 (g) SO 3 (g) Adding equations for reactions S(s) + O 2 (g) SO 2 (g) Manipulating K: adding reactions NET EQUATION SO 2 (g) + 1/2 O 2 (g) SO 3 (g) [SO 3 ] [SO 2 ][O 2 ] 1/2 K 2 = K 1 = [SO 2 ] / [O 2 ] [SO 3 ] [O 2 ] 3/2 K tot = ADD REACTIONS  MULTIPLY K

20. 27 Oct 97Chemical Equilibrium20 Changing direction S(s) + O 2 (g) SO 2 (g) SO 2 (g) S(s) + O 2 (g) Manipulating K: Reverse reactions

21. 27 Oct 97Chemical Equilibrium21 Chemistry of Sulfur Elemental S : stable form is S 8 (s) Oxides of S : SO 2 (g) and SO 3 (g) - significant in atmospheric pollution Industrially: Oxides generated as needed; ‘stored’ as the hydrate SO 3 (g) + H 2 O (l)  H 2 SO 4 (aq) Sulfuric acid is HIGHEST VOLUME chemical (fertilizers, refining, manufacturing) sources: desulfurizing natural gas roasting metal sulfides

22. 27 Oct 97Chemical Equilibrium22 Concentration Units We have been writing K in terms of mol/L. These are designated by K c But with gases, P = (n/V)RT = conc RT P is proportional to concentration, so we can write K in terms of PARTIAL PRESSURES. These constants are called K p. K c and K p have DIFFERENT VALUES (unless same number of species on both sides of equation) Manipulating K : K p for gas rxns

23. 27 Oct 97Chemical Equilibrium23 Concentration of products is much greater than that of reactants at equilibrium. The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. K= p P (H 2 O) 2 P (H 2 ) 2 P (O 2 ) = 1.5 x 10 80 2 H 2 (g) + O 2 (g) 2 H 2 O (g) The reaction is strongly product-favored. K >> 1

24. 27 Oct 97Chemical Equilibrium24 What about the reverse reaction ? This reaction is strongly reactant-favored. Conc. of products is much less than that o f reactants at equilibrium. Meaning of K: AgCl rxn K c = K << 1 Ag + (aq) + Cl - (aq) AgCl(s) K rev = K c -1 = 5.6x10 4. It is strongly product-favored. AgCl(s) Ag + (aq) + Cl - (aq) [Ag + ] [Cl - ]= 1.8 x 10 -5

25. 27 Oct 97Chemical Equilibrium25 CH 3 —C —C —CH 3 H H H H n-butane H CH 3 —C—CH 3 iso-butane CH 3 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. Meaning of K : butane isomerization K = [iso] [n] = 2.5 If [iso] = 0.35 M and [n] = 0.25 M, is the system at equilibrium? If not, which way does the rxn “shift” to approach equilibrium?

26. 27 Oct 97Chemical Equilibrium26 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. Reaction is NOT at equilibrium. [Iso] must INCREASE and [n] must DECREASE. Q = 1.4 which is LESS THAN K =2.5 To reach EQUILIBRIUM [iso] [n] Q has the same form as K,... but uses existing concentrations For n-Butane iso-Butane

27. 27 Oct 97Chemical Equilibrium27 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

28. 27 Oct 97Chemical Equilibrium28 Examples of equilibrium questions Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calculate equilibrium concentrations. H 2 (g) + I 2 (g) 2 HI(g) b. From an initial non-equilibrium condition, what are the concentrations at equilibrium?

29. 27 Oct 97Chemical Equilibrium29 H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Initial 1.001.000 DEFINE x = [H 2 ] consumed to get to equilibrium. Change-x-x+2x At equilibrium1.00-x1.00-x2x Step 1. Set up table to define EQUILIBRIUM concentrations in terms of initial concentrations and a change variable [H 2 ][I 2 ][HI]

30. 27 Oct 97Chemical Equilibrium30 Step 2 Put equilibrium concentrations into K c expression. 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

31. 27 Oct 97Chemical Equilibrium31 [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M Step 3. Solve for x. 55.3 = (2x) 2 /(1-x) 2 In this case, take square root of both sides. 7.44= 2x 1.00x- H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Solution gives: x = 0.79 Therefore, at equilibrium

32. 27 Oct 97Chemical Equilibrium32 “...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.

33. 27 Oct 97Chemical Equilibrium33 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

34. 27 Oct 97Chemical Equilibrium34 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

35. 27 Oct 97Chemical Equilibrium35 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

36. 27 Oct 97Chemical Equilibrium36 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

37. 27 Oct 97Chemical Equilibrium37 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?

38. 27 Oct 97Chemical Equilibrium38 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)

39. 27 Oct 97Chemical Equilibrium39 EQUILIBRIUM AND EXTERNAL EFFECTS Add catalyst ---> no change in K A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

40. 27 Oct 97Chemical Equilibrium40 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)