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Chapter 1: Chemistry and Measurement Problems: 2, 4-5, 7-10, 14, 16, 17, 24-26, 27a, 28-44, 47-52, 55-56, 59-77, 79-88, 93-96, 105-112, 115-132 science: methodical study of nature followed by a logical explanation of the observations Chemistry: 1.1 study of matter and its properties, the changes that matter undergoes, and the energy associated with those changes MODERN CHEMISTRY In what fields is chemistry used today? And how is it used? 1.2 EXPERIMENT AND EXPLANATION Consider the following: On the first day of school, you get in your car and turn the key on the ignition, but nothing happens. What could be the problem? The Scientific Method 1. Perform experiments and record observations on system 2. Analyze the data, and propose a hypothesis to explain observations 3. Conduct additional experiments to test hypothesis. If initial hypothesis holds up to extensive testing, the hypothesis becomes a theory. theory: a tested qualitative explanation of basic natural phenomena If all or part of the hypothesis does not hold up to testing, then it is adjusted or a new hypothesis is proposed to explain the observations. law: a simple statement or mathematical equation about a fundamental relationship CHM 151: Chapter 1 page 1 1.5 MEASUREMENT AND SIGNIFICANT FIGURES measurement: a number with attached units When a measurement is recorded, all the numbers known with certainty are given along with the last number, which is estimated. All the digits are significant because removing any of the digits changes the measurement's uncertainty. Ruler A 0 1 2 3 4 5 0 1 2 3 4 5 Ruler B Ruler C 4.1 Ruler 4.2 4.3 Measurement/quantity 4.4 # of sig figs A ____________ ____ B ____________ ____ C ____________ ____ Which ruler gives the most exact measurement? _______________ Guidelines for Sig Figs (if measurement is given): Count the number of digits in a measurement from left to right: 1. When there is a decimal point: – For measurements greater than 1, count all the digits (even zeros). – 62.4 cm has 3 sig figs, 5.0 m has 2 sig figs, 186.000 g has 6 s.f. – For measurements less than 1, start with the first nonzero digit and count all digits (even zeros) after it. – 0.011 mL and 0.00022 kg each have 2 sig figs 2. When there is no decimal point: – Count all non-zero digits and zeros between non-zero digits – 125 g has 3 sig figs, 107 mL has 3 sig figs CHM 151: Chapter 1 page 2 How many significant digits do the following numbers have? # of sig figs of sig figs # of sig figs a. 165.3 _____ c. 90.40 _____ e. 0.19600 _____ b. 105 _____ d. 100.00 _____ f. 0.0050 _____ SCIENTIFIC NOTATION Some numbers are very large or very small difficult to express. For example, Avogadro’s number = 602,000,000,000,000,000,000,000 an electron’s mass = 0.000 000 000 000 000 000 000 000 000 91 kg Also, it's not clear how many sig figs there are in some measurements. For example, Express 100.0 g to 3 sig figs: ___________ Express 100.0 g to 2 sig figs: ___________ Express 100.0 g to 1 sig fig: ___________ To handle such numbers, we use a system called scientific notation. All numbers can be expressed in the form: where N 10n N =digit term= a number between 1 and 10, so there can only be one number to the left of the decimal point: #.#### n = an exponent = a positive or a negative integer (whole #). To express a number in scientific notation: Count the number of places you must move the decimal point to get N between 1 and 10. Moving decimal point to the right (if # < 1) negative exponent. Moving decimal point to the left (if # > 1) positive exponent. CHM 151: Chapter 1 page 3 Express the following numbers in scientific notation (each with 3 sig figs): 555,000 __________________ 0.000888 __________________ 602,000,000,000,000,000,000,000 ______________________ ROUNDING OFF NONSIGNIFICANT DIGITS How do we eliminate nonsignificant digits? • If first nonsignificant digit < 5, just drop ALL nonsignificant digits • If first nonsignificant digit 5, raise the last sig digit by 1 then drop ALL nonsignificant digits For example, express 72.58643 with 3 sig figs: last significant digit 72.58643 g first nonsignificant digit 72.58643 to 3 sig figs _______________________ Express each of the following with the number of sig figs indicated: a. 376.276 to 3 sig figs _______________________ b. 500.072 to 4 sig figs _______________________ c. 0.00654321 d. 1,234,567 e. 2975 to 3 sig figs to 5 sig figs to 2 sig figs _______________________ _______________________ _______________________ Express measurements in scientific notation when necessary to make it clear how many sig figs there are in the measurement. CHM 151: Chapter 1 page 4 ADDING/SUBTRACTING MEASUREMENTS When adding and subtracting measurements, your final value is limited by measurement with the largest uncertainty—i.e. the number with the fewest decimal places. Ex 1: 106.61 + 0.25 + 0.195 = 107.055 107.055 to the correct number of sig figs: _______________ Ex 2: 725.50 – 103 = 622.50 622.50 to the correct # of sig figs: ____________________ MULTIPLYING/DIVIDING MEASUREMENTS When multiplying or dividing measurements, the final value is limited by the measurement with the least number of significant figures. Ex 1: 106.61 0.25 0.195 = 5.1972375 5.1972375 to the correct # of sig figs: ________________ Ex 2: 106.61 91.5 = 9754.815 w/ correct sig figs: _______________ MULTIPLYING/DIVIDING WITH EXPONENTIAL NUMBERS: When multiplying or dividing measurements with exponents, use the digit term (N in “N 10n”) to determine number of sig figs. 1033 Ex. 1: (6.02 1023)(4.155 109) = 2.50131 How do you calculate this using your scientific calculator? Step 1. Enter “6.02 1023” by pressing: 6.02 then EE or EXP (which corresponds to “ 10”) then 23 Your calculator should read or something similar. Step 2. Multiply by pressing: CHM 151: Chapter 1 page 5 Step 3. Enter “4.155 109” by pressing: 4.155 then EE or EXP (which corresponds to “ 10”) then 9 Your calculator should now read Step 4. Get the answer by pressing: Your calculator should now read: or something similar indicating 2.50131 33 or 2.50131 E 33 10 33 2.50131 Thus, the answer with the correct sig figs = ________________________ Be sure you can do exponential calculations with your calculator. Many of the calculations we do in chemistry involve very large and very small numbers with exponential terms. Ex. 2:(3.75 1015)(8.6 104) = 3.225 1020 with the correct sig figs: ___________________ Ex. 3: (3.75 1015) (8.605 104) = 4.357931435 1010 with the correct sig figs: ___________________ EXACT NUMBERS Although measurements can never be exact, we can count an exact number of items. For example, we can count exactly how many students are present in a classroom, how many M&Ms are in a bowl, how many apples in a barrel. 1.8 UNITS AND DIMENSIONAL ANALYSIS (FACTOR-LABEL METHOD) UNIT EQUATIONS AND UNIT FACTORS Unit equation: Simple statement of two equivalent values Unit conversion factor = unit factor = equivalents: - Ratio of two equivalent quantities Unit equation 1 dollar = 10 dimes CHM 151: Chapter 1 Unit factor 10 dimes 1 dollar or 1 dollar 10 dimes page 6 Equivalents are exact if we can count the number of units equal to another or the units are in the same system (metric or English). For example, the following unit factors and unit equations are exact: 365 days 1 year 7 days 1 week 12 inches 1 foot and 1 yard 3 feet Exact equivalents have an infinite number of sig figs never limit number of sig figs! Note: When the relationship between two units or items is exact, the “” (meaning “equals exactly”) is used instead of the basic “=” sign. Other equivalents are inexact or approximate because they are measurements or approximate relationships, such as 1.61 km 1 mile 65 mi hour speed of light = 3.00 108 m s Approximate equivalents do limit the sig figs for the final answer. UNIT (DIMENSIONAL) ANALYSIS PROBLEM SOLVING 1. Write the units for the answer. 2. Determine what information to start with. 3. Arrange all other unit factors—showing them as fractions—with correct units in the numerator and denominator, so all units cancel except for the units needed for the final answer. 4. Check for correct units and number of sig figs in the final answer. CHM 151: Chapter 1 page 7 Example 1: If a marathon is 26.2 miles, how many inches are in the marathon? (1 mile 5280 feet) Example 2: The speed of light is about 3.00 x 108 meters per second. Express this speed in miles per hour. (1.609 km = 1 mile, 1000 m 1km) 1.6 SI Units (from French "le Système International d’Unités") – standard units for measurement Metric system: A unified decimal system of measurement with a basic unit for each type of measurement CHM 151: Chapter 1 quantity basic unit symbol length meter m mass gram g volume liter L time second s page 8 Metric Prefixes Multiples or fractions of a basic unit are expressed as a prefix Each prefix = power of 10 The prefix increases or decreases the base unit by a power of 10. Prefix Symbol Multiple/Fraction mega M 106 kilo k 103 deci d 0.1 = 10-1 centi c 0.01 = 10-2 milli m 0.001 = 10-3 micro (Greek “mu”) 10–6 nano n 10–9 pico p 10–12 KNOW these metric units above (highlighted in Table 1.3)! Metric Conversion Factors Ex. 1 Ex. 2 Complete the following unit equations: a. 1 kg = __________ g c. 1 cm = __________ m b. 1 s = __________ ns d. 1 L = __________ mL Write 2 unit factors for each of the unit equations above. CHM 151: Chapter 1 page 9 3.3 Metric-Metric Conversions: Solve the following using dimensional analysis. Ex. 1 Convert 75 miles per hour into units of meters per second. Ex. 2 Convert 12.0 kilograms into milligrams. Metric-English Conversions English system: Our general system of measurement. Scientific measurements are exclusively metric. However, most Americans are more familiar with inches, pounds, quarts, and other English units. A method of conversion between the two systems is necessary. Know these conversions! Quantity English unit Metric unit English–Metric conversion length 1 inch (in) 1 cm 1 in. = 2.54 cm (exact) mass 1 pound (lb) 1g 1 lb = 453.6 g (approximate) volume 1 quart (qt) 1 mL 1 qt = 946.4 mL (approximate) All other metric-English conversions will be provided on quizzes and exams. Ex. 1 What is the mass in kilograms of a person weighing 215 lbs? CHM 151: Chapter 1 page 10 Ex. 2 What is the volume in cups for a 2.0-L bottle? Ex. 3 A light-year (~5.88x1012 miles) is the distance light travels in one year. Calculate the speed of light in meters per second. (1 mile=1.61 km approx.) Temperature: — A measure of the average energy of a single particle in a system. The instrument for measuring temperature is a thermometer. Temperature is generally measured with these units: Fahrenheit degree (°F) Celsius degree (°C) References English system Metric system freezing pt for water 32°F 0°C boiling pt for water 212°F 100°C Phoenix Summer day 113°F 45°C Conversion between Fahrenheit and Celsius scales: C = CHM 151: Chapter 1 (F - 32) 1.8 F = (C 1.8) + 32 page 11 Ex. 1 In Europe, temperature is generally reported in the Celsius scale. A European tourist asks you how hot it is. What is the equivalent Celsius temperature if on that cool Summer day it’s only 95°F? Kelvin Temperature Scale There is a third