Aqa a level chemistry

Alternative transcript:

Inorganic Reactions & Ionic Equations. Unit 2 - Inorganic Chemistry (Paper 1).... 2.1 Periodicity.. Periodic Table & Periodic Trends. 2.2 Group 2 The Alkaline Earth Metals. Group 2 Alkaline Earth Metals....... Dr. C. Boes 3 7 .7 7 9 11 .11 .13 .15 .17 .19 21 .21 .23 .25 .27 .29 28282 w w w 33 .33 .35 .37 38 .39 .41 43 .43 .45 47 445 .47 .49 .51 53 53 53 55 .55 alevelchemistryrevision.co.uk 2.3 Group 7 The Halogens. Group 7 Halogens & Water Treatment.. Halides...... Test for Ions & Displacement Reaction & Solubility. List of Anions & Naming Salts. 2.4 - Investigative and Practical Skills. Accuracy and Reliability & Uncertainty & Atomic Models. Acids & Bases........ Acids and Bases Preparation & Preparing a Standard Solution.....69 Titrations.............. Unit 3 - Organic Chemistry (Paper 2) 3.1 Introduction to Organic Chemistry. Naming Rules & Types of Organic Formulae. Functional Groups. Three Main Types of Organic Reactions & Mechanisms. Types of Isomers. 3.2 Alkanes...... Alkanes & Terms Crude Oil Fuel & Greenhouse Effect. 3.3 Haloalkanes.. Radical Substitution in Alkanes Ozone Layer & CFCs. Haloalkanes. 3.4 Alkenes.. Alkenes - Reaction with Halogens.. Alkenes - Addition of Hydrogen Halides and Water. Alkenes - Hydration and Fermentation. Addition Polymers. 3.5 Alcohols Alcohols Oxidation of Alcohols & Aldehydes/Ketones.. 3.6 Organic Analysis...... Mass Spectrometry & IR Spectroscopy.. Separating Funnel & Reflux Apparatus. Distillation Apparatus & Drying with Anhydrous Salts. Appendix: Periodic Table of Elements............ Dr. C. Boes 57 .57 .59 .61 .63 65 .65 .67 4 .71 .73 .73 .73 .75 .77 .79 .81 .81 .83 .85 ..85 .87 .89 .91 23 .91 .93 .95 .97 99 .99 .101 103 .103 .105 .107 .109 alevelchemistryrevision.co.uk How to use these notes These revision notes are organized in chapters according to the current AQA Year 1 & AS syllabus (from Sept 2015). Each chapter contains individual revision cards covering all the necessary topics. Everything in italic is optional knowledge, aimed at students who want to excel or want to continue to year 2. Bold represents important keywords or key definitions. 'Data sheet' indicates information which will be provided on the data sheet or in the question during the exam and does not need to be memorized. Important information and exam- specific tips are highlighted in yellow. How to memorize: The revision cards are introduced by their titles and keywords on a separate page. After reading the title you should try to write down the content of the card without looking at the next page. The keywords give you hints about the content. Write down everything you remember, even if you are not sure. Then check if your answers are correct; if not, rewrite the incorrect ones. At the beginning, when you are still unfamiliar with the cards, it might help to read them a few times first. If they contain a lot of content, you can cover the revision card with a piece of paper and slowly reveal the header and sub content. While you uncover it try to remember what is written in the covered part, e.g. the definition for a term you just uncovered. This uncovering technique is for the early stages, later you should be able to write down the whole content after just reading the header. If this is the case, move to the next card. If not, bookmark the card and memorize it repeatedly. Do at least four to five sessions per week until you know all the cards of one chapter word-perfectly. When you have memorized a revision card apply your new knowledge by answering topic questions. Then move on to the next section. Generally it is better to do shorter sessions more often than longer sessions less frequently. An even better option is to ask somebody to check your knowledge by reading the header aloud and comparing your answer to the content. Exam techniques Begin with a quick look through the exam. How is it structured; what topics are coming up and how many questions are there? Then work systematically through it from the beginning, but keep an eye on the time. When you fall behind shorten your answers and leave difficult topics for the end. Underline or highlight the important information/data in the question. If just names for compounds are given, write the chemical formula above it (e.g. sulfuric acid -> H₂SO4). Circle the functional groups in an organic formula and name them. Draw the carbons and hydrogens in skeletal formulae or displayed formulae if the structural formulae are given. Dr. C. Boes 5 alevelchemistryrevision.co.uk Make sure you read the question thoroughly and be aware what actions are expected from you from the command words used. Identify the topic of the question and mentally bring up the flashcards associated to the topic. They will help you with the answer. If you have problems understanding the question, read it again slowly and also read through the follow up sub-questions (a, b, c etc.) sometimes the topic and the initial question becomes clearer. If you still do not understand the question or cannot come up with all the answers, do not spend any more time on it. Write down your best answers or just standard keywords/phrases from the flashcard. Writing something is better than writing nothing. You might still get some marks for it. Circle the question and come back to it at the end of the exam. If you do calculations, write down a list of the data given (time permitting - otherwise just underline) and the formulae/equations which you are using (even if your calculation is wrong, you might get a mark for the correct formula). Always show your workings and do the unit calculations. This means writing the units next to the numbers and cancelling or multiplying them accordingly. You should get the correct unit for your final answer. If not, you might not have converted them correctly (e.g. cm³ into dm³) or have used the wrong equation. After writing down the final answer check if it