The Basics • Elements, Molecules, Compounds, Ions • Parts of the Periodic Table • How to Name 1 Classification of Elements •Metalloids semi-metals) – along ••Nonmetals Metals – found –(or found on the on the left-side right-hand of thethe sidestair-step of line the Periodic Periodic Table Table Properties are intermediate between metals and nonmetals 2 Elements , Molecules, Compounds, and Ions • Element – one single type of atoms – Al Cu He – Naturally occurring elements that are Diatomic are still elements – N2 O2 F2 Cl2 Br2 I2 H2 – How many elements are in Mn(SO4)2 ? 3 – How many Atoms? 9 • Molecule - smallest electrically neutral unit of a substance that still has the properties of that substance, 2 or more different elements 3 Elements , Molecules, Compounds, and Ions • Compounds – groups of atoms – Ionic and Molecular • Molecular Compounds – share electrons typically 2 or more non-metals (hydrocarbons) • Example H2S CO2 C5H10 • Ionic Compound (salts) – transfer electrons typically metal and non metal (watch for poly atomic ions) • FeS Mg(OH)2 (NH4)3PO4 4 Ions • Ions have either lost or gained electrons • Typically Metals lose electrons to become positive • Example cations • Mono-valent Mg -> Mg2+ + 2e– Group 1A = Metal1+, 2A =Metal2+ and 3A = metal3+ – Multi-valent Fe --> Fe2+ + 2e- and Fe --> Fe3+ + 3e- • Non-metals gain electrons - anion • S + 2e- -> S25 Poly Atomic Ions • A molecular compound with a charge • NH4+ • Ammonium • CO32- • Carbonate • SO42- • Sulfate • NO3- • Nitrate • OH- • Hydroxide • H3O+ • Hydronium 6 Acids and Bases From Ions H+ or OH• Acids look for hydrogen up front (HA) or as COOH • • • • • Example HF H3PO4 C4H6COOH Strong Acids HCl, H2SO4, HBr, HI, HClO3,HNO3 Base look for hydroxide or NH group Example KOH C4H6NH2 7 Naming Compounds 1. Ionic or Covalent 2. Ionic – two ions or Poly atomic ions Covalent 2 non-metals Or a hydrocarbon Type of Metal Mono-valent metals groups 1A 2A 3A Multi-valent Metal Transitions metals and under the stairs Name the Metal Name the Nonmetal + ide (if PAI use its name) Find the Charge on the Metal To make the compound neutral Write the Charge with roman numeral Name nonmetal + ide (if PAI use its name) CaF2 RbNO3 Al(OH)3 calcium fluoride rubidium nitrate aluminum hydroxide Ni+ClPb2+SO42Pb4+(SO4)22- nickel(I) chloride lead(II) sulfate lead(IV) sulfate 8 Ionic naming • Name to Formula– Final compound must be neutral based on subscripts and charges • Magnesium Fluoride Mg2+ + F- MgF2 • Ammonium Sulfide NH4+ + S2- (NH4)2S • Tin(II) Carbonate Sn2+ + CO32- SnCO3 • Iron(III) Oxide Fe3+ + O2- Fe2O3 • Iron(II) Oxide Fe2+ + O2- FeO 9 Ionic or Covalent Ionic – two ions or Poly atomic ions Covalent 2 non-metals Or a hydrocarbon Use the prefix to tell how many of each atom there are Mono is never used with the first element Example PBr3 CCL4 P2O5 CO Phosphorous tribromide Carbon tetrachloride diphosphorous pentoxide carbon monoxide Hydrogen up front Most likely an Acid you should have memorized Hydrocarbons Look for how many carbons HCl - Hydrochloric acid HI - Hydroiodic acid HBr - Hydrobromic acid HNO3 Nitric Acid H2SO4 - Sulrufic acid HClO - Hypochlorous acid One – methane CH4 Two – ethane C2H6 Three – propane C3H8 Four – butane C4H10 10 Molecular Compounds • Name to formula – charge does not matter this time, just use the prefixes or memorize • Tetraarsenic hexoxide As4O6 • Sulfur hexafluroide SF6 • Butane C4H10 • Nitric acid HNO3 11 Lewis Dot Structures • Rules 1. Fewest number of atoms goes in the middle or C if present 2. Connect remaining elements with single bonds 3. Make sure all elements have 8 electrons (H only 2) 4. Count the number of electrons in structure 5. Add up valence electrons from PT – Too many e- in structure: remove 2 adjacent pairs fill in with one bond – Too few e- in structure: add to central atom 12 EXAMPLES: • • • • • • CH4 1. (1) C + (4) H (1)(4e-) + (4)(1e-) = 8e2. Spatial order H H C H 3. Draw bonds H 4. Octet rule satisfied? 5. # of e- match? 13 EXAMPLES: • • • CO2 1. (1) C (1)(4e-) 2. Spatial order + + • 3. Draw Bonds • • 4. Octet rule satisfied? 