scale for measuring temperature: the Kelvin scale. The unit for temperature in the Kelvin scale is Kelvin (K, NOT °K!). The Kelvin scale assigns a value of zero kelvins (0 K) to the lowest possible temperature, which we call absolute zero and corresponds to –273.15°C. (The term absolute zero is used because this is the theoretical lowest limit.) Conversion between °C and K: K = ˚C + 273.15 ˚C = K – 273.15 Ex. 1 The average temperature in Reykjavik, Iceland during the summer is about 15˚C. What is the equivalent Kelvin temperature? What is the equivalent Fahrenheit temperature? Ex. 2 In the movie Terminator 2, a tanker crashes and pours out liquid nitrogen. This freezes the T-1000 because liquid nitrogen has a temperature of 77 K. What is the equivalent temperature in degrees Celsius? CHM 151: Chapter 1 page 12 1.7 DERIVED UNITS Volume: Amount of space occupied by a solid, gas, or liquid. – measured using graduated cylinder, a buret, a pipet, a volumetric flask, etc. – generally in units of liters (L), milliliters (mL), or cubic centimeters (cm3) – Know the following: 1 L 1 dm3 1 mL 1 cm3 These are both exact! DETERMINING VOLUME Volume is determined in three principal ways: 1. Volume of any liquid can be measured directly using calibrated glassware (graduated cylinder, pipets, burets, etc.) 2. Volume of a solid with a regular shape (rectangular, cylindrical, uniformly spherical or cubic, etc.) can be determined by calculation. – e.g. volume of rectangular solid = length x width x thickness 3. Volume of an irregular solid is found indirectly by the amount of liquid it displaces. This technique is called volume by displacement. Volume By Displacement a. Fill a graduated cylinder halfway with water, and record the initial volume. b. Carefully place the object into the graduated cylinder so as not to splash or lose water. c. Record the final volume. d. Volume of object = final volume – initial volume CHM 151: Chapter 1 page 13 density: The amount of mass in a unit volume of matter density = mass volume or d = m V generally in units of g/cm3 or g/mL For water: 1.00 g of water occupies a volume of 1.00 cm3 d= m 1.00 g = = 1 . 00 g /cm 3 3 V 1.00 cm Applying Density as a Unit Factor Given the density for any matter, you can always write two unit factors. For example, the density of ice is 0.917 g/cm3. Two unit factors would be: 0.917 g cm 3 or cm 3 0.917 g Ex. 1 Aluminum has a density of 2.70 g/cm3. What is the mass of a piece of aluminum with a volume of 0.525 cm3? Ex. 2 Ethanol is used in alcoholic beverages and has a density of 0.789 g/mL. What volume of ethanol (in liters) would have a mass of 500.0 mg? CHM 151: Chapter 1 page 14 Ex. 3 A piece of silver metal weighing 194.3 g is placed in a graduated cylinder containing 242.0 mL of water. The volume of water now reads 260.5 mL. Calculate the density of silver. 1.3 CONSERVATION OF MASS matter: anything that has mass and volume mass: A measure of the amount of matter an object possesses. – measured with a balance and NOT AFFECTED by gravity MASS WEIGHT = MASS x Acceleration due to gravity Mass is not affected by gravity! EARTH mass = 68 kg weight = 150 lbs MOON mass = 68 kg weight = 25 lbs SPACE mass = 68 kg weight = 0 lbs law of conservation of mass: Matter is neither created nor destroyed reaction (rxn): REACTANTS (starting materials) (substances before rxn) PRODUCTS (substances after rxn) Mass of the product(s) in a reaction must = mass of the reactant(s)! For example: Example: CHM 151: Chapter 1 11 g hydrogen + 89 g oxygen = 100 g water A 95-g sample of sodium chloride, NaCl, contains 58 g of chlorine. Determine the mass of sodium in the sample. page 15 1.4 MATTER: PHYSICAL STATE AND CHEMICAL COMPOSITION Matter exists in one of three physical states: solid, liquid, gas (See Fig. 1.11) solid: Has definite shape, fixed volume – At molecular level, particles are packed closely, in a fixed position liquid: Fixed volume, but shape can change—takes shape of container – Particles are packed closely together but can move past each other gas (or vapor): Volume is variable, particles are widely spaced If volume expands, particles move apart If volume decreases, particles, move closer together Takes shape of container ELEMENTS, COMPOUNDS, AND MIXTURES PHYSICAL & CHEMICAL PROPERTIES Physical Properties: physical state (solid, liquid, gas) color density melting and boiling points electrical & heat conductivity solubility hardness odor Chemical Properties: how a substance reacts with other substances PHYSICAL & CHEMICAL CHANGES physical change: – a process that does not alter the chemical composition of a substance – eg. changing shape, changing physical state, dissolving – eg. boiling water, melting ice, hammering gold into foil chemical change or reaction: – a process that changes the chemical composition (and thus the chemical formula) of starting materials (reactants) — eg. oxidation of matter (burning or rusting) release of gas bubbles (fizzing) formation of insoluble solid (precipitation) release of heat or light CHM 151: Chapter 1 page 16 pure substances: – Matter having constant composition, definite and consistent properties Two types of pure substances: elements: — consist of only one type of atom — cannot be broken down by chemical reaction — eg. carbon (C), hydrogen (H2), sulfur (S8), copper wire (Cu) compounds: — consist of more than one type of atom and has a specific formula — can be broken down by chemical reaction — eg. ethanol (C2H5OH)can be broken down to C, H, & O Two or more pure substances can combine to form mixtures: mixtures: — consist of various compounds and/or elements, with no specific formula — Matter having variable composition with either definite or varying properties depending on the sample — can be broken down into individual components — eg. Any alloy like brass, steel, 10-K to 18-K gold; course mixtures like sea water, carbonated soda, salt and iron fillings; air consists mostly of nitrogen and oxygen gases Also, be able to identify elements, compounds, and mixtures given images at the atomic/molecular level. (See Concept Check 1.1 on p. 14) CHM 151: Chapter 1 page 17 Chapter 2: Atoms, Molecules, and Ions Problems: 2.1 2.1, 2.5-2.6, 2.8,2.11-2.14, 2.17-2.19, 2.21, 2.26, 2.29-2.32, 2.35-2.40, 2.43-2.54, 2.57-2.94, 2.99-2.118 ATOMIC THEORY OF MATTER Postulate from John Dalton’s Model of the Atom 1. All matter is composed of indivisible atoms. 2. An element is composed of only one type of atom – All atoms of one type of element always behave the same way. – Atoms of different elements do not behave the same way. 3. Two or more elements combine to form compounds. law of constant composition: A compound always has the same elements in the same proportion by mass – i.e. a compound always has the same formula law of multiple proportions: Two elements can combine to form different compounds (e.g. C and O can combine to form CO and CO2), and the masses of each element for each compound are always in a fixed ratio (because always whole #’s of atoms of each element) 4. A chemical reaction involves only the separation, combination, or rearrangement of atoms—NEVER their creation or destruction. – This is also called the law of conservation of mass. Ex. 1: If 2.50 g of iron powder react with 1.44 g of yellow sulfur powder, what is the mass of the iron sulfide product? Ex. 2: If 10.11 g of limestone (calcium carbonate) decomposes with heat to 8.51 g of calcium oxide and CO2 gas, what mass of CO2 is produced? CHM 151: Ebbing Chapter 2 Notes page 1 2.2 THE STRUCTURE OF THE ATOM Discovery of the Electron J. J. Thomson (1897) – demonstrated that cathode rays are composed of tiny, negatively charged subatomic particles electrons (e–) Eugen Goldstein (late 1880s) – canal rays are composed of positively charged subatomic particle protons (p+) And decades later: – James Chadwick (1935), Nobel Prize winner for his discovery neutron (n) = neutral subatomic particle Nuclear Model of the Atom Ernest Rutherford's Alpha-Scattering Experiment (Figure 2.8) – Alpha () particles shot at a thin gold foil – A detector set up around the foil to determine what happens to the particles – Most of a particles went straight through, some deflected, a few bounced back Rutherford’s interpretation of the scattering experiment results –Most alpha () particles pass through foil Atom is mostly empty space with electrons moving around the space –Some particles deflected or bounce back Atom must also contain a dense region, and particles colliding with this region are deflected or bounce back towards source region = atomic nucleus (contains atom’s protons and neutrons) Why this is called the “Nuclear Model of the Atom” Rutherford also estimated the size of the atom and its nucleus: nucleus (d~10-15 m) atom (diameter ~10-10 m) If nucleus = size of a small marble, then atom BankOne Ballpark! CHM 151: Ebbing Chapter 2 Notes page 2 Properties of Protons, Neutrons, and Electrons Subatomic Particle Charge Location Mass (amu) proton +1 inside nucleus 1.00728 neutron 0 inside nucleus 1.00866 electron -1 outside nucleus 0.00055 2.3 NUCLEAR STRUCTURE; ISOTOPES Nuclide Symbol (also called Atomic Notation): – shorthand for keeping track of protons and neutrons in the nucleus atomic number (Z): whole number of p+ = number of e– in neutral atom mass number (A): whole number sum of protons and neutrons in an atom – Note electrons contribute almost no mass to an atom mass number=A atomic number=Z E element symbol Ex. 1: Write the atomic notation for carbon-14. How many neutrons are in each neutral carbon-14 atom? Ex. 2: Write the atomic notation for uranium-235 (Z=92). How many neutrons are in each neutral uranium-235 atom? Isotopes Each element always has the same number of protons, but the number of neutrons may vary. Atoms with different numbers of neutrons are called isotopes. – e.g. Carbon exists as carbon-12, carbon-13, and carbon-14 where each carbon atom has 6 protons but 6, 7, or 8 neutrons – Isotopes are identified with an element name followed by the mass number – eg. uranium-235 (U-235), carbon-12 (C-12), cobalt-60 (Co-60), etc. CHM 151: Ebbing Chapter 2 Notes page 3 2.4 ATOMIC WEIGHTS Atoms are too small to weigh directly – eg. one carbon atom has a mass of 1.99 x 10-23 g–too inconvenient! need more convenient unit for mass atomic mass unit (amu) Carbon-12 was chosen and given a mass value of 12 amu 1 amu = 1/12 the mass of carbon-12 Mass of all other atoms measured relative to mass of carbon-12 Average Atomic Mass of an Element – The mass of carbon on the Per. Table is 12.01 amu, NOT 12.00 amu–WHY?! – Atomic masses reported on the Periodic Table are weighted averages of all naturally occurring isotopes for each element. Ex. 1 If 98.89% of carbon exists as carbon-12, which has a mass of 12.00000, while 1.11% exists as carbon-13, which has a mass of 13.00335, calculate the average atomic mass for carbon. average atomic mass = (0.9889)(12.00000 amu) + (0.0111)(13.00335 amu) = ____________________ Ex. 2 The atomic masses of the two stable isotopes for boron, B-10 (19.78%) and B-11 (80.22%), are 10.0129 amu and 11.0093 amu, respectively. Calculate the average atomic mass for boron. 