makes sense (is the number in the expected range; does it have the correct sign in front, e. g. – for an exothermic reaction etc.). Calculation answers should always be given in decimals, never fractions. Furthermore make sure you have answered all the questions and everything asked for (e.g. state symbols, significant figures etc.) For multiple choice questions: read all answers and strike through the ones which are definitely wrong. Choose the correct or most likely of the remaining ones. If you have no clue take the longest answer. Always choose an answer -> you have at least a 25 % chance to get it right :-). Do not spend too much time on a question. Rule of thumb is 1 min per mark. If you are unsure, circle the question and come back to it at the end of the exam. More tips about how to plan your revision and how to prepare for exams can be found on my website: https://www.alevelchemistryrevision.co.uk Disclaimer: Due to the changing nature of mark schemes it cannot be guaranteed that answering according to these notes will give you full marks. These notes constitute only one part of a full revision program and work alongside other methods, like practising past papers. They have been created with great care; however, errors or omissions cannot be excluded. They are designed for the final stage of revision and require a thorough understanding of the topics. Dr. C. Boes 6 alevelchemistryrevision.co.uk Unit 1 - Physical Chemistry (Paper 1 & 2) 7404 1.1 Atomic Structure (Paper 1 & 2) Dr. C. Boes Basic Definitions Atom Element Isotopes Atomic number Mass number Ion Relative atomic mass with equation Relative isotopic mass Molar mass Relative molecular mass 7 alevelchemistryrevision.co.uk Basic Definitions Atom: smallest unit of an element -> consist of electrons organized in orbitals/shells and a nucleus made from protons and neutrons Subatomic Particle Proton Neutron Electron Element: same kind of atoms (same atomic number) Isotopes: atoms with same number of protons but different number of neutrons (same element: same atomic number but different mass number) Atomic number: number of protons (equals number of electrons). Mass number: protons + neutrons Ion: charged particle (different numbers of protons and electrons) positive -> cation (more protons than electrons) negative -> anion (more electrons than protons) => formed when an atom gains or loses electrons to get a full outer shell a%: b%: Relative Atomic mass Ar Is the average (weighted) mass of an element's isotopes (atoms) relative to 1/12 the mass of a ¹2C atom [no unit] A₁: A₂: Relative mass 1 1 1/2000 A₁ = (a% x A₁) + (b% x A₂) 100 percentage of Isotope 1 percentage of Isotope 2 Relative Isotopic mass of Isotope 1 Relative Isotopic mass of Isotope 2 Charge +1 0 -1 => A, and Relative Isotopic Abundance can be worked out from a Mass Spectrum (A₁, A₂ from x-axis and a%, b% from y-axis of each peak) -> see revision card 'Mass Spectrometry' Molar mass M Mass of one mole of a substance [g mol^¹] Dr. C. Boes Relative isotopic mass Is the mass of an atom of an isotope relative to 1/12 the mass of a ¹2C atom [no unit] Relative molecular Mass Mr (relative formula mass) of a compound is the sum of the relative atomic masses of all its atoms [no unit] 8 alevelchemistryrevision.co.uk Dr. C. Boes Electron Configuration Function of Electron Configuration Definition of orbital Four rules Equation for number of electrons per shell Meaning of the numbers & letters Box diagram for Nitrogen (Shapes of s- and p-orbitals) Table of subshells (s, p, d, f) 9 alevelchemistryrevision.co.uk Electron Configuration -> Distribution of electrons across the shells and subshells of an atom Orbitals Definition: Region of space in which electrons are most likely to be found; one orbital can hold up to two electrons, which must have opposite spins. Electrons fill up the lowest energy shells/subshells first Orbitals of the same subshell are filled individually first (electrons repel) 4s get emptied and filled before 3d (see Y2 rev. card 'Transition Metals') For cations take off electrons, for anions add electrons according to charge Number of electrons per shell: 2n² n: shell number [He] 2s²2p³ ● N: 1s² 2s²2p³ 2: shell (principal quantum number) p: subshell 3: number of electrons (in subshell) Box diagram: Energy Dr. C. Boes Sphere Subshells S P d f Electronic configuration of N 2p³ Shapes of orbitals (year 2): s-orbital: 2s² 1s² ↑↓ ↑↓ Orbitals 1 3 5 7 ↑↑↑ 10 p-orbital: Dumbbell Electrons 2 6 10 14 alevelchemistryrevision.co.uk 1.2 Amount of Substance (Paper 1 & 2) Mole & Molar Gas Volume Dr. C. Boes Mole: Definition and equation Equation for calculating number of particles Molar gas volume at standard conditions Equation for calculating volume of gas from moles Converting cm³ into dm³ Ideal gas equation with SI units Two Tips 11 alevelchemistryrevision.co.uk Mole Def.: 1 mole = 6.02 x 1023 particles (atoms, molecules, ions, electrons etc.) -> Avogadro's number NA, see data sheet n = m Mr n: number of moles [moles] m: mass [g] M₁: molar mass [g/mol] []: units Number of particles N = n x NA N: number of particles [no unit] NA: Avogadro's number 6.02 x 1023 [mol¹] Molar Gas Volume Volume of 1 mole of any gas = 24 dm³ (data sheet) -> at standard conditions (298 K/25 °C, 100 kPa) Vx = n x 24 dm³ mol-¹ Vx: unknown volume [dm³] n: number of moles [moles] Converting cm³ into dm³: x cm³- = X dm³ 1000 Ideal gas equation: PV = nRT Dr. C. Boes p: pressure [Pa] V: volume [m³] 1 m³ = 1000 dm³ R: 8.31 J K¹ mol ¹ (gas constant) -> data sheet T: temperature [K] (K = 273 + °C) -> make sure you are using correct SI units - see [ ] -> the ratios of the volumes of reactants and products can be used to work out the mole equation (reaction equation): Volumes are proportional to moles 12 alevelchemistryrevision.co.uk Empirical and Molecular Formulae Definitions of molecular and empirical formulae Relationships between these formulae for salts and molecules