5. # of e- match? (2) O (2)(6e-) = 16e- O C O 14 EXAMPLES: • • NH3 1. • (1) N + (1)(5e-) + 2. Spatial order (3)H (3)(1e-) = • 3. Draw bonds • • 4. Octet rule satisfied? 5. # of e- match? 8e- H N H H 15 EXAMPLES: • • CCl4 1. • (1) C + (4) Cl (1)(4e-) + (4)(7e-) = 32e2. Spatial Order Cl Cl C Cl 3. Draw bonds Cl • • 4. Octet rule satisfied? 5. # of e- match? • 16 EXAMPLES: • • NH4+ 1. (1) N + (4) H • (1)(5e-)+ (4)(1e-) 2. Spatial order - (1)(+) (1)(1e-) = 8eH H N H H [ • 3. Draw bonds • • 4. Octet rule satisfied? 5. # of e- match? + ] 17 EXAMPLES: • • • • • • SO421. (1) S + (4) O + (2)(-) (1)(6e-)+ (4)(6e-) + (2)(1e-) = 32e2. Spatial Order O 2O S O 3. Draw bonds O [ 4. Octet rule satisfied? 5. # of e- match? ] 18 EXAMPLES: • • CN1. (1) C + (1) N + • (1)(4e-) + (1)(5e-)+ 2. Spatial order • 3. Draw Bonds • • 4. Octet rule satisfied? 5. # of e- match? (1)(-) (1)(1e-) = 10e- [C N ] 19 EXAMPLES: • • • CO321. (1) C + (3) O + (2)(-) (1)(4e-)+ (3)(6e-) + (2)(1e-) = 24e2. Spatial Order 2O C O 3. Draw bonds O • • 4. Octet rule satisfied? 5. # of e- match? • [ ] 20 VSEPR: • Regions of electron density (where pairs of electrons are found) can be used to determine the shape of the molecule. • CO2 O C O • Here there are two regions of electron density. • The regions want to be as far apart as possible, so it is linear. 21 EXAMPLES: 4 1 H • CH4 H C H • There are four electron pairs. 2 H • You would expect that the bond angles 3 would be 90° but… • Because the molecule is three-dimensional, the angles are 109.5°. • The molecule is of tetrahedral arrangement. 22 EXAMPLES: • NH3 4 H N H • Four regions of electron density H • But one of the electron pairs is a3 lone pair • The shape is called trigonal pyramidal 1 2 23 EXAMPLES: • • • • 4 1 H2O H O H Four regions of electron density But two are lone pairs 3 This structure is referred to as bent 2 24 EXAMPLES: • CO32- 1 3 [ O C O O 2- ] • Three regions of electron density • This structure is referred to as trigonal planar 2 25 Practice determining molecular shape: • H2S – 4 regions of e- density – 2 lone pairs – bent H S H H S H 26 Practice determining molecular shape: • SO2 – 3 areas of e- density – 1 lone pair – bent O S O O S O 27 Practice determining molecular shape: • CCl4 – 4 areas of e- density – tetrahedral Cl Cl C Cl Cl 3d 28 Practice determining molecular shape: • BF3 – 3 areas of e- density – trigonal planar F B F F F F B F 29 Practice determining molecular shape: • NF3 – 4 areas of e- density – 1 lone pair – pyramidal F N F F 3d 30 16.3 Polar Bonds and Molecules • In covalent bonds, the sharing of electrons can be equal • or it can be unequal. 31 Nonpolar Covalent Bonds • Nonpolar covalent bond - This is a covalent bond in which the electrons are shared equally. • EXAMPLES: • H2 • Br2 • O2 • N2 • Cl2 • I2 Br Cl H N FIO OBr Cl H N FI • F2 32 Polar Bonds and Molecules • If the sharing is unequal, the bond is referred to as a dipole. • A dipole has 2 separated, equal but opposite charges. • “∂” means partial + _ 33 Polar Bonds and Molecules • Polar covalent bond - a covalent bond that has a dipole • It usually occurs when 2 different elements form a covalent bond. • EXAMPLE: • H + Cl H Cl 34 Polar Bonds and Molecules • Electronegativity - This is the measure of the attraction an atom has for a shared pair of electrons in a bond. • Electronegativity values increase across a period and up a group. 35 Examples: • Identify the type of bond for each of the following compounds: • HBr Br = 2.8 H = 2.1 .1 < 0.7< 1.9 Polar Covalent H Br 36 Examples: • NaF F Na • N2 N N = = = = 4.0 0.9 3.1 > 2 3.0 3.0 0.0 Na F Ionic Non-Polar Covalent N N 37 Molecular Polarity • If there is only one bond in the molecule, the bond type and polarity will be the same. • If the molecule consists of more than 2 atoms, you must consider the shape. To determine its polarity, consider the following: – Lone pairs on central atom • If so… it is polar – Spatial arrangement of atoms • Do bonds cancel each other out (symmetrical)? – If so… nonpolar • Do all bonds around the central element have the same difference of electronegativity? – If so… nonpolar 38 Polar Molecules • If the molecule is symmetrical it will be nonpolar. – Exception hydrocarbons are nonpolar • If the molecule is not symmetrical it will be polar, with a different atom or with lone pair(s) Br Cl C Cl Cl 39 Attractions Between Molecules • Van der Waals forces – the weakest of the intermolecular forces. These include London dispersion and dipole-dipole forces. – London dispersion forces – between nonpolar molecules and is caused by movement of electrons 40 Attractions Between Molecules • van der Waals forces(cont.)