普通化学(全英文)教学大纲 Syllabus for general chemistry () ClassWebpagehttp://www.chemistry.fudan.edu.cn/mnetteacher/gfzheng/teaching.html 1. Introduction( Chapter 1-3) 1. 1. Elements, compounds and mixtures 12. Atoms. molecules and ions 1.3. SI units and converting units nd intensive prope 1.5. Amount of substance 16. Law of conservation of matter 1.7. Mass relationships in chemical reactions 18. Balancing the chemical reaction 1.9. Limiting reactants 1. 10. Empirical formula 2. Electronic Structures and the Periodic table(chapter 7-8) 2.1. Atomic structure, Isotope 2.2. The period ic table (a). Main groups, metals and non-metals (b). S-block, p-block, d-block elements 23. Electron cloud and Atomic orbitals 2 4. Electron configurations (a). 4 quantum numbers: n, l,m,ms (b). Pictures of the orbitals: S, px, py, pz (c). How to fill electrons in different orbitals (d). Be able to draw the electron configuration of the first 20 elements (e). Unpaired electrons, Spin, Paramagnetic and Diamagnetism 2.5. Comparing the atomic orbitals with Bohr's model of atoms( Classical but incorrect) 2.6. Electron's states a). Each state is related to specific energy (b). e" will transit between states when absorbing/giving off energy, in a form of" light (c). Ground state, Excited states, the first excited state (d). Energy of light": relationship among wavelength, frequency, and photon's energy (e). Absorption spectrum and Emission spectrum (f). Red shift and blue shift Materials color
普通化学（全英文）教学大纲 Syllabusfor General Chemistry (I) Class Webpage: http://www.chemistry.fudan.edu.cn/m_netteacher/gfzheng/teaching.html 1．Introduction (Chapter 1—3) 1.1．Elements, compounds, and mixtures 1.2．Atoms, molecules and ions 1.3．SI units and converting units 1.4．Extensive and intensive properties 1.5．Amount of substance 1.6．Law of conservation of matter 1.7．Mass relationships in chemical reactions 1.8．Balancing the chemical reaction 1.9．Limiting reactants 1.10．Empirical formula 2．Electronic Structures and the Periodic Table (Chapter 7—8) 2.1．Atomic structure, Isotope 2.2．The periodic table (a)．Main groups, metals and non-metals (b)．s-block, p-block, d-block elements 2.3．Electron cloud and Atomic orbitals 2.4．Electron configurations (a)．4 quantum numbers: n, l, m, ms (b)．Pictures of the orbitals: s, px, py, pz (c)．How to fill electrons in different orbitals (d)．Be able to draw the electron configuration of the first 20 elements (e)．Unpaired electrons, Spin, Paramagnetic and Diamagnetism 2.5．Comparing the atomic orbitals with Bohr’s model of atoms (Classical but incorrect) 2.6．Electron’s states (a)．Each state is related to specific energy (b)．e - will transit between states when absorbing/giving off energy, in a form of “light” (c)．Ground state, Excited states, the first excited state (d)．Energy of “light”: relationship among wavelength, frequency, and photon’s energy (e)．Absorption spectrum and Emission spectrum (f)．Red shift and blue shift (g)．Materials’ color
2.7. Atomic and ionic rad i 2.8. Ionization energy: the first, second,.. ionization energy 2.9. Electron affinity (Note here we use a different definition of the electron affinity from the textbook.) 2.10. Exothermic and Endothermic reactions 3. Chemical Bonds, Valence Bond, and Molecular Shape(Chapter 9) 31. Valence electrons 3.2. Ionic bond 33. Covalent bond 3. 