Work out empirical formulae from masses Empirical Formulae from percentages Molecular formulae from empirical formulae and molecular mass Example Calculation Dr. C. Boes 13 alevelchemistryrevision.co.uk Empirical and Molecular Formulae Molecular formula: Actual number of atoms (of each type of element) in a molecule, e.g. Ethene C₂H4 => closer to reality; generally used in equations. Empirical formula: Smallest whole number ratio of atoms (of each element) in a compound; CH₂ => Empirical formula is used when trying to find out the molecular formula of an unknown organic compound by burning it. For Salts: chemical formula is identical to empirical formula For Molecules: molecular formula is a multiple of empirical formula Work out empirical formulae from grams (burning hydrocarbon): ● Use n=m/M to calculate moles of each element; take mole ratios into account, e.g. H in H₂O 2:1 Divide by smallest mole number to get ratio (empirical formula) How to work out empirical formulae from percentages: • Change percentages into grams, then same as above Work out molecular formulae from empirical formulae and molecular mass Divide molecular M, by empirical M, to get factor Multiply empirical formula with factor to get molecular formula Example calculation When an unknown hydrocarbon with M₁ = 70 g/mol is burnt in excess oxygen, we get 6.6 g of CO₂ and 2.7 g of H₂O (elemental analysis). What is the empirical and molecular formula of this compound? Calculate empirical formula first Moles CO₂: n = 6.6 g/44 g/mol = 0.15 mol (1 mole C in 1 mole CO₂ -> factor 1) => Moles C: n = 0.15 mol Moles H₂O: n = 2.7 g/18 g mol¹ = 0.15 mol (2 mole H in 1 mole H₂O -> factor 2) => Moles H: n = 2 x 0.15 mol = 0.3 mol Divide by smallest mole number: C 0.15/0.15 = 1, H 0.3/0.15 = 2 Ratio 1:2 C:H Empirical formula: C₂H₂ => CH₂ Calculate Molecular formula from Mr and empirical formula Unknown hydrocarbon M₁ = = 70 g mol-¹ Empirical molar mass CH₂ M₁ = 14 g mol-¹ Factor: 70 g mol¹/14 g mol™¹= 5 5 x CH₂ => C5H10 Molecular formula: => Unknown Hydrocarbon was Pentene Dr. C. Boes 14 alevelchemistryrevision.co.uk Water of Crystallisation Dr. C. Boes & Types of Crystal Structures Two terms Characteristics (three points) Experiment to determine water of crystallisation (two points) Example Calculation Four types of crystal structures 15 alevelchemistryrevision.co.uk Water of crystallisation Terms ● Anhydrous salt -> no water Hydrated salt -> water incorporated in lattice ● Water of crystallisation is expressed with dot and mole number in chemical formula Experiment to determine water of crystallisation when heated hydrated salts lose water of crystallisation and become lighter (anhydrous) -> heat to constant mass difference in mass is due to the water lost and can be used to calculate the number of moles of water of crystallisation in the chemical formula (X): -> Calculate moles H₂O and moles anhydrous salt by using n = m/M -> Divide moles of H₂O by moles of anhydrous salt to get X -> Similar to calculating empirical formulas -> water is included in molar mass M, and belongs to the compound. => the number of moles H₂O per mole salt is written in the formula: Na₂CO3 10H₂O M₁ = 286 g/mol (2x23 + 12 + 3x16 + 10x18) Example Calculation When 6.42 g of hydrated magnesium sulphate, MgSO4 XH₂O is heated 3.14 g of anhydrous magnesium sulphate is left. What is the formula of the hydrated salt, e.g. X? Water lost: 6.42 g -3.14 g = 3.28 g Moles H₂O: n = 3.28 g/18 g mol¹ = 0.182 moles Moles MgSO4: n = 3.14 g/120.4 g mol¹ 1 = 0.0261moles X = 0.182 mole / 0.0261 mole = 6.95 = 7 => MgSO4.7H₂O (Epsom salt) Types of Crystal Structures ● Ionic (NaCl) ● Metallic (Mg) ● ● Macromolecular - giant covalent (diamond, graphite) Molecular (I2, ice) Dr. C. Boes 16 alevelchemistryrevision.co.uk Mole Equations - Calculate Masses & Percentage Yield Dr. C. Boes Calculation Steps Rule for rounding Rules for significant figures Equation for percentage yield Units Tip Four reasons for loss 17 alevelchemistryrevision.co.uk Mole Equations – Calculate Masses 1) Underline or highlight all data given in the exam question. Calculate moles for the given compound by using n = m/M. 2) 3) Circle or highlight mole numbers in front of related compounds (given and unknown). 4) 5) 6) 7) 8) Determine mole ratio for unknown compound. Get mole factor by dividing both mole numbers by the same number (here: 4) so that the given compound becomes 1. Calculate mass of unknown compound using m = n x M. 9) Do not round whilst still calculating. Carry as many digits through the calculation as possible (at least 3) until you reach the final answer. Write the answer with the appropriate number of significant figures: if the data are given in 3 significant figures then the answer should also be given in 3 significant figures (see below). 10) If the data are given in 2 and 3 significant figures then the answer should be given in 2 significant figures (always the lowest one). Multiply moles of given compound with factor to get moles of unknown compound. 11) The first significant figure is the first digit which is not zero: 0.0109. 