* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Click chemistry wikipedia , lookup
Electrolysis of water wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Molecular orbital diagram wikipedia , lookup
Atomic orbital wikipedia , lookup
Acid–base reaction wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
Hydrogen-bond catalysis wikipedia , lookup
Process chemistry wikipedia , lookup
Oxidation state wikipedia , lookup
Bond valence method wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Isotopic labeling wikipedia , lookup
Periodic table wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Chemical element wikipedia , lookup
Metallic bonding wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Resonance (chemistry) wikipedia , lookup
Lewis acid catalysis wikipedia , lookup
Electrochemistry wikipedia , lookup
Chemical reaction wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Biochemistry wikipedia , lookup
Transition state theory wikipedia , lookup
Rate equation wikipedia , lookup
Molecular dynamics wikipedia , lookup
Photosynthetic reaction centre wikipedia , lookup
Electronegativity wikipedia , lookup
History of chemistry wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Electron configuration wikipedia , lookup
Bioorthogonal chemistry wikipedia , lookup
Hypervalent molecule wikipedia , lookup
Atomic nucleus wikipedia , lookup
Homoaromaticity wikipedia , lookup
Chemistry: A Volatile History wikipedia , lookup
Strychnine total synthesis wikipedia , lookup
Metalloprotein wikipedia , lookup
Organosulfur compounds wikipedia , lookup
Extended periodic table wikipedia , lookup
Chemical bond wikipedia , lookup
History of molecular theory wikipedia , lookup
IUPAC nomenclature of inorganic chemistry 2005 wikipedia , lookup
Matter & Measurement EIT Review F2007 Dr. J.A. Mack Matter is the physical material of the universe. Matter is anything that occupies space (volume) and has mass. www.csus.edu/indiv/m/mackj/ Matter is made up of relatively few elements. Matter consists of atoms and molecules. Part 1 Each element is a unique atom. Atoms combine to form molecules 1 3 5 7 Classifying Matter: 1 • Physical Properties can be determined without changing the chemical make–up of the sample. STATES OF MATTER • SOLIDS — have rigid shape, fixed volume. • Examples of physical properties are: External shape can reflect the atomic and molecular arrangement. –Melting Point,Boiling Point, Density, Mass, Touch, Taste, Temperature, Size, Color, Hardness, Conductivity. • LIQUIDS — have no fixed shape and may not fill a container completely. • Examples of physical changes are: • GASES — expand to fill their container. –Melting, Freezing, Boiling, Condensation, Evaporation, Dissolving, Stretching, Bending, Breaking 8 • Physical Properties can be determined without changing the chemical make–up of the sample. 9 Chemical Properties are those that do change the chemical make–up of the sample. • Examples of physical properties are: –Melting Point,Boiling Point, Density, Mass, Touch, Taste, Temperature, Size, Color, Hardness, Conductivity. Examples of chemical properties are: Burning, Cooking, Rusting, Color change, Souring of milk, Ripening of fruit, Browning of Apples, • Examples of physical changes are: photography, Digesting food –Melting, Freezing, Boiling, Condensation, Evaporation, Dissolving, Stretching, Bending, Breaking 10 Note: Chemical properties are actually chemical changes. 