– Dipole interactions – between polar molecules and is caused by a difference in electronegativity. 41 Attractions Between Molecules • Hydrogen bonds – attractive forces in which hydrogen, covalently bonded to a very electronegative atom (N, O, or F) is also weakly bonded to an unshared (lone) pair of electrons on another electronegative atom. H O H H H O H O H 42 Attractions Between Molecules • Ionic Bonding-occurs between metals and nonmetals when electron are transferred from one atom to another. • These bonds are very strong. 43 Summary of the Strengths of Attractive Forces Ionic bonds hydrogen bonds dipole-dipole attractions LDF 44 Writing and Balancing Chemical Equations Example: Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water. 45 Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water. 1. Write the formulas of all reactants to the left of the arrow and all products to the right of the arrow. Sodium + water sodium hydroxide + hydrogen Translate the equation and be sure the formulas are correct. Na + H2O NaOH + H2 46 Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water. 2. Once the formulas are correctly written, DO NOT change them. Use coefficients (numbers in front of the formulas), to balance the equation. DO NOT CHANGE THE SUBSCRIPTS! _____Na + _____H2O ____NaOH + _____H2 47 3. Begin balancing with an element that occurs only once on each side of the arrow. Ex: Na 2 2 2O ____NaOH 2 _____Na + _____H + _____H2 When you are finished, you should have equal numbers of each element on either side of the equation Na 2 Na 2 H 4 H 4 O 2 O 2 48 4. To determine the number of atoms of a given element in one term of the equation, multiply the coefficient by the subscript of the element. Ex: In the previous equation (below), how many hydrogen atoms are there? 4 2 2 2 ____Na + _____H 2O ____NaOH + _____H2 49 • Balance elements one at a time. • Balance polyatomic ions that appear on both sides of the equation as single units. (Ex: Count sulfate ions, not sulfur and oxygen separately) (NH•4)2Balance SO4 (aq) +HBaCl (aq) BaSO (s) +one 2NHthat and2 O last. Save4 the is in 4Cl (aq) the most places for last… • Use Pencil! 50 Practice: • Balance the equation for the formation of magnesium nitride from its elements. Mg2+ N3Mg3N2 3 ____Mg + ____N2 ____Mg3N2 51 Ex: NH3 + O2 NO2 + H2O • H can be balanced by placing a 2 in front of NH3 and a 3 in front of H2O. Then put a 2 in front of NO2 for nitrogen to balance. 2 _____NH 3 + _____O2 3 2 ____NO 2 + ____H 2O 52 2 7/2 2 _____NH 3 + _____O 2 3 2O ____NO 2 + ____H • Now all that is left to balance is the oxygen. There are 2 O on the reactant side and 7 on the product side. Our only source of oxygen is the O2. Any whole number we place in front of the O2 will result in an even number of atoms. The only way to balance the equation is to use a coefficient of 7/2. 53 Stoichiometry – study of calculations of quantities in chemical reactions using balanced chemical equations. 2Mg + 2 moles Mg O2 + 2MgO 1mole O2 2 moles MgO 54 2Mg + O2 2MgO • The mole ratios can be obtained from the coefficients in the balanced chemical equation. • What are the mole ratios in this problem? • Mole ratios can be used as conversion factors to predict the amount of any reactant or product involved in a reaction if the amount of another reactant and/or product is known. 55 What’s that mean? Well, a stoichiometry problem gives you an amount of one chemical and asks you to solve for a different chemical. To get from one type of chemical to another, a MOLE RATIO must be found between the two chemicals. You get the MOLE RATIO from the BALANCED CHEMICAL EQUATION! 