Week 1 Lecture 1 - Introduction to chemistry MACROSCOPIC VS SUBMICROSCOPIC 2KI (aq) + Pb(NO3)2 (aq) 2KNO3 (aq) + PbI2 (s) - Pb2+ (aq) and I- (aq) combine to form ppt (yellow) Ppt is insoluble in water PbI2 is ionic – forms a 3D structure CLASSIFYING MATTER - - Physical State: Gas Solid Liquid Composition: Element Compound Mixture PHYSICAL STATES OF MATTER - Gas, liquid and solid Can also exist in supercritical state between gas and liquid Mobility of atoms decreases from Gas>Liquid>Ice Density usually decreases from Solid>Liquid>Gas (Ice has Lower density than water because of arrangement) COMPOSITION OF MATTER - Substance = matter of uniform composition throughout sample Substance divided into two categories: Element – Atoms or molecules consisting of one kind of atom Compounds – molecules consisting of two pr more different kinds of atoms MIXTURES - Two or more substances together Exhibit properties of the substances that comprise it Either: Heterogeneous – variation in composition throughout sample Homogeneous (also called solution) – constant composition throughout sample CLASSIFICATION OF MATTER – BASED ON COMPOSITION DIFFERENT PROPERTIES OF MATTER PROPERTY Physical Chemical Intensive Extensive DESCRIPTION Observed w/out changing a substance into another substance When a substance is changed into another substance Independent of amount of substance that is present – important when identifying a substance Depend on amount of substance present EXAMPLE(S) Colour, odour, density, melting point, boiling point and hardness Flammability Density, boiling point or colour Mass, volume or energy SEPARATION OF MIXTURES – METHODS - Filtration: Separation of solid and liquid Solid filtered out of liquid - Distillation: Separates liquid from liquid Uses different boiling point of substances Boils solution to vaporize substance 1 Substance 1 then condenses and collects in receiving flask After substance 1 is boiled then only substance 2 remains in flask - Chromatography: Mixture dissolved in the solvent Solvent is added throughout the process Components separate Each component is collected as it reaches bottom of the chromatography column Lecture 2 – Introduction to Chemistry SI BASE UNITS USED: - Length = Meter (m) Mass = Kilogram (Kg) Temperature = Kelvin (K) Time = Second (s) Amount of Substance = Mole (mol) Electric Current = Ampere (A) Luminous Intensity = Candela (cd) METRIC SYSTEM PREFIXES: PREFIX GigaMegaKiloDeciCentiMilliMicroNanoPicoFemto- ABBREVIATION G M k d c m µ n p f TEMPERATURE - Celsius scale – based on properties of water 0ᵒC = freezing point of water 100ᵒC = boiling point of water MEANING 109 – billion 106 – million 103 – thousand 10-1 – one tenth 10-2 – one hundredth 10-3 – one thousandth 10-6 – one millionth 10-9 – one billionth 10-12 10-15 - - Kelvin scale – based on properties of gases No negative temperatures Lowest possible temperature = absolute zero (0 K) ᵒC + 273 = K TOOLS TO MEASURE VOLUMES OF LIQUID - - Deliver variable volumes: Graduated cylinder Syringe Burette Deliver specific volume Pipette Hold a specific volume: Volumetric flask Volume (m3) is not an SI unit because it is derived from length LENGTH, VOLUME AND MASS - Litre (L) = cube 1dm long on each side (dm 3) Millilitre (mL) = cube 1cm long on each side (cm 3) Microlitre (µL) = cube 1mm long on each side (mm3) 1L = 1000cm3 1kg = 1000g 1g = 1000mg DENSITY - Density = mass per unit volume (g/cm3) Densities of select substances @25ᵒC SUBSTANCE Air Balsa wood Ethanol Water Ethylene glycol Table sugar Table salt Iron Mercury Gold Osmium DENSITY (G/CM3) 0.001 0.16 0.79 1.00 1.09 1.59 2.16 7.87 13.53 19.32 22.59 UNCERTAINTY IN MEASUREMENTS - Different measuring devices = different uses and degrees of accuracy All - measured numbers have some degree of inaccuracy Last digit measured is considered reliable but not exact (e.g. 28.88mL) PRECISION AND ACCURACY EXPERIMENTAL ERROR Error∈measurement =Experimentally determined value−accepted value Percentage error= Error∈ Measurement X 100 % Accepted Value DETERMINATE VS INDETERMINATE ERRORS Sume of the Squares of the Deviations for each Measuremnet √ ¿ the Avereage Standard Deviation= ¿ One Less thanthe Number of Measurements SIGNIFICANT FIGURES – THE ROLE OF ZEROS SIGNIFICANT FIGURES – IN CALCULATIONS - Rule 1 – when adding or subtracting, the number of decimal places in answer is equal to the number of decimal places in number with fewest digits after decimal point Rule 2 – in multiplication or division, the number of sig.fig in answer is determined by quantity with fewest sig.fig Rule 3 – number rounded off, last digit to be retained is increased by one only if following digit 5 or greater Lecture 3 – Introduction to Chemistry ATOMIC STRUCTURE Particle Charge Proton 1+ Neutron Neutral Electron 1- Unified atomic mass unit (u): 1.66054 x 10 -24 g - Electron charge = -1.6022 x 10-19C ATOMIC MASS - Carbon – standard Mass (amu) 1.0073 1.0087 5.486 x 10-4 - 12 C mass = exactly 12u 1u = 1.66054 x 10-24 g Hydrogen was standard with mass of 1 Different isotopes of carbon are taken into account – leads to average atomic mass of 12.01u MASS NUMBERS - Mass number = protons + neutrons Atomic number = protons or electrons 23 11Na – 11e, 11p and 12n (12n = 23-11) ISOTOPES - Atoms of same element with different masses, due to them having different numbers of neutrons but the same number of protons AVERAGE ATOMIC WEIGHT Atomic Weight= Isotope 1 % Abundance Isotope 2 ( Mass of Isotope 1 )+( )(Mass of Isotope 2) (% Abundance ) 100 100 AVERAGE ATOMIC WEIGHT MEASUREMENT - Spectrometer measures atomic and molecular weights As well as % abundancies of different isotopes PERIODIC TABLE OF ELEMENTS - Systematic organization of elements - Arranged in order of atomic number Periods – Horizontal rows Elements – order of increasing atomic number Groups – vertical columns (elements with similar properties) Non-metals and metals – divided by steplike line 19 – Atomic Number K – Atomic Symbol 39.0983 – Atomic Weight - Repeating pattern of properties and reactivity: Element nonreactive gas soft reactive metal element etc. SOME GROUPS IN PERIODIC TABLE GROUP 1 2 16 17 18 IONS AND IONIC COMPOUNDS NAME Alkali Metals Alkaline Earth Metals Chalcogens Halogens Noble Gases ELEMENTS Li, Na, k, Rb, Cs, Fr Be, Mg, Ca, Sr, Ba, Ra O, S, Se, Te, Po F, Cl, Br, I, At He, Ne, Ar, Kr, Xe, Rn COMMON CATIONS COMMON ANIONS IONIC COMPOUNDS - Charge on cation becomes subscript on anion Charge of anion becomes subscript on cation If these subscripts are not in lowest whole number ratio – divide them by greatest common factor MOLECULES AND MOLECULAR COMPOUNDS CHEMICAL FORMULAS - Molecular – structure E.g. H2O2 - Empirical – simplest form E.g. HO Week 2 – Lectures and Slides Lecture 4 – Introduction to Chemistry NOMENCLATURE – IONIC COMPOUNDS NAMES OF CATIONS - - - Cations of metal – same name as metal Na+ = Sodium ion Zn2+ = Zinc ion Al3+ = aluminium ion Transition metals can form cations with different charges Roman numerals indicate charges Exception = Ag+ and Zn2+ Cu+ = copper (I) ion and Cu2+ = copper (II) ion Co2+ = cobalt (II) ion and Co3+ = cobalt (III) ion Cations of Non-metals – names end in —ium NH4+ = ammonium ion