13 2 Units of Measure: Temperature: English Customary Weights and Measures The International System of Units (SI) distance (phase change) meter inch Temperature is a measure of particle motion foot = 12 inches micrometer = 10−6 meters yard = 3 feet centimeter = 10−2 meters mile = 5280 feet kilometer = 103 meters (all motion stops) Units with in a system can be represented by units within the system. Units within the metric system are related by powers of 10 14 15 What are Significant Figures? Significant figures communicate the uncertainty in a measurement. Given the mass: 45. 8724g With the uncertainty: ± 0.001g Which numbers are certain ? (significant) You need to memorize these values and have the ability to convert between. 16 17 3 The error (±) tells us which digit is uncertain: Counting Significant Figures 45. 8724g 1. All non zero numbers are significant ± 0.001 2. All zeros between non zero numbers are significant The uncertainty occurs at the thousands place: 3. Leading zeros are NEVER significant. (Leading zeros are the zeros to the left of your first non zero number) Therefore, 4. Trailing zeros are significant ONLY if a decimal point is part of the number. (Trailing zeros are the zeros to the right of your last non zero number). 45. 8724 These digits are certain The blue digits are significant, This digit is uncertain or we say there are 5 significant figures 18 Determine the number of Sig. Figs. in the following numbers zeros written behind the decimal are significant… 1256 4 sf 1056007 7 sf 0.000345 3 sf 0.00046909 5 sf 1780 3 sf 770.0 4 sf 0.08040 4 sf 19 Scientific notation: Now count each step as a power of ten to find the exponent. not trapped by a decimal place. 563,490. Five steps: the exponent is 5 5.63490 × 105 20 21 4 Multiplication and Division: Sig. Figures in Calculations Determine the correct number of sig. figs. in the following calculation, express the answer in scientific notation. Multiplication/Division The number of significant figures in the answer is limited by the factor with the smallest number of significant figures. 4 sf 2 sf 4 sf 23.50 ÷ 0.2001 × 17 Addition/Subtraction The number of significant figures in the answer is limited by the least precise number (the number with its last digit at the highest place value). NOTE: Defined numbers (numbers from tables and references) never limit calculations. The sf in the result is limited to the number with the least amount of sf. The answer must be rounded to 2 sf. 22 Sig. Figs. Addition and Subtraction 23.50 ÷ 0.2001 × 17 from the calculator: 1996.501749 How many sig. figs. are allowed in the following calculation? 10 sf 391 - 12.6 +156.1456 Your calculator knows nothing of sig. figs. !!! in sci. not.: 1.996501749 x 103 Rounding to 2 sf: 2.0 x 103 23 To determine the correct decimal to round to, align the numbers at the decimal place: 391 -12.6 +156.1456 no digits here One must round the calculation to the least significant decimal. 24 25 5 391 -12.6 +156.1456 534.5456 one must round to here Combined Operations: When there are both addition / subtraction and multiplication / division operations, the correct number of sf must be determined by examination of each step. (answer from calculator) Example: Complete the following math mathematical operation and report the value with the correct # of sig. figs. round to here (units place) Answer: 535 (26.05 + 32.1) ÷ (0.0032 + 7.7) = ??? 26 27 3 sf (26.05 + 32.1) ÷ (0.0032 + 7.7) = ??? 26.05 (26.05 + 32.1) = (0.0032 + 7.7) 1st determine the correct # of sf in the numerator (top) (26.05 + 32.1) = (0.0032 + 7.7) 2nd determine the correct # of sf in the denominator (bottom) The result will be limited by the least # of sf (division rule) 28 + 32.1 0.0032 + 7.7 2 sf The result may only have 2 sf 29 6 (26.05 + 32.1) = (0.0032 + 7.7) The Representation of Matter: 58.150 In chemistry we use chemical formulas and symbols to represent matter. 