4. Lewis structures (a). Octet rule (b). Single, double, and triple bonds (c). Bond order, bond length, and bond energy Be able to calculate the heat of a reaction from the given bond energies (d). Lone pair electrons, Coordinate bond (e). Be able to write Lewis structures of common molecules ( f. Exception of the octet rule: more than 8 e"; fewer than 8e"; odd number of e (g). Resonance structures (h ). Formal charges -a way to select the most effective Lewis structure(s) (. Failure of Lewis structure: octet rule exception; resonance structure; paramagnetism 3.5. Electronegativity 3.6. Nonpolar and polar covalent bonds 3.7. Valence Bond (VB)method (a). Definition of the valence bond method (b). g bond andπbond How o bond and T bond are made from S, px, py, pz orbitals? How o bond and bond relate to single/double/triple bonds (c). Hybrid orbitals(only applied to central atoms ): sp, sp, sp, spd (d ). Bond angle and the shape of molecules (e). Cis-and Trans-isomerism(sp- hybrid f). VSEPR method: predict the electron pair shapes and molecular shapes How to calculate the hybrid type of the central atom(s) How to arrange lone pair electrons and bonding electrons (g). Advantages and disad vantages of the Vb method 3.8. Molecule's polarity: sum of all bonds' dipoles 3.9. Brief introduction to organic molecules
2.7．Atomic and ionic radii 2.8．Ionization energy: the first, second, … ionization energy 2.9．Electron affinity (Note here we use a different definition of the electron affinity from the textbook.) 2.10．Exothermic and Endothermic reactions 3．Chemical Bonds, Valence Bond, and Molecular Shape (Chapter 9) 3.1．Valence electrons 3.2．Ionic bond 3.3．Covalent bond 3.4．Lewis structures (a)．Octet rule (b)．Single, double, and triple bonds (c)．Bond order, bond length, and bond energy: Be able to calculate the heat of a reaction from the given bond energies (d)．Lone pair electrons, Coordinate bond (e)．Be able to write Lewis structures of common molecules (f)．Exception of the octet rule: more than 8 e- ; fewer than 8 e- ; odd number of e- (g)．Resonance structures (h)．Formal charges – a way to select the most effective Lewis structure(s) (i)．Failure of Lewis structure: octet rule exception; resonance structure; paramagnetism 3.5．Electronegativity 3.6．Nonpolar and polar covalent bonds 3.7．Valence Bond (VB) method (a)．Definition of the valence bond method (b)．σ bond and π bond How σ bond and π bond are made from s, px, py, pz orbitals? How σ bond and π bond relate to single/double/triple bonds? (c)．Hybrid orbitals (only applied to central atoms): sp, sp2 , sp3 , sp3d (d)．Bond angle and the shape of molecules (e)．Cis- and Trans- isomerism (sp2 hybrid) (f)．VSEPR method: predict the electron pair shapes and molecular shapes How to calculate the hybrid type of the central atom(s) How to arrange lone pair electrons and bonding electrons (g)．Advantages and disadvantages of the VB method 3.8．Molecule’s polarity: sum of all bonds’ dipoles 3.9．Brief introduction to organic molecules
4. Molecular Orbital(MO) Method( Chapter 10) 41. Introduction to the mo method (a). Bonding, Antibond ing, and Nonbonding MOs (b). g bond andπbond Compare" o bond and T bond"in VB method and in MO method 4.2. Bond orders, unpaired electrons (a). Compare"Bond order"definitions in Lewis structure, VB, and MO method (b). Compare"Lone pair electrons"in Lewis structure, VB, and MO method (c). Predict the stability of molecules/ions (d). Predict paramagnetism/diamagnetism of a molecule 4.3. Homonuclear diatomic molecules(1st and 2nd period elements) 44. Heteronuclear d atomic molecules 4.5. Conjugated II bond and electron delocalization (a). Be able to judge whether a molecule has a conjugated II bond (b). Draw the molecular shape, and jud ge the p-orbitals and p-electrons (c). Calculate the IInm 46. HOMO and LUMO 4.7. Band theory of bonding in solids (a). Valence band, conduction band, band gap (b). Metal, semiconductor and insulator 48. Inter-molecular interactions (a). van der Waals interaction: polar VS. nonpolar molecules (b). Hydrogen bond (c). Bond energy: chemical bond > hydrogen bond> van der Waals interaction (d). Relationship of Inter-molecular interactions with Melting, boiling point, hardness (e). Hydrophobic and hydrophilic interaction 4.9. Introduction to proteins and nucleic acids (a). Amino acid Peptide Protein: which interactions are involved in each step? (b). Oligonucleotide, Single strand DNA,Double strand DNA: which interactions?