2.5 PERIODIC TABLE OF THE ELEMENTS A vertical column is called a group or family. – Elements belonging to the same group exhibit similar chemical properties – e.g. Li, Na, and K all react vigorously with water; He, Ne, and Ar do not react with any other substances A horizontal row is called a period or series. CHM 151: Ebbing Chapter 2 Notes page 4 Main-Group (Representative or A Group) Elements Those elements in groups 1, 2, 13, 14, 15, 16, 17, 18 (or IA to VIIIA) – Group 1 or IA: alkali metals – Group 2 or IIA: alkaline earth metals – Group 17 or VIIA: halogens – Group 18 or VIIIA: noble gases (because they are all unreactive gases) Transition Metals (or B Group Elements) – Elements in Groups 3 to 12 (middle of the Periodic Table) Dimitri Mendeleev proposed that elements display recurring properties according to increasing atomic mass first Periodic Table arranged elements according to increasing atomic mass H.G.J. Moseley’s high-energy X-ray radiation experiments of atomic nuclei Repeating properties of elements more clearly reflected by the arrangement of elements according to increasing atomic number Periodic Table’s arrangement today – Trends for increasing atomic mass are identical with those for increasing atomic number, except for Ni & Co, Ar & K, Te & I. NAMES & SYMBOLS OF THE ELEMENTS – Every element has an individual name, symbol, and number. Convention for writing chemical symbols – Use first letter (capitalized) of element name: hydrogen H, carbon C – If symbol already used, include second letter (in lower case) of name: helium He, calcium Ca, cobalt Co – Some symbols come from Latin names lead=plumbum Pb gold= “shining dawn”=aurum Au Know the first 56 elements of the periodic table, as well as lead (Pb) and uranium (U) for Exam #1. You will be given a periodic table with only the symbols written. Spelling counts! CHM 151: Ebbing Chapter 2 Notes page 5 SOLIDS, LIQUIDS, AND GASES KNOW the physical state of each element! In their natural state under room temperature conditions: – Only mercury (Hg) and bromine (Br) are liquid – H, N, O, F, Cl, and all Noble gases (group VIIIA) are gases – All other elements are solids METALS, NONMETALS, & METALLOIDS (or SEMIMETALS) metals shiny appearance malleable, ductile conduct heat & electricity react with nonmetals e.g. aluminum, copper, gold nonmetals dull appearance brittle nonconductor react with metals and nonmetals e.g. carbon, oxygen, sulfur metalloids (or semimetals): Have properties of metals and nonmetals Organization of the nonmetals, semimetals, metals on the Periodic Table – Nonmetals (except H) are concentrated on the top-right of the Periodic Table – Semimetals are along the stair-step – All remaining elements are metals Diatomic Molecules: consist of two atoms – Recognize that some elements exist as diatomic molecules – H2, O2, N2, Cl2, F2, I2, Br2 – These are often called “the magic seven” since there are seven of them, and six of them form a 7 on the periodic table 2.6 CHEMICAL FORMULAS: MOLECULAR AND IONIC SUBSTANCES molecule: a compound of two or more nonmetal atoms, which are held together by covalent bond A molecular formula indicates how many atoms are actually present – e.g. in water, H2O, there are 2 H atoms and 1 O atom CHM 151: Ebbing Chapter 2 Notes page 6 chemical formulas: – Symbolically express the number of atoms of each element in a compound – Number of atoms is indicated by a subscript following the element’s symbol (If there is no subscript, only one atom of that element is in the compound.) – Some chemical formulas use parentheses more than one of that subunit present in the compound –e.g. C2H4(OH)2 has 2 C, 6 H, and 2 O Example: How many atoms of each element are in TNT: C7H5(NO2)3? ___ C atoms, ___ H atoms, ___ N atom, ___ O atoms IONIC SUBSTANCES – When atoms lose or gain electrons, they form charged particles called ions. Metals lose e–s positively charged ions = cations Nonmetals gain e–s, negatively charged ions = anions Main-group elements generally form ions--i.e. gain or lose electrons--to get the same number of electrons as a noble gas Ions from representative metals are usually isoelectronic with–i.e. have the same number of electrons as–one of the noble gases! Charges shown as superscripts – Group IA elements +1 charge: Li+, Na+, K+, etc. (“+” = “+1”) – Group IIA elements +2 charge: Mg+2, Ca+2, Sr+2, Ba+2,etc. – Group IIIA elements +3 charge: Al+3 – Group IVA elements +4 charge: Sn+4, Pb+4 – Group VA elements –3 charge: N–3, P–3 – Group VIA elements –2 charge: O–2, S–2, Se–2 – Group VIIA elements –1 charge: F–, Cl–, Br–, I–,etc. monoatomic ions: from a single atom (eg Na+, Cl–, O2–) polyatomic ions: from 2 or more atoms (eg. OH–, MnO4–, SO42–) CHM 151: Ebbing Chapter 2 Notes page 7 ionic compound: a compound composed of a metal + nonmetal(s), which are held together by an ionic bond (electrostatic attraction) An ionic compound is actually a network of ions, with each cation surrounded by anions, and vice versa. A 2-dimensional slice would look like this: Why ionic compounds have the highest melting points compared to other substances – to melt the crystal solid, all the bonds between ions have to be broken formula unit: simplest unit of an ionic compound--e.g. NaCl, Fe2O3, Fe2(SO4)3 – represent the ratio of ions present in the compound, not the actual number of ions NAMING IONIC COMPOUNDS CATIONS: positively charged ions – Metal atoms lose valence electrons to form cations. I. Groups IA, IIA, IIIA elements, silver (Ag), zinc (Zn) and cadmium (Cd) form only one type of ion each: – Group IA elements +1 charge always (e.g. Li+=lithium ion) – Group IIA elements +2 charge always (e.g. Mg+2=magnesium ion) – – Group IIIA elements +3 charge always (e.g. Al+3=aluminum ion) + 2+ 2+ – silver ion = Ag ; zinc ion = Zn ; cadmium ion = Cd CHM 151: Ebbing Chapter 2 Notes page 8 II. The Stock system is used to name most transition metals, Sn, Pb, and other metals that form more than one ion: +2 +3 – e.g. iron (Fe), a transition metal, forms 2 different ions: Fe and Fe , +2 +4 – e.g. lead (Pb), in Group IVA, forms 2 different ions: Pb and Pb When a metal can form more than one ion, each ion is named: element name (charge in Roman numerals) + ion Cu + = copper (I) ion Cu 2+= copper (II) ion Pb 2+= lead (II) ion Pb 4+= lead (IV) ion + Na = ________________________ 3+ Co = ________________________ 2+ = ________________________ Ba = ________________________ 3+ = ________________________ Cd = ________________________ Ni Al Fe 2+= iron (II) ion Fe 3+= iron (III) ion 2+ 2+ ANIONS: formed only by nonmetals When a nonmetal forms an ion, it is named: element stem name + -ide suffix + ion e.g. S 2– O2– = oxide ion N = nitrogen atom N3– = nitride ion O = oxygen atom = ________________________ – F = ________________________ – Br = ________________________ – Cl = ________________________ POLYATOMIC IONS – Know the formulas and names of the following polyatomic ions: NH4+ = ammonium ion Hg22+ = mercury (I) ion MnO4– = permanganate ion C2H3O2– = acetate ion PO43– = phosphate ion CN– = cyanide ion CHM 151: Ebbing Chapter 2 Notes CrO42– = chromate ion Cr2O72– = dichromate ion SO42– = sulfate ion SO32– = sulfite ion NO3– = nitrate ion NO2– = nitrite ion OH– = hydroxide ion CO32– = carbonate ion HCO3– = hydrogen carbonate ion – ClO = hypochlorite ion ClO2– = chlorite ion ClO3– = chlorate ion ClO4– = perchlorate ion page 9 WRITING CHEMICAL FORMULAS formulas of compounds: Cation + anion symbols and number of each Compounds should be neutral +ve charges = -ve charge Simple techniques for writing chemical formulas: 1. If both ions have charges that are exactly opposite (+1 & -1, 2+ & -2, etc.), formula contains one of each (This also applies for polyatomic ions.) Na+ + Cl– NaCl and Na+ + CN– NaCN K+ + Cl– ___________ K+ + NO3– ___________ Mg2+ + O2– ___________ Mg2+ + SO42– ___________ Al3+ + N3– ___________ Al3+ + PO43– ___________ 2. For monatomic ions with different charges, use the crossover rule: Make the negative charge the subscript of cation, and make positive charge Ba Al 2+ 3+ – + Cl 2– + O 3. For polyatomic ions, where ions have different charges, also use the crossover rule — Express more than one polyatomic ion with subscripts and parentheses. Ba Al 2+ 3+ – + NO3 + CO3 2– EXCEPTION FOR CROSSOVER RULE: Ions with +4 and 2– charges! Pb +4 2– + O Ionic compound formulas must reflect lowest ratio of elements Instead of Pb2O4, it should be PbO2 ! CHM 151: Ebbing Chapter 2 Notes page 10 Also applies to polyatomic ions! Pb +4 + SO4 2– NAMING COMPOUNDS Given known charges for some elements get charge on transition metals! I. If one of each (cation & anion) present same but opposite signs (a)CuCl: Cl ion has what charge? ______ Cu ion must have what charge? ______ and what formula? ______ What is the name for the Cu ion in this compound? _______________ What is the name of this compound? _______________ (b)FeS: S ion has what charge? ______ Fe ion must have what charge? ______ and what formula? ______ What is the name for the Fe ion in this compound? _______________ What is the name of this compound? _______________ Also applies to polyatomic ions: (a)CuSO4: SO4 ion has what charge? ______ Cu ion must have what charge? ______ and what formula? ______ What is the name for the Cu ion in this compound? _______________ What is the name of this compound? _______________ (b)FePO4: PO4 ion has what charge? ______ Fe ion must have what charge? ______ and what formula? ______ What is the name for the Fe ion in this compound? _______________ What is the name of this compound? _______________ CHM 151: Ebbing Chapter 2 Notes page 11 II. If ions have different charges reverse crossover, making subscripts the charges with positive sign on cation and negative sign on anion. (Also applies to polyatomic ions.) Get the individual ions for each compound below: Co2O3 Fe2(CO3)3 EXCEPTION FOR CROSSOVER RULE: Ions with +4 and 2– charges PbS2 Sulfide ion must have -2 charge, so 2 sulfide ions total charge = –4 To get overall charge of zero on compound, Pb must have +4 charge! Naming Ionic Compounds: 1. Get the individual ions for each compound 2. CATION NAME + ANION NAME, minus “ion” Name of compound Ex. CaCl2 = ______________ ______________________ individual ions name of compound Fe2S3 = ______________ ______________________ BaSO4 = ______________ ______________________ Co(NO3)3 = ______________ ______________________ Na3P = ______________ ______________________ Cu2CO3 = ______________ ______________________ TiO2 = ______________ ______________________ CHM 151: Ebbing Chapter 2 Notes page 12 Given the name of a compound, predict the formula: — KNOWING charges on ions formed by representative elements! — KNOWING how to use polyatomic ions and their charges when given to you! cesium bromide: ____________________ _____________ individual ions formula of compound copper (I) sulfide: _____________ _______________________ barium chloride: _______________________ _____________ lead (IV) phosphide: _____________ _______________________ iron (III) nitrite: _____________ _______________________ silver hydroxide: _____________ _______________________ tin (II) permanganate: _____________ _______________________ calcium phosphate: _____________ _______________________ Binary Molecular Compounds: composed of 2 or more nonmetals NAMING: # of atoms of