7.70032 = 7.5516 Why? We are “macroscopic”: large in size on the order of 100’s of cm = 7.6 2 sf Atoms and molecules are “microscopic”: on the order of 10-12 cm 30 Where do we begin… 31 The modern periodic table is defined by: The Periodic Table Groups (families) (columns down) Periods (rows across) Dmitri Mendeleev (1834 - 1907) 32 33 7 Which elements do we need to know? Chemical Symbols and Formula: Elements: Start here as a minimum… H = hydrogen O = oxygen C = carbon Molecules: H2 = hydrogen O2 = oxygen H2O = water UhUh-Oh! this is confusing… As many as possible! Yes it is… Get over it and get used to it! CO2 = carbon dioxide 34 1 35 The Groups are labeled: The Periods are labeled: 1A 7A 8A 2A 2 3 3A 4A 5A 6A 3B 4B 5B 6B 7B The “A” refers to the “main group elements” 36 8B 1B 2B The “B” refers to the transition metal elements. 37 8 Periodic Table: Metallic arrangement 1 IA 1 18 VIIIA 2 IIA 13 IIIA 14 IVA 2 3 4 5 Periodic Table: Metallic arrangement 3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 9 VIIIB 10 11 IB 12 IIB Metals 15 VA 16 VIA 17 VIIA 1 IA 1 s al et m on N 18 VIIIA 2 IIA 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA 2 3 3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 9 VIIIB 10 11 IB 12 IIB 4 5 6 6 7 7 somewhere in between metals and non-metals Metalloids 38 Atomic Number, Z Atom: The smallest divisible unit of an element Compound: A substance made of two or more atoms An element’s identity is defined by the number of protons in the nucleus: Z 13 Al 26.981 39 Ion: A charged atom or molecule Cation: Positive ion Anion: Negative ion Atomic number NOMENCLATURE Format for naming chemical compounds using prefixes, suffixes, and other modifications of the names of elements which constitute compounds. Atom symbol Atomic weight 40 41 9 Ion Charges: inert or “noble” gasses: +1 +3 -3 -2 -1 +2 Variable Metals form Cations non-metals form anions Metalloids can do both 42 43 Binary Compounds: Metal & non-Metal Compounds fall into one of two classes: Inorganic Salts Molecules metal cation non-metal + + non-metal or polyatomic anion non-metal (no cations or anions) Metal of fixed oxidation (charge) state combined with a non-metal. non-metal takes on “ide” suffix Examples: The two use different formalisms for naming… 44 Cation Anion Formula Name K+ Cl− KCl Potassium chloride Ca2+ O2- CaO Calcium Oxide Na+ S2- Na2S Sodium sulfide Al3+ S2- Al2S3 Aluminum sulfide 45 10 Metals of variable charge (transition) with a non-metal Examples: Cation Pb2+ Some common polyatomic ions: modify transition metal name with roman numeral Anion Formula Name Cl− PbCl2 lead (II) chloride pronounced: lead - two - chloride Pb4+ Cl− PbCl4 lead (IV) chloride Fe3+ O2− Fe2O3 Iron (III) oxide NH4+ H3O+ CO32– HCO32– NO2– NO3– SO42– SO32– PO43– C2H3O2– ammonium hydronium carbonate hydrogencarbonate or bicarbonate nitrite nitrate sulfate sulfite phosphate acetate 46 Ternary Compounds: Those with three different elements Type A: metal of fixed charge with a complex ion Cation Anion Formula 47 Metal of variable charge transition) with a complex ion Cation Anion Formula Name Fe3+ NO3− Fe(NO3)3 Iron (III) nitrate Fe2+ NO −2 Fe(NO2)2 Iron (II) nitrite Name K+ OH− KOH Potassium hydroxide Ca2+ OH− Ca(OH)2 Calcium hydroxide Na+ SO 24− Na2SO4 Sodium sulfate Al3+ SO 24− Al2(SO4)2 Aluminum sulfate 48 49 11 Examples: Formula NonNon-metal with a nonnon-metal When non-metals combine, they form molecules. They may do so in multiple forms: CO CO2 Because of