H3O+ = hydronium ion NAMES OF ANIONS - - - - - Monoatomic anions: Replace end of name with –ide H- = hydride ion O2- = oxide ion N3- = nitride ion Few simple polyatomic anions also end in –ide OH- = hydroxide ion CN- = cyanide ion N3- = azide ion O22- = peroxide ion Oxyanions – end in –ate or –ite (–ite oxyanion has same charge as –ate, but one less oxygen) NO3- = nitrate ion and NO2- = nitrite ion SO42- = sulfate ion and SO32- = sulphite ion ClO3- = chlorate ion and ClO2- = chlorite ion Note when to use prefixes per—and hypo— ClO4- = perchlorate ion ClO3- = chlorate ion ClO2- = chlorite ion ClO- = hypochlorite ion Oxyanions with H+ - add prefix hydrogen or dihydrogen PO43- = phosphate ion HPO32- = hydrogen phosphate ion H2PO2- = dihydrogen phosphate ion This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:54:49 GMT -06:00 https://www.coursehero.com/file/143247979/Lectures-and-Slides-Week-2docx/ NAMES AND FORMULAE OF IONIC COMPOUNDS Ionic Compound CaBr2 NaHSO4 Mg(OH)2 TiCl2 Co2O3 FeO Fe2O3 Ions Involved Ca2+ and 2 BrNa+ and HSO4Mg2+ and 2 OHTi2+ and 2 Cl2 Co3+ and 3 O2Fe2+ and O22 Fe3+ and 3 O2- Name Calcium Bromide Sodium Hydrogen Sulfate Magnesium Hydroxide Titanium(II) Chloride Cobalt(III) Oxide Iron(II) Oxide Iron(III) Oxide ACID NOMENCLATURE NOMENCLATURE – MOLECULAR COMPOUNDS - - Name of element farther to the left in periodic table or lower in same group – written first Prefix used to denote number of atoms of each element in the compound (mono- not used on first element listed) Ending of second element = —ide CO2 = carbon dioxide CCl4 = carbon tetrachloride If prefix ends with a or o and name of element begins with vowel – two successive vowels often elided into one N2O5 = dinitrogen pentoxide CO = carbon monoxide This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:54:49 GMT -06:00 https://www.coursehero.com/file/143247979/Lectures-and-Slides-Week-2docx/ Lecture 1 – Stoichiometry BALANCING CHEMICAL EQUATIONS - CH4 (g) + O2 (g) CO2 (g) + H2O (g) Balanced version = CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) LHS = 1C, 4H, 4O & RHS = 1C, 4H, 4O TYPES OF REACTIONS ACCORDING TO CHEMICAL REACTIVITY - Combination – Reactants combined to form one product A + B AB - Decomposition – Reactant is split into two or more products AB A + B - Combustion – Reactant reacts with oxygen A + O2 CO2 + H2O COMBINATION 2Mg (s) + O2 (g) 2MgO (s) - Ribbon of Mg metal surrounded by oxygen gas in the air Intense white flame produced as Mg atoms react with O 2 Reaction forms MgO (white, ionic solid) DECOMPOSITION - Inflation of car air bags: 2NaN3 (s) 2Na (s) + 3N2 (g) - Production of Lime: CaCO3 (s) CaO (s) + CO2 (g) COMBUSTION - O2 from air is used as a reactant CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) FORMULA AND MOLECULAR WEIGHT - Formula Weight = sum of atomic weights for atoms in chemical formula E.g. for Na the formula weight is the atomic weight (23.0 u) If substance is a molecule – formula weight also called molecular weight Molecular Weight = sum of atomic weights of atoms in a molecule E.g. for C6H12O6 (glucose) the molecular weight is 180.0 u PERCENTAGE COMPOSITION OF A COMPOUND This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:54:49 GMT -06:00 https://www.coursehero.com/file/143247979/Lectures-and-Slides-Week-2docx/ % Element = - ( Number of atoms)( Atomic weight) x 100 Formula weight of compound Example: % C ∈C 6 H 12 O6= ( 6 ) ( 12.0 u ) 72.0 u x 100= x 100=40.0 % 180.0u 180.0 u AVOGADRO’S NUMBER AND MOLAR MASS - NA = 6.02x1023 mol-1 1 mole = number of atoms found in exactly 12g of 12C 6.02x1023 (Avogadro’s Number) carbon atoms in 12g 12C Therefore: 1 mol 12C atoms = 6.02x1023 atoms 1 mol CO2 molecules = 6.02x1023 molecules 1 mol CO22- ions = 6.02x1023 ions MOLAR MASS - Molar Mass (Mr) = mass of 1mol of a substance (i.e. g/mol) Molar mass of an element = atomic weight (from periodic table) Formula weight (in u) = Molar mass (in g/mol) – are the same number CONVERTING MASS, MOLES AND NUMBERS - Mass (g) > (Use molar mass) > Moles (mol) > (Use NA) > Formula units (-) (Mass / Molar mass) = Moles Number of molecules = NA x Moles This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:54:49 GMT -06:00 https://www.coursehero.com/file/143247979/Lectures-and-Slides-Week-2docx/ Powered by TCPDF (www.tcpdf.org) Week 3 Lecture 2 – Stoichiometry Empirical Formulas from Analyses - EXAMPLE: o 73.9% Hg and 26.1% Cl by weight – can work out molar ratio o Means that 100g of the substance contains 73.9g Hg and 26.1g Cl o Covert masses to moles: o Hg, 1 mol = 200.6g o Cl, 1 mol = 35.5g o Molar ratio: o Hg: 73.9g x 1mol/200.6g = 0.368 mol o Cl: 26.2g x 1 mol/35.5g = 0.735 mol o Hg: 0.368 mol/0.368 mol = 1.00 o Cl: 0.735 mol/0.368 mol = 1.997 ~ 2.00 o Empirical formula = HgCl2 Finding Molecular formula from Empirical formula - Calculate empirical formula weight – formula weight calculated from empirical formula Obtain a whole number multiple by diving molecular weight by empirical formula weight Use this multiple to multiply through subscripts in empirical formula EXAMPLE – MESITYLENE: o C3H4 w/ molecular weight of 121.0 amu o 3 x 12 + 4 x 1 = 40 amu o (Experimental M. wt) / (empirical formula weight) = 121 amu/40 amu = 3.02 ~ 3.0 o Multiple = 3.0, so molecular formula = C9H12 Lecture 3.1 – Stoichiometry Quantitative Information from Balanced Equations - Coefficients in balanced equation represent both rea=lative numbers of molecules and realative numbers of moles Therefore, they also represent relative masses 2H2 + O2 2H2O = 2 moles of H, 1 mole of O and 2 moles of H2O = 2x2.0g + 32.0g 2x18.0g This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:55:56 GMT -06:00 https://www.coursehero.com/file/143252532/Lectures-and-Slides-Week-3docx/ Multiplying an Equation Through by a Common Multiple - Can multiply coefficients in a balanced equation by any multiple – still has correct ratio of moles EXAMPLE: o Zn (s) + 2HCL (aq) ZnCl2 (aq) + H2 (g) o 1 mole + 2 moles 1 mole + 1 mole IF WE HAVE S MOLES OF ZN (X2): o 2 moles + 4 moles 2 moles + 2 moles IF WE HAVE 0.5 MOLES OF ZN (X0.5): o 0.5 moles + 1 mole 0.5 moles + 0.5 moles Masses of products and Reactants in Balanced Equation - Can work out how many grams of water will be produced by burning given amounts of H 2 and O2 together EXAMPLE – COMBUSTION OF BUTANE: 2 C4H10 (l) + 13 O2 (g) 8CO2 (g) + 10 H2O (l) 2 moles + 13 moles 8 moles + 10 moles 2 x 58.0 g(m.m. C4H10) 8 x 44.0 g(m.m.CO2) General Plan for Stoichiometry Calculations - MASS REACTANT (n = M/Mr) MOLES REACTANT (Stoichiometric Factor) MOLES PRODUCT (x molar mass) MASS PRODUCT How Many Grams of CO2 will be Obtained by Burning 1.00g of C4H10? 