4．Molecular Orbital (MO) Method (Chapter 10) 4.1．Introduction to the MO method (a)．Bonding, Antibonding, and Nonbonding MOs (b)．σ bond and π bond Compare “σ bond and π bond” in VB method and in MO method 4.2．Bond orders, unpaired electrons (a)．Compare “Bond order” definitions in Lewis structure, VB, and MO method (b)．Compare “Lone pair electrons” in Lewis structure, VB, and MO method (c)．Predict the stability of molecules/ions (d)．Predict paramagnetism/diamagnetism of a molecule 4.3．Homonuclear diatomic molecules (1st and 2nd period elements) 4.4．Heteronuclear diatomic molecules 4.5．Conjugated Π bond and electron delocalization (a)．Be able to judge whether a molecule has a conjugated Π bond (b)．Draw the molecular shape, and judge the p-orbitals and p-electrons (c)．Calculate the Πnm 4.6．HOMO and LUMO 4.7．Band theory of bonding in solids: (a)．Valence band, conduction band, band gap (b)．Metal, semiconductor and insulator 4.8．Inter-molecular interactions (a)．van der Waals interaction: polar vs. nonpolar molecules (b)．Hydrogen bond (c)．Bond energy: chemical bond >> hydrogen bond > van der Waals interaction (d)．Relationship of Inter-molecular interactions with Melting, boiling point, hardness (e)．Hydrophobic and hydrophilic interaction 4.9．Introduction to proteins and nucleic acids (a)．Amino acid → Peptide → Protein: which interactions are involved in each step? (b)．Oligonucleotide → Single strand DNA → Double strand DNA: which interactions?
5. Acids and Bases( Chapter 15, 16.1--16.4) 5. 1. The Bronsted-Lowry definitions(H, OH transfer) (a). Strong acid(base)and Weak acid(base) (b). Conjugated acid-base pair Note in each acid-base reactions, there are two conjugated acid-base pairs (c). Poly-protic acid(base), and amphoteric substance 5.2. Lewis acid-base definition (transfer of lone pair electrons to an empty orbital) 5.3. lon pro ater,Kw=[H3O]x [OH-]=10-4(@25C (a). Kw is a constant in all aqueous solution(under a constant temperature) (b).[H3O]>10M→ acidic;田H3O=107M→ neutral;田H3O+10-5M? (b). weak acid only Consider the Kal only( for poly-protic acid) If Cacid/Ka>100>[H30]=v(Cacid X Ka) If Cacid/Ka need to solve the quadratic equation (c). one strong acid+ one weak acid Let the strong acid dissociate completely first, and then setup the" ICE equation of the weak acid's dissociation (d). one weak acid its conjugated base(it is a buffer Ina buffer [H30]=Ka(Cacid/Cbase) (e). one acid one bas Let the acid-base reaction go into completion first. (Need to find out the limiting reactants! Then the remaining species will belong to one of the 5.5. Calculation of oH-l in base solution(s) 5.6. Hydrolysis of salt (a). How to quickly judge whether a salt is acidic, neutral or basic (b). Calculate the pH value of a salt solutio 5.7. Buffer solutions Note for a good buffer, its acid/base ratio should be 0. 1-10 5.8.Tt (a). Titration curve: use a strong base to titrate a strong acid (b). Titration curve: use a strong base to titrate a weak acid
5．Acids and Bases (Chapter 15, 16.1—16.4) 5.1．The Bronsted-Lowry definitions(— H+ , OHtransfer) (a)．Strong acid (base) and Weak acid (base) (b)．Conjugated acid-base pair Note in each acid-base reactions, there are two conjugated acid-base pairs (c)．Poly-protic acid (base), and amphoteric substance 5.2．Lewis acid-base definition (— transfer of lone pair electrons to an empty orbital) 5.3．Ion product of water, Kw = [H3O+ ] × [OH- ] = 10-14 (@ 25 oC) (a)．Kw is a constant in all aqueous solution (under a constant temperature) (b)．[H3O+ ] > 10-7 M → acidic; [H3O+ ] = 10-7 M → neutral; [H3O+ ] 10-5 M? (b)．weak acid only Consider the Ka1 only (for poly-protic acid) If Cacid/Ka > 100 → [H3O+ ] = √(Cacid × Ka) If Cacid/Ka ≤ 100 → need to solve the quadratic equation (c)．one strong acid + one weak acid Let the strong acid dissociate completely first, and then setup the “ICE” equation of the weak acid’s dissociation. (d)．one weak acid + its conjugated base (it is a buffer!) In a buffer: [H3O+ ] = Ka × (Cacid / Cbase) (e)．one acid + one base Let the acid-base reaction go into completion first. (Need to find out the limiting reactants!) Then the remaining species will belong to one of the previous categories. 5.5．Calculation of [OH- ] in base solution(s) 5.6．Hydrolysis of salt (a)．How to quickly judge whether a salt is acidic, neutral or basic (b)．Calculate the pH value of a salt solution 5.7．Buffer solutions Note for a good buffer, its acid/base ratio should be 0.1~10. 5.8．Titration (a)．Titration curve: use a strong base to titrate a strong acid (b)．Titration curve: use a strong base to titrate a weak acid