this we need to specify the number of each atom by way of a prefix. BCl3 boron trichloride SO3 sulfur trioxide NO nitrogen monoxide we don’t write: 1 = mono 5 = penta 2 = di 3 = tri 6 = hexa 4 = tetra N2O4 7 = hepta Name: nitrogen monooxide or mononitrogen monoxide dinitrogen tetraoxide 50 51 Type I Acids: Acids derived from –ide anions. D) Writing formulas for acids and Bases •An acid is a substance that produces H+ when dissolved in water. •Certain gaseous molecules become acids when dissolved in water. •A base produces OH− when dissolved in water. •Bases often are Group I and Group II hydroxide salts. 52 The names for these acids follows the formula: “hydro” + the root of the ide anion + ic “acid” Anion: Acid: Name: chloride HCl hydrochloric acid fluoride HF hydrofluoric acid 53 12 Examples: H+ and S2- Anion: it takes 2 H+ to cancel one S2- Acid: Name: (nitrate) NO3− HNO3 nitric acid (sulfate) SO24 − H2SO4 sulfuric acid (acetate) C2 H 3O2− HC2H3O2 acetic acid vinegar H2S hydro sulfuric acid 54 55 Practice: CO2 carbon dioxide BaCO3 carbon monoxide CO FeO Iron (II) oxide P2O5 diphosphorous pentaoxide zinc phosphide Zn3P2 nitrogen trihydride NH3 NiBr2 nickel (II) bromide N2SO4 sodium sulfate barium carbonate (ammonia) 56 57 13 Common Names: H2O water ammonia NH3 please go to my chem. 1A web site and download a nomenclature practice worksheet if you need more review. CH4 methane http://www.csus.edu/indiv/m/mackj/chem1a/chem1A_lab.html NO nitric oxide The link to the worksheet is at the top of the page. N2O nitrous oxide 58 59 The Composition of an Atom: Modern Atomic Theory: •Atoms are made of protons, neutrons and electrons. •The nucleus of the atom carries most of the mass. The atom is mostly empty space • It consists of the protons and neutrons surrounded by a cloud of electrons. The charge on the electron is –1 The charge on the proton is +1 •protons and neutrons in the nucleus. There is no charge on the neutron •the number of electrons is equal to the number of protons. The Atomic Number or number of protons in the nucleus defines an element. •electrons in space around the nucleus. 60 61 14 Isotopes, Atomic Numbers, and Mass Numbers •Atomic number (Z) = number of protons in the nucleus. •Mass number (A) = total number of nucleons in the nucleus (i.e., protons and neutrons). 75 protons + neutrons •One nucleon has a mass of 1 amu 75 34 Se (Atomic Atomic Mass Unit) Dalton” Unit a.k.a “Dalton 34 protons •Isotopes have the same Z but different A. •The elements are arranged by Z on the periodic table. protons = 34 By convention, for element X, we write A Z X electrons = 34 neutrons = 75-34 = 41 62 63 Molar Masses Avagadro’s Number Since we can equate mass (how much matter) with moles (how many particles) we now have a conversion that relates the two. Since one mole of 12C has a mass of 12g (exactly), 12g of 12C contains 6.022142 x 1023 12C-atoms. But carbon exists as 3 isotopes: C-12, C-13 &C-14 mols The average atomic mass of carbon is 12.011 u. molar mass (g/mol) = grams The Molar Mass of a substance is the amount of matter that contains one-mole or 6.022 × 1023 particles. From this we conclude that 12.011g of carbon contains 6.022142 x 1023 C-atoms Is this a valid assumption? × factor aka: Avogadro's number (NA) Yes, since NA is so large, the statistics hold. The atomic masses on the Periodic Table also represent the molar masses of each element in grams per mole (g/mol) 64 65 15 Molar Masses (Molecular Weights) of Compounds: So if you have 12.011g of carbon… you have 6.022×1023 carbon atoms! The molar mass of a molecular compound is the sum of the molar masses of its atoms. So if you have 39.95g of argon… you have 6.022×1023 argon atoms! Example: The molar mass of CO2 is: if you have a mole of dollar bills… you are Bill Gates… you have 6.022×1023 bucks! 