1. GRAMS OF REACTANT MOLES OF REACTANT Conversion factor: 1 = 1 mol/58.0 g 1.00g x 1mol/58g = 0.0172 mol 2. MOLES OF REACTANT MOLES OF PRODUCT 2 moles C4H10 8 moles CO2 Moles CO2 = 0.0172 x 8/2 = 0.0688 moles 3. MOLES OF PRODUCT GRAMS OF PRODUCT 0.0688 moles CO2 = 0.0688 mol x 44.0 g/1 mol 4. GRAMS OF CO2 = 3.03g This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:55:56 GMT -06:00 https://www.coursehero.com/file/143252532/Lectures-and-Slides-Week-3docx/ Lecture 3.1 – Stoichiometry Limiting Reactants - - - Limiting reactant = Reactant that is completely used up in a reaction EXAMPLE: o 2 Bd + 1 Ch = Bd2Ch o If we have 12Ch and 8Bd, we can only make 4Bd 2Ch o Bd = limiting reactant EXAMPLE: o 2H2 + O2 2H2O o 2 mol + 1 mol 2 mol o If we have 4mol H2 and 1 mol O2 – only make 2mol H2O with 2 mol H2 left over o O2 = limiting reactant Same general stoichiometry calculations plan applies to limiting reactants Reaction to Study - - 2Al + 3Cl2 Al2Cl6 QUESTION: mix 5.40 g of Al with 8.10g of Cl2 – what mass of Al2Cl6 can form? Mass of reactants Moles of reactant Moles of product Mass of products 1. COMPARE ACTUAL MOLE RATIO OF REACTANTS TO THEORETICAL MOLE RATIO: If (mol Cl2)/(mol Al) = 3/2 – even No limiting reagent, because both are used up in reaction If (mol Cl2)/(mol Al) > 3/2 – more Cl than Al, Cl in excess Al = limiting Reagent If (mol Cl2)/(mol Al) < 3/2 – more Al, Al in excess Cl = limiting Reagent 2. CALCULATE MOLES OF EACH REACTANT: 5.40g Al = 1mol/27.0g = 0.200 mol Al 8.10g Cl2 = 1mol/70.9g = 0.114 mol Cl2 3. FIND MOLE RATIO OF REACTANTS: (mol Cl2)/(mol Al) = (0.114 mol)/(0.200 mol) = 0.57 0.57 < 2/3 4. LIMITING REAGENT = Cl2 Grams of Cl2 Moles of Cl2 (1mol Al2Cl6/3mol Cl2) moles of Al2Cl6 Grams of Al2Cl6 1. 2. 1 mol Al 2Cl 6 =0.0380 mol Al 2 cl 6 3 mol Cl 2 266.4 g Al 2 Cl 6 0.0380 mol Al 2Cl 6 X =10.1 g Al 2 Cl 6 mol 0.114 mol Cl 2 X 3. 10.1 g of Al2Cl6 Lecture 3.2 – Stoichiometry This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:55:56 GMT -06:00 https://www.coursehero.com/file/143252532/Lectures-and-Slides-Week-3docx/ Limiting Reactants Continued - Cl2 = limiting reactant AL WAS PRESENT IN EXCESS, BUT HOW MUCH? Find how much Al was required Find how much Al is in excess 1. 2Al + 3Cl2 products 2. 0.200 mol of Al and 0.114 mol for Cl2 (LR) 3. 0.114 mol Cl 2 X 2mol Al =0.0760 mol Al required 3 mol Cl 2 4. Excess Al = Al available – Al required 5. = 0.200 mol – 0.0760 mol 6. = 0.124 mol Al in excess Percentage Yield Percentage yield= - Actual yeild ( g ) X 100 % Theorectical yield( g) Theoretical yield = quantity of product that forms if all limiting reagent reacts – calculated quantity Actual yield = less than theoretical yield – measured quantity Find Theoretical yield: with moles, Mr and n = m/Mr This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:55:56 GMT -06:00 https://www.coursehero.com/file/143252532/Lectures-and-Slides-Week-3docx/ Powered by TCPDF (www.tcpdf.org) Week 4 – Solution Stoichiometry Lecture 1 – Solution Stoichiometry Introduction Aqueous Reactions and solution Chemistry - Chemical reactions in water are important – life occurs in water and chemical reactions in the environment occur mainly in water General properties of aqueous Solutions - Electrolytic Properties: NaCl dissolved in water produces solution that conducts electricity, whereas table sugar (sucrose C12H22O11) does not. Demonstration of electrical conductivity of an electrolyte: Bulb glows in presence of electrolyte Does not glow in pure water Circuit contains a cathode and an anode in solution (Na + towards cathode and Cltowards anode) Electrolyte produces ions that conduct electric current Distribution of Charge on the Water Molecule - Water molecule = polar Means that hydrogens have partial (+) charges and oxygen has partial (–) charge Oxygen is more electronegative than hydrogen – therefore angular shape ( V ) Due to shape of the molecule H and O get charges (+/-) Partial positive charges (δ+) attracted to negative charges Partial negative charges (δ-) attracted to positive charges Structure of water around anions and cations - Each anion surrounded by partial positive (+) charges from waters Each cation surrounded by partial negative (-) charges from waters O attracted to cation H attracted to anion Ionic Compounds - Ionic compounds dissolve in water – tend to dissociate completely NaCl (s) + H2O (l) Na+ (aq) (Cation) + Cl- (aq) (Anion) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:04 GMT -06:00 https://www.coursehero.com/file/143250292/Lectures-and-Slides-Week-4docx/ - side surrounds Na – attracted to + H side surrounds Cl – attracted to – Starts from edge of solid and water molecule attracts ion by ion until whole solid dissolved Molecular Compounds - - Most do not dissociate when dissolved in water Methanol (molecular compound) + H2O Methanol dissolved in water Methanol molecule not ionized Homogenous and spread throughout water + and – attracts slight but not complete (?) Intermolecular interacts between + part of water and – part of Methanol molecule Molecular Compounds that ionize in water - Some molecular compounds dissociate (ionize) in water – mainly acids Strong acids: Hydrochloric acid HCl (aq) H+ (aq) + Cl- (aq) Dissociate completely – Strong electrolyte Single arrow shows equilibrium well to the right Weak acids: Acetic acid HC2H3O2 (aq) H+ (aq) + C2H3O2- (aq) Dissociate only partially – Weak electrolyte Double arrow shows equilibrium does not lie completely to right Strong and Weak Electrolytes - - Strong electrolytes = completely ionized in water E.g. NaOH NaOH (aq) Na+ (aq) + OH- (aq) Single arrow used to indicates complete ionization Weak electrolytes = partially ionized in water E.g. Acetic Acid – only ionizes to small extent HC2H3O2 (aq) H+ (aq) + C2H3O2- (aq) Indicated by double arrow Lecture 2 – Precipitation Reactions - Ppt = insoluble solid formed by a reaction in solution E.g. Mix clear AgNO3 and NaI solutions produces yellow AgI (s) ppt Soluble Ionic Compounds - Table given in tests COMPOUNDS CONTAINING EXCEPTIONS This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:04 GMT -06:00 https://www.coursehero.com/file/143250292/Lectures-and-Slides-Week-4docx/ NO3C2H3O2ClBrISO42- None None Ag+ Hg22+ Pb2+ Ag+ Hg22+ Pb2+ Ag+ Hg22+ Pb2+ Sr2+ Ba2+ Hg22+ Pb2+ Insoluble Ionic Compounds COMPOUNDS CONTAINING S2OHCO32PO43- EXCEPTIONS Alkali metal cations: NH4+ Ca2+ Sr2+ Ba2+ Alkali metal cations: NH4+ Ca2+ Sr2+ Ba2+ Alkali metal cations: NH4+ Alkali metal cations: NH4+ Ionic Equations - MOLECULAR EQUATION – do not show as ions Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq) COMPLETE IONIC EQUATIONS – show all ions Pb2+ + 2NO3- + 2H+ + 2I- Pbl2 (s) + 2K+ + 2NO3NET IONIC EQUATION – omit ions that do not change during reaction In e.g. those ions are NO3- and K+ Pb2+ (aq) + 2I- (aq) PbI2 (s) EXAMPLES: What is the net ionic reaction that corresponds to the molecular reaction below? Ca(NO3)2 (aq) + Na2CO3 (aq) CaCO3 (s) + 2 NaHO3 (aq) 1. Write out complete ionic equation and then remove spectator ions Ca2+ + 2NO3- + 2Na+ + CO32- CaCO3 (s) + 2Na+ + 2NO32. Net ionic equation = Ca2+ (aq) + CO32- (aq) CaCO3 (s) Acid-Base Reactions ACIDS - Acids = substance that ionize aq solutions to form protons (H+ (aq) ions) Proton donator HCl, HNO3 are monoprotic acids – they ionize to give only one proton per acid molecule: This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:04 GMT -06:00 https://www.coursehero.com/file/143250292/Lectures-and-Slides-Week-4docx/ - HNO3 (aq) H+ (aq) + NO3- (aq) H2SO4 is a diprotic acid – gives two protons H2SO4 (aq) 2H+ (aq) + SO42- (aq) H2SO4 – only first ionization is complete BASES - Substance that accept protons Proton acceptor Common bases are