Outline for general Chemistry (lD) 6A. Solution( Chapter 4) 1 class 61. Solute solvent. and solution 6.2. Solubility(S)and molarity(C) a). Solubil ity: grams of solute /100 g solvent; Molarity: moles of solute /1 L solution (b). Be able to convert between solubility and molarity of a given solution 6.3. Electrolyt (a). Strong electrolyte: fully dissociate into cations and anions in solution or melt (b). Weak electrolyte: partially dissociate(with a dissociation factor between 0-1), and majority remains as molecules 64. Acid base and salt Note that"insoluble salt" is still a strong electrolyte 6.5. Reaction/calculation between ions in solution (a). Water solubility of different acids, bases and salts (b). Find out the limiting reactants for calculation 6.6. Ionic equation 6B. Precipitate(Chapter 16.5--16.8) I class 6.7. Solubility product (a). Ksp=[cation]m x [anion]n,(where m, n are coefficients of ions in the precipitate's dissociation reaction) (b). Convert between solubility product(Ksp)and solubility(s) 6.8. Pred ict the solubility formation Compare the magnitude of ion product, [cation]m x [anion]n, with the Ksp (a). cation]m x [anion under saturated solution, No precipitate (b). [cation]m x [anion]n=Ksp> saturated solution, Start to precipitate (c). [cation]m x [anion]n>Ksp)Over saturated solution, Precipitate is formed 6.9. The common ion effect 6.10. Calculation of ion concentrations in solution When more than one precipitate can be formed, which one will precipitat When the 2nd precipitate starts to form, what is the remaining ion concentration for the 1st ion? 7. Oxidation-Reduction Reaction( Chapter Il) 2 class 7.1. Oxidation numbers(Oxidation state) 7.2. Rules for assigning the oxidation number Apply the 5 steps in order
Outline for General Chemistry (II) 6A．Solution (Chapter 4) 1 class 6.1．Solute, solvent, and solution 6.2．Solubility (S) and molarity (C) (a)．Solubility: # grams of solute / 100 g solvent; Molarity: # moles of solute / 1 L solution (b)．Be able to convert between solubility and molarity of a given solution 6.3．Electrolyte (a)．Strong electrolyte: fully dissociate into cations and anions in solution or melt (b)．Weak electrolyte: partially dissociate (with a dissociation factor between 0~1), and majority remains as molecules 6.4．Acid, base, and salt Note that “insoluble salt” is still a strong electrolyte 6.5．Reaction/calculation between ions in solution (a)．Water solubility of different acids, bases and salts (b)．Find out the limiting reactants for calculation 6.6．Ionic equation 6B．Precipitate (Chapter 16.5—16.8) 1 class 6.7．Solubility product: (a)．Ksp = [cation]m × [anion]n , (where m, n are coefficients of ions in the precipitate’s dissociation reaction) (b)．Convert between solubility product (Ksp) and solubility (S) 6.8．Predict the solubility formation – Compare the magnitude of ion product, [cation]m × [anion]n , with the Ksp (a)．[cation]m × [anion]n Ksp → Over saturated solution, Precipitate is formed 6.9．The common ion effect 6.10．Calculation of ion concentrations in solution When more than one precipitate can be formed, which one will precipitate first? When the 2nd precipitate starts to form, what is the remaining ion concentration for the 1st ion? 7．Oxidation-Reduction Reaction (Chapter 11) 2 class 7.1．Oxidation numbers (Oxidation state) 7.2．Rules for assigning the oxidation number Apply the 5 steps in order