56 A balanced chemical equation tells the quantity of reactants and products as well as what they are. 2 mol 1 mol 2 mol *the coefficients are* 57 The MOLE RATIO is your mechanism of transition between the chemical that is your starting given and the chemical you are solving for. The MOLE RATIO is the bridge between the two different chemicals moles moles (given) given (want) Mole Ratio ? want 58 EXAMPLE • How many grams of KClO3 must decompose to produce KCl and 1.45 moles O2? 2KClO3 1.45 moles O2 → 2KCl + 3O2 2 mol KClO3 122.6 g KClO3 3 mol O2 1 mol KClO3 mol-mol ratio = 119 g KClO3 GFM 59 • CaCO3, limestone, is heated to produce calcium oxide, CaO, and CO2. What mass of limestone is required to produce 156.0 g of CaO? CaCO3 (s) CaO (s) + CO2 (g) 156.0 g CaO GFM 1 mol CaO 1 mol CaCO3 100.1 g CaCO3 56.1 g CaO 1 mol CaO 1 mol CaCO3 mol-mol ratio GFM = 278.4 g CaCo3 60 EXAMPLE • Calculate the number of joules released by the oxidation of 5.00 moles of Na completely react with oxygen gas. ΔH = -416 kJ/mol 4Na + O2 2Na2O 1000 J 5.00 mol Na 2 mol Na2O -416 kJ 4 mol Na 1 mol Na2O 1 kJ mol-mol ratio enthalpy 1.04x106 J 61 Atomic orbital – the region in space where the electron is likely to be found A quantum mechanical model of a hydrogen atom, which has one electron, in its state of lowest energy. The varying density of the spots indicates the relative likelihood of finding the 62 electron in any particular region. Electrons can be described by a series of 4 quantum numbers. You must be familiar with all of these! 63 1. Principle quantum number (n) -describes the principal energy level an electron occupies -values of 1,2,3,4,etc 1 2 3 4 5 6 7 64 2. Azimuthal quantum number (l) -describes the shape of atomic orbitals -s orbitals are spherical, p orbitals are peanut shaped, d orbitals are daisies and f orbitals are fancy -designates a sublevel -values of 0 up to and including n-1 0=s, 1=p, 2=d, 3=f 65 66 67 68 69 Where do I find the orbital shapes? 70 3. magnetic quantum number (ml) -Designates the spatial orientation of an atomic orbital in space -values of -l to +l so s has 1 orbital p has 3 orbitals (x, y, and z) d had 5 orbitals (xy, xz, yz, x2-y2 and z2) f has 7 orbitals 71 4. spin quantum number -describes the orientation of the individual electrons; values of +1/2 and -1/2 -each orbital can hold 2 electrons with opposite spins 72 Orbitals Electrons Symbol Shape s sphere 1 2 p peanut 3 6 d Daisy/ donut 5 10 f fancy 7 14 73 Now, try letsthe try three to do in anyour electron notes. configuration for carbon. Begin He 1s2with the first quantum number and use the periodic table to write the configuration. Na 1s2 2s2 2p6 3s1 Ti 1s2 2s2 2p6 3s2 3p6 4s2 3d2 Quick and Easy Electron Configuration 1 2 3 4 5 6 7 3 4 5 6 4 5 74 For a shorter way to write electron configuration, write the nearest noble gas and then continue. AKA “Shorthand Notation”. Ex: Ti can be written as OR 1s2 2s2 2p6 3s2 3p6 4s2 3d2 [Ar] 4s2 3d2 75 Electron Configuration of Ions • For the loss of an electron remove electrons from the last orbital Ge 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2 Ge2+ 1s2 2s2 2p6 3s2 3p6 4s2 3d10 For the gain of electrons add them in the last orbital filled At [Xe] 6s2 4f14 5d10 6p5 At- [Xe] 6s2 4f14 5d10 6p6 76 Increasing Atomic Radius Atomic Radius Increasing Atomic Radius 77 • Ion Size • Anions are larger than the atoms from which they were formed. • The negative charge means more electrons are present causing the size of the ion to be larger. • Cations are smaller than the atoms from which they were formed. • The positive charge means fewer electrons are surrounding the nucleus, thus pulling the existing electrons closer and causing the ion to be smaller. • Trend: ionic radius increases from right to left and top to bottom 78 Increasing Ionic Radius Ionic Radius Increasing Ionic Radius 79 • Ionization energy - energy required to overcome the attraction of the nuclear charge and remove an electron from a gaseous atom • 1st ionization energy: the energy required to remove the first electron • 2nd ionization energy: the energy required to remove the second electron • 3rd ionization energy: the energy required removing the third electron • Trend: ionization energy increases from left to right and bottom