OH- compounds for group 1 cations (NaOH, KOH) or heavier Group 2 (Ba(OH)2) These produce hydroxide (OH- (aq)) ion in solution NaOH (aq) Na+ (aq) + OH- (aq) Bases react with proton (net ionic reaction below) H+ (aq) (acid) + OH- (aq)(base) H2O (l)(water) Lecture 3 – Concentrations Concentration of Solutions - Molarity – used to express concentrations of solutions Molarity= - Moles Solute( substance dissolved ∈ solution) Volume of Solution∈litres Units = mole/litre = M EXAMPLE: WHAT STRENGTH SOLUTION IS OBTAINED IF WE DISSOLVE 0.250 MOL CUSO 4 IN A 250ML FLASK 250 ml = 0.250 litre Molarity= - Moles Solute 0.250 moles = =1.00 M Volume of Solution∈litres 0.250 litres EXAMPLE: WHAT MOLARITY SOLUTION DO WE OBTAIN IF WE DISSOLVE 4.82 G CUSO 4 IN A 200ML FLASK 200ml = 0.200 litre Molar mass CuSO4 = 4.82 g / 159.5 g = 0.0302 moles Molarity= 0.0302moles =0.151 M 0.200 litres Expressing the concentration of an Electrolyte - - A 1.00 M SOLUTION OF NaOH IS ALSO 1.00 M IN Na+ IONS AND OH- IONS NaOH (aq) Na+ (aq) + OH- (aq) 1 Molar 1 Molar 1 Molar A 1.00 M SOLUTION OF H2SO4 IS 1.00 M IN SO42- BUT 2.00 M IN H+ H2SO4 (aq) 2H+ (aq) + SO42- (aq) 1 molar 2 molar 1 molar This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:04 GMT -06:00 https://www.coursehero.com/file/143250292/Lectures-and-Slides-Week-4docx/ - A 1.00 M SOLUTION OF Na3PO4 IS 1.00 M IN PO43- BUT 3.00 M IN Na+. WHAT ABOUT A 0.1 M SOLUTION Na3PO4 (aq) 3 Na+ (aq) + PO43- (aq) 1 M x 0.1 3 M x 0.1 1 M x 0.1 Interconverting Molarity, Moles and Volume - - EXAMPLE: CALCULATE NUMBER OF MOLES HNO3 IN 2.0L OF 0.200 M HNO3 1. Molarity = moles/litres 2. 0.200 M = moles/2.0 L 3. Moles = (0.200 mol/1 L) x 2.0 L = 0.40 moles EXAMPLE: HOW MANY GRAMS OF Na2SO4 ARE REQUIRED TO MAKE 0.350L OF 0.500 M Na2SO4? 1. Moles = 0.350 L x 0.500 mol/L = 0.175 moles 2. 0.175 moles Na2SO4 3. Molar Mass Na2SO4 = 142.0 g/mol 4. Grams = 0.175 mol x 142.0 g/mol = 24.9 g Dilution - - Take volume of solution and place it in a larger container and make it more dilute by adding water - MOLES = MOLARITY x LITRES In diluting a solution – number of moles of dissolved substance stays the same Moles before dilution = moles after dilution - M= MCONC x VCONC = MDIL x VDIL Example: How many ml of 3.00 M H2SO4 needed to make 450 ml of 0.100 M H2SO4 1. Mconc x Vconc = Mdil x Vdil 2. 3.0 M x Vconc = 0.100 M x 450 ml 3. Vconc = 15.0 ml Moles Volume Lecture 4 – Solution Stoichiometry - Able to convert from gras to moles and use M = moles / volume to convert from moles to molarity EXAMPLE: 1 LITRE OF 0.5 M Cu SO4 (0.5 mol / 1 litre) x 1 litre = 0.5 mol Example: 79.8 g of CuSO4 79.8 g x (1 mole / 159.5 g) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:04 GMT -06:00 https://www.coursehero.com/file/143250292/Lectures-and-Slides-Week-4docx/ General Plan for Stoichiometry Calculations – Solutions - Mass Reactant > Moles Reactant > Moles Product > Mass Product or Volume Product EXAMPLE: HOW MANY GRAMS OF CA(OH)2 NEEDED TO NEUTRALIZE 25.0 ML OF 0.100 M HCL 1. 2HCl + Ca(OH)2 CaCl2 + 2H2O 2 mol 1 mol 1 mol 2 mol Will need 1 mole of Ca(OH)2 to neutralize 2 moles HCl 2. Mole HCl = 0.100 M x 0.025 L = 0.0025 moles 2HCl + Ca(OH)2 CaCl2 + 2H2O 2 mol 1 mol 1 mol 2 mol 3. Factor = moles we have / moles in equation = 0.0025 / 2 = 0.00125 Moles of Ca(OH)2 = 0.00125 x 1 = 0.00125 mol 4. Grams = 74.2g x 0.00125 mol = 0.0926 g EXAMPLE: WHAT VOLUME OF 0.30 M HCL SOLUTION IS NEEDED TO COMPLETELY REACT 3.5 G OF CA(OH)2 1. 2HCl + Ca(OH)2 CaCl2 + 2H2O 2. Number of moles of Ca(OH)2 = 3.5 g / 74.1 g = 0.047 mol Moles of HCl = 0.047 x 2 = 0.094 mol 3. Molarity = moles / volume 0.30 = 0.094 / volume 4. Volume = 0.094 / 0.30 = 0.3 L (310ml) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:04 GMT -06:00 https://www.coursehero.com/file/143250292/Lectures-and-Slides-Week-4docx/ Powered by TCPDF (www.tcpdf.org) Lecture 1 – Thermodynamics 14.1: The Nature of Energy - Energy = capacity to do work (w) or transfer heat (q) Work = energy used to cause an object with mass to move Heat = energy that causes temperature of an object to increase FORMS OF ENERGY - Different forms of energy are interconvertible Kinetic Energy 1 2 Ek= m v 2 - m = mass, v = velocity Potential Energy Ep=mgh - m = mass, h = height relative to reference point, g = gravitional constant (9.81 ms -2) Electrostatic Potential Energy Eel= - kq 1 q 2 r k = constant (8.99 x 109 JmC-2), r = distance between charges (q1 & q2) UNITS OF ENERGY - SI unit for energy is joule (J) 1 cal = 4.184 J 1000 cal = 1 x 103 cal = 1 kcal 1 kcal = 4184 J = 4.184 kJ REACTION SYSTEM AND SURROUNDINGS - System = just reaction and contents Surroundings = exterior environment and glassware (excluding of contents) Universe = whole thing System + Surroundings = Universe IS MATTER TRANSFERABLE? IS ENERGY TRANSFERABLE? OPEN Yes CLOSED No ISOLATED No Yes Yes No 14.2: The First Law of Thermodynamics - Also known as Law of Conservation of Energy 1st Law = energy is conserved Energy cannot be created or destroyed – rather there is a set amount in universe which can be converted from one from to another, or transferred from one place to another This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:13 GMT -06:00 https://www.coursehero.com/file/143249408/Lectures-and-Slides-Week-5docx/ INTERNAL ENERGY (U) - Sum of all kinetic and potential energy Not possible to measure absolute value of U – possible to measure change in U (ΔU) ∆ U =U Final −U Initial - Internal energy = example of state function: values of Uinitial and Ufinal are fixed and not dependent on how change comes about Initial State = reactants Final State = products System: ΔU > 0 Net GAIN of energy Surroundings – Loss of energy ΔU < 0 Net LOSS of energy Surroundings – Gain in energy RELATING ΔU TO HEAT AND WORK - System undergoes chemical or physical change – change in internal energy can be determined from extent of heat and work exchanged with the surroundings Δ U =q+ w - Q = heat transferred, w = work performed Sign conventions (with respect to system): q + Heat GAINED - Heat LOST - + Energy enters - - Energy leaves w + Work done ON system - Work done BY system ENDOTHERMIC AND EXOTHERMIC PROCESSES - Endothermic = system gains (absorbs) heat (+q) Exothermic = system loses heat (-q) 14.3: Enthalpy - Many reactions described as isobaric Isobaric = constant pressure of atmosphere, Patm Work (w) that accompanies such reactions is generally mechanical in nature – resulting in change in volume of system This is pressure-volume (P-V) work - ΔV Vf < Vi + ΔV Vf > Vi System COMPRESSED System EXPANDS Work done ON system Work done BY system This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:13 GMT -06:00 https://www.coursehero.com/file/143249408/Lectures-and-Slides-Week-5docx/ +w -w w=−P ∆ V - Where P-V is the only form of work – transfer of heat is accounted for by Enthalpy (H): ∆ H=∆ U + P ∆ V ∆ H= ( q+ w ) +(−w) ΔH = qp - Sign Conventions: + ΔH + qp - ΔH - qp Heat GAINED by system Heat LOST by system Process is ENDOTHERMIC Process is EXOTHERMIC Lecture 2 – Thermodynamics 14.4 Enthalpy of Reaction - Enthalpy (H) is also a state function – therefore enthalpy of rection (ΔrxnH) depends only one initial and final states - Thermochemical