element indicated by Greek prefix before element name 1. For first element, Greek prefix + element name 2. For second element, Greek prefix + element name stem + "ide" — If only one atom present, “mono-” is generally omitted, except in a few cases (eg. CO=carbon monoxide) # of atoms Greek prefix # of atoms Greek prefix 1 mono 6 hexa 2 di 7 hepta 3 tri 8 octa 4 tetra 9 nona 5 penta 10 deca CHM 151: Ebbing Chapter 2 Notes page 13 Examples: CO2= carbon dioxide P4O10= _________________________________ CCl4= _________________________________ SF6= _________________________________ Cl2O5= _________________________________ Some binary molecular compounds also have common names –e.g. everyone knows (or should know) H2O is water Common compounds and names you need to know: NH3 = ammonia CH4 = methane H2O2 = hydrogen peroxide ACIDS: Aqueous solutions of a compound that releases H+ ions – usually have H in front, in water indicated by aqueous (aq) – naming depends on the ion from which the acid forms add # of H's equal to negative charge HF (aq) = hydrofluoric acid F– = fluoride ion add # of H's equal to negative charge – HNO2 (aq) = nitrous acid NO2 = nitrite ion – add # of H's equal to negative charge HNO3 (aq) = nitric acid NO3 = nitrate ion For some acids, the stem name changes: 2– add # of H's equal to negative charge 2– add # of H's equal to negative charge H2SO3 (aq) = sulfurous acid SO3 = sulfite ion H2SO4 (aq) = sulfuric acid SO4 = sulfate ion Exercises: add # of H's equal to negative charge – ________ (aq) = __________________ Cl = ___________ add # of H's equal to negative charge 2– ________ (aq) = _________________ CrO4 = ___________ - add # of H's equal to negative charge _______ (aq) = _______________ C2H3O2 = ___________ 3– add # of H's equal to negative charge ________ (aq) = ________________ PO4 = ___________ CHM 151: Ebbing Chapter 2 Notes page 14 Hydrates: Has a specific # of water molecules bonded per formula unit. CuSO4·5 H2O (s) hydrate (contains 5 "waters of hydration") Naming a hydrate: name of anhydrous salt CuSO4 (s) + 5 H2O (l) anhydrous salt (has no waters) + Greek prefix indicating number of waters + "hydrate" CuSO4·5 H2O: ____________________________ CaCl2·6 H2O: ____________________________ 2.9 WRITING CHEMICAL EQUATIONS (EQNS) chemical equation: formulas and symbols describing a chemical rxn A + B reactants starting materials C + D products substance(s) resulting from chemical rxn Physical state of all reactants and products are indicated using subscripts: (s) = solid (l) = liquid (g)= gas (aq) = aqueous (ions or compounds in solution) Example: HCl (aq) + NaHCO3 (s) NaCl (aq) + H2O (l) + CO2 (g) 2.10 BALANCING CHEMICAL EQUATIONS coefficient: Whole #s in front of each reactant or product, indicating how many of each is present subscript: Whole # following each element in a compound, indicating the # of each element present CHM 151: Ebbing Chapter 2 Notes page 15 Balancing by Inspection GUIDELINES 1. Count the # of elements on both sides of the equation 2. Change the coefficients (NEVER the subscripts) to get the same # of elements on both sides of the equation – Balance the equation using the following order: – Metals – Polyatomic ions – Balance as a whole! – Hydrogen – Carbon – Oxygen – All other atoms Examples: ______ H2 (g) + ______ Cl2 (g) ______ Al (s) + ______ O2 (g) ______ C5H12 (l) + ______ O2 (g) ______ HCl (g) ______ Al2O3 (s) ______ H2O (g) + ______ CO2 (g) Treat polyatomic ions as ONE UNIT—Do not break them up into atoms! ______ Ca(C2H3O2) 2 (aq) + ______ K3PO4 (aq) ______ Ca3(PO4) 2 (s) + ______ Al2(SO4) 3 (aq) + CHM 151: Ebbing Chapter 2 Notes ______ KC2H3O2 (aq) ______ Ba(NO3) 2 (aq) ______ BaSO4 (s) + ______ Al(NO3) 3 (aq) page 16 Chapter 3: Calculations with Chemical Formulas and Equations Problems: 3.2-3.3, 3.6-3.7, 3.10, 3.13, 3.17-3.94, 3.97-3.106 3.2 The Mole Concept Avogadro’s Number (NA) = 6.02 x 1023 (to 3 sig figs) 1 mole (abbreviated mol) = 6.02 x 1023 entities Similar to: 1 dozen = 12 entities: 1 dozen doughnuts = 12 doughnuts 1 mole of doughnuts = 6.02 x 1023 doughnuts How many eggs are in 3 dozen eggs? ___________________ How many eggs are in 3 moles of eggs? How many C atoms are in 3 mol of C atoms? Atomic weights and molar masses: — The mass of 1 C atom (on average) is 12.01 amu — The mass of 1 mole of C atoms is 12.01 g (or 12.01 g/mol) 1 mole (6.02 x 1023) is the amount of atoms of any element that has a mass in grams equal to the mass of ONE atom in amu. The atomic masses reported for each element in the Periodic Table gives the atomic weight (or molecular/formula weight for compounds) in amu and the molar mass in g/mol. Ex. What is the molar mass for each of the following? (Use the atomic masses reported for each in the Periodic Table.) a. Mg c. Ar b. Si d. Sn CHEM 151 Chapter 3 page 1 of 14 Molar mass (MM): Mass in grams of 1 mole of any element/compound – To obtain, multiply the molar mass of each element by the number of each present, then add up all the constituent parts. Ex. Determine the molar mass of each of the following compounds: a. O 2: 2 (molar mass of O) = 2 (16.00 g/mol) = 32.00 g/mol b. H3PO4: c. Al2(SO4)3: Mole Calculations Ex. 1 How many moles of Ne are in 0.500 g Ne? Ex. 2 How many Ne atoms are in 0.500 g of Ne? Ex. 3 How many moles of CO2 are in 0.500 g of CO2? CHEM 151 Chapter 3 page 2 of 14 Ex. 4 How many CO2 molecules are in 0.500 g of CO2? Ex. 5 How many oxygen atoms are in 0.500 g of CO2? Molar Volume: Volume occupied by 1 mole of any gas Avogadro's Law: At the same temperature and pressure, equal volumes of gases contain the same number of molecules Standard temperature and pressure (STP): T=0˚C and P=1.00 atm At STP, 1 mole of gas occupies 22.4L! (3 sig figs) Molar Volume Calculations Ex. 1 How many moles of He occupy a volume of 3.36L at STP? Ex. 2 What mass of SO3 occupies a volume of 2.05 L at STP? CHEM 151 Chapter 3 page 3 of 14 Ex. 3 What is the volume occupied by 5.000 g of NH3 at STP? Ex. 4 How many Ne atoms are present in 0.124 L of Ne gas at STP? 3.3 Mass Percentages from the Formula mass percentage: The mass of one element in a compound divided by the mass of the entire compound Steps to determine percentage composition: 1. Calculate the mass of each individual element in the compound 2. Add up all the masses of each element to get the total mass of compound 3. Divide the mass of each individual element with the total mass of compound Ex. 1 What is the percent composition of C and O in CO2? Ex. 2 What is the percent composition of Fe and O in rust, Fe2O3? Ex. 3 What is the mass of iron in 5.00 g of rust? CHEM 151 Chapter 3 page 4 of 14 3.4 Elemental Analysis; Percentages of Carbon, Hydrogen, & Oxygen When a hydrocarbon (i.e. a compound containing only carbon and hydrogen) or a hydrocarbon derivative (containing carbon, hydrogen, and oxygen) is burned, the products are carbon dioxide and steam: CxHyOz (g) + O2 (g) CO2 (g) + H2O (g) Thus, if you have an unknown hydrocarbon is burned, we can use the amount of carbon dioxide and steam formed to determine the mass percentage of carbon and hydrogen in the original hydrocarbon. What to keep in mind: 1. All the C in the CO2 came from original compound mass of C in the CO2 equals mass of C present in the original compound 2. All the H in the H2O came from original compound mass of H in the H2O equals mass of H present in the original compound 3. If there is O in the original compound: mass of O in original = mass of original – mass of C – mass of H 4. Divide the mass of each component by the mass of the compound to get the mass percentage Ex. 1 Combustion of an 11.5-mg sample of ethanol produces 22.0 mg of CO2 and 13.5 mg of H2O. What is the mass percentage of carbon, hydrogen, and oxygen in ethanol? CHEM 151 Chapter 3 page 5 of 14 Ex. 2 Cumene is a hydrocarbon—i.e., it contains only carbon and hydrogen, and it is used in the production of acetone and phenol. Combustion of 47.6 mg of cumene produces 156.8 mg of CO2 and 42.8 mg of water. Calculate the mass percentage of carbon and hydrogen in cumene. 3.5 Determining Formulas Empirical Formula: Simplest whole-number ratio of atoms in a compound Molecular Formula: Chemical formula of a compound that expresses the actual number of atoms present in one molecule. – The molecular formula will either be exactly the same as or some multiple of the empirical formula! What is the empirical formula of benzene, C6H6? __________ What is the molecular formula of benzene, C6H6? __________ What is the empirical formula of ascorbic acid, C6H8O6? _____________ What is the molecular formula of ascorbic acid, C6H8O6? ____________ Guidelines for Determining the Empirical Formula of a Compound 1. Find the # mols of each element in the compound 2. Divide each # mols of each element by smallest # mols to get ratio of atoms 3. Get a whole number ratio for all atoms in the compound: – If within 0.1 of a whole number, round to that whole number – If any ratio ends close to .5 multiply ALL subscripts by 2 – If any ratio ends close to .33 or .66 multiply ALL subscripts by 3 CHEM 151 Chapter 3 page 6 of 14 Empirical Formula from Composition Ex. 1: Determine the empirical formula for iron oxide if a sample of iron oxide consists of 3.497 g of Fe and 1.503 g of O? Determining Molecular Formula 1. Follow the same steps for determining empirical formula. 2. For the Molecular Formula, you need to be given the molar mass of the compound. Find the molar mass of the empirical formula. 3. Divide the molar mass of the compound by the molar mass of the empirical formula to get the factor with which to multiply each subscript in the empirical formula. Ex. 2: Acetylene is a hydrocarbon, a compound that consists only of carbon and hydrogen. A 4.500-g sample of acetylene was analzyed and found to contain 4.151 g of carbon. If acetylene has a molar mass of 26.0 g/mol, determine both the empirical formula and the molecular formula for acetylene. CHEM 151 Chapter 3 page 7 of 14 Empirical and Molecular Formulas from Percent Composition: 1. Assume 100.0 g of the compound change percentages to grams! 2. Follow the same steps for determining empirical formula. 3. Divide the molar mass of the compound by the molar mass of the empirical formula to get the factor with which to multiply each subscript in the empirical formula. Ex. 3: Quinine is used as an antimalarial drug. An analysis of quinine Indicates the compound consists of 74.03% carbon, 7.47% hydrogen, 8.64% nitrogen, and 9.86% oxygen. If quinine's molar mass is 325 g/mol, determine the empirical and molecular formulas for quinine. 3.4 Molar Interpretation of a Chemical Equation A + B products starting materials substance(s) resulting from chemical reaction Interpreting a Chemical Equation H2 (g) + Cl2 (g) 1 molecule 1 molecule 2 H2 (g) + ____ molecule(s) CHEM 151 Chapter 3 C + D reactants 2 HCl (g) 2 molecules O2 (g) ____ molecule(s) page 8 of 14 2 H2O (g) ____ molecules It follows that any multiples of these coefficients will be in same ratio! 