to top 80 Increasing Ionization Energy Increasing Ionization Energy Ionization Energy 81 • Electronegativity – the tendency for the atoms of the element to attract electrons when they are chemically combined with atoms of another element • Note: Noble gases don’t have values for electronegativity because their outer orbitals are full and they do not need to gain or lose electrons to be stable. • Trend: electronegativity increases from left to right and bottom to top 82 Increasing Electronegativity Increasing Electronegativity Electronegativity 83 We can measure not only the length of each wave of light, but also its frequency of occurrence. Parts of a wave: Amplitude amplitude- height of the wave from the origin to the crest wavelength - - distance between the crests frequency - - the number of wave cycles to pass a given point per unit of time. The units of frequency are 1/s, s-1, or Hertz (Hz) = c/ where c = speed of light c= 3.00 x 108 m/s As increases, decreases. Ex. A certain wavelength of yellow 16 -1 light has a frequency of 2.73 x 10 s . Calculate its wavelength. = c/ = c/ = 3.00 x 108m/s 2.73 x 1016s-1 = 1.10 x 10-8 m Spectrum- series of colors produced when sunlight is separated by a diffraction gradient. ROY G. BIV Red: has the longest wavelength, lowest frequency Lowest energy Violet: has the shortest wavelength, highest frequency highest energy Kinetic theory- The tiny particles in all forms of matter are in constant motion. 1. A gas is composed of particles, usually molecules or atoms. We treat them as, Hard spheres, Insignificant volume, and Far from each other 2. The particles in a gas move rapidly in constant random motion. 3. All collisions are perfectly elastic. 92 Gas Laws • • • • One single set of conditions PV = nRT Two Sets of conditions P1 V1 = P2 V2 T1 T2 93 Ex. If a helium-filled balloon has a volume of 3.40 L at 25.0oC and 120.0 kPa, what is its volume at STP = 0°C, 1 ATM? V1 = 3.40L V2 = ? P1 = 120.0 kPa P2 = 1 atm = 101.3 kPa T1 = 25.0oC + 273 = 298K T2 = 0oC + 273 = 273K P1V1 = P2V2 (120.0 kPa)(3.40L) = (101.3 kPa)V2 T1 T2 298K 273K V2 = 3.69L 94 Ex. A 5.0 L flask contains 0.60 g O2 at a temperature of 22oC. What is the pressure (in atm) inside the flask? P=? V= 5.0L R = 0.0821 (L.atm/mol.K) T = 22°C +273= 295K n = 0.60g O2 1 mol O2 = 0.01875 mol 32.0g O2 PV = nRT P(5.0L) = (0.01875mol)(0.0821Latm/molK)(295K) P = 0.091 atm 95 Factors affecting solubility: • The nature of the solute and the solvent – “like dissolves like” • Miscible - liquids that are soluble in each other – Ex. ethanol and water • Immiscible- liquids that are not soluble in each other – Ex. oil and water Molarity(M) = moles of solute liters of solution • A 1 M solution contains 1 mole of solute per 1 L of solution. A 0.5 M NaCl solution has 0.5 mol NaCl in 1 L total solution. EXAMPLES: • What is the concentration in molarity of a solution made with 1.25 mol NaOH in 4.0 L of # mol solution? M= M=? #mol = 1.25 mol V = 4.0 L M = 1.25 mol 4.0 L #L = 0.3125 M = 0.31 M Exothermic reactions have - H Endothermic reactions have + H 100 Enthalpy (H)the amount of heat in a system at a given temperature Enthalpy change: H =q=mCT 101 Ex. A 25 g sample of a metal at 75.0oC is placed in a calorimeter containing 25 g of H2O at 20.0oC. The temperature stopped changing at 29.4oC. What is the specific heat of the metal? 103 Standard Heat of Formation of a compound ( Hfo) * Hfo of a free element in its standard state is zero. H = Hfo products - Hforeactants 104 Ex. 1. Calculate H for the following reaction: (endo or exo thermic?) CaCO3(s) CaO(s) + CO2(g) ∆Hfo values: CaCO3 = -1207.0 kJ/mol CaO = -635.1 kJ/mol CO2 = -393.5 kJ/mol ∆H = [-635.1 + (-393.5)] – [-1207.0] ∆H = 178.4 kJ endothermic, heat going in 105 2. Calculate the heat of reaction for the following reaction: (endo or exo thermic?) 