equation = balanced equation when Δ rxnH takes place Example: ΔrxnH = Hproducts – Hreactants 2H2 (g) + O2 (g) 2H2O (g) ΔH = - 483.6 kJ 1. Enthalpy is an extensive proterty H2 (g) + 0.5 O2 (g) H2O (g) ΔH = - 241.8 kJ 2. Reverse equation = opposite sign 2H2O (g) 2H2 (g) + O2 (g) ΔH = + 483.6 kJ 3. States of matter influence magnitude 2H2 (g) + O2 (g) 2H2O (l) ΔH = - 571.6 kJ Lecture 3 – Thermodynamics 14.5: Calorimetry - Value of ΔH can be determined experimentally Calorimeter = experimental device used to do this HEAT CAPACITY AND SPECIFIC HEAT CAPACITY - All substances change temperature when subjected to heat transfer Magnitude of this change differs from substance to substance Temperature change experienced by an object = determined by heat capacity (C) Heat capacity = amount of heat required to change temperature (ΔT) of object by 1 K or 1 ᵒC For pure substances – also consider molar heat capacity (C m) and Specific Heat Capacity (C) Specific Heat Capacity = amount of heat required to change the temperature of 1 gram of a substance by 1 K or 1 ᵒC This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:13 GMT -06:00 https://www.coursehero.com/file/143249408/Lectures-and-Slides-Week-5docx/ - C = J / (g.ᵒC) Magnitude of ΔT is independent of units ᵒC K = +273 Specific heat capacity helps limit evaporation in water, so biological lifeforms are not harmed by sharp increases/decreases in temperature CONSTANT PRESSURE CALORIMETRY - Pressure controlled (kept constant) – ΔH can be readily measured (ΔH = qp) Simple coffee-cup calorimeter (CCC) used in lab to obtain informative results Example of an isolated system – Styrofoam prevents (limits) heat exchange with greater surroundings qrxn = -qsoln qsoln = Csoln x Msoln x ΔT - Small part of the result in solution ΔT is partly from surroundings (minimal) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:13 GMT -06:00 https://www.coursehero.com/file/143249408/Lectures-and-Slides-Week-5docx/ Powered by TCPDF (www.tcpdf.org) Week 6 Lecture 4 14.6 Hess’s Law - Enthalpies of reaction for which experimental data is not available can be calculated using Hess’s Law ∆ H 1=( ∆ H 2+∆ H 3+ ∆ H 4 ) ∆ H=∆ Hf −∆ Hi Lecture 5 14.7 Enthalpies of Formation - Standard state – form of a substance when P = 1bar and T = 298K (25ᵒC) When all reactants and products are in their standard states – most stable state Graphite = most stable form of carbon Standard enthalpy change indicated as ΔHᵒ Enthalpy of formation (ΔfH) = enthalpy change associated with formation of a substance from its constituent elements This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:25 GMT -06:00 https://www.coursehero.com/file/143249070/Lectures-and-Slides-Week-6docx/ - Standard enthalpy of formation of a compound (ΔfHᵒ) – defined as enthalpy change for the formation of 1 mole of a substance from its constituent elements with all species in their standard states 2C (graphite) + 3H2 (g) + ½ O2 (g) C2H5OH (l) ΔfHᵒ = -277.7 kJmol-1 Important considerations: - - If an element exists in more than one form under standard conditions – most stable form is selected E.g. thus O2 is used instead of O or O3 and graphite instead of diamond By definition, formation reactions involved formation of 1 mole of a substance; enthalpies of formation are therefore reported in units of kJ per mole (kJmol -1) USING ENTHALPIES OF FORMATION TO CALCULATE ENTHALPIES OF REACTION - - Can apply Hess’s Law order to determine ΔrxnHᵒ - need to use the individual formation reactions for each species Can make use of following relationship – which holds since any reaction can be broken down into a series of formation reactions ΔrxnHᵒ = ∑nΔfHᵒ (products) - ∑mΔfHᵒ (reactants) ∑nΔfHᵒ (products) = sum of enthalpies of formation of products ∑mΔfHᵒ (reactants) = sum of enthalpies of formation of reactants Quantity and amount of moles needs to be taken into consideration Lecture 6 14.8 Spontaneous Processes - Can calculate ΔU and ΔH Important consideration is the extent of a reaction (how likely is it to take place) Spontaneous process = one in which change occurs in a particular direction and without ongoing external intervention - - Occurs under given set of conditions Reactions may be described as spontaneous irrespective of rate at which they occur Gas expanding into a vacuum – instantaneous Nail rusting – days to weeks Temperature and pressure can influence whether or not a reaction is spontaneous Examples: falling brick, rusting nail, gas expanding into a vacuum 2Na(s) + Cl2(g) 2NaCl(s) Non-spontaneous = reverse of spontaneous and do require work - Does not occur under given set of conditions This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:25 GMT -06:00 https://www.coursehero.com/file/143249070/Lectures-and-Slides-Week-6docx/ REVERSIBLE AND IRREVERSIBLE PROCESSES - - - Reversible No net change to surroundings Restoring system causes requires no work Irreversible Change in the surroundings Restoring system rewires work Under given set of conditions = spontaneous processes are irreversible and take place in a particular direction Entropy – important for determining whether reaction will be spontaneous or not Entropy = characteristic system that gives us indication of extent to which energy is disordered Entropy ΔS = amount of disorder in a system 14.10 The molecular Interpretation of Entropy MOLECULAR MOTIONS AND ENERGY - - Molecules can undergo three kinds of motion: Translational – entre molecule moves Vibrational – atoms and bonds move Rotational – molecule spins When molecule heated – motion (and Ek) increases Refer to molecules as having motional energy BOLTZMANN’S EQUATION AND MICROSTATES - - Microstate is a single arrangement of positions and kinetic energies of a given set of molecules (system) For any system, entropy (S) related to total number of microstates (W) S = K ln W ΔS = k ln (Wfinal)/(Winitial) Number of microstates available in system increases with increase in volume, temperature and number of independent particles MAKING QUALITIES PREDICTIONS ABOUT ΔS This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:25 GMT -06:00 https://www.coursehero.com/file/143249070/Lectures-and-Slides-Week-6docx/ Increase in disorder / entropy (+ΔS) THIRD LAW OF THERMODYNAMICS - If temperature if a system decreased to such a point S L (W - S Solvent that all molecular motion is halted G then a single microstate = 1) is essentially achieved Solution 2NH (g) N (g) + 3H (g) 3 2 2 Temperature at which this is achieved = Absolute zero (0K) - Entropy of a pure crystalline system at 0K is zero (S = 0) - Pure crystalline solid heated – entropy increases from 0: 1. At std temperature (298K) Sᵒ for pure elements is not zero 2. In general Sᵒ (g) > Sᵒ (L) > Sᵒ (s) 3. Value of Sᵒ usually increases with increasing molar mass 4. Value of Sᵒ usually increase with increasing number of atoms This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:25 GMT -06:00 https://www.coursehero.com/file/143249070/Lectures-and-Slides-Week-6docx/ Week 7 Lecture 7 14.9 Entropy and the Second law of Thermodynamics SECOND LAW OF THERMODYNAMICS - Spontaneous (irreversible) process characterized by overall increase in entropy of universe