7.3. Oxidation numbers and the period ic table Highest possible oxidation number = of its valence e"s Lowest possible oxidation number = of e"s needed to gain for an octet 74. Oxidation-reduction reaction (a). Any oxidation number change Red-ox reaction (b). Oxidation half-reaction(for Reducing agent ) Oxidation# t, lose e"s, being oxidized (c). Reduction half-reaction(for Oxid izing agent): Oxidation#l, gain e"s, being reduced 7.5. Balance the Red-ox reactions(Half-reaction method Find out species that are oxidized or reduced( based on their oxidation #s); Balance each half reaction If in a basic solution add oH-to each side to neutralize h Double check the conservation of mass and charges 8. Changes in States(Chapter 5, 12, 13.5--13.7) 3 classes 8. 1. State of Matters-Gas, liquid, and solid 8.2. Gas and Ideal Gas 2 assumptions for the ideal gas molecules: (1)no molecule size; (2)no intermolecular interaction 83. Ideal gas law: PV=nRT Select the correct value for the gas constant R, based on the units 8.4. Avogadro's law a).@ same T P, 1 mole of any ideal gas occupies the same volume (b).@ Standard temperature and pressure, I mole of any ideal gas takes 22. 4 liter 8.5. Mole fraction(Xi Xi=ni/ntotal= Pi /Ptotal=Vi/vtotal Note that all the n' s here only inchude gas speci Ies 8.6. Partial pressure(Pi): pressure of a gas component if it solely occupies the whole volume ◆ Ptotal:=PA+PB+.+P1 Pi= Pt 8.7. Partial volume(Vi): Volume of a gas component under t and Ptotal Vtotal=VA+Vb+ 8.8. Phase, Phase change, and Phase equilibrium (a). Liquid evaporation and the vapor pressure Estimate the relative magnitude of different molecules'vapor pressure, based on their molecular structures (b). Change of the vapor pressure with T: In(P2/P1)=-AHvap/[R(1/) (c). Boiling temperature(Tb): Tb, vapor pressure= 1 atm (d). Melting Freezing curves Understand the molecular view of melting freezing process Why the temperature does not increase when a solid melts, or a liquid boils? Why a liquid can go lower than its freezing point before it freezes? (e). Sublimation deposition
7.3．Oxidation numbers and the periodic table Highest possible oxidation number = # of its valence es; Lowest possible oxidation number = # of es needed to gain for an octet 7.4．Oxidation-reduction reaction (a)．Any oxidation number change → Red-ox reaction (b)．Oxidation half-reaction (for Reducing agent): Oxidation # ↑, lose es, being oxidized (c)．Reduction half-reaction (for Oxidizing agent): Oxidation # ↓, gain es, being reduced 7.5．Balance the Red-ox reactions (Half-reaction method) Find out species that are oxidized or reduced (based on their oxidation #s); Balance each half reaction; If in a basic solution, add OH- to each side to neutralize H+ ; Double check the conservation of mass and charges 8．Changes in States (Chapter 5, 12, 13.5—13.7) 3 classes 8.1．State of Matters – Gas, liquid, and solid 8.2．Gas and Ideal Gas 2 assumptions for the ideal gas molecules: (1) no molecule size; (2) no intermolecular interaction 8.3．Ideal gas law: PV = nRT Select the correct value for the gas constant R, based on the units. 8.4．Avogadro’s law (a)．@ same T & P, 1 mole of any ideal gas occupies the same volume; (b)．@ Standard temperature and pressure, 1 mole of any ideal gas takes 22.4 liter. 8.5．Mole fraction (Xi): Xi = ni / ntotal = Pi / Ptotal = Vi / Vtotal Note that all the n’s here only include gas species 8.6．Partial pressure (Pi): pressure of a gas component if it solely occupies the whole volume Ptotal = PA + PB + …… + Pi Pi = Ptotal × Xi 8.7．Partial volume (Vi): Volume of a gas component under T and Ptotal Vtotal = VA + VB + …… + Vi Pi × Vtotal = Ptotal × Vi 8.8．Phase, Phase change, and Phase equilibrium (a)．Liquid evaporation and the vapor pressure Estimate the relative magnitude of different molecules’ vapor pressure, based on their molecular structures (b)．Change of the vapor pressure with T: ln (P2/P1) = – ΔHvap /[R (1/T2 – 1/T1)] (c)．Boiling temperature (Tb): @ Tb, vapor pressure = 1 atm (d)．Melting & Freezing curves: Understand the molecular view of melting & freezing process: Why the temperature does not increase when a solid melts, or a liquid boils? Why a liquid can go lower than its freezing point before it freezes? (e)．Sublimation & deposition