1 x (12.01 g/mol) + 2 x (16.00 g/mol) and if you have 6.022×1023 avocados… = 44.01 g/mol a “guacamole” you have… 66 How many oxygen atoms are there in 25.1g of chromium (III) acetate? Cr(C2H3O2)3 step 2: calculate the molar mass… Question: What is the weight % of each element in C2H6? 229.13 g/mol First determine the molar mass of C2H6: 1 mol of C 2 H 6 has a mass of 30.07 g step 3: use dimensional analysis to solve the problem… 25.1g Cr(C 2 H3O 2 )3 × 1mol Cr(C 2 H3O 2 )3 229.13g Cr(C2 H 3O 2 )3 × 6.022 × 1023 O − atoms = 1mol O 3.96 × 1023 Percent Composition: The relative amounts of each atom in a molecule or compound can be represented fraction of the whole. step 1: write the correct chemical formula… Cr3+ & C2H3O2− 68 (2×12.01 + 6 ×1.008) g/mol Next determine the mass of hydrogen in 1 mol of the compound: 6mol O × 1mol Cr(C 2 H 3O 2 )3 1 mol C2 H 6 × O-atoms 69 6 mol H 1.0079 g H × = 6.047 g H 1 mol C2 H 6 1 mol H 70 16 Determining a Formula from Percent Composition: Now relate the mass of H in one mol of the compound to the molar mass of the compound Given the relative percentages of each element in a compound, 1 × 100 = 20.11% H 6.047 g H × 30.07g C2 H 6 10 % X, Since there is only C as the remaining element: % C = 100 − % H 20 % Y, 30 % Z etc… one can find the empirical formula of the compound. = 79.89 % C The empirical formula of a compound or molecule represents the simplest ratio of each element in 1 mol of the compound or molecule. The compound C2H6 is 20.11% H & 79.89% C 71 Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? Solution: Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound? 2. Convert the grams of each element to moles. determine X, Y & Z in (CXHYOZ) (g element X → mole X etc…) 1. Since the percentages for each element sum to 100%, if one equates % to grams (g), the sum of the masses must be 100g. (i.e. one can assume 100g of the compound) 64.82 g C 21.59 g O 72 13.59 g H 73 64.82g C × 1 mol C 12.011 g C = 5.397 mol C 21.59g O × 1 mol O 16.00 g O = 1.349 mol O 13.59g H × 1 mol H = 13.48 mol H 1.0079 g H 74 17 3. Divide each of the individual moles by the smallest number of moles to gain the molar ratios for each element in the compound. These are the formula subscripts. (X2Y3 etc…) Subscript for C Subscript for H **If the ratios are fractional (0.5, 1.5 or 0.333) multiply each ratio by a whole number to get even number formula subscripts. 0.5 × 2 = 1 X = 4.001 = 4 Y = 9.992 =10 Z = 1.000 = 1 Subscript for O 13.48 1.349 5.397 = 9.992 Z = = 1.000 X= = 4.001 Y = 1.349 1.349 1.349 Examples: Rounding to the nearest whole numbers: 0.25 × 4 = 1 C4H10O The empirical formula is: The results of this calculation tells us only about the empirical formula of the compound. To determine the molecular formula, we need more information. This will be shown in a later example. 0.333 × 3 = 1 75 76 Balancing Chemical Reactions: Example Chemical Equations: C2H6 + 2 C’s & 6 H’s Mass is conserved in a chemical reaction. → O2 2 O’s balance last CO2 + H2O 1 C & 2 O’s 2 H’s & 1 O balance H first Total mass of reactants = Total mass of products 2 2H6 ___C + → O2 CO2 63 H2O ___ + balance C next Chemical equations must therefore be balance for mass 2C2H6 + balance O The number and type of atoms on either side of the equation must be equal! 