2H2(g) + O2(g) 2H2O(g) ∆Hfo values: H2O(g) = -241.8 kJ/mol ∆H = [2(-241.8)] – [2(0) + 0] ∆H = -483.6 kJ Remember to multiply heat values by coefficients!!! Exothermic Heat leaving 106 Entropy Calculation Example: Mn(s) + 2O2(g) MnO4(s) S° J mol-1 K-1 Mn (s) = 32.8 O2 = 205.0 MnO4 = 120.5 ∆S = [S° Products] – [S° Reactants] ∆S = [120.5] – [2(205.0) + 32.8] ∆S = -322.3 J/mol K 107 Entropy, S - a measure of randomness or disorder • associated with probability (There are more ways for something to be disorganized than organized.) • Entropy increases going from a solid to a liquid to a gas. • Entropy increases when solutions are formed. • Entropy increases in a reaction when more atoms or molecules are formed. • The entropy of a substance increases with temperature. 108 Gibbs free energy, G • Energy available to do work • Go = standard free energy change – change in free energy that occurs if the reactants in their standard states are converted to products in their standard states Go =Ho -T So When Go is negative the reaction is spontaneous is the forward direction When Go is positive the reaction is nonspontaneous is the forward direction 109 • What temperature would a reaction be spontaneous if ΔH = 9500 J/mol and ΔS = 6.5J/mol K? Go =Ho -T So 9500 = T(6.5) T = 1461 K Above 1461 K this reaction will be spontaneous and below it will not 110 • Will this reaction be spontaneous at 100°C? ΔH = -18 KJ/mol and ΔS = 94.3 J/mol K Go =Ho -T So Yes at all temperatures Go =(Ho) -T (So) Go =(-) - (+)(+) Go = can only be negative Don’t believe me try putting in the values and craze high/low number for temperature 111 Go =Ho -T So • A spontaneous reaction has a negative G. For example, when ice melts H is positive (endothermic), S is positive and G = 0 at 0oC. • If... Entropy, ΔS Enthalpy, ΔH Spontaneity Positive Positive Yes at high temp Negative Positive Never spontaneous Positive Negative Always spontaneous Negative Negative Yes at low temp 112 Collision Theory: • In order to react, two or more particles must collide with sufficient energy (called the activation energy) and with the proper molecular orientation. If the colliding particles do not have either of these two prerequisites, no product is formed. 113 Factors that affect reaction rate: • Temperature- Reactions go faster at higher temperatures. Particles have more kinetic energy. More colliding particles have enough energy to overcome the activation energy barrier. 114 Factors that affect reaction rate: • Concentration- Increasing the concentration of reactants usually increases the reaction rate. If there are more particles to collide, there should be a greater number of collisions that produce products. 115 Factors that affect reaction rate: • Catalysts- A catalyst is a substance that speeds up a reaction by lowering the activation energy barrier. It is not a product or reactant and it is not used up or changed itself. 116 Rate Law • equation that is written that expresses how the reaction rate of a particular reaction is dependent upon the concentrations of its reactants. • For the reaction aA + bB cC + dD, the general form of the rate law would be: Rate = k [A]a[B]b 117 Rate = k x y [A] [B] • Rate is usually expressed as mol/L∙time. • k is the specific rate constant. It is constant for a given reaction at a given temperature. The faster a reaction, the larger the k value. • [A] and [B] represent the concentrations of reactants A and B in moles per liter (M). • x and y are the order of the reactant. They can only be determined by analyzing experimental data. These exponents are usually positive integers. 118 EXAMPLES 2A + B 2C Ex. [A] [B] Rate 1 0.1 0.2 0.10 2 0.1 0.4 0.20 3 0.2 0.4 0.80 Determine the rate law: 3 [A] x 0.1 0.1 ( 0.2) = ( 0.8) [B] x 0.1 0.2 ( 0.4) = 0.2 23 =8 1 rate = k [A] [B] 119 Half-life Based on how much time does it take ½ of the substance to change into products x = time / half-life (number of half-lives) mf = (mi)(1/2)x 120 If you start with 2.00 g of nitrogen-13 how many grams will remain after 4 half lives? mf = 2.00 (1/2)4 mf =0.125 g Phosphorous-32 has a half-life of 14.3 yr. How many grams remain after 57.2 yr from a 4.0 g sample mf = 4.00 (1/2)(57.2/14.3) mf = 0.25 g 121 BrØnstedLowry • Acid: hydrogen-ion donor (proton donor) • Base: hydrogen-ion acceptor (proton acceptor) NH3 + H2O NH4+ + OHbase acid BrØnstedLowry • Acid: hydrogen-ion donor (proton donor) • Base: hydrogen-ion acceptor (proton acceptor) NH3 + H2O NH4+ + OHbase acid Working Ka