ΔunivS = [ΔsysS + ΔsurrS] > 0 - If no change in entropy of universe – dealing with reversible process: ΔunivS = [ΔsysS + ΔsurrS] = 0 - To determine whether reaction is spontaneous or not – consider ΔunivS ΔsysS = readily determined using this equation: ΔrxnHᵒ = ∑nΔfHᵒ (products) - ∑mΔfHᵒ (reactants) - For isothermal (constant temperature) process at equilibrium (forward and backwards rates are the same) following relationship useful alternative to calculate ΔsysS ΔsysS = (qrev) / T - qrev = heat transferred if process reversible T = temperature (K) at which process occurs ENTROPY CHANGES IN SURROUNDINGS - To calculate ΔsurrS: ΔsurrS = (-qsys) / T = (-ΔHsys) / T - Surroundings and system = equal magnitude and opposite signs If we know ΔH of system we can find ΔS of surroundings 14.11 Entropy Changes in Chemical Reactions - Can measure ΔH using calorimetry No easy method to measure ΔS Absolute values of entropy (S) can be determined Standard molar entropies (Sᵒ) for substances in their standard state reported in units of Jmol-1K-1 Examples in table 4.6 Observations: At standard temperature (289K) Sᵒ for pure elements is not zero In general, Sᵒ (g) > Sᵒ (l) > Sᵒ (s) Value of Sᵒ usually increases with increasing molar mass Value of Sᵒ usually increases with increasing number if atoms This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:33 GMT -06:00 https://www.coursehero.com/file/143250134/Lectures-and-Slides-Week-7docx/ - Entropy is a sate function – ΔrxnS = Sfinal – Sinitial Change in entropy for a chemical reaction calculated by: ΔSᵒ = ∑nSᵒ (products) - ∑mSᵒ (reactants) - These are the absolute entropies of the products and reactants ΔH = kJ and is therefore much larger than ΔS (J) Lecture 8 14.12 Gibbs Free Energy - Reaction spontaneity depends on both enthalpy and entropy Expressed in state function: Gibbs Free Energy (G) G = H – TS ΔG = ΔH – TΔS - T = constant under isothermal conditions From entropy considerations: ΔunivS = ΔsysS + ΔsurrS - Substitute so RHS is only in terms of system ΔunivS = ((-ΔsysH) / T) + ΔsurrS - Cross multiply by -T: -TΔunivS = ΔsysH – TΔsurrS - Compare this to ΔG = ΔH – TΔSthat - Therefore: ΔG = -TΔunivS - ΔG Forward reaction is spontaneous Backward reaction is not spontaneous Reaction is in equilibrium Forward reaction is not spontaneous Backward reaction is spontaneous ΔG = 0 + ΔG - Change in Gibbs free energy for a reaction can be calculated if changes in enthalpy and entropy for same reaction are known Change in Gibbs free energy for a reaction can also be calculated, as for enthalpy, using standard Gibbs free energies of formation: - ΔGᵒ = ∑nΔfGᵒ (products) - ∑mΔfGᵒ (reactants) ΔGᵒ = 0 – substance that only contains one atom type and is already in standard stable state (e.g.O2) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:33 GMT -06:00 https://www.coursehero.com/file/143250134/Lectures-and-Slides-Week-7docx/ Lecture 9 14.13 Gibbs (Free) Energy and Temperature ΔG = ΔH – TΔS +ΔH – Endothermic -ΔH – Exothermic +ΔS – increase in disorder ΔG = +ΔH – (T(+ΔS)) ΔG = + Spontaneous at HIGH values of T ΔG = -ΔH + TΔS ΔG = Spontaneous at ALL values of T -ΔS – Decrease in disorder ΔG = +ΔH – (T(-ΔS)) ΔG = + Not spontaneous at ALL values of T ΔG = -ΔH – (T(-ΔS)) ΔG = Spontaneous at LOW values of T Gibbs (Free) Energy and the Equilibrium Constant ΔG - (Spontaneous) 0 + (Not spontaneous) K >1 =1 <1 Species Favoured Products 50:50 (equally favoured) Reactants This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:33 GMT -06:00 https://www.coursehero.com/file/143250134/Lectures-and-Slides-Week-7docx/ Powered by TCPDF (www.tcpdf.org) Week 8 Kinetics - Lecture 1 - Change = transformation of reactants to products Reaction can be classified as spontaneous regardless of how long it takes (ms or days/months/years) Chemical Kinetics = area of chemistry that deals with speed (rate) of a reaction 15. 1 Factors That Affect Reaction Rates - Reactions involve breaking and forming bonds Rate at which these take place depend on nature of reactants Factors influence rate of chemical reaction: o Physical state of reactants Larger surface area = faster reaction Smaller surface area = slower reaction o Concentration of reactants More atoms = more likely to collide Higher concentration = quicker reaction o Reaction temperature Higher temperature = speeds up reactions – more kinetic energy and movement Lower temperature = slows down reactions o Presence of a catalyst Lowers activation energy Catalyst is not used up during reaction Rate increases 15.7 Catalysts - - Homogenous catalysis = same phase as reacting molecules E.g. 2H2O2 (aq) 2H2O (l) + O2 (g) o In absence of catalyst this reaction = extremely slow o Including bromide: o 2Br- (aq) + 2H+ (aq) + 2H2O2 (aq) Br2 (aq) + 2H2O (l) o Br2 (aq) + 2H2O2 (aq) 2Br- (aq) + 2H+ (aq) + O2 (g) o Bromide ions (together with H+) catalyse this reaction o Br2 (aq) = intermediate – produced in reaction 1 and consumed in reaction 2 Heterogenous catalysis = different phase than reacting molecules o Catalyst is usually a solid o Reactants may be gas or liquid o Reactant molecules first absorb to catalyst surface (which is highly reactive) o Then reaction proceeds to yield product molecules – which later desorb from surface o Absorption Reaction Desorption This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:38 GMT -06:00 https://www.coursehero.com/file/143250270/Lectures-and-Slides-Week-8docx/ 15.2 Reaction Rates - - Speed = extent of change in a given time interval Measurable change with time = concentration (M/s or M s -1) AB A used up and B formed At the end, there is the same number of total mol in reaction vessel as at the start of the reaction Reaction rate can be expressed relative to disappearance of reactant (A) or the appearance/formation of product (B) −Change∈concentraion of A Change∈time Change∈concentraion of B Average rate of appearance of B= Change∈time Average rate of disapperance of A= Rates are always expressed as positive quantities CHANGE IN RATE WITH TIME - E.g. in lecture notes Average rate decreases with increasing time Concentration of reactants decreases – resulting in a decrease in the number of collisions Can determine the instantaneous rate – rate of reaction at a particular time Instantaneous Rate= ΔConcentration( M ) Δ time Instantaneous rate= tangent of concentration(Y) against time(X) graph REACTION RATES AND STOICHIOMETRY - AB o Reaction stoichiometry is 1:1 o - −∆[ A] ∆ [B] = ∆t ∆t 2A B + C o Reaction stoichiometry is 2:1 (A:B or A:C) o o - Rate= −∆ [ A ] ∆ [ B] ∆ [ C] =2× ∨2 × ∆t ∆t ∆t −1 ∆ [ A ] 2 ∆ [ B ] 2 ∆ [ C ] Rate= = ∨ 2 ∆t 2 ∆t 2 ∆t Rate= aA+bBcC+dD o Rate= −1 ∆ [ A ] −1 ∆ [ B ] 1 ∆ [ C ] 1 ∆ [ D ] = = = a ∆t b ∆t c ∆t d ∆t Kinetics - Lecture 2 15.3 Concentration and Rate Laws This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:38 GMT -06:00 https://www.coursehero.com/file/143250270/Lectures-and-Slides-Week-8docx/ - Rate law = equation that shows how rate of reaction depends on concentration of all reactants For general equation: a A + b B c C + d D Rate Law is: Rate = k[A]m[B]n k = rate constant/coefficients Exponents (m and n) = reaction orders REACTION ORDERS: EXPONENTS IN THE