12) If the last non-significant figure is 1 - 4 round down, if 5 - 9 round up. 13) 0.001 = 1 x 10-3 -> use standard form for scientific calculations. Example 1) Calculate amount of O₂ (in grams) produced if 3.24 g of iron(III) nitrate is heated 4Fe(NO3)3 -> 2Fe₂O3 + 12NO₂ + 30₂ 2) Moles Fe(NO3)3 : n = 3.24 g/241.8 g/mol = 0.0134 moles See mole equation above 3) 4) Ratio: 3:4 5) Factor for O₂: 3/4 = 0.75 (Fe(NO3)3 : 4/4 = 1) 6) Moles O₂: 0.75 x 0.0134 moles = 0.01005 moles 7) Mass O₂: m = n x M = 0.01005 mol x 32 g/mol = 0.322 g Percentage Yield percentage yield = actual yield theoretical yield units of actual/theoretical yield of products: [grams] or [moles] -> make sure actual and theoretical yield of products are in the same unit Reasons for loss ● ● Reaction not complete (clumps instead of powder) Loss of product (sticking to vessel, evaporation of liquids) ● By-products Impurities of reactants Dr. C. Boes x 100 18 alevelchemistryrevision.co.uk Atom Economy & Concentrations Dr. C. Boes Equation for atom economy Two rules Benefits of high atom economy (four points) Example calculation Equation for mole concentration Converting dm³ in cm³ Equation for mass concentration Converting mole concentration into mass concentration 19 alevelchemistryrevision.co.uk Atom Economy % atom economy = Mr desired product x 100 EM, all reactants ● Σ: sum 100 % for addition reactions (no waste products) multiply M, with mole numbers from chemical equation Environmental and economic benefits of high atom economy Avoiding waste High sustainability (less raw material) More profitable (raw material and removal of waste products expensive) High efficiency Example calculation (NH4)2SO4(s) + 2NaOH(aq) -> 2NH3(g) + Na₂SO4(aq) + 2H₂O) Calculate the percentage atom economy for the production of ammonia % atom economy = x 100 Concentration Mole concentration: c=n Mass concentration: n: moles [mol] V: volume [dm³] c: concentration [mol dm³³] 1 dm³ = 1000 cm³ = Cm = m V Dr. C. Boes 2 x 17 2x17 + 142 + 2x18 16.0 % m: mass [g] C: mass concentration [g dm³³] Convert mole concentration into mass concentration: Cm = M₁x c 20 alevelchemistryrevision.co.uk 1.3 Bonding (Paper 1 & 2) Ionic and Covalent Compounds and Bonds Definition of compound Names and Characteristics of ionic compounds (four points) Three physical properties Definition of ionic bond Draw lattice Names and Characteristics of covalent compounds (four points) Three physical properties Definition of covalent bond Definition & symbol of dative covalent bond Dr. C. Boes 21 alevelchemistryrevision.co.uk Ionic and Covalent Compounds and Bonds Compound: Atoms of different elements bonded together Ionic compounds - Salts: Metal/Non-metal Consist of ions ● ● ● @jo Na 'Dot-and Cross' diagram: square bracket with charge around ion Form giant ionic lattice with alternating charges (see diagram below) Chemical formula gives ratio (ion charges have to be balanced) Physical properties: - - high melting points (strong ionic bonds, lots of energy needed) - soluble in water - conduct electricity in solution or when molten (ions can move) Examples: NaCl, MgCl₂ Ionic bond: electrostatic attraction between oppositely charged ions Na Cl Na Cl Na Cr CI Na Cl Na Cl Nat Na Cl Na Cl Nat Cr CI Na Cl Na Cl Na+ Covalent compounds - Molecules: Non-metals Collection of atoms 'Dot-and Cross' diagram: overlapping circles for bonds Different shapes Formula tells which atoms are directly connected to each other Physical properties: - low melting point (weak intermolecular forces, less energy needed) - not soluble in water - not conducting electricity Examples: CH4, H₂O Covalent bond: Strong electrostatic attraction between two nuclei (+) and the shared pair of electrons (-) -> sharing a pair of electrons Dative covalent bond: both electrons of the covalent bond come from one atom (arrow instead of dash) e.g. NH4* H Dr. C. Boes H + N. H CI H 22 alevelchemistryrevision.co.uk Shapes of Molecules I Theory (five points) Shapes of molecules: One central atom with 2 partners One central atom with 3 partners One central atom with 4 partners One central atom with 3 partners & 1 lone pair of electrons One central atom with 2 partners & 2 lone pairs of electrons Dr. C. Boes 23 alevelchemistryrevision.co.uk Shapes of Molecules Electron-Pair Repulsion Theory: • Negative charges of electron pairs in covalent bonds and in lone pairs repel They try to get as far apart as possible • Lone pairs of electrons on the central atom repel more (closer to atom) => smaller bond angle (increment - 2.5° per electron pair) • Shape depends on number of charge clouds (bonds, lone pairs) A double bond counts as a single bond (one charge cloud) ● One central atom with 2 partners (use the term ‘charge clouds'): O => linear, 180°, CO₂, BeCl₂ One central atom with 3 partners: F. 109.5° 180° H => trigonal planar, 120°, BF3 One central atom with 4 partners: H H 'В H Dr. C. Boes 120° CH 104.5° H => tetrahedral, 109.5°,CH, NH4* (draw straight-line-bonds next to each other) One central atom with 3 partners & 1 lone pair of electrons: H 107° H => trigonal pyramidal, 107°, NH3, SO3²- One central atom with 2 partners & 2 lone pairs of electrons: H H => bent/non-linear, 104.5°, H₂O + H 24 alevelchemistryrevision.co.uk Shapes of Molecules II One central atom with 3 partners and 2 lone pairs of electrons One central atom with 4 partners and 1 lone pair of electrons One central atom with 4 partners & 2 lone pair of electrons One central atom with 5 partners One central atom with 6 partners Dr. C. Boes How to get other examples How to get number of lone electrons 25 alevelchemistryrevision.co.uk One central atom with 3 partners and 2 lone pairs of electrons F => t-shaped, 88, CIF 3 One central atom with 4 partners and 1 lone pair of electrons: F 87° 90° => seesaw, 102° & 87°, SF4 One central atom with 4 partners and 2 lone pairs of electrons: => square planar, XeF4, 90° -> see Y2 rev. card 'Stereoisomerism in complexes' CI -F 88⁰ One central atom with 5 partners: CI F F F 102° => trigonal bipyramidal, 120° & 90°, PC15 One central atom with 6 partners: F Dr. C. Boes CI F F => octahedral, 90°, SF 120° 90° To get other examples: use a different element from the same group as the central atom and keep same partners, e.g. PH3 for NH3 Number of lone electrons = number of outer electrons – number of bonds 26 alevelchemistryrevision.co.uk Electronegativity & Ionisation Energy Dr. C. Boes Definition Electronegativity Properties of electronegativity (five points) Definition of 1st Ionisation Energy Reactivity (two points) 1st and 2nd ionisation energies Difference between 2nd and 1st ionisation energies (three points) Big jump Drops in 1st ionisation energies with reasons Trends across period (three points) Trends down group (three points) 27 