2 H2 (g) + O2 (g) 2 H2O(g) 1000 _____ molecule(s) _____ molecule(s) _____ molecule(s) N _____ molecule(s) _____ molecule(s) _____ molecule(s) Since N = Avogadro’s # = 6.021023 molecules = 1 mole 2 H2 (g) ___ mole(s) + O2 (g) ___ mole(s) 2 H2O(g) ___ mole(s) Thus, the coefficients in a chemical equation give the mole ratios of reactants and products in a chemical equation. Stoichiometry (STOY-key-OM-e-tree): calculation of the quantities of reactants and products involved in a chemical reaction 3.7 Amounts of Substances in a Chemical Reaction Consider the following: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) 1. Use unit factors to determine how many moles of O2 are needed to completely react with 2.25 moles of CH4. 2. How many moles of CO2 form when 5.25 moles of O2 completely react? 3. How many moles of H2O form when 7.50 moles of CO2 form? CHEM 151 Chapter 3 page 9 of 14 Mass-Mass Stoichiometry Problems MASS OF KNOWN Ex. 1: Molar Mass MOLES OF MOLE-MOLE Ratio KNOWN MOLES OF UNKNOWN Molar Mass MASS OF UNKNOWN Chlorofluorocarbons (CFCs) have a negative impact on the environment. One of the most common CFCs is CFC-12 or freon-12, which has the formula CCl2F2 and can be prepared from the reaction between hydrogen fluoride gas, HF, and carbon tetrachloride, CCl4. 2 HF (g) + CCl4 (l) CCl2F2 (g) + 2 HCl (g) 1. Calculate the mass of CCl4 necessary to react completely with 50.0 g of HF. 2. Calculate the mass of CCl2F2 produced when 50.0 g of HF reacts completely. Mass-Volume Stoichiometry Problems MASS OF KNOWN Molar Mass MOLES OF MOLE-MOLE Ratio KNOWN Molar Volume MOLES OF of a GAS = UNKNOWN 22.4 L/mol VOLUME OF UNKNOWN MOLES OF UNKNOWN MASS OF UNKNOWN OR Molar Volume VOLUME OF of a GAS = 22.4 L/mol KNOWN CHEM 151 Chapter 3 MOLES OF KNOWN MOLE-MOLE Ratio page 10 of 14 Molar Mass Ex. 1: Calculate the volume (in liters) of hydrogen gas produced when 0.165 g of aluminum metal reacts completely with dilute HCl at STP, 2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g) Ex. 2: An automobile airbag inflates when N2 gas results from the explosive decomposition of sodium azide (NaN3), 2 NaN3 (s) spark 2 Na (s) + 3 N2 (g) Calculate the mass of NaN3 needed to produce 50.0 L of N2 gas at STP. 3.8 LIMITING REACTANT (LIMITING REAGENT) In practice, reactants will not always be present in the exact amounts necessary for all reactants to be converted completely into products. Only in a limited supply of the some reactants (usually the more expensive) are present, so these are completely used up “limiting reagent(s)” since they limit the amount of product that can be made Some reactants (usually the least expensive) are present in larger amounts and are never complete used up “reactant(s) in excess” GUIDELINES FOR SOLVING LIMITING REAGENT PROBLEMS: 1. Calculate the mass of any product that can be made from each reactant – Use mass-mass or mass-volume conversions 2. Whichever reactant produces the smaller amount of product limiting reagent 3. All other reactant(s) reactant in excess CHEM 151 Chapter 3 page 11 of 14 PANCAKES MAKING BISQUICK 2 cups Bisqui ck 1 cup milk 2 eggs 14 pancakes 2 cups of Bisquick + 1 cup milk + 2 eggs 14 pancakes Ex. 1 If you have 10 cups of Bisquick, 10 cups of milk, and 14 eggs, how many pancakes can you make? Indicate the limiting reagent(s) and the reagent(s) in excess. Consider: 1 lb Bisquick= 6 cups of Bisquick 1 gallon of milk = 16 cups milk If you have 3.0 lbs of Bisquick, 2.0 gallon of milk, and a dozen eggs, a. How many dozen pancakes can you make? b. What is the limiting reagent? c. What are the reagents in excess? CHEM 151 Chapter 3 page 12 of 14 Consider the reaction to produce ammonia: N2 (g) + 3 H2 (g) 2 NH3 (g) Ex. 1 How many moles of ammonia form if 10.0 moles of N2 reacts with 20.0 moles of H2? (Indicate the limiting reagent and reagent in excess.) Ex. 2. If 50.0 g of N2 reacts with 10.0 g of H2, a. What mass of ammonia is produced? b. Indicate the limiting reagent and the reactant in excess. c. What mass of the reactant in excess remains after the reaction? CHEM 151 Chapter 3 page 13 of 14 Experimental yield; Percent Yield Percent yield = actual yield 100% theoretical yield theoretical yield: Amount of product one should get based on the chemical equation and the amount of reactants present – One generally calculates this in grams from info given actual yield: Amount of product one actually obtains – Generally smaller than the theoretical yield because of impurities and other adverse conditions in the lab – This value is generally given for a problem Example: For the reaction of 50.0 g of N2 with 10.0 g of H2, we determined the theoretical yield of anmmonia to be what? (See Ex. 2 on p. 13) theoretical yield = ________________________ If 49.6 g of ammonia was actually produced, calculate the percent yield for the reaction. percent yield = CHEM 151 Chapter 3 page 14 of 14 Chapter 4: Chemical Reactions Problems: 1-3, 5-6, 11, 18, 20-40, 43b, 44a, 46, 47-56, 59-72, 79-92, 95-98, 101-104 solution: composed of a solute dissolved in a solvent solute: component present in smaller amount solvent: component present in greater amount aqueous solution: solution where water is the dissolving medium (the solvent) Evidence for Chemical Reactions 1. A gas is produced—indicated by bubbles 2. A precipitate (ppt) is formed when 2 solutions are combined 3. Heat energy change is noted – exothermic reaction: releases heat (reaction vessel feels hot) – endothermic reaction: absorbs heat (reaction vessel feels cold) Types of Chemical Reactions Precipitation Reactions (also called double-replacement reactions) Acid-Base Neutralization Reaction Oxidation-Reduction (Redox) Reactions (also called combination, decomposition, and single-replacement reactions) 4.3 PRECIPITATION REACTIONS soluble = ions stay in solution (no solid/precipitate) insoluble = solid/precipitate forms Solubility Rules: Indicate if an ionic compound is soluble/insoluble in water. Solubility Rules for Ionic Compounds in Water Soluble if the ionic compound contains: Insoluble if the ionic compound contains: 1. Li+, Na+, K+, NH4+ (ALWAYS!) 6. Carbonate ion (CO32-), but Li+, Na+, K+, NH4+ 2. Nitrate ion (NO3–) 7. Chromate ion (CrO42-), but Li+, Na+, K+, NH4+ 3. Acetate ion (C2H3O2–) 8. Phosphate ion (PO43-), but Li+, Na+, K+, NH4+ 4. Halide ions (X–): chloride (Cl–), bromide (Br–), or iodide ion (I–), but AgX, PbX2, and Hg2X2 are insoluble 9. Sulfide ion (S2–), but compounds with Li+, Na+, K+, NH4+, and CaS, SrS, and BaS are soluble. (SO42-), but CaSO4, BaSO4, 5. Sulfate ion and PbSO4 are insoluble. CHM 151: Chapter 4 Notes 10. Hydroxide ion (OH–), but compounds with Li+, Na+, K+, NH4+, Ca(OH)2, Sr(OH)2, and Ba(OH)2 are soluble. page 1 of 18 Ex. 1 Use the Solubility Rules to predict whether the following ionic compounds are soluble or insoluble. Indicate the physical state as (aq) for soluble compounds and (s) for insoluble compounds. a. NaCl e. CaS i. K2CO3 b. CuS f. Li2CrO4 j. Ag3PO4 c. Na2SO4 g. Mg(OH)2 k. Li2S d. KOH h. BaSO 4 l. (NH4)2S In a precipitation reaction, two solutions react to form a precipitate: AX (aq) + BZ (aq) AZ (s) + BX (aq) To balance and complete the following reactions: 1. Exchange the anions, writing the formulas for the products based on the charges of the ions! 2. Use the Solubility Rules to determine if each product is soluble or insoluble – If one (or more) is/are insoluble, a precipitate reaction has occurred, so write the formulas for the products, indicating the precipitate as (s), then balance the equation. – If both products are soluble—both (aq)—then write NR=no reaction. 1. MgSO4 (aq) 2. CaCl2 (aq) 3. KC2H3O2 (aq) CHM 151: Chapter 4 Notes NaOH (aq) + AgNO3 (aq) + + LiNO3 (aq) page 2 of 18 4.4 ACID-BASE (NEUTRALIZATION) REACTIONS: Arrhenius Definitions: acid: A substance that releases hydrogen ions (H+) when dissolved in water – Some acids are monoprotic (release only H+ per molecule) – e.g. HCl, HBr, HI, HNO3, HClO4 – Some acids are polyprotic (release more than on H+ per molecule) – e.g. H2SO4 and H2CO3 are both diprotic, H3PO4 is triprotic base: A substance that releases hydroxide ions (OH–) when dissolved in water In an acid-base reaction, – Hydrogen ions (H+) from acid react with the hydroxide ions (OH–) from base water, H2O – The cation (M+) from the base combines with the anion from the acid (X–) the salt. A general equation for an acid-base neutralization reaction is shown below: HX (aq) + MOH (aq) acid H2O (l) + MX (aq) base water salt Because water is always produced, an acid always reacts with a base! Ex. 1 Complete and balance each of the equations below: NaOH (aq) a. HCl (aq) + b. H2SO4 (aq) + KOH (aq) c. H3PO4 (aq) + Ca(OH)2 (aq) CHM 151: Chapter 4 Notes page 3 of 18 Brønsted-Lowry Definitions: acid: A substance that donates a proton (H+) base: A substance that accepts a proton (H+) – It need not contain hydroxide ion (OH–). According to Brønsted-Lowry, an acid-base reaction simply involves a proton transfer, not necessarily the formation of water and a salt. Consider the following acid-base reaction: HCl (aq) + NH3 (aq) NH4+ (aq) B-L acid + Cl– (aq) B-L base Ex. 1 Indicate the Brønsted-Lowry acid and base in each of the following: a. NH3 (aq) b. NH3 (aq) + H2O (l) + NH4+ (aq) HNO3 (l) NH4+ (aq) OH– (aq) + + NO3– (aq) Acid-Base Reactions with Gas Formation – Some acid-base reactions involve the formation of carbon dioxide gas, CO2 (g), in addition to water and a salt. – When the base contain carbonate ion (CO32–) or hydrogen carbonate ion (HCO3–), then the products of the acid-base reaction are water, carbon dioxide gas, and a salt. The general equations for the unbalanced acid-base reactions are below: HX (aq) + MCO3 (aq) acid base HX (aq) + MHCO3 (aq) acid base H2O (l) + CO2 (g) + MX (aq) water carbon dioxide salt H2O (l) + CO2 (g) + MX (aq) water carbon dioxide salt Because water is always produced, an acid always reacts with a base! CHM 151: Chapter 4 Notes page 4 of 18 Ex. 1 Complete and balance each of the equations below: Na2CO3 (aq) a. HCl (aq) + b. HNO3 (aq) + CaCO3 (aq) c. H2SO4 (aq) + KHCO3 (aq) 4.1 Ionic Theory of Solutions and Solubility Rules Strong Acids Strong Bases HCl, HBr , HI, HNO3, HClO4, H2SO4 LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 strong electrolytes: substances that are strong/good conductors of electricity – strong acids, strong bases, all soluble salts – These substances dissociate to produce many ions in water many ions present to move electrons/conduct electricity strong electrolyte weak electrolytes: substances that are weak/poor conductors of electricity – weak acids, weak bases, insoluble salts – These substances barely dissociate to produce only a few ions in water few ions to move electrons/conduct electricity weak electrolyte nonelectrolytes: substances that cannot conduct electricity – sugar (e.g. sucrose), ethanol (C2H5OH), and other molecules that are not acids – These molecules do not break down into ions. – these compounds remain intact as neutral molecules that have no charge no ions to move electrons/conduct electricity. CHM 151: Chapter 4 Notes page 5 of 18 4.2 MOLECULAR AND IONIC EQUATIONS molecular equation: chemical equation showing reactants and products are compounds total/complete ionic equation: – shows strong electrolytes as individual ions while all solids, liquids, gases, and weak electrolytes remain intact as compounds spectator ions: ions that do not form solids, liquids, gases, or weak electrolytes – appear on both sides of total ionic equation as ions net ionic equations: shows only solids, liquids, gases, weak electrolytes (weak acids and weak bases), and ions undergoing reaction – excludes spectator ions Guidelines for Writing Net Ionic Equations 1. Balance the chemical equation. 