RATE LAW - Exponents (m and n) – indicate how reaction rate is affected by concentration of each reactant Values of m and n must be determined experimentally Exponent (m or n) 0 Meaning Zero order 1 First order 2 Second order - Interpretation Rate independent of reactant concentration Rate directly proportional to concentration Rate changes as square of concentration Overall reaction order = sum of orders UNITS OF RATE CONSTANTS - Rate constant does not depend on concentration It is affected by temperature and presence of catalyst Units of rate constant, k, depend on overall order of reaction Rate Equation Rate = k Rate = k[A] Rate = k[A][B] Rate = k[A]2 Rate = k[A][B]2 Overall Reaction Order 0 1 2 Units of k (t in seconds) Ms-1 s-1 M-1 s-1 3 M-2 s-1 This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:38 GMT -06:00 https://www.coursehero.com/file/143250270/Lectures-and-Slides-Week-8docx/ Powered by TCPDF (www.tcpdf.org) Week 9 Kinetics – Lecture 3 15.4 The Change of Concentration with Time (Integrated Rate Equations) - Integrated form of rate equation – used to find reaction rate at different times AB o Differential form of reaction rate (for disappearance of A) is: o Rate= −d [ A ] dt ZERO-ORDER REACTIONS - Rate independent of concentration Rate= t −d [ A ] 0 k [ A ] =k dt t ∫ d [ A ] =∫−k dt 0 0 [ A ]t −[ A ] 0=−k (t−0) [ A ]t =−kt + [ A ] 0 - Linear graph (y = mx + c) FIRST-ORDER REACTIONS - Rate directly proportional to concentration Rate= −d [ A ] 1 k [ A ] =k [ A ] dt t t ∫ [ A1 ] d [ A ] =∫ −k dt 0 0 ln [ A] t−ln [ A ]0 =−k (t−0) ln [ A] t=−kt+ ln [ A ]0 - Linear graph (y = mx + c) – ln[A] Slope – [A] ln [ A ]t =−kt [ A ]0 [ A ]t =[ A ]0 e−kt This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:43 GMT -06:00 https://www.coursehero.com/file/143250236/Lectures-and-Slides-Week-9docx/ SECOND-ORDER REACTIONS - Rate directly proportional to square of concentration Rate= −d [ A ] 2 =k [ A ] dt t t ∫ 1 2 d [ A ] =∫ −k dt 0 [A] 0 −1 −1 − =−k (t−0) [ A ]t [ A] 0 1 1 =−kt+ [ A ]t [ A ]0 - Linear graph (y = mx + c) HALF-LIFE - Half-life of a reaction (t1/2) = time required for concentration of reactant to reach one-half if its initial value 1 [ A ]t 1 = [ A] 0 2 2 - Zero-Order o [A]t = -kt + [A]0 o t1 = 2 - 2k First-Order o Half-life is constant o ln [ A ] t=−kt+ ln [ A ] t o t1 = 2 - [ A ]0 ln 2 k Second-Order o 1 1 =kt+ [ A ]t [ A ]0 This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:43 GMT -06:00 https://www.coursehero.com/file/143250236/Lectures-and-Slides-Week-9docx/ o t1 = 2 1 k [ A ]0 15.5 Temperature and Rate - Rates of most chemical reactions increase with increasing temperature The faster rate at higher temperature – due to increase in rate constant with increasing temperature Exponential relationship Reaction rates affected by both concentration of reactants and by temperature Collison model – molecules must collide in order to react Greater the number of collisions = greater reaction rate Not all collisions lead to reaction Molecules also need to be orientated in a certain way for collisions to result in reaction – Orientation Factor ARRHENIUS EQUATION - Increase in reaction rate with increasing temperature does not show a linear dependence Exponential relationship given by Arrhenius equation: k = Ae-Ea/RT - k = rate constant Ea = Activation energy R = gas constant (8.314 J/K mol) T = temperature (K) A = frequency factor – relates to frequency of favourably-orientated collisions (changes very little with temperature – can be considered a constant) Arrhenius equation predicts that as activation barrier increases, the reaction rate decreases DETERMINING ACTIVATION ENERGY - - Arrhenius equation can be re-worked into a linear form o ln k = −E a + ln A RT o ln k = −E a 1 ∙ + ln A – Linear form (negative (-) slope) R T Two different reaction temperatures: o Assume A is constant This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:43 GMT -06:00 https://www.coursehero.com/file/143250236/Lectures-and-Slides-Week-9docx/ o ln k 1−ln k 2=ln k 1 Ea 1 1 = ( − ) k2 R T 2 T 1 This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:43 GMT -06:00 https://www.coursehero.com/file/143250236/Lectures-and-Slides-Week-9docx/ Powered by TCPDF (www.tcpdf.org) Week 10 Electronic Structure Limitations of the Bohr Model - Only works for atoms/ions with single electron (e.g. H or He +) Cannot account for more numerous lines in multi-electron atoms However is does make two important postulates: o Electrons exist only in discrete energy levels o Energy is involved in moving electron from one level to another – energy absorbed as photons to excite electron from one level to a higher energy level and energy is emitted as photons when an electron drops to a lower energy level The wave Behaviour of Matter - Louis de Broglie – matter also had a wave property For any particle: λ= - - h mv v – velocity v – frequency λ – wavelength m – mass h – Planck’s constant (6.626 x 10-34) How to get to this equation: 1. E = hv 2. E = mv2 3. hv = mv2 4. λ = v/v 5. λ = hv/mv2 Quantity mv = momentum Electron thus is not only a particle, but also a wave Confirmed in that electrons can be diffracted by crystals Since electrons are waves – inappropriate to treat them as particles (as in Bohr model) Schrodinger Wave Equation Ӈψ = Eψ - - Same account of e in H atom as Bohr Gives correct account of heavy atoms ψ2 – measure of e- density in various regions around the nucleus Orbitals and Quantum Numbers - Solution to Schrodinger wave equation leads to set of wavefunctions that yields 4 types of quantum numbers Single quantum number yielded by Bohr model This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ - - - These numbers are: O PRINCIPLE QUANTUM NUMBER (n) Values 1, 2, 3,… This corresponds to quantum number in Bohr model O AZIMUTHAL/ ANGULAR MOMENTUM QUANTUM NUMBER (l) Values of 0 to n-1 for each value of n Corresponding orbitals = 0 – s; 1 – p; 2 – d; 3 – f O MAGNETIC QUANTUM NUMBER (ml) Values -l to +l for each value of l O SPIN QUANTUM NUMBER (ms) Values + ½ or – ½ Each value of ml – there are two values of ms Leads to occupation of each orbital by two electrons of opposite spin These quantum numbers lead to the shells (different n values) and subshells (different values of l) that lead to our modern understanding of chemistry Number of orbitals in each sub-shell = determined by m l Only two electrons of opposite spin can occupy each orbital = determined by m s This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ Representation of Orbitals This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ Many-electron Atom - PAULI EXCLUSION PRINCIPLE – no two electrons in an atom can have all four quantum numbers (n, l, ml, ms) the same Result of this is that each orbital can hold a max of two electrons, with m s = + ½ or – ½ Failure of Bohr Model for Many-electron Atoms - For H-atom – all subshells within same shell have the same energy Bohr model works for H-atom – generates single quantum number (n), which predicts only energy of shells and not the existence of subshells Many-electron atom – subshells within single shell do not have same energy Cause of failure of the Bohr model when applied to