alevelchemistryrevision.co.uk Electronegativity Definition: Ability of an atom to attract the electron pair in a covalent bond Fluorine most electronegative element ● Elements closer to F: more electronegative ● ● Ionisation Energy Definition of 1st Ionisation energy: Energy required to remove one electron from one mole of gaseous atoms [kJ/mol] Large differences in electronegativity result in ionic compounds Small differences result in covalent compounds Type of bond (ionic/covalent) can be predicted from the electronegativity difference Reactivity ● ● decreases down Halogen group (want to gain one electron) increases down Alkali group (want to lose one electron) 1st Ionisation energy: 2nd Ionisation energy: 2nd Ionisation energy larger than 1st Ionisation energy • increasingly positive ion • less repulsion amongst remaining electrons smaller ions Big jump when new shell is broken into => closer to the nucleus Drop in 1st ionisation energy between Groups 2 (Be) and 3 (B) -> start of p-subshell, which is further away from nucleus -> full 2s² subshell provides additional shielding -> despite increased nuclear charge Og) >0¹(g) + e 0+ *(g) -> 0²(g) + e* Drop between Groups 5 (N) and 6 (0) -> one orbital of the 3 p-orbitals gets filled with two electrons, which repel each other Periodic Trends of Electronegativity & Ionisation Energy Increases across period: - more protons - same shell (smaller radius) => stronger nuclear attraction Decreases down group: - more shells, more distance - more shielding - despite more protons => less nuclear attraction Dr. C. Boes 28 alevelchemistryrevision.co.uk Metallic Bonding & Simple and Giant Covalent Structures Definition of metallic bonding Trends of melting point across period (three points) Trends of melting point down group (two point) Characteristics of metals Diagram of structure Dr. C. Boes Simple covalent structures for three elements Definition and properties of giant covalent structures Definition of allotropes Two allotropes of Carbon with properties 29 alevelchemistryrevision.co.uk Metallic Bonding Definition: Electrostatic attraction between metal cations and delocalized electrons from the outer shell (sea of electrons) -> high melting points (large energy required to overcome strong attraction) Across the period higher melting points: more delocalized electrons (negative charges) charges of cations increase: Na, Mg²+, Al³+ smaller ions => higher charge density: ratio of charge to volume Down the group lower melting points: more shells greater distance Characteristics of metals: electrical & thermal conductors (free moving electrons) malleable & ductile; dense, shiny & soft - only alloys are hard Giant metallic lattice ● cations -> Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ delocalized-00 OO electrons Simple Covalent Structures (Molecules) Sulphur: yellow solid, simple molecules like Sg rings or amorphous Phosphorus: white solid, simple molecules like P4 Iodine: black solid, I₂ -> molecular lattice (Van der Waals forces) -> Low melting points (less energy to overcome weak intermolecular forces) Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ 8000 00 20 OO 90 OO Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+) Giant Covalent Structures Def.: Network of covalently bonded atoms (Macromolecular Structures) High melting points (Large energy needed to break covalent bonds) C, Si => four covalent bonds (Group 4) Allotropes: pure forms of the same element that differ in structure Allotropes of carbon: Diamond: 4 bonds, tetrahedral -> hard, cold, high melting point, not conducting electricity, not soluble Graphite: 3 bonds, trigonal planar, sheets of hexagons -> slippery, single electrons are delocalized and conduct electricity, strong, lightweight, insoluble Dr. C. Boes 30 alevelchemistryrevision.co.uk Intermolecular Forces - Boiling Points Function of intermolecular forces Relationship between intermolecular forces and boiling points Three types of intermolecular forces Cause of polar bonds Why polar bonds do not always make polar molecules Causes of temporary and induced dipoles Van der Waals depends on... (three points) Three elements which form hydrogen bonds Behaviour of the Hydrogen atom Strength of hydrogen bond compared to other forces Drawing hydrogen bonds Anomalous properties of water (two points) Dr. C. Boes 31 alevelchemistryrevision.co.uk Intermolecular Forces keep molecules together -> model to explain physical states of molecules: Melting & Boiling points The stronger the intermolecular forces the higher the melting/boiling points -> more energy is required to overcome these forces of attraction I) Permanent dipole-dipole interactions Permanent dipoles contain polar bonds due to different electronegativities -> shift in electron density 8+ 8- H - CI 6 means slightly (or partially) Molecules with polar bonds can be permanent dipoles (polar molecules) Polar bonds do not always make polar molecules (Dipoles) Symmetric arrangement where polar bonds cancel each other out results in non polar molecule, e.g. CO₂ II) Induced dipole-dipole interactions: Van der Waals Non-polar molecules contain non polar bonds (equal electronegativities: Cl₂) Temporary/instantaneous dipole Electron cloud moving randomly (uneven distribution) This temporary dipole can induce another dipole => induced dipole Weakest intermolecular force, but found in all atoms and molecules <HI) ● Can override permanent dipole-permanent dipole interactions (HCI < Can hold molecules together in a molecular lattice (1₂) ● ● ● Depends on: ● number of electrons (size of molecule) surface area & points of contact -> shape (how close molecules can come together) ● III) Hydrogen Bonds O, F, N (very electronegative) with an H (high charge density) bonded directly to these atoms: -NH₂, -OH, HF • Strongest inter-molecular force (H forms weak bonds with lone pair of e) Responsible for high solubility in water Write partial charges on top and draw