2. Convert the molecular equation to total ionic equation – Leave solids, liquids, gases, and weak acids and bases as compounds – Break down strong acids, strong bases, all aqueous salts—show as (aq) 3. Cancel spectator ions to get net ionic equation – If canceling spectator ions eliminates all ions NO REACTION (NR) – If coefficients can be simplified, do so to get the lowest ratio. 4. Make sure total charges (+ve and –ve) are equal on both sides of equation. Write the net ionic equation for each of the following: a. H2SO4 (aq) + CHM 151: Chapter 4 Notes NaOH (aq) H2O (l) + Na2SO4 (aq) page 6 of 18 b. H3PO4 (aq) + c. Na2SO4 (aq) + d. KBr (aq) + KOH (aq) H2O (l) + K3PO4 (aq) NaNO3 (aq) + Ba(NO3)2 (aq) Cu(C2H3O2)2 (aq) CuBr2 (aq) + BaSO4 (s) KC2H3O2 (aq) 4.10 OXIDATION-REDUCTION (REDOX) REACTIONS Combination Reactions: A + Z AZ – Usually, a metal and a nonmetal react to form a solid ionic compound: metal + nonmetal where ionic compound (s) indicates the reactants are also heated. Ex. 1 Complete and balance each of the equations below: a. Na (s) + Cl2 (g) b. Al (s) + O2 (g) c. Zn (s) + S8 (s) CHM 151: Chapter 4 Notes page 7 of 18 Decomposition Reactions: AZ A + Z Balance the following decomposition reactions: a. _____ Ca(HCO3)2 (s) _____ CaCO3 (s) + _____ H2O (l) + _____ CO2 (g) b. _____ KClO3 (s) MnO 2 , _____ KCl (s) + _____ O2 (g) ACTIVITY SERIES: Relative order of elements arranged by their Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn > Fe > Cd > Co > Ni > Sn > Pb > (H) > Cu > Ag > Au Note: The Activity Series will be given to you on quizzes and exams. Displacement Reactions: A + BZ metal A + aqueous soln B AZ + B aqueous soln A + metal B To balance and complete the following rxns: – Check the Activity Series to see which metal is more active. The more active metal will prefer to be in solution (aq). 1. Mg (s) + CdSO4 (aq) 2. Cd (s) + CuSO4 (aq) 3. Cu (s) + ZnSO4 (aq) CHM 151: Chapter 4 Notes page 8 of 18 To balance and complete the following rxns: – Check the Activity Series to see which metal is more active, the metal or H. The more active metal will prefer to be in solution (aq). metal A + aqueous acid solution 1. Zn (s) + 2. Cu (s) + aqueous solution A + H2 (g) HCl (aq) HCl (aq) ACTIVE METALS: Li > K > Ba > Sr > Ca > Na — React directly with water active metal + H2O (l) metal hydroxide + H2 (g) 1. Na + H2O (l) 2. Fe + H2O (l) COMBUSTION RXNS: 1. C3H8 (g) + 2. C6H6O (l) + 3. C2H2 (g) + CHM 151: Chapter 4 Notes CxHy + O2 (g) H2O (g) + CO2 (g) CxHyOz + O2 (g) H2O (g) + CO2 (g) O2 (g) O2 (g) O2 (g) page 9 of 18 OXIDATION NUMBERS: actual or hypothetical charge of an atom in a compound if it existed as a monatomic ion Guidelines for Assigning Oxidation Numbers 1. The oxidation number of an element in its natural form is 0. – e.g. the oxidation number is zero for each element in H2, O2, Cl2, P4, Na, etc. 2. The oxidation number of a monatomic ion is the charge on the ion. – e.g. the oxidation number of Na in Na+ is +1; the oxidation number of N in N3– is -3; the oxidation numbers for Al2O3 are +3 for Al and –2 in O. 3. In a compound or polyatomic ion, – Group I elements are always +1. – Group II elements are always +2. – Fluorine is always –1. – Oxygen is usually –2 (except in the peroxide ion, O22–, when O is –1) – Hydrogen is usually +1 (except when it is with a metal, like NaH or CaH2, then it is –1) 4. In a compound, the sum of all oxidation numbers must equal 0. In a polyatomic ion, the sum of all oxidation numbers must equal charge. Example: Determine the oxidation number for each element in the following: a. H2SO4: H: _____, S: _____, O: _____ b. KClO3: K: _____, Cl: _____, O: _____ c. CaCr2O7: Ca: _____, Cr: _____, O: _____ d. C2O42–: C: _____, O: _____ e. Ni(OH)2: Ni: _____, O: _____, H: _____ Oxidation: lose electrons (oxidation number goes up) Reduction: gain electrons (oxidation number goes down) In a redox reaction – One reactant Loses Electrons/is Oxidized (LEO) – Another reactant Gains Electrons/is Reduced (GER) An easy way to remember is “LEO the lion goes GER!” – The element or reactant that is oxidized is the reducing agent. – The element or reactant that is reduced is the oxidizing agent. CHM 151: Chapter 4 Notes page 10 of 18 For each of the following reactions, 1. Balance the equation. 2. Identify the reactant that is oxidized and the reactant that is reduced. 3. Identify the oxidizing agent and the reducing agent. a. Zn (s) + b. Mn (s) + O2 (g) c. Ca (s) + H2O (l) d. C2H2 (g) + e. KClO3 (s) f. H2O2 (aq) AgNO3 (aq) O2 (g) Zn(NO3)2 (aq) + MnO2 (s) KCl (s) + Ag (s) H2O (l) + Ca(OH)2 (aq) + CO2 (g) + H2 (g) H2O (g) O2 (g) O2 (g) Example f is a disproportionation reaction, where an element in one oxidation state is simultaneously oxidized and reduced. CHM 151: Chapter 4 Notes page 11 of 18 WORKING WITH SOLUTIONS 4.7 Molar Concentration = Molarity Molarity = moles of solute (reported in units of M=molar) liters (L) of solution Ex. 1 Find the molarity of a solution prepared by dissolving 0.100 mol of NaCl in 250.0 mL of solution: Ex. 2 Find the molarity of a solution prepared by dissolving 1.25 g of KOH in 250.0 mL of solution: Ex. 3 Indicate the molarity of each ion in the solutions indicated below: Ex. 4 a. [Cl–] = _____________ in 0.500 M CaCl2 (aq) b. [Na+] = _____________ in 0.125 M Na3PO4 (aq) c. [NO3–] = _____________ in 1.500 M Mg(NO3)2 (aq) d. [SO42–] = _____________ in 1.250 M Al2(SO4)3 (aq) Circle the solution in each set with the highest [H+]: a. 0.100 M HF (aq) b. 0.100 M H2CO3 (aq) Ex. 5 0.100 M HCl (aq) 0.100 M H2SO3 (aq) 0.100 M HNO2 0.100 M H2SO4(aq) Explain why the hydroxide ion concentration, [OH–], in a 1.00 M NH4OH solution is not 1.00 M. CHM 151: Chapter 4 Notes page 12 of 18 4.8 Diluting Solutions Dilution Equation: M1 V1 = M2 V2 where M1=initial molarity, V1=initial volume, M2=final molarity, V2=final volume Ex. 1: What is the molarity of a HCl solution prepared by diluting 15.0 mL of 6.00 M HCl to give a total volume of 100.0 mL? Ex. 2: What is the molarity of a NaOH solution prepared by diluting 12.5 mL of 0.500 M NaOH to give a total volume of 50.0 mL? Writing Molar Concentration Unit Factors and Molarity Calculations: Ex. 1 Write 2 unit factors for each of the following: a. 6.00 M HCl solution b. 0.125 M NaCl solution Ex. 2 Calculate the number of moles of HCl present in 50.0 mL of 6.00 M HCl. Ex. 3 Calculate the mass of NaOH in 25.0 mL of a 0.500 M NaOH solution. CHM 151: Chapter 4 Notes page 13 of 18 Ex. 4 What volume (in L) of a 0.250 M NaCl solution contains 5.00 g of NaCl? Solution Stoichiometry Ex1. One important property of oxalic acid (H2C2O4) is its ability to remove rust (Fe2O3), as shown in the following equation: Fe2O3 (s) + 6 H2C2O4 (aq) 2 Fe(C2O4)3–3(aq) + 3 H2O (l) + 6 H+(aq) Calculate the mass of rust in grams that can be removed with 175 mL of a 0.250 M oxalic acid solution. Ex2. Barium hydroxide and sodium sulfate react to form barium sulfate precipitate. 2 AgNO3 (aq) + CaCl2 (aq) 2 AgCl (s) + Ca(NO3)2 (aq) Calculate the amount of precipitate formed when 22.75 mL of 0.820 M silver nitrate reacts with excess calcium chloride. CHM 151: Chapter 4 Notes page 14 of 18 4.6 Volumetric Analysis standard solution: an acid or base solution where the concentration is known, generally to 3 sig figs — used to analyze properties of substances, such as neutralizing power of commercial antacids, tartness of wine, etc. acid-base indicators: – Solutions that are pH sensitive & change color – Generally have their color change occurring for pH7 since reactions monitored are neutralization reactions, which are complete at pH=7 titration: The gradual addition of standard solution to another solution of unknown concentration until the reaction between the two is complete, as signaled by an indicator changing color endpoint: When one reactant has completely reacted with the other reactant, as evidenced by an indicator changing color Ex1. Find the molarity of a HCl solution if 25.50 mL of HCl is required to neutralize 0.375 g of Na2CO3 as shown in the following equation: 2 HCl (aq) + Na2CO3 (aq) 2 NaCl (aq) + H2O (l) + CO2 (l) Ex2. Find the molarity of a NaOH solution if 42.15 mL of NaOH is required to neutralize 0.424 g of oxalic acid, H2C2O4, as shown in the following balanced equation: 2 NaOH (aq) + H2C2O4 (aq) CHM 151: Chapter 4 Notes Na2C2O4 (aq) + 2 H2O (l) page 15 of 18 Ex 3. A 10.0 mL sample of vinegar (or acetic acid, HC2H3O2) requires 37.55 mL of a 0.255 M NaOH solution for complete neutralization. Calculate the molarity of the acetic acid solution if the balanced equation for the reaction is: HC2H3O2 (aq) + NaOH (aq) H2O (l) + NaC2H3O2 (aq) Ex 4. A 10.0 mL sample of battery acid (H2SO4) is titrated with 0.275 M NaOH. If the acid concentration is 0.555 M, what volume of NaOH is required for the titration? H2SO4 (aq) + 2 NaOH (aq) 2 H2O (l) + Na2SO4 (aq) Ex 5. Citric acid (abbreviated H3Cit) is a triprotic acid—ie. it has three H+ ions that can react to produce water. If 36.10 mL of 0.223 M NaOH is used to neutralize a 0.515 g sample of citric acid, what is the molar mass of the acid? H3Cit (aq) + 3 NaOH (aq) CHM 151: Chapter 4 Notes Na3Cit (aq) + 3 H2O (l) page 16 of 18 solution: composed of a solute dissolved in a solvent solute: component present in smaller amount solvent: component present in greater amount MASS PERCENT CONCENTRATION (M/M%) M/M% = mass of solute mass of solute 100% 100% mass of solution mass of solute + mass of solvent Ex1. What is the mass percent concentration of a solution made by dissolving 25.0 g of HCl in 65.0 g of water? (What is the solute, and what is the solvent?) Ex2. A person accused of a DUI violation submitted a 5.00 g sample of blood for alcohol content analysis. The analysis determined the presence of 4.59 mg of alcohol in the blood. If a person with a blood alcohol content (mass percent of alcohol in the blood) of 0.08% is considered legally impaired, was this person driving while impaired—i.e., was the blood alcohol content greater than or equal to 0.08%? Ex3. Intravenous injections of glucose are sometimes administered to patients with low blood sugar. If a normal glucose solution is 5.00%, what is the mass of solution that contains 12.7 g of glucose? CHM 151: Chapter 4 Notes page 17 of 18 Ex4. Intravenous saline injections are sometimes administered to restore electrolyte balance in trauma patients. What is the mass of water required to dissolve 2.00 g of NaCl for a 0.90% saline solution? CHM 151: Chapter 4 Notes page 18 of 18 CHAPTER 5: GASES AND THE KINETIC-MOLECULAR THEORY Problems: 1, 10, 20, 23-46, 51-81, 83-94, 101-103, 105-107, 109, 113-114 AN OVERVIEW OF THE PHYSICAL STATES OF MATTER At "normal atmospheric conditions" (25°C and 1 atm) – Elements that are gases: H2, N2, O2, F2, Cl2, ozone (O3), all noble gases – No ionic compounds exist as gases – Some molecules are gases (CO, CO2, HCl, NH3, CH4); most are solids or liquids Physical characteristics of gases – Gases assume the volume and shape of their containers – Gases are the most compressible of the