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Chemical Bonding and Molecular Structure (Chapter 9)

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

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Chemical Bonding and Molecular Structure (Chapter 9)

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  1. 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 Chemical Equilibrium

  2. 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.) Chemical Equilibrium

  3. Bond Polarity (2) • Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond. BOND ENERGY “pure” bond 339 kJ/mol calculated real bond 432 kJ/mol measured Difference 92 kJ/mol. This difference is the contribution of IONIC bonding It is proportional to the difference in ELECTRONEGATIVITY, c. Chemical Equilibrium

  4. Electronegativity, c c 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) Chemical Equilibrium

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

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

  7. Symmetric molecules Molecular Polarity • Molecules such as HCl and H2O 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” Chemical Equilibrium

  8. Molecular Polarity: H2O Water is polar because: a) O-H bond is polar b) water is non-symmetric The dipole associated with polar H2O is the basis for absorption of microwaves used in cooking with a microwave oven Chemical Equilibrium

  9. HBF2 is polar BF3 is NOT polar Molecular Polarity in NON-symmetric molecules B—F bonds are polar molecule is NOT symmetric Atom Chg.  B +ve 2.0 H +ve 2.1 F -ve 4.0 B +ve F -ve B—F bonds are polar molecule is symmetric Chemical Equilibrium

  10. Fluorine-substituted Ethylene: C2H2F2 C—F bonds are MUCH more polar than C—H bonds. (C-F) = 1.5, (C-H) = 0.4 CIS isomer • both C—F bonds on same side  molecule is POLAR. TRANS isomer • both C—F bonds on opposite side  molecule is NOT POLAR. Chemical Equilibrium

  11. CHEMICAL EQUILIBRIUMChapter 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) Chemical Equilibrium

  12. Co(H2O)6Cl2 (aq) Co(H2O)6Cl2 (aq) + 2 H2O Properties of an Equilibrium Equilibrium systems are • DYNAMIC (in constant motion) • REVERSIBLE • can be approached from either direction 16_CoCl2.mov (16z01vd1.mov) Pink to blue Co(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2 Chemical Equilibrium

  13. FeCl3 (aq) NaSCN(aq) Fe3+ + SCN- FeSCN2+ FeSCN (aq) 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. 16_FeSCN.mov 16m03an1.mov Chemical Equilibrium

  14. Chemical Equilibria CaCO3(s) + H2O(l) + CO2(g) Ca2+(aq) + 2 HCO3-(aq) At a given T and pressure of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT. Chemical Equilibrium

  15. a A + b B c C + d D THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type the following is a CONSTANT (at a given T) : If K is known, then we can predict concentrations of products or reactants. Chemical Equilibrium

  16. 2 NOCl(g) 2 NO(g) + Cl2(g) 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] [Cl2] Before 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 Chemical Equilibrium

  17. 2 (0.66) (0.33) = 0.080 K = 2 (1.34) Calculate K from equil. [ ] 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Before 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 Chemical Equilibrium

  18. S O O S(s) + O2(g) SO2(g) NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq) Writing and ManipulatingEquilibrium Expressions Solids and liquids NEVER appear in equilibrium expressions. Chemical Equilibrium

  19. S(s) + O2(g) SO2(g) [SO3] SO2(g) + 1/2 O2(g) SO3(g) K2 = [SO2][O2]1/2 [SO3] S(s) + 3/2 O2(g) SO3(g) [O2]3/2 Manipulating K: adding reactions K1 = [SO2] / [O2] Adding equations for reactions NET EQUATION Ktot = ADD REACTIONS  MULTIPLY K Ktot = K1 x K2 Chemical Equilibrium

  20. S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g) Manipulating K: Reverse reactions Changing direction Chemical Equilibrium

  21. Chemistry of Sulfur Elemental S : stable form is S8 (s) sources: desulfurizing natural gas roasting metal sulfides Oxides of S : SO2 (g) and SO3 (g) - significant in atmospheric pollution Industrially: Oxides generated as needed; ‘stored’ as the hydrate SO3(g) + H2O (l) H2SO4(aq) Sulfuric acid is HIGHEST VOLUME chemical (fertilizers, refining, manufacturing) Chemical Equilibrium

  22. Manipulating K : Kp for gas rxns Concentration Units We have been writing K in terms of mol/L. These are designated by Kc 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 Kp. Kc and Kp have DIFFERENT VALUES (unless same number of species on both sides of equation) Chemical Equilibrium

  23. 2 H2(g) + O2(g) 2 H2O (g) P(H2O)2 K = p P(H2)2P(O2) The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. = 1.5 x 1080 K >> 1 Concentration of products is much greater than that of reactants at equilibrium. The reaction is stronglyproduct-favored. Chemical Equilibrium