H-bond as dotted line between H and lone pair of electrons (bond angle 180°): 8+ 8- 8+ 8- H-F:----- H-F Dr. C. Boes 8+ 8- 32 HI Anomalous properties of water -> high boiling point -> lower density of ice because of more H-bonds than liquid water : alevelchemistryrevision.co.uk 1.4 Energetics (Paper 1 & 2) Enthalpy Changes - Definitions & Bond Enthalpies Dr. C. Boes Definitions for exothermic/endothermic Definition for enthalpy change Definitions for standard enthalpy change of: Reaction Formation Combustion Neutralisation (Atomisation) Two equations to calculate AH, from bond enthalpies Definition of mean bond enthalpy Two rules Exothermic/endothermic in relation to bond enthalpies 33 alevelchemistryrevision.co.uk Enthalpy Changes - Definitions Exothermic -AH, negative: energy released into surroundings -> heat up Endothermic +AH, positive: energy taken from surroundings -> cool down Enthalpy change: Heat change in a reaction at constant pressure [KJ mol™¹] Standard enthalpy change º: under standard conditions: 100 kPa, 298 K ~ of reaction AH,: the enthalpy change when the reaction occurs in the molar quantities shown in the chemical equation, under standard conditions in their standard states ~ of formation AH,: the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions 2C(s) + 3H2(g) + 1/2O2(g) -> C₂H5OH(1) or ~ of combustion AH: the enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, all reactants and products being in their standard states ~ of neutralisation AHneut: the enthalpy change when 1 mole of water is formed from the neutralisation of hydrogen ions (H*) by hydroxide ions (OH) under standard conditions H+ + OH -> H₂O(1) (aq) (aq) Bond Enthalpies ● of atomisation AHat: is the enthalpy change when I mole of gaseous atoms is formed from the element in its standard state (Year 2) 1//2Cl2(g) -> Cl(g) ΔΗ, = ΣΗ bonds broken – ΣΗ bonds formed AH, = 2H bonds (reactants) - EH bonds (products) Mean bond enthalpy: average enthalpy required to break one mole of a covalent bond over different compounds in the gaseous state [kJ mol¹]. -> always positive (also for bonds formed) Bond forming releases energy Bond breaking requires energy If EH bonds formed > EH bonds broken => exothermic If EH bonds formed < EH bonds broken => endothermic Dr. C. Boes 34 alevelchemistryrevision.co.uk Dr. C. Boes Calorimeter Purpose Equation for enthalpy change Five reasons for underestimate Extrapolation (Diagram of calorimeter) Example calculation Three tips 35 alevelchemistryrevision.co.uk Calorimeter (-> to measure enthalpy change of a reaction) (Practical) q=mcAT enthalpy change [Joules] C: mass of water [g] (ignore masses of salts added to water) specific heat capacity of water (4.18 Jg¹ K¹ ->data sheet) AT: temperature change [°C] (A°C = A°K) q: m: Reasons why enthalpy change is underestimated by experiment Heat absorbed by container & Heat lost to surroundings Incomplete combustion & Evaporation of volatile fuel Non-standard conditions ● ● -> record initial temperature for a suitable time before starting the reaction -> Use extrapolation method to account for heat loss => extrapolate to start time to get minimum/maximum Temperature Thermometer Water Stirrer Total volume of solution: Density of water: Mass of water: w Air Dr. C. Boes Example Calculation - Neutralisation 30 cm³ of 0.80 mol dm³ HCl were neutralised with 20 cm³ of 1.2 mol dm-³ NaOH. The temperature of the solutions increased from 20 °C to 26 °C. What is the standard enthalpy change of neutralization for this reaction? HCl + NaOH -> NaCl + H₂O Insulation Standard enthalpy change of neutralization: -> divide by the number of moles Combustion Chamber Reactant/Fuel 20 cm³ + 30 cm³ = 50 cm³ 1 g cm³ 50 cm³ x 1 g cm³ = 50 g q=mcAT q= 50 g x 4.18 Jg¹¹ K¹ x 6 K = 1254 J Moles water formed: n = 0.03 dm³ x 0.8 mol dm³ = 0.024 moles 36 density p = m/V Hneut q/n = 1254 J/0.024 moles = -52 kJ mol-¹ If temperature has increased -> exothermic reaction => put - sign in front If temperature has decreased -> endothermic reaction => put + sign in front alevelchemistryrevision.co.uk Dr. C. Boes Hess's Law Definition Purpose Equation Two rules Three rules for triangles Example Calculation 37 alevelchemistryrevision.co.uk Hess's law The total enthalpy change for a reaction is independent of the route taken (as long as the initial and final conditions are the same) -> To calculate enthalpy changes for unknown reactions from known reactions (e.g. reaction enthalpies from formation enthalpies) ● AHr,c=EAH₁ (products) - ZAH, (reactants) AH, Enthalpy change of reaction or combustion [kJ mol™¹] AH: Enthalpy change of formation [kJ mol¹] AH, Standard enthalpy change of reaction Elements like O₂ have no formation enthalpy (zero) multiply formation enthalpies with mole numbers from reaction Triangles/arrows Add enthalpies (AH₂ AH3) while going to the same products (endpoint) of the unknown enthalpy AH, by using the alternative route: going along arrow -> positive sign for enthalpy going against arrow -> negative sign for enthalpy ΔΗ, C₂H5OH + 302 ΔΗ;° (CO2) = AH₁ (H₂O) = AH, (C₂H5OH) = ΔΗ, (02) = ΔΗ, ΔΗ, Θ Dr. C. Boes ΔΗ, = . - AH₂ + AH3 Example Calculation for standard enthalpy change of combustion for ethanol: Ꮎ AH₂ Ꮎ 2C + 3H₂ + 3.50₂ = -394 kJ mol-¹ -286 kJ mol -277 kJ mol-¹ 0 kJ mol-¹ ZAH, (products) = [(2x-394) + (3x-286)] = -1369 kJ mol™¹ 2CO₂ + 3H₂O 38 AH3 ZAH, (reactants) [-277] | kJ mol-¹ alevelchemistryrevision.co.uk 1.5 Kinetics (Paper 2) Rates of Reactions Dr. C. Boes Rates depend on... (four points) Definition & equation Graph shows... Four measurement methods Collision theory (four points) Definition of activation enthalpy Maxwell-Boltzmann Distribution for temperature increase Maxwell-Boltzmann Distribution with and without catalyst Tip Area below curve represents... Restriction Effect of catalyst on activation energy 39 alevelchemistryrevision.co.uk Rates of Reactions (Practical) Rates (speed) depend on temperature surface