multi-electron atom Electron Configurations - Electron configuration = way in which electrons are distributed among various orbitals Orbitals filled in order of increasing energy – two electrons of opposite spin per orbital Orbital diagram: o Each orbital represented by a box o Half arrow up (spin up) = positive spin (m s = + ½ ) o Half arrow down (spin down) = negative spin (m s = - ½ ) o Electrons Indicates number of having electrons in sub-shell opposite spins are spin-paired when in the same orbital Hund’s Rule - Orbitals that have same energy = degenerate - HUND’S RULE = for degenerate orbitals, lowest energy attained when number of electrons with same spin is maximized This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ Uncertainty Principle - If electron is a wave – cannot state position with any accuracy HEISENBERG postulated: UNCERTAINTY PRINCIPLE – we cannot precisely know exact momentum of an electron and also its location in space Condensed Electron Configurations - Abbreviate electron configuration by only writing noble gas before element (e.g. [Ne]) and electrons in outermost occupied shell (valence shell) Transition Metals - - Row ended by Ar marks beginning of 4th row (n = 4) In this row = first row of transition elements This is where the 3d orbital is being filled 4s = lower in energy than 3d orbital Therefore 3d orbital does not fill right away First two electrons for K and Ca go into this orbital K = [Ar] 4s1 Ca = [Ar] 4s2 After 4s is filled then 3d starts to fill up E.g. Mn = [Ar] 4s2 3d5 Lanthanoids and Actinoids - Filling up f orbital Although energies of 4f and 5d are very close – electrons also occupy 5d orbitals La = [Xe] 6s2 5d1 Lu = [Xe] 6s2 4f14 Electron Configurations and The Periodic Table - Each group of elements in periodic table has characteristic electronic configuration E.g. Halogens have characteristic configuration of ns2 np5 of highest energy occupied shell Valence electrons determine chemical properties of elements and occupy highest energy shell Electron Configuration of ions - Add or subtract n number of electrons from last orbital depending on +n or –n Electron Configurations of Transition Metal Cations This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ - S orbital filled before d-subshell However, when cations formed – d-subshell is lower in energy and therefore any valence electrons left occupy the d-subshell Atomic Numbers (Z) - Each element produces X-rays of a distinct frequency This frequency increases with atomic number Atomic number = number of protons in a nucleus This concept allowed correct placement of certain elements on periodic table E.g. Ar (Z = 18) and K (Z = 19) but atomic weight Ar > K Core and Valence Electrons - Valance electrons in 3s shell experience small net positive charge Core electrons shield valence electrons from nucleus Effective Nuclear Charge - Many-electron atom – each electron is simultaneously attracted to nucleus and repelled by other electrons Generally – valence electrons screened from nucleus by core electrons (~S, where S is screening constant = number of core electrons) Effective nuclear charge = approximate charge experienced by valence electrons Thus, effective nuclear charge (Zeff) experienced by valence electrons is given by: Zeff = Z – S E.g. Na: Zeff = 11 – 10 ~ 1+ (because of electrons in higher shells are less good at shielding nucleus – charge experienced by Na+ is actually +2.5) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:48 GMT -06:00 https://www.coursehero.com/file/164129940/Lectures-and-Slides-Week-10docx/ Powered by TCPDF (www.tcpdf.org) Week 11 Periodic Properties of Elements Development of Periodic Table - Elements in same group generally have similar chemical properties Effective Nuclear Charge - Zef - Nuclear charge that electron experiences depends on both factors - Effective nuclear charge = Zeff = Z – S - Many-electron atom – electrons both attracted to nucleus and repelled by other electrons Z = atomic number S = screening constant (close to number of inner e -s Zef increases across a period – nuclear charge increases while number of core electrons remains the same Zef increases only slightly down a group – larger electron cores are less able to screen outer electrons Atomic and Ionic Size - Bonding atomic radius = half of distance between covalently bonded nuclei Size increases down a group – number of shells and electrons are increasing Size decreases across a period – increase in Zef causes electrons to pull closer to nucleus This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:53 GMT -06:00 https://www.coursehero.com/file/164129780/Lectures-and-Slides-Week-11docx/ Ion Sizes - Radius when cation formed – cations are smaller because fewer electrons relative to protons (repulsions reduced) - Radius when anion is formed – anions are larger because more electrons relative to protons (repulsions increased) - For series of ions which are isoelectronic (same number of electrons) – radius of ion decreases as nuclear charge increases (electrons more strongly attracted to nucleus) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:53 GMT -06:00 https://www.coursehero.com/file/164129780/Lectures-and-Slides-Week-11docx/ Ionization Energy - IE = energy required to remove an electron from an atom in the gas phase - More energy required to remove each successive electron - 1st IE = energy required to remove first electron 2nd IE = energy required to remove second electron When all valence electrons removed – ionisation energy increases drastically IE decrease down a group – outer electrons further from nucleus and are easier to remove IE increases across a period – Zef increases, therefore electrons are held more tightly and difficult to remove EXCEPTIONS: o o Group 3: IE lower (electrons easier to remove) – these elements have only one electron in p subshell Group 6: IE lower – these elements have 4 electrons in p subshell (repulsion of paired electrons is unfavourable and makes these electrons easier to remove) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:53 GMT -06:00 https://www.coursehero.com/file/164129780/Lectures-and-Slides-Week-11docx/ Electron Affinity - EA = energy change when an atom gains an electron in the gas phase EA does not change significantly down a group EA increases across a period (becomes more -ve/exothermic) – Zef increases, therefore increased attraction for extra electron EXCEPTIONS: o o Group 2: +ve EA – electrons would have to go into previously empty p subshell Group 5: EAs less -ve than expected – half filled p subshell and would have to start pairing electrons Main Group Ions - Elements from groups 1 and 2 – tend to form cations with noble gas configuration (i.e. oxidation number = group number) This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:53 GMT -06:00 https://www.coursehero.com/file/164129780/Lectures-and-Slides-Week-11docx/ - Elements from group 6 and 7 – tend to form anions with noble gas configuration This study source was downloaded by 100000824697111 from CourseHero.com on 02-27-2023 12:56:53 GMT -06:00 https://www.coursehero.com/file/164129780/Lectures-and-Slides-Week-11docx/ Powered by TCPDF (www.tcpdf.org)