2C2H6 + O2 → 7 O2 ____ 4 C’s 12 H’s 14 O’s 77 4 CO2 ___ → + 4CO2 + 6H2O 6H2O 4 C’s 12 H’s 14 O’s 78 18 Reduction and Oxidation Reactions: RedOx Oxidation Numbers +1 +2 +3 -3 -2 -1 Oxidation involves an atom or compound losing electrons Variable Reduction involves an atom or substance gaining electrons Niether process can occur alone… that is, there must be an exchange of electrons in the process. Metals take on their formal charge: Non-metals do the same The substance that is oxidized is the reducing agent The substance that is reduced is the oxidizing agent Chemists use oxidation numbers to account for the transfer of electrons in a RedOx reaction. 79 Electrochemistry: Oxidation numbers In the compound potassium bromate (KBrO3), the oxidation number of bromine (Br) is? The compound is neutral so the sum of the oxidation numbers should be zero. 5 +1 3×(-2) = -6 ?? KBrO3 1+ ?? + (-6) = 0 80 Balancing REDOX reactions: Fe + O2 → Fe2O3 oxidation states: 0 0 +3 -2 oxidation half reaction: {Fe → Fe+3 reduction half reaction: {O2 + 3e-} x 4 + 4e- → 2O2-} x 3 Balance electrons transferred then sum the half RXN’s: 4Fe + 3O2 + 12e- → 2Fe2O3 + 12e- ?? = 5 4Fe + 3O2 → 2Fe2O3 81 82 19 Quantitative calculations: Mass and moles 2HCl (aq) + Ba(OH)2 (aq) Æ 2H2O (l) + BaCl2(aq) Consider the following reaction: HCl (aq) + Ba(OH)2 (aq) Æ H2O (l) + BaCl2(aq) g BaCl2 Æ mol BaCl2 Balacing: Æ mol HCl 2 HCl (aq) + Ba(OH)2 (aq) Æ 2 H2O (l) + BaCl2(aq) How many moles of HCl are consumed if 1.50 g of BaCl2 are produced assuming that Ba(OH)2 is in excess? Solution: convert to g BaCl2 then convert Æ using the Molar mass of BaCl2 mol BaCl2 1.50 g BaCl 2 × 1 mol BaCl 2 2 mol HCl × = 0.0144 mol HCl 208.24 g BaCl 2 1 mol BaCl 2 to Æ Molar mass BaCl2 mol HCl Molar ratio for equation using the Molar ratio for equation 83 84 Reaction Yields: Stoichiometry: Conversions between masses & moles in chemical reactions. The theoretical yield is the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry. Limiting Reactant: ÖWhen one reactant is present in an amount such that it is completely consumed before all other reactants, we say that it limits the reaction. The actual yield is the experimentally measured amount of products that results upon completion of the reaction. Ö The other reactants are said to be in excess. excess Ö The Theoretical Yield is determined by the stoichiometry of the limiting reactant. Ö The limiting reactant can only be determined through molar ratios. It cannot be identified by mass. The percent yield is a measure of the extent of the reaction in terms of the actual vs. the theoretical yield. % Yield = 85 Actual Yield (in grams or moles) Theoretical Yield (in grams or moles) × 100 86 20 Example: A solution of sodium phosphate is added to a solution of aqueous barium nitrate. A white ppt is observed. The Behavior of Solutes: Strong Electrolytes: complete dissociation into ions Unbalanced Equation: 1 M Na3PO4(aq) Æ 3M Na+ (a) + 1M PO43– Na3PO4 (aq) + Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + NaNO3 (aq) Molecular: 4M overall in ions Non–Electrolytes: 2Na3PO4 (aq) + 3Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + 6NaNO3 (aq) no dissociation into ions 1M CH3OH (aq) methanol Ionic: 1M overall in molecules / / 6Na+ (aq) + 2PO43- (aq) + 3Ba2+ (aq) + 6NO3- (aq) / 1M HC2H3O2(aq) Æ H+ (aq) + / → Ba3(PO4)2 (s) + 6Na+ (aq) + 6NO3- (aq) Weak Electrolytes: partial dissociation into ions Net Ionic: C2H3O2–(aq) 2PO43- (aq) + 3Ba2+ (aq) → Ba3(PO4)2 (s) between 1 & 2 M overall 87 How many grams of sodium phosphate are in 35.0 mL of a 1.51 M Na3PO4 solution? Solutions and Concentration Molarity: 88 Moles of solute per liter of solution. mL solution Æ L Æ mols Na3PO4 Æ g Na3PO4 moles of solute