area (size of particles) catalyst concentrations of reactants (or pressure for gases) Rate of reaction: change of concentration (product or reactant) over time t r = Ac At Graph: shows increase/decrease over time Measurement ● Collision Theory ● ● ● Activation energy E₂: Minimum amount of energy needed to begin breaking reactant bonds and start a chemical reaction (always positive) ● Precipitation (measure time until marker disappears) Change in mass when gas given off (balance) Volume of gas given off (syringe) Titration (put samples on ice to stop reaction) Maxwell-Bolzmann Distributions for Temperature Increase & with Catalyst Higher temperature -> faster => more successful collisions: Ekin > Ea ● Higher concentration -> successful collisions more likely Larger surface area -> particles can access more area Catalyst -> see revision card 'Catalyst' Number of particles Temperature Increase Dr. C. Boes Lower Temperature Kinetic Energy Higher Temperature E₁ Particles with Ekin > E Number of particles 40 With and Without Catalyst Most Likely Energy Kinetic Energy Mean Energy E₂' with catalyst Graphs should not touch y-axis. Area below curve represents total number of molecules (stays the same). Maxwell Boltzmann distribution applies to gases only. Catalyst reduces activation energy -> more molecules exceeding Ea. Particles with Ekin > E₂ alevelchemistryrevision.co.uk E₂ Catalyst Definition (Homogeneous and heterogeneous catalysts with examples) Effect of catalyst on equilibrium Catalysts save... (Catalyst poison) Mechanism of solid catalyst (four steps) Diagram Enthalpy profile diagram with and without catalyst Dr. C. Boes 41 alevelchemistryrevision.co.uk Catalyst Definition: Increases rate of reaction by lowering activation energy (more successful collisions) and providing an alternative reaction pathway. Catalysts are not used up during the reactions (unchanged). ● ● ● ● ● ● homogenous catalyst => same physical state as the reactants: enzymes/substrate (aq/aq) or H₂SO4 (aq/aq) in organic chemistry (Year 2) heterogeneous catalyst => different physical states (Year 2) catalytic converter Pt/Rh (s/g) in car burns hydrocarbons and converts: 2CO+2NO -> 2CO₂ + N₂ -> CO & NO are toxic other examples: Ni (Hydrogenation), Fe (Haber), V₂O5 (Contact process) Contact process: V₂O5 + SO₂ -> V₂O4 + SO3 V₂O4 + 1/2O₂ -> V₂05 -> Regeneration (Year 2) does not change chemical equilibrium catalysts save energy and costs by lowering reaction temperature/pressure and increasing reaction speed Catalyst poison: binds stronger than reactant -> blocking catalyst surface Mechanism for solid catalyst: H₂ + Cl₂ -> 2HC1 adsorption (not absorption!) of reactants (H₂, Cl₂) at catalyst surface - this weakens bonds in reactants (lowers activation energy) - - new bonds of products (H-Cl) are formed desorption of products (HCI) after reaction. CI---CI Enthalpy Reactants H-H Dr. C. Boes H H-CI CI Enthalpy profile diagram of exothermic reaction with and without catalyst ← Catalyst ← Catalyst Enthalpy Profile Heterogeneous Catalyst Reaction Pathway E, Activation Energy E' Activation Energy with catalyst 42 -AH,: Enthalpy Change Products alevelchemistryrevision.co.uk 1.6 Equilibria (Paper 1 & 2) Equilibrium and Reversible Reactions Definition of reversible reaction Definition for dynamic equilibrium (three points) Application Le Chetallier Haber process with catalyst and change of conditions Effect of catalyst on chemical equilibrium Equilibrium law for a general reaction equation Dr. C. Boes Equilibrium constant (seven points): The larger K.... Equilibrium constant depends on... When temperature is increased... For exothermic reactions... Units of Ke How does change of concentrations affect Ke Solids and liquids... 43 alevelchemistryrevision.co.uk Equilibrium and Reversible Reactions Reversible reaction: products convert back to reactants (reaction goes both ways) -> incomplete reaction => draw double arrow: Dynamic equilibrium (steady state) -> rates of the forward and reverse reactions are equal ● -> Important for yield in chemical industry Le Chetallier: When the conditions of a system at equilibrium changes, the position of the equilibrium shifts in the direction that opposes (counteracts) the change both reactions still going on; balance => low percentage of product concentrations of reactants and products remain constant only in closed system Ammonia manufacture (Haber-process): iron catalyst (200 atm, 450 °C) N2(g) + 3H2(g) 2NH3(g) AH-93 kJ/mol (exothermic) + pressure + heat ● => Catalyst does not change equilibrium but increases rate of both reactions (equilibrium is reached faster) ● + concentration reactants - concentration products Equilibrium Equation aA + bB Equilibrium law -> shifts to the side with less moles gas (here to the right) -> shifts in the direction of endothermic process (here to left) -> shifts to right -> shifts to right Kc = [C] [D]d [A]ª[B]b []: concentration Ke: equilibrium constant ● the larger K, the more yield (Kc> 1 more products, K. < 1 more reactants) Ke only temperature dependent when temperature is increased then K, increases for endothermic reactions and decreases for exothermic reactions for exothermic reactions: compromise between rate and yield needed ● need to calculate units of Ke by cancelling out units of concentration if concentration of [C] or [D] is increased (numerator), concentration of [A] and [B] must also increase (denominator) to keep K, constant (equilibrium moves to the left) => Ke stays constant solids (and liquids) do not appear in the equation for Ke Dr. C. Boes cC + dD 44 alevelchemistryrevision.co.uk Equilibrium Concentrations Calculation How to approach an equilibrium question (two points) Example calculation Dr. C. Boes 45 alevelchemistryrevision.co.uk Equilibrium Concentrations Calculation -> highlight all data given (bold) in the question and write down mole ratios -> create a table as specified below 0.90 moles of Nitrogen dioxide were thermally decomposed at 450 °C in a container of 23 dm³. 0.40 moles of Oxygen were found in the equilibrium mixture. Calculate K, for this reaction. 