and Kb problems • 1st: Write the equation • 2nd: Set up a reaction diagram (RICE diagram) • 3rd: Set up Ka or Kb expression • 4th: Substitute values into Ka expression • 5th: Solve Ka expression for X. • 6th: Calculate pH from H+ or OHconcentration. R = reaction I = initial C = change E = equilibrium Example: Calculate the pH of a 0.10 M solution of acetic acid. The Ka for acetic acid is 1.8 x 10-5. R HC2H3O2 H+ + C2H3O2I .1 0 0 -x +x +x C E .1- x x x Ka = [H+] [C2H3O2-] [HC2H3O2] [x] [x] = [0.1-x] = 1.8 x 10-5 [x] [x] -5 1.8 x 10 = [.1-x] x2 = 1.8 x 10-5 .1-x x2 -5 1.8 x 10 = .1 x2 = 1.8 x 10-6 x = 1.3 x 10-3 = [H+] pH = -log [H+] pH = -log(1.3 x 10-3) pH= 2.87 Example: Calculate the pH of a 0.25 M solution of HCN. The Ka for HCN is 6.2 x 10-10. R HCN H+ + CNI .25 0 0 C -x +x +x E .25 - x x x +][CN-] [H [x][x] -10 = Ka = 6.2 x 10 = [HCN] .25 – x Ka = 6.2 x 10-10 = [H+][CN-] = [x][x] [HCN] .25 – x X2 = 6.2 x 10-10 .25 X2 = 1.55 x 10-10 x = 1.24 x 10-5 = [H+] pH = -log [H+] pH = -log(1.24 x 10-5) pH= 4.9 • Calculate the pH of a 0.15 M solution of ammonia. The Ka of ammonia is 1.8 x 10-9. R I C E NH3 + H2O (l) NH4+ + OH.15 0 0 -x +x +x .15 –x Ka = [NH4+] [OH-] [NH3] x [x][x] = .15 – x x Ka = [NH4+] [OH-] = [x][x] [NH3] .15 – x pH = -log (H+) X2 = 1.8 x 10-9 .15 X2 = 2.7 x 10-10 X = 0.000016= [H+] pH = -log (0.000016) pH = 4.78 LeChatelier's Principle • When a stress is applied to a system, the equilibrium will shift in the direction that will relieve the stress. Henry Louis Le Chatelier Changes in concentration • An increase in concentration of: – a reactant will cause equilibrium to shift to the right to form more products. – a product will cause equilibrium to shift to the left to form more reactants. • A decrease in concentration of: – a product will cause equilibrium to shift to the right to form more products. – a reactant will cause equilibrium to shift to the left to make more reactants. A+BC+D • • • • Remove A or B Add C or D Remove C or D Add A or B ← ← → → Changes in temperature • Treat energy as a product or reactant and temperature changes work just like changes in concentration! • An increase in temperature of an exothermic reaction (H is negative) will cause equilibrium to shift to the left. A decrease in temperature of an exothermic reaction will cause equilibrium to shift to the right. • An increase in temperature of an endothermic reaction (H is positive) will cause equilibrium to shift to the right. A decrease in temperature of an endothermic reaction will cause equilibrium to shift to the left. Example: 181 kJ + N2 + O2 2NO H = 181 kJ (endothermic) • addition of heat • lower temperature Example: 2SO2 + O2 2SO3 H= -198 kJ (exothermic) • increase temperature • remove heat + 181 kJ Example: CaCO3 + 556 kJ CaO + CO2 • decrease temperature Changes in pressure • Changes in pressure only affect equilibrium systems having gaseous products and/or reactants. • Increasing the pressure of a gaseous system will cause equilibrium to shift to the side with fewer gas particles. • Decreasing the pressure of a gaseous system will cause equilibrium to shift to the side with more gas particles. Addition of a catalyst • Adding a catalyst does not affect equilibrium. Catalysts speed up the forward and reverse reactions equally. Example: P4(s) + 6Cl2(g) 4PCl3(l) • increase container volume – Shifts to side with more gas • decrease container volume – Shifts to side with less gass • add a catalyst N/C – Inert gases have no effect on equilibrium Example: Consider the reaction: 2NO2(g) N2(g) + 2O2(g) + HEAT which is exothermic • NO2 is added • N2 is removed • The volume is halved N/C • He (g) is added • The temperature is increased N/C • A catalyst is added Mechanisms • Uni molecular A→B AB → A + B A2 → 2A • Bi molecular A + B → AB AB + C → AC + B Only one thing changing at a time 142 Overall reactions NO2 + F2 NO2F + F (slow step) NO2 + F NO2F (fast step) Simplify F on both sides 2NO2 + F2 NO2F Intermediate F Rate = k[NO2]1[F2]1 Rate depends on both NO2 and F2 143 Overall reactions PO2 + Cl PClO2 (fast step) PClO2 + PO2 P2O4 + Cl (slow step) Simplify Cl and PClO2 on both sides 2PO2 P2O4 Intermediate PClO2 