states of matter – Gases will mix evenly and completely when confined to the same container – Gases have much lower densities than liquids and solids; in units of g/L 5.1 GAS PRESSURE AND ITS MEASUREMENT – Because gas molecules are in constant motion, gases exert pressure on any surface they encounter Atmospheric Pressure: – pressure exerted by column of air on an area exposed to Earth's atmosphere – depends on location, temperature, and weather conditions – decreases as altitude increases because air becomes thinner ! About 760 mmHg at sea level barometer: instrument that measures atmospheric pressure Units of Pressure Standard Atmospheric Pressure (1 atm): 760 mmHg at 0°C at sea level 1 atm ! " 760 mmHg " 760 torr = 101.325 kPa Ex. 1: If the atmospheric pressure is measured to be 725 mmHg on a given day in Phoenix, express this atmospheric pressure in torr, atm, and kPa. CHM 151: Chapter 5 Notes Page 1 of 10 5.2 EMPIRICAL GAS LAWS Boyle’s Law: Pressure-Volume Changes – When T and n are constant, V is indirectly proportional to pressure of the gas Charles’ Law: Volume-Temperature Changes – When P and n are constant, V is directly proportional to the absolute T Gay-Lussac’s Law: Pressure-Temperature Changes – When V and the n are constant, P is directly proportional to the absolute T Combined Gas Law – Consider when there are changes in P,V, and T for a gas, but n and R remain constant Exercises: If a value (P,V,n, or T) is not given, that value is constant. Ex. 1. If 25.0 mL of hydrogen gas are heated from 225K to 675K, calculate the new volume. CHM 151: Chapter 5 Notes Page 2 of 10 Ex. 2. A 250.0-mL sample at 1.20 atm is compressed to 125.0 mL. Calculate the new pressure. Ex. 3. A sample of CO2 gas at 795 torr is cooled from 25°C until the new pressure is 125 torr. Calculate the new temperature of the gas. Ex. 4. A sample of krypton gas at –80.0°C and 1245 torr occupies 50.5 mL. What is the volume at STP (1.00 atm and 0.00°C)? 5.3 The Ideal Gas Law: PV=nRT where P=pressure (in atm), V=volume (in L), n=# of moles of gas, L ! atm T=temperature (in K), and the ideal gas constant, R = 0.0821 mol ! K Standard temperature and pressure (STP): 0°C and 1 atm Ex. 1: Calculate the volume for 1.00 mole of gas at STP. CHM 151: Chapter 5 Notes Page 3 of 10 Ideal Gas Calculations NOT at STP Ex. 2: How many moles of NO2 gas occupy a volume of 5.00 L at 50.00°C and 735 torr? Ex. 3. Calculate the mass of nitrogen gas that occupies a volume of 75.0 L at 35.00°C and 2.50 atm. (Remember that nitrogen exists as a diatomic molecule.) Further Applications of the Ideal Gas Law: Density and Molar Mass Ex. 1: Calculate the density (in g/L) of H2S (g) at STP. Ex. 2: An unknown gas having a mass of 2.041g occupies a volume of 1.15 L at 740 torr and 20.0°C. Calculate the molar mass of the unknown gas. Ex. 3: Calculate the density of ammonia (NH3) in g/L at 645 mmHg and 65°C. – Use R and solve using unit analysis. CHM 151: Chapter 5 Notes Page 4 of 10 5.4 STOICHIOMETRY PROBLEMS INVOLVING GAS VOLUMES – The first step in every stoichiometry problem is getting moles! At STP, use PV 22.4L , and for other conditions, start with: n = mol RT Ex. 1 Magnesium nitride reacts with water to give ammonia gas: Mg3N2 (s) + 6 H2O (l) " 3 Mg(OH)2 (s) + 2 NH3 (g) What volume of ammonia would be produced at STP given 5.00 g of magnesium nitride and excess water? Ex. 2 When heated, NH4NO3 decomposes to give off steam and nitrogen gas: NH4NO3 (s) " 2 H2O (g) + N2 (g) How many grams of ammonium nitrate are required to produce 6.37 L of steam at 25.00°C and 720.0 torr? Ex. 3 The active agent in many hair bleaches is hydrogen peroxide, H2O2. The amount of hydrogen peroxide present can be determined by titration with a standard permanganate, MnO4–, solution: 2 MnO4– (aq) + 5 H2O2 (aq) + 6 H+ (aq) " 5 O2 (g) + 2 Mn+2 (aq) + 8 H2O (l) Calculate the molarity of hydrogen peroxide if 28.75 mL of hydrogen peroxide produced 695 mL of oxygen gas at 0.950 atm and 315 K? CHM 151: Chapter 5 Notes Page 5 of 10 The Volume-Amount Relationship: Avogadro's Law Avogadro's Law: At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas present. " When the reactants and products in a chemical equation are gases, we can relate the amounts of each gas to each other in terms of a volume-to-volume ratio, just like the mole-to-mole ratio. Example: 2 SO2 (g) + O2 (g) " 2 SO3 (g) 2 mol 1 mol 2 mol 2L 1L 2L All Gaseous Species in a Reaction – If all the reactants and products involved in a problem are gases, use volumevolume ratios since Avogradro’s Law applies. Ex. 1: Consider the reaction to produce sulfur trioxide gas: 3 H2 (g) + N2 (g) " 2 NH3 (g) Calculate the volume of ammonia formed if 5.00 L of hydrogen gas reacts with 3.00 L of nitrogen gas. 5.5 GAS MIXTURES; LAW OF PARTIAL PRESSURES partial pressure: pressures of individual gas components in a mixture Dalton's Law of Partial Pressure: – total pressure of a mixture of gases is the sum of the partial pressures of the gases present PTotal = P1 + P2 + P3 + ... Example: A mixture of gases contains nitrogen (N2), oxygen (O2), and trace gases. Given the following partial pressures for the gases: P N2 = 0 .78 atm P O2 = 0 .19 atm P trace = 0 .05 atm Determine the total pressure for the mixture. CHM 151: Chapter 5 Notes Page 6 of 10 mole fraction: ratio of the number of moles of one component to sum total of all the moles of all components X A !" # of moles of A total # of moles of all gases in mixture We can use mole fraction to calculate the partial pressure of a gas in a mixture – For system with more than many gases, the partial pressure of nth component: Pn = Xn PTotal Ex. 1 A mixture of gases contains 4.46 mol of neon, 0.74 mol of argon, and 2.15 mol of xenon. Calculate the partial pressures of the gases if the total pressure is 2.00 atm at a given temperature. 5.6 KINETIC THEORY OF AN IDEAL GAS 1. Particle volume: Gas molecules are so small they occupy no volume – i.e. "points" that have mass but negligible volume 2. Particle motion: Gas molecules move in constant, random, straight-line motion. 3. Particle attraction: Gas molecules do not attract or repel one another – i.e. they have no effect on the motion of other particles except upon collision 4. Particle collisions: Collisions are perfectly elastic – i.e. energy totally transferred from one molecule to another, and overall energy of the system remains the same; no energy lost to friction 5. The average kinetic energy (KE or energy of motion) of the molecule is proportional to the temperature of the gas in Kelvins " at higher temperatures, gas molecules move more quickly " at lower temperatures, gas molecules move more slowly CHM 151: Chapter 5 Notes Page 7 of 10 5.7 MOLECULAR SPEEDS; DIFFUSION AND EFFFUSION kinetic energy: energy associated with the motion of an object KE = 1 (mass)(speed)2 2 Molecular Speeds – In 1860, Maxwell formulated an equation to determine the speed of molecules at a given instant Combining ideal gas equation and KMT, we can get root-mean-square speed, urms: urms = where R=0.0821 3RT MM L ! atm , T in Kelvins, MM=molar mass mol ! K This equation indicates that the higher the molar mass of a particle, the slower the particle moves. Same Gas at Two Different Temperatures Example: Given two distribution curves corresponding to N2 at 0°C and 500°C, match each curve with the corresponding temperature. # of molecules molecular speeds Note: At higher temperatures, the curve for any gas flattens out because more molecules have higher molecular speeds. CHM 151: Chapter 5 Notes Page 8 of 10 Different Gases at the Same Temperature: Relationship Between Molar Mass and Molecular Speed Example: Given the distribution curves of equal samples of H2 and N2 at STP, match each curve with the corresponding gas, H2 or N2. # of molecules molecular speeds Note: Heavier molecules move slower than lighter molecules Diffusion and Effusion Diffusion: gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties Example: Consider a tube where gas molecules can be injected into both ends. If a sample of neon gas and a sample of fluorine gas are injected into opposite ends of the tube at the same time, predict where along the tube the two gases would meet. Explain why. Ne F2 Effusion: process of a gas under pressure escaping from a container via a small opening. Example: Consider a container sealed with a small opening. A mixture of several gases is added to the container. Assuming the gases do not react, indicate the order that the gases escape out of the container, starting with the gas that escapes the fastest. Explain why. CHM 151: Chapter 5 Notes Ar Ne Kr N2 O2 Page 9 of 10 Graham’s Law of Effusion – rate of effusion of molecules " # For 2 gases: 1 MMg as (same temp. and pressure) MMGas2 rate of effusion of Gas1 ! rate of effusion of Gas2 MMGas1 Ex. 1 Calculate the ratio of effusion rates for molecules of methane and carbon dioxide. Ex. 2 If it takes 10.0 s for a sample of neon to escape from a hole in a container, how long would it take for an equal amount of nitrogen gas, N2, to escape from the hole in the same container? CHM 151: Chapter 5 Notes Page 10 of 10 Chapter 6: Thermochemistry Problems: 1, 6, 8-11, 13, 18-22, 25-28, 33-60, 63-78, 85-94, 97-102 heat: form of energy that is transferred from a body at a higher temperature to one at a lower temperature – "heat flow" means heat transfer thermodynamics: study of heat and its transformations thermochemistry: study of heat flow that accompanies chemical reactions 6.1 ENERGY AND ITS UNITS energy: potential or capacity to move matter kinetic energy (KE): energy associated with an object’s motion – e.g. a car moving at 75 mph has much greater KE than the same car moving at 15 mph ! Greater damage if the car crashes at 75 mph than at 15 mph KE = 1 mv2 2 potential energy (PE): energy associated with an object’s position or its chemical bonds – A 10-lb bowling ball has higher PE when it’s 10 feet off the ground as opposed to 10 inches off the ground ! Greater damage on your foot if it hit your foot after falling 10 feet than if it hit after falling only 10 inches – In terms of chemical bonds, the stronger the bond ! more energy required to break the bond ! higher the potential energy of the bond joule (J): 1J= 1 kg " m2 s2 – SI (i.e. standard) unit of energy – To appreciate the size of a joule, note that 1 watt = 1 ! So a 100-watt bulb uses 100 J every second. J s – Heat can also be reported in kilojoules (kJ), where 1 kJ = 1000 J CHM 151: Chapter 6 Thermochemistry page 1 of 12 Example: Calculate the kinetic energy (in joules) Randy Johnson fastball if the baseball has a mass of 143 g and travels at 95 miles per hour. (1 mph = 0.4469 m/s approximately) calorie (cal): unit of energy used most often in the U.S. – amount of energy required to raise the temperature of 1 g of water by 1˚C 1 cal ! 