2NO₂ 2NO Initial moles n₁ Moles reacted 5 Equilibrium moles Equilibrium conc. [mol dm³ Calculate equilibrium moles: 0.4 moles O₂ -> X² Equilibrium Concentrations: c=n V + NO₂ 0.9 0.8 0.1 (ner) 0.0043 Dr. C. Boes 0₂ Divide all equilibrium moles by volume, e.g. -3 [NO₂] = 0.1 mol = 0.0043 mol dm 23 dm NO 2:1 ratio between NO and O₂ => 2 x 0.4 moles = 0.8 moles NO in equilibrium 1:1 (2:2) ratio between NO₂ and NO -> 0.8 moles NO₂ have reacted -> x x: reacted moles of reactant (NO₂) (related to x' by mole-ratios) Equilibrium moles of reactant (NO₂): ner= n₁-x ner = 0.9 mole - 0.8 mole = 0.1 moles => 0.1 moles NO₂ left in equilibrium 46 0.8 (x) 0.035 0₂ 0.4 (x²) 0.017 -3 Kc = [NO]² x [0₂] = (0.035)² x 0.017 mol dm-³ = 1.1 mol dm` [NO₂]² (0.0043)² alevelchemistryrevision.co.uk 1.7 Redox Reactions (Paper 1) Redox Reactions - Oxidation States & Disproportionation Dr. C. Boes Definitions of Oxidation and Reduction Purpose of oxidation states Six rules for oxidation states Oxidising/Reducing agent Tip Definition of disproportionation 47 alevelchemistryrevision.co.uk Redox reactions - Oxidation states Oxidation Is Loss of electrons -> oxidation state becomes more positive Reduction Is Gain of electrons -> oxidation state becomes more negative => OILRIG Oxidation states/numbers -> help to determine chemical formula of compound Def: identical to the charge of ions in a salt; Rules: ● the charge an element would have, in a molecule, if it were a salt. always zero for pure element in its basic state (uncombined element) group number indicates maximum oxidation state the more electronegative element gets a negative -, the less electronegative element a positive oxidation number the sum of all oxidation states for neutral compounds is zero (2x+3)+(3x-2) = 0 AL2₂03 +3 the sum of all oxidation states for ions equals their charge SO₂2- (+6) + (4x-2) = -2 +6 -2 most common oxidation states in compounds: Halogens -1 Oxygen -2 (-1) Hydrogen +1 (-1) Metals positive (alkali metals +1, alkaline earth metals +2) Oxidising agent: gets reduced (helps other element to get oxidised) -> electron acceptor Reducing agent: gets oxidised (helps other element to get reduced) -> electron donor Oxidising/Reducing agent is the whole compound not just the element which changes oxidation state Disproportionation Definition: same element gets reduced and oxidized -> Special kind of Redox reaction Cu₂O + 2H+ -> Cu²+ + Cu + H₂O +1 +2 0 -> see also revision card 'Water Treatment' Dr. C. Boes 48 alevelchemistryrevision.co.uk Balancing Redox Equations Balancing a simple redox reaction (two points) Balancing a complicated redox reaction (two points) Two important rules for balancing redox reactions Half equations Dr. C. Boes 49 alevelchemistryrevision.co.uk Balancing Redox Equations How to balance a simple redox equation First write the chemical formula of the product to the right side of the arrow. Balance this formula with subscript numbers according to oxidation states (group number) by using lowest common multiple. Then balance equation by putting numbers in front of reactants and products (never change subscript numbers, as this would create a different substance) ● ● ● If not just elements reacting (ions/molecules): First balance electrons, starting with the biggest difference in oxidation states (between left and right side of equation) ● ● ● ● 4A1+30₂ -> 2A1₂203 0 0 +3 -2 ● ● (lowest common multiple: 6) then balance all the other reactants and products (single elements like sulphur first; oxygen before hydrogen) 8I + 8H* + H₂SO4 -> 41₂ +H₂S +4H₂O -1 +6 0 -2 Half-Equations split up redox reaction in two separate oxidation and reduction reactions => half equations use oxidation states to determine number of transferred electrons number of electrons transferred must be the same in overall equation - > multiply to get lowest common multiple (here: 6) Biggest difference: S: +6 -> -2 I: -1 ->0 => loses le => 8I needed (-> 41₂) number of electrons lost and gained must be the same! charges have to be balanced as well (81* = 8H¹) 3Zn + 2Fe³+ -> 2Fe + 3Zn²+ 0 +3 0 +2 half equations: Dr. C. Boes Fe³+ + 3e -> Fe Zn -> => gains 8e Zn²+ + 2e Sometimes H or H₂O have to be added to half-equation to balance them see also Year 2 revision card 'Redox Equations' 50 |x 2 |x 3 alevelchemistryrevision.co.uk Two Main Types of Inorganic Reactions & Ionic Equations Dr. C. Boes Two main types Tip Two minor types Four state symbols Tip Definition of spectator ion 51 alevelchemistryrevision.co.uk Two Main Types of Inorganic Reactions 1) Redox (Reduction/Oxidation) -> transfer of electrons (Oxidation states change) 4A1+30₂ -> 2Al2O3 0 0 +3 -2 -> Displacement & Disproportionation are special cases of Redox reactions You can quickly spot a redox reaction if an element becomes an ion or part of a compound and visa versa in a reaction equation (see above) 2) Acid-Base (Neutralisation) (Year 2) -> transfer of protons H* (Oxidation states do not change) HCl + NaOH -> NaCl + H₂O +1 -1 +1 -2 +1 +1 -1 +1 -2 Minor reaction types ● Thermal decomposition CaCO3(s)-> CaO(s) + CO2(g) Precipitation Ag (aq) + Cl(aq) -> AgCl(s) -> No change of oxidation states State Symbols s = solid, 1 = liquid, g = gas, aq = aqueous (solution in water) If the melting or boiling point is below room temperature (25 °C, 298 K) then the substance is a liquid or a gas respectively. Ionic Equations -> only show the reacting particles Spectator Ions: Ions which do not take part in the reaction They can be removed from the full equation to give an ionic equation Full equation: Cl₂ + 2NaBr -> Br₂ + 2NaCl 0 +1 -1 0 +1 -1 Na does not change oxidation states -> spectator ion, removed from equation: Ionic equation: Cl₂ + 2Br -> Br₂ + 2C1* 0 -1 0 -1 Dr. C. Boes End of free Sample! 52 alevelchemistryrevision.co.uk