8.9. Phase d iagrams Understand the meaning of different zones, lines and points in a phase Be able to use a phase diagram to explain different phenomena 8.10. Two additional concentration representation Mole fraction(X): percentage of moles(between solute and solvent) 8.11. Colligative properties: only depend on the number of solute particles (a). Vapor pressure lowering-Raoult's law Understand the reason of vapor pressure lowering from molecular point of vIew For nonvolatile solute: P= Po solvent X Xsolvent: P= Po solvent x Xsolute For two volatile components: Ptotal =P1,0 XI+ P2,0x X2 Calculate the ratios of two volatile components in both liquid and gas phases the one with higher vapor pressure will be enriched in the gas phase Distillation (b). Boiling point elevation ◆ Increase of Tb:△Tb=Tb-Tb=Kb×m;( Kb is a constant for the solvent,m is the molality of solute) Using the boiling point elevation to calculate the solute's Molar mass (c). Freezing point depr Decrease ofTE. ATf= Tf-Tf= KfX m;(Kf is a constant for the solvent, m is the molality of solute) (d ). Osmotic pressure II=c R*T Using the osmotic pressure to calculate the solute's Molar mass Summary and Exercise Class #3 Midterm Exam I class
8.9．Phase diagrams Understand the meaning of different zones, lines and points in a phase diagram Be able to use a phase diagram to explain different phenomena 8.10．Two additional concentration representation: Molality (m): # of solute moles per solvent mass in kg Mole fraction (X): percentage of moles (between solute and solvent) 8.11．Colligative properties: only depend on the number of solute particles (a)．Vapor pressure lowering – Raoult’s law Understand the reason of vapor pressure lowering from molecular point of view For nonvolatile solute: P = P0, solvent × Xsolvent; ΔP = P0, solvent × Xsolute For two volatile components: Ptotal = P1,0 × X1 + P2,0 × X2 Calculate the ratios of two volatile components in both liquid and gas phases – the one with higher vapor pressure will be enriched in the gas phase → Distillation (b)．Boiling point elevation: Increase of Tb: ΔTb = Tb ’ – Tb = Kb × m; (Kb is a constant for the solvent, m is the molality of solute) Using the boiling point elevation to calculate the solute’s Molar mass. (c)．Freezing point depresstion Decrease of Tf: ΔTf = Tf ’ – Tf = Kf × m; (Kf is a constant for the solvent, m is the molality of solute) (d)．Osmotic pressure Π = Csolute * R * T Using the osmotic pressure to calculate the solute’s Molar mass. Summary and Exercise Class #3 1 class Midterm Exam 1 class
9. Chemical Thermodynamics and thermochemistry( Chapter 6) classes 9.I. System, Universe Open system, Closed system, Isolated system 9.2. State and State function (a). State functions changes that only depends on the initial state and the final state, not on the path of ch es (b). Internal energy and Internal energy change (U, AU (c). Heat(Q)and Work (W)-they are NoT state functions Q=n C.(Tfinal- Tinitial), where C(Cp or Cv) is the molar heat capacity 9.3. Thermodynamics First Law: AU=Q-W ). Signs of△U,Q,andW (b). Calculation of the volume work (i.e. expansion, compression) Under constant external pressure(Pex):W=Pex. AV Under vacuum(Pex=0):W=0·△v=0 9.4. Enthalpy and Enthalpy change(H, AH) Relationship between△Hand△U:△H=AU+△ nasri 9.5. Thermal Chemical Equation AH and Q (heat) are extensive properties, and relate to moles of a reaction Exothermic(q>0): AH0 9. 6. Hess law When a reaction can be written as combination of a few other reactions. those thermodynamic state functions,(e.g.△U,△H,△S,△G, K) can be expressed as corresponding combination How to find out the combination coefficients of different sub-reactions? 