Catalyst Cl Rate = k[PO2]2 Rate depends on only PO2 144 Oxidation Reduction Reactions • Oxidation-reduction reactions- chemical changes that occur when electrons are transferred between reactants. • Also called REDOX reactions Oxidation Reduction Reactions • • • • • • Oxidation Modern definition - loss of electrons Examples 4Fe + 3O2 2Fe2O3 (rusting of iron) C + O2 CO2 (burning of carbon) C2H5OH + 3O2 2CO2 + 3H2O (burning of ethanol) To help remember these definitions, use one of these mnemonic devices: Oxidation Is Loss Reduction Is Gain LEO (Lose Electrons-Oxidation) the lion goes GER (Gain Electrons-Reduction) Formation of Ions • Ex. 2Na + S Na2S • Sodium goes from the neutral atom to the 1+ ion. Therefore, it has lost an electron (It was oxidized). Sulfur goes from the neutral atom to the 2- ion. Therefore, it has gained two electrons. (It was reduced) Question • Lead loses four electrons. It take on a charge of ___. Does this mean that it is oxidized or reduced? REDOX Reaction Examples • Identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent for each of the following: +4 -2 +1 -1 +2 -1 0 +1 MnO2 + 4HCl MnCl2 + Cl2 + 2H2O Mn +4 +2 Gained 2 e- -2 Reduced If it was Reduced, then the reactant that contains Mn acts as the “Oxidizing Agent”. (MnO2) Cl -1 0 Lost 1 e- Oxidized If it was Oxidized, then the reactant that contains Cl acts as the “Reducing Agent”. (HCl) Batteries • Electrochemical cells that convert chemical energy into electrical energy are called voltaic cells. • The energy is produced by spontaneous redox reactions. • Voltaic cells can be separated into two half cells. • A half cell consists of a metal rod or strip immersed in a solution of its ions. Batteries • We write half reactions to show what happens in each part of the cell. – Example Write the half reactions that occur in the Fe2+/Ni2+ cell. – Oxidation Fe Fe2+ + 2e– Reduction Ni2+ + 2e- Ni Diagram of voltaic cell for the reaction of zinc and copper. • Diagram of voltaic cell for the reaction of zinc and copper. • Oxidized: Zn Zn2+ + 2e• Reduced: Cu2+ + 2e- Cu Direction of electron flow From anode to cathode anode cathode Solution of Zinc ions Solution of Copper ions Half-Cells • The half cells are connected by a salt bridge. A salt bridge is a tube containing a solution of ions. • Ions pass through the salt bridge to keep the charges balanced. • Electrons pass through an external wire. • The metal rods in voltaic cells are called electrodes. • Oxidation occurs at the anode and reduction occurs at the cathode. (An Ox and Red Cat) • The direction of electron flow is from the anode to the cathode. (FAT CAT ) Calculating the Charge of a Battery • The potential charge of a battery can be calculated with a set of values from a table of reduction potentials. – To do this, write the oxidation and reduction half reactions. – Look up the cell potentials from the data table. – Flip the sign of the cell potential for oxidation. – Add the potentials together. Example A common battery is made with nickel and cadmium. What is the cell potential of this battery? (E0Cd = -0.40V, E0Ni = -0.25V) Oxidation Reduction Cd Cd2+ + 2eE0 = 0.40V Ni2+ + 2e- Ni + E0 = -0.25V Total = E0 = 0.15V Spontaneous Electrochemical Reactions ΔG = -nFE Recall the NiCd battery Total = E0 = 0.15V Positive Voltage, -ΔG, spontaneous reaction Negative Voltage, +ΔG, nonspontaneous reaction 157 The word, polymer, implies that polymers are constructed from pieces (monomers) that can be easily connected into long chains (polymer). When you look at the above shapes, your mind should see that they could easily fit together. There are two types of polyethylene polymers (plastics). One is when the polyethylene exists as long straight chains. The picture here shows the chains of one carbon with two hydrogen atoms repeating. The chain can be as long as 20,000 carbons to 35,000 carbons. This is called high density polyethylene (HDPE). Low density polyethylene (LDPE) is made by causing the long chains of ethylene to branch. That way they cannot lie next each other, which reduces the density and strength of the polyethylene. This makes the plastic lighter and more flexible.