use M as a conversion factor {units: mol/L} Molarity (M) = L of Solution 35.0mL × molarity is a conversion factor that transforms units of volume to mole and vise–versa 163.94g L 1.51 mol Na 3 PO4 × × 1 mol Na 3 PO 4 10 mL L 3 = 8.66g Na 3 PO 4 89 90 21 The pH Scale: Dilutions: old Molarity × old Volume 1 pH = log + H New Molarity = new Volume 1 Since log = − log(x) x M1 × V1 M2 = Rearranging: [H+] = Molarity of H+ (or anything for that matter…) pH = -log[H+] In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC V2 pH = - log [H+] = -log (1.00 x 10-7) = - [0 + ((-7)] M1 × V1 = M2 × V2 = 7 91 What is the pH of the 0.0515 M HCl solution that Jane made? pH = -log[H+] 92 Reaction Quotient & Equilibrium Constant HCl (aq) → H+(aq) + Cl−(aq) Strong electrolyte! Equilibrium Established Therefore [H+] = [HCl] pH = − log[H + ] = − log(0.0515) H2 + I2 → 2 HI H 2 + I 2 U 2HI = 1.29 93 94 22 The Reaction Quotient, Q THE EQUILIBRIUM CONSTANT In general, ALL reacting chemical systems are For any type of chemical equilibrium of the type aA + bB→cC + dD the following is a CONSTANT (at a given T) characterized by their REACTION QUOTIENT, Q. aA + bB→cC + dD conc. of products K = [C]c [D]d [A]a [B]b equilibrium constant conc. of reactants If Q < K, then system will shift to the right, convert reactants to products. If Q > K, then system will shift to the left, products convert to to reactants. If Q = K, then system is at equilibrium. If K is known, then we can predict concentrations of products or reactants. 95 Forms of Chemical Bonds • There are 2 extreme forms of connecting or bonding atoms: 96 When there exists a difference in the Electronegativity (EN) between the two bonding atoms, (∆EN) the bond is said to be polar. Example: ∆EN Bond Br2 0 electrons from one atom to another HCl 0.9 polar-covalent • Covalent—some valence electrons NaF 3.1 ionic • Ionic—complete transfer of 1 or more non-polar shared between atoms • Most bonds are somewhere in between. 97 98 23 Electronegativity: Rank the following bonds by increasing strength: The measure of the tendency for a given atom to polarize the electrons in a covalent bond is called the Electronegativity (EN) of an atom. Electronegativity is related to the ionization energy and the electron affinity of an atom. H−F, H−Cl and H−I Each bond has hydrogen in common so we have a basis for comparison. From the periodic table, in terms of electronegativity: F > Cl > I EN increases Periodic Table Bond Strength H−I < H−Cl < H−F inc. EN EN decreases One concludes that the H-F bond is the most polar, followed by H-Cl then H-I. 99 Multiple bonds in covalent molecules: Bond Strength and Bond Properties: The oxygen and nitrogen that makes up the bulk of the atmosphere also exhibits covalent bonding in forming diatomic molecules. : : → O=O “double bond” Covalent bond strength increases with increasing ∆EN example: HCl bond is stronger than the HBr bond Covalent bond strength increases with increasing bond order : : N. O: : : + : : : : O: 100 + N⋅ → :N≡N: “triple bond” : : example: O=O bond in stronger than O–O bond triple > double > single Polyatomic Molecules (More than two atoms) Bond Order: 3 2 1 Bond length decreases with increasing bond order (Strength) : : : : Carbon dioxide: CO2 O=C=O example: 101 O=O bond is shorter than O–O bond 102 24 Heat Transfer: Surroundings heat in heat out System H–bonding in water brings about a network of interactions which explain phenomena such as: capillary action surface tension q<0 (–) ∆E < 0 q>0 (+) ∆E > 0 why ice floats 103 Thermochemistry: Energy in E Final Ef > Ei Heat and the Specific Heat Capacity: Energy out Ef < Ei When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. E initial work out energy work in 104 The amount of heat (q) transfer is related to the mass and temperature by: q = m × C × ∆T q = heat lost or gained (J) q in E Final ∆Esystem > 0 (+) ∆Esystem < 0 (–) J C = the Specific Heat Capacity of a compound o g⋅ C q out E initial m = mass of substance (g) 105 ∆T is the temperature change in degrees Celsius or Kelvin’s 106 25 Enthalpy: ∆H Ideal Gas Law How many kJ of energy are released when 128.5 g of methane (CH4(g)) is combusted? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) 128.5g CH 4 × P×V = constant for a set number of gas moles n×T ∆H = -802 kJ 802 kJ mol CH 4 × = 6.42×103 kJ 1mol CH 4 16.04 g P×V = n×R ×T How many hours would this power a 100 W light bulb if one could use all of this energy? 100 W = 100 Js–1 6.42×103 103 J × s × min × hr = kJ × 100J 60s 60 min kJ R = "gas constant" = 0.08206 L ⋅ atm mol ⋅ K PV = nRT 17.8 hrs 107 Problems: How many grams of krypton (Kr) will it take to exert a pressure of 11.2 atm in an 18.5 L at 28.2 oC. PV = nRT PV n= = RT 108 What wavelengths correspond to FM radio (93.5 MHz) signals? λ×υ = c c λ= υ L ⋅ atm R = 0.082057 mol ⋅ K 3.00 × 108 m 11.2 atm × 18.5 L 0.08206 L ⋅ atm × ( 28.2 + 273.15) K mol ⋅ K 8.3789 mols Kr × 83.80 g = 702 g Kr mol = 8.3789 mols Kr λ= 93.5 MHz × s 106 Hz 1s −1 × 1 MHz Hz = 3.21 m (3sf) 109 110 26 What wavelengths correspond to FM radio (93.5 MHz) signals? λ×υ = c c λ= υ 3.00 × 108 m λ= 93.5 MHz × s 106 Hz 1s −1 × 1 MHz Hz = 3.21 m 111 112 We can use the periodic table to determine the electron configuration by counting: Each element’s outermost electrons (valence) are related to the elements position on the periodic table. N: 7 electrons 1 3 2 4 5 1: 1s1 All of the subshells below the valence are full. 113 2: 1s2 3: 2s1 4: 2s2 1s2 2s2 5: 2p1 6 6: 2p2 7 7: 2p3 2p3 114 27 Electron Configurations cont… Periodic Trends Orbital box notation: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 1s 2s 2p 3s ↑ 3p ↑ 3p ↑↓ Aluminum: Al (13 electrons) Electron Configuration 3s ↑↓ ↑↓ ↑↓ ↑↓ 2p 2s 1s2 2s2 2p6 3s2 3p1 ↑↓ 1s 115 116 Kinetics: Rate Law & Reaction Order Reaction Orders: The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms−1). (It also has deeper significance, which will be discussed later) For the general reaction: A reaction order can be zero, or positive integer and fractional number. a A + b B → cC + d D Rate = k [A]x [B]y x and y are the reactant orders determined from experiment. x and y are NOT the stoichiometric coefficients. 117 Order 0 1 2 0.5 1.5 0.667 Name zeroth first second one-half three-half two-thirds 118 28 Recognizing a first order process: AÆproducts Determining the Rate constant for a first order process Taking the log of the integrated rate law for a first order process we find: Whenever the conc. of a reactant falls off exponentially, the kinetics follow first order. ln ([A] = [A]o e − kt ) ln[A] = ln[A]o − k × t A plot of ln[A] versus time (t) is a straight line with slope -k and intercept ln[A]o [A] = [A]o e − kt 119 A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s? 120 A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s? Using the integrated rate law, substituting in the value of k and 900s we find: Step 1: Determine the magnitude of the rate constant, k. k = 0.00385 s-1 t 1 = ln 2 = 0.693 2 k k k= [A] = e − kt [A]o ln 2 ln 2 = = 0.00385s −1 t1 180s −1 [A] = e −0.00385 s × 900 s [A]o = 0.0312 Since the ratio of [A] to [A]0 represents the fraction of [A] that remains, the % is given by: 2 100 × 0.0312 = 3.12% 121 122 29