9.7. Standard States(XT) For a pure solid or liquid they are standard states, concentration=1 For a gas>Standard states: Partial =1 atm For an ion in solution Standard states: [ion]=1 M If non-stand ard states, we need to convert,e.g.△Gr=△G°r+ R.In Q 9.8. Calculation of△H (a). Using the enthalpy of formation(△H°):△H=∑(△H° products)-x(△H° reactants) (b). Using the enthalpy of combustion(△H°):△H°=∑(△H。 reactants)-∑(△H。 products) (c). Using the bond energy(B.E.).△H°=∑(BE. reactants)-∑(B.E. products 9.9. Entropy and Entropy Change(s, As (a). Factors that affect entropy: State of matter, Temperature, Pressure, Molecular weight, molecular symmetry (b). Qualitatively, if gas molecules#↑→ΔS↑ 9.10. Thermodynamics Second Law:△ Sisolated=△ Ssystem+△ Surroundings>0 △ Sisolated>0÷ spontaneous;△ Sisolated=0 equilibrium, ISolated<0 non-spontal 9. 11. Gibbs free energy and change(GT, AGT)
9．Chemical Thermodynamics and Thermochemistry (Chapter 6) 4 classes 9.1．System, surroundings, and universe System + Surroundings = Universe Open system, Closed system, Isolated system 9.2．State and State function (a)．State functions: changes that only depends on the initial state and the final state, not on the path of changes (b)．Internal energy and Internal energy change (U, ΔU) (c)．Heat (Q) and Work (W) – they are NOT state functions!! Q = n ∙ C ∙ (Tfinal – Tinitial), where C (Cp or Cv) is the molar heat capacity 9.3．Thermodynamics First Law: ΔU = Q – W (a)．Signs of ΔU, Q, and W (b)．Calculation of the volume work (i.e. expansion, compression) Under constant external pressure (Pex): W = Pex ∙ ΔV Under vacuum (Pex = 0): W = 0 ∙ ΔV = 0 9.4．Enthalpy and Enthalpy change (H, ΔH) Relationship between ΔH and ΔU: ΔH = ΔU + ΔngasRT 9.5．Thermal Chemical Equation ΔH and Q (heat) are extensive properties, and relate to moles of a reaction. Exothermic (q > 0): ΔH 0 9.6．Hess Law – When a reaction can be written as combination of a few other reactions, those thermodynamic state functions, (e.g. ΔU, ΔH, ΔS, ΔG, K) can be expressed as corresponding combination. How to find out the combination coefficients of different sub-reactions? 9.7．Standard States (Xo T): For a pure solid or liquid → they are standard states, concentration = 1 For a gas → Standard states: Ppartial = 1 atm For an ion in solution → Standard states: [ion] = 1 M If non-standard states, we need to convert, e.g. ΔGT =ΔGo T + R∙T∙ln Q 9.8．Calculation of ΔH (a)．Using the enthalpy of formation (ΔHf o ): ΔHo = ∑ (ΔHf o products) – ∑(ΔHf o reactants) (b)．Using the enthalpy of combustion (ΔHc o ): ΔHo = ∑ (ΔHc o reactants) – ∑(ΔHc o products) (c)．Using the bond energy (B.E.): ΔHo = ∑ (B.E. reactants) – ∑(B.E. products) 9.9．Entropy and Entropy Change (S, ΔS) (a)．Factors that affect entropy: State of matter, Temperature, Pressure, Molecular weight, molecular symmetry (b)．Qualitatively, if gas molecules # ↑ →ΔS ↑ (c)．ΔS o = ∑ (So products) – ∑(So reactants) 9.10．Thermodynamics Second Law: ΔSisolated = ΔSsystem + ΔSsurroundings > 0 ΔSisolated > 0 → spontaneous; ΔSisolated = 0 → equilibrium; ΔSisolated < 0 → non-spontaneous 9.11．Gibbs free energy and change (GT, ΔGT)
(a).G=H-T·S,△G=△H-T·△s,△G°=△H°-T·△S° (b). AG is strongly temperature dependent; (AH and AS are not strongly affected by t) (c). Using△G(△OT4G°") to predict changes:△G>0÷ spontaneous;△G=0 equilibrium; AG 0→ spontaneous@all △H>0,△S0,△S>0→ spontaneous@ high T(T>△H△S (e). Calculation of△G298K:△G298K°=∑(△Gr,298K° products)-∑(△Gf,298K° reactants) (f). Calculation of△Gr°:△G°r=△H°298K-T·△S298K,( assume△H,△ S are not ependent) (g). Calculation of△ g from△G°:△Gr=△G"r+ R. Q,(Q=[ products]/ reactants]) 9. 12. Thermodynamics Third Law: for all materials, SoK=0 10. Chemical Equilibrium( Chapter 14, 17) 3 classes 10. 1. Introduction to chemical equilibrium 10. 