  24. AgCl(s) Ag+(aq) + Cl-(aq) Ag+(aq) + Cl-(aq) AgCl(s) Meaning of K: AgCl rxn Kc = [Ag+] [Cl-] = 1.8 x 10-5 K << 1 Conc. of products is much less than that of reactants at equilibrium. This reaction is strongly reactant-favored. What about the reverse reaction ? Krev = Kc-1 = 5.6x104. It is strongly product-favored. Chemical Equilibrium

  25. n-butane iso-butane H H H CH3—C—CH3 CH3 —C —C —CH3 H H CH3 [iso] = 2.5 K = [n] Meaning of K : butane isomerization 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. 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? Chemical Equilibrium

  26. [iso] 0.35 For n-Butane iso-Butane Q = = = 1.40 [n] 0.25 Q - the reaction quotient Q has the same form as K, . . . but uses existing concentrations All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. Q = 1.4 which is LESS THAN K =2.5 Reaction is NOT at equilibrium. To reach EQUILIBRIUM [Iso] must INCREASE and [n] must DECREASE. Chemical Equilibrium

  27. Q/K 1 Q  Q = K Typical EQUILIBRIUM Calculations 2 general types: a. Given set of concentrations, is system at equilibrium ? Calculate Q compare to K IF: Q > K or Q/K > 1  REACTANTS Q < K or Q/K < 1  PRODUCTS Q=K at EQUILIBRIUM Chemical Equilibrium

  28. H2(g) + I2(g) 2 HI(g) Examples of equilibrium questions b. From an initial non-equilibrium condition, what are the concentrations at equilibrium? Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calculate equilibrium concentrations. Chemical Equilibrium

  29. H2(g) + I2(g) 2 HI(g) Kc = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations in terms of initial concentrations and a change variable Initial 1.00 1.00 0 DEFINE x = [H2] consumed to get to equilibrium. Change -x -x +2x At equilibrium 1.00-x 1.00-x 2x [H2] [I2] [HI] Chemical Equilibrium

  30. H2(g) + I2(g) 2 HI(g) Kc = 55.3 Step 1 Define equilibrium condition in terms of initial condition and a change variable [H2] [I2] [HI] At equilibrium 1.00-x 1.00-x 2x Step 2 Put equilibrium concentrations into Kc expression. Chemical Equilibrium

  31. 2x 7.44 = 1.00 - x H2(g) + I2(g) 2 HI(g) Kc = 55.3 Step 3. Solve for x. 55.3 = (2x)2/(1-x)2 In this case, take square root of both sides. Solution gives: x = 0.79 Therefore, at equilibrium [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M Chemical Equilibrium

  32. 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. “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” Chemical Equilibrium

  33. Shifts in EQUILIBRIUM : Concentration • 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. ADDING PRODUCTS - equilibrium shifts to REACTANTS ADDING REACTANTS - equilibrium shifts to PRODUCTS - GAS-FORMING; PRECIPITATION REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION Chemical Equilibrium

  34. n-Butane Isobutane [iso] = 2.5 K = [n] Effect of changed [ ] on an equilibrium Solution A. Calculate Q with extra 1.50 M n-butane. INITIALLY: [n] = 0.50 M [iso] = 1.25 M CHANGE: ADD +1.50 M n-butane What happens ? 16_butane.mov (16m13an1.mov) Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63 Q < K . Therefore, reaction shifts to PRODUCT Chemical Equilibrium

  35. B A Butane/Isobutane B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane] Initial 0.50 + 1.50 1.25 Change - x + x Equilibrium 2.00 - x 1.25 + x Solution x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane. Chemical Equilibrium

  36. N2O4(g) 2 NO2(g) Effect of Pressure (gas equilibrium) Increase P in the system by reducing the volume. Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT PN2O4 increases 16_NO2.mov (16m14an1.mov) PNO2 decreases See Ass#2 - question #6 Chemical Equilibrium

  37. CO2(g) + H2O(liq) CO2(aq) + heat LOWER T HIGHER T EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change  change in K • Consider the fizz in a soft drink Kc = [CO2(aq)]/[CO2(g)] • Change T: New equilib. position? New value of K? • Increase T • Equilibrium shifts left: [CO2(g)]  [CO2 (aq)] • K decreases as T goes up. • Decrease T • [CO2 (aq)] increases and [CO2(g)] decreases. • K increases as T goes down Chemical Equilibrium

  38. N2O4 + heat 2 NO2 (colorless) (brown) Temperature Effects on Chemical Equilibrium Kc = 0.00077 at 273 K Kc = 0.00590 at 298 K Horxn = + 57.2 kJ Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction 16_NO2RX.mov (16m14an1.mov) Chemical Equilibrium

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

  40. CHEMICAL EQUILIBRIUMChapter 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) Chemical Equilibrium

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