2. Equilibrium constants(K)-K only depends on temperature 10.3. Calculations involving equilibrium Set up the chemical equation with "ICE(initial, change, and equilibrium concentrations, pressures, or moles, then solve the equation ofK 10.4. Relationship between K and△G°r:△Gr=- RTIn K 10.5. Predict the reaction direction by comparing the magnitude of Q and K:(QK) 10.6. Hess law for Ag and K Multiple ionic equilibria in solution(using Hess Law to find out combination Pay attention to K: e.g. if (reaction 1)=2*(reaction 2)-3*(reaction 3) →△G1=2*△G2-3*△G3.K1=K2/K3 10.7. Shift of equilibrium (a). Qualitatively, use Le Chatelier's principle to predict the shift direction (b). Quantitatively, set up"ICE"equation to solve the equation ofk (c). Change of K with T: In(K2/K1=-AH/[R(1/T2-1TDI 10.8. Learn how to plot Concentration- Time graphs 11. Chemical Kinetics( Chapter 18) 3 classes 11.1. Rates of Reactions. Rate and concentration v=△c/At=dC/dt for aa+bb dd:v=-dA/adt=-dB/b t=dD/ddt 112. Rate laws How to find rate laws based on reactants ' concentrations and reaction rates Differential rate laws:v=k·[AP·[B Integrated rate laws: C-t
(a)．G = H – T ∙ S, ΔG = ΔH – T ∙ ΔS, ΔGo = ΔHo – T ∙ ΔS o (b)．ΔG is strongly temperature dependent; (ΔH and ΔS are not strongly affected by T) (c)．Using ΔG (NOT ΔGo !!) to predict changes: ΔG > 0 → spontaneous; ΔG = 0 → equilibrium; ΔG 0 → spontaneous @ all T ΔH > 0, ΔS 0, ΔS > 0 → spontaneous @ high T (T > ΔH/ΔS) (e)．Calculation of ΔG298Ko : ΔG298Ko = ∑ (ΔGf, 298Ko products) – ∑(ΔGf, 298Ko reactants) (f)．Calculation of ΔGT o : ΔGo T = ΔHo 298K – T ∙ ΔS o 298K , (assume ΔH, ΔS are not temperature dependent) (g)．Calculation of ΔG from ΔGo : ΔGT =ΔGo T + R∙T∙ln Q , (Q = [products] / [reactants]) 9.12．Thermodynamics Third Law: for all materials, S0K = 0 10．Chemical Equilibrium (Chapter 14, 17) 3 classes 10.1．Introduction to chemical equilibrium 10.2．Equilibrium constants (K) – K only depends on temperature! 10.3．Calculations involving equilibrium Set up the chemical equation with “ICE” (initial, change, and equilibrium) concentrations, pressures, or moles, then solve the equation of K. 10.4．Relationship between K and ΔGo T: ΔGo T = – R∙T∙ln K 10.5．Predict the reaction direction by comparing the magnitude of Q and K: (Q K) 10.6．Hess law for ΔG and K Multiple ionic equilibria in solution (using Hess Law to find out combination coefficients) Pay attention to K: e.g. if (reaction 1) = 2 * (reaction 2) – 3 * (reaction 3) → ΔG1 = 2 *ΔG2 – 3 *ΔG3, K1 = K2 2 / K3 3 10.7．Shift of equilibrium (a)．Qualitatively, use Le Châtelier’s principle to predict the shift direction (b)．Quantitatively, set up “ICE” equation to solve the equation of K. (c)．Change of K with T: ln (K2/K1) = – ΔHo / [R (1/T2 – 1/T1)] 10.8．Learn how to plot Concentration ~ Time graphs 11．Chemical Kinetics (Chapter 18) 3 classes 11.1．Rates of Reactions, Rate and concentration v = Δc/Δt = dC/dt for aA + bB → dD: v = – dA/a∙dt = – dB/b∙dt = dD/d∙dt 11.2．Rate laws How to find rate laws based on reactants’ concentrations and reaction rates Differential rate laws: v = k ∙ [A]m ∙ [B]n Integrated rate laws: C ~ t
11.3. Zero-order. First-order. Second-order reaction Be able to write rate laws for Zero-order and First-order reactions Half-life(tin): Att=tin, C=0.5 C For the first-order reaction, ti/ is independent of initial concentration (In 2)/k Units of rate constant k for d ifferent order of reactions (a). Ele (b). Change ofk with T: In(k2/k1)=-Ea/[R(1/T2-1/Tu)I (c). If there is one slowest step, this step is the rate determining step (d ). If there is no slowest step, use" Steady-state approximation "of any intermediates Summary and Exercise Class #4 Final exam
11.3．Zero-order, First-order, Second-order reaction Be able to write rate laws for Zero-order, and First-order reactions Half-life (t1/2): At t = t1/2, C = 0.5 C0 For the first-order reaction, t1/2 is independent of initial concentration: t1/2 = (ln 2) / k Units of rate constant k for different order of reactions 11.4．Reaction mechanism (a)．Elementary reactions (b)．Change of k with T: ln (k2/k1) = – Ea / [R (1/T2 – 1/T1)] (c)．If there is one slowest step, this step is the rate determining step. (d)．If there is no slowest step, use “Steady-state approximation” of any intermediates Summary and Exercise Class #4 1 class Final Exam