4.184 J (Note: This is EXACT!) Example: What is the kinetic energy in calories of the Randy Johnson fastball described above? law of conservation of energy: 6.2 energy is neither created nor destroyed but converted from one form to another HEAT OF REACTION system: that part of the universe being studied surroundings: the rest of the universe outside the system Direction and Sign of Heat Flow Let q = heat flow, q is + when heat flows into the system from the surroundings q is – when heat flows out of the system into the surroundings CHM 151: Chapter 6 Thermochemistry page 2 of 12 heat of reaction (qreaction): heat associated with a chemical reaction Where Does the Heat of Reaction Come From? Bond Energy – energy required to break a particular bond in 1 mol of gaseous molecules – always positive since breaking a bond always requires energy – a quantitative measure of the strength of a bond (i.e. stability of molecule) Breaking and Forming Bonds – energy is absorbed by reactants when their bonds are broken, and energy is released by products when their bonds are formed If ! energy released $ ! energy required $ & # & # & > # when products' & # to break && ## && ## " bonds are formed% "reactants' bonds % ' endothermic reaction If ! energy released $ ! energy required $ & # & # & < # when products' & # to break && ## && ## " bonds are formed% "reactants' bonds % ' exothermic reaction Endothermic reaction: qreaction = + – Heat flows from surroundings to reaction system ' surroundings feel cooler – e.g. water evaporating is endothermic H2O (l) ' H2O (g) qreaction = +ve Exothermic reaction: qreaction = – – Heat flows from reaction system to surroundings: ' surroundings feel hotter – e.g. propane burning is exothermic C2H5OH (l) + 3 O2 (g) ' 3 H2O (g) + 2 CO2 (g) qreaction = –ve 6.6 MEASURING HEATS OF REACTION Heat Capacity and Specific Heat specific heat (s): amount of heat necessary to raise the temperature of 1 gram of any substance by 1°C; has units of J/g°C – water has relatively high specific heat (4.184 J/g°·C) – because the earth is covered by so much water ' why temperatures on earth do not vary by large degree CHM 151: Chapter 6 Thermochemistry page 3 of 12 heat capacity (C): amount of heat necessary to raise the temperature of a given amount of any substance by 1°C; in units of J/°C molar heat capacity: heat capacity per mole of a substance (in J/mol·°C) Use the following equations to solve problems: q = heat capacity ! ! T or q = (specific heat) ! (mass) ! ! T where !T=change in temperature Ex. 1. How much heat is released by a 150.0-g sample of copper that cools from 100.0°C to 25.0°C? (The molar heat capacity for copper is 24.6 J/mol·°C) Ex. 2 To raise the temperature from 23.5°C to 100.0 °C for a 15.5-g sample of silver, 279.9 J is required. What is the specific heat of silver? Ex. 3 A beaker with 250.1 g of water is heated from 25.0°C to its boiling point. If the specific heat of water is 4.184 J/g·°C, how much heat is required to heat the water? CHM 151: Chapter 6 Thermochemistry page 4 of 12 Ex. 4 When drinking an ice-cold beverage, a person must raise the temperature of the beverage to 37.0°C (normal body temperature). One argument for losing weight is to drink ice-cold beverages since the body must expend about 1 calorie per gram of water per degree Celsius—i.e. the specific heat of water = 1.00 cal/g·°C —to consume the drink. a. Calculate the amount of energy expended (in cal) to consume a 12oz beer (about 355 mL) if the beer is originally at 4.0°C. Assume the drink is mostly water and its density is 1.01 g/mL. b. If the label indicates 103 Calories (where 1 nutritional Calorie (Cal) equal 1 kcal), what is the net calorie gain or loss when a person consumes this beer? Is this a viable weight loss alternative? Measurement of Heat of Reaction calorimeter: – an instrument that measures heat changes for physical and chemical processes – insulated, so the only heat flow is between reaction system and calorimeter Coffee-Cup Calorimeter – also called constant-pressure calorimeter since under atmospheric pressure – polystyrene cup partially filled with water – since polystyrene is good insulator, very little heat lost through cup walls – heat evolved by a reaction is absorbed by water, and the heat capacity of calorimeter is the heat capacity of the water heat of reaction: CHM 151: Chapter 6 Thermochemistry qreaction = – mwater ! 4.184 J !""T g !˚C page 5 of 12 Ex. 1 A 28.2 g sample of nickel is heated to 99.8°C and placed in a coffee cup calorimeter containing 150.0 g of water at 23.5°C. After the metal cools, the final temperature of metal and water is 25.0°C. Calculate the specific heat of the metal assuming no heat is lost to the surroundings or the calorimeter. 6.2 Enthalpy and Enthalpy Change Enthalpy (H): refers to heat flow into and out of a system under constant pressure Under constant pressure (usually the case since under atmospheric pressure), qreaction = !H = Hproducts – Hreactants 6.4 For an endothermic reaction: ! H = +ve For an exothermic reaction: ! H = –ve Thermochemical Equations: thermochemical equation: shows both mass and enthalpy relationships Consider: Ice melting to form water at 0°C and 1 atm — requires energy (6.01 kJ/mol) We can represent the melting of 1 mol of ice as an equation: H2O (s) " H2O (l) !H = +6.01 kJ Note: ! H is +ve since ice must absorb heat to form water CHM 151: Chapter 6 Thermochemistry page 6 of 12 Consider: The formation of water releases heat: 2 H2 (g) + O2 (g) ! 2 H2O (g) "H = –571.6 kJ Note: " H is –ve since heat is lost to the surroundings Guidelines for Thermochemical Equations 1. Sign of "H indicates if reaction is exothermic ("H<0) or endothermic ("H>0). 2. The coefficients in the chemical equation represent the numbers of moles of reactants and products for the "H given. – e.g. " H is –185 kJ for 1 mol H2 + 1 mol Cl2 to form 2 mols of HCl 3. The physical states must be indicated for each reactant and product. – e.g. For water, the solid and the liquid states vary by 6.01 kJ 4. "H generally reported for reactants and products at 25°C. 6.5 APPLYING STOICHIOMETRY TO HEATS OF REACTION Rule 1. The magnitude of " H is directly proportional to the amount of reactant or product. ! Consider heat as you would a reactant or product in a mole-tomole ratio, where "H is the heat released or absorbed for the moles of reactants and products indicated in the equation Example: Consider H2 (g) + Cl2 (g) ! 2 HCl (g) "H = –185 kJ a. What is "H for when 5.00 mol of hydrogen gas reacts completely? b. What is "H for the formation of 1.00 g of hydrogen chloride gas? CHM 151: Chapter 6 Thermochemistry page 7 of 12 Rule 2. For a reverse rxn, ! H is equal in magnitude but opposite in sign. If H2O (s) " H2O (l) then H2O (l) " ! H fusion): heat of fusion (! H2O (s) !H = +6.01 kJ !H = –6.01 kJ heat associated with a solid melting, in kJ/mol; i.e. solid " liquid !H!"!!Hfusion ! H vapor): heat of vaporization (! heat associated with a liquid vaporizing, in kJ/mol; i.e. liquid " gas !H!"!!Hvapor Example: If a 24.6-g ice cube requires 8.19 kJ of heat to melt completely, calculate !Hfusion for ice (in kJ/mol)? Rule 3: If all the coefficients in a chemical equation are multiplied by a factor n " !H is multiplied by factor n: If H2O (s) " H2O (l) !H = +6.01 kJ " 2 [H2O (s) " H2O (l)] = 2 H2O (s) "2 H2O (l) !H = 2(+6.01 kJ) = 12.0 kJ " 1 1 1 [H2O (s) " H2O (s)] = H2O (s) " H2O (l) 2 2 2 !H = 6.7 1 (+6.01 kJ) =3.01kJ 2 Hess’s Law Hess’s Law of heat summation: The value of !H for a reaction is the same whether it occurs in one step or in a series of steps ! H = ! H1 + ! H2 + ! H3 + ... CHM 151: Chapter 6 Thermochemistry page 8 of 12 Example 1: Cgraphite (s) + 2 H2 (g) ! CH4 (g) Here are the reactions involved in the formation of CH4: (a) Cgraphite (s) + O2 (g) ! CO2 (g) "H = –393.5 kJ (b) 2 H2 (g) + O2 (g) ! 2 H2O (l) "H = –571.6 kJ (c) CH4 (g) + 2 O2 (g) ! CO2 (g) + 2 H2O (l) "H = –890.4 kJ Rearranging the data allows us to calculate "H for the reaction: (a)Cgraphite (s) + O2 (g) (b)2 H2 (g) + O2(g) (c) ! ! CO2 (g) "H = –393.5 kJ 2 H2O (l) "H = –571.6 kJ C O2 (g) + 2 H2O (l) ! Cgraphite (s) + 2 H2 (g) ! CH4 (g) + 2 O2 (g) CH4 (g) "H = 890.4 kJ "H = –74.7 kJ If the coefficients in an equation can be simplified, multiply by the necessary factors to cancel all intermediate compounds and get the correct coefficients for your final equation. Ex. 2 Rearrange the following data: (a) Cgraphite (s) + O2 (g) ! CO2 (g) "H = –393.5 kJ (b) 2 CO (g) + O2 (g) "H = 566.0 kJ ! 2 CO2 (g) to calculate the enthalpy change for the reaction: 2 Cgraphite (s) + O2 (g) ! 2 CO (g) CHM 151: Chapter 6 Thermochemistry page 9 of 12 Ex. 3 From the following data: (a)N2 (g) + 3 H2 (g) ! 2 NH3 (g) "H = –92.6 kJ (b)N2 (g) + 2 O2 (g) ! 2 NO2 (g) "H = (c) "H = –571.6 kJ 2 H2 (g) + O2 (g) ! 2 H2O (l) 67.70 kJ Calculate the enthalpy change for the reaction: 4 NH3 (g) + 7 O2 (g) ! 4 NO2 (g) + 6 H2O (l) 6.6 Standard Enthalpies of Formation ( "Hf! ) Definitions of standard state (or reference form) 1. A gas at 1 atm 2. An aqueous solution with a concentration of 1 M at a pressure of 1 atm 3. Pure liquids and solids 4. The most stable form of elements at 1 atm and 25°C allotrope: one or 2 or more forms of an element in the same physical state – e.g. diamond and graphite are allotropes of carbon; O 2 (g) and ozone, O3 (g) are allotropes of oxygen standard enthalpy of formation ( "H!f ) – enthalpy change for formation of 1 mole of an element or compound from its elements in their standard states (i.e. naturally occurring form at 1 atm and 25˚C) Note: "H!f =0 for any element in its naturally occurring (most stable) form CHM 151: Chapter 6 Thermochemistry page 10 of 12 Examples of !H!f are as follows: 1 Ag (s) + Cl2 (g) " AgCl (s) 2 "" 1 N2 (g) + O2 (g) " NO2 (g) 2 !H = –127.1 kJ " !H!f (AgCl, s) = -127.1 kJ !H = +33.2 kJ """ !H!f (NO2, g) = +33.2 kJ Example: Explain for each of the following if !H˚ is a heat of formation. a. N2 (g) + 3 H2 (g) " 2 NH3 (g) b. H2O (l) " H2O (g) c. Cgraphite (s) + 2 H2 (g) d. 2 CO (g) + O2 (g) e. CO2 (s) " CO2 (g) CHM 151: Chapter 6 Thermochemistry " CH4 (g) " 2 CO2 (g) page 11 of 12 Calculation of ! H° – superscript ° denotes taken under 1 atm and 25°C For the reaction: aA + bB " cC + dD where a,b,c,d = stoichiometric coefficients ! = # n !H!f (products) – # m !H!f (reactants) !Hrxn = [c !H!f (C) + d !H!f (D)] – [a !H!f (A) + b !H!f (B)] ! Use Table 6.2 to solve for !Hrxn for each of the following: Ex. 1: C graphite (s) + O2 (g) " CO2 (g) ! = (1mol) !H!f (CO2, g) – [(1mol) !H!f (Cgraphite, s) + (1mol) !H!f (O2, g)] !Hrxn = Ex. 2: 2 CH3OH (l) + 5 O2 (g) " 2 CO2 (g) + 4 H2O (g) ! = !Hrxn Ex. 3 Consider the reaction for the combustion of glucose: C6H12O 6 (s) + 6 O2 (g) " 6 CO2 (g) + 6 H2O (g) ! !Hrxn = –2803 kJ Calculate the enthalpy of formation for glucose using Table 6.2. CHM 151: Chapter 6 Thermochemistry page 12 of 12 Chapter 7: Quantum Theory of the Atom Problems: 1, 8-11, 15, 18, 25, 27-40, 49-54, 63-66, 71-72, 83-84 7.1 The Wave Nature of Light – Light travels through space as a wave – Waves have the following characteristics (see Fig. 7.4) ! =Greek “lambda”): distance between successive peaks wavelength (! != distance ; generally in units of m, cm, nm wave " =Greek “nu”): number of waves passing by a given point in 1 s frequency (" "= wave 1 cycle ; generally in hertz (Hz) = or time s s Electromagnetic (EM) spectrum:!! continuum of radiant energy (Fig. 7.5) visible region:!! ! portion of the EM spectrum that we can perceive as color For example, a "red-hot" or "white-hot" iron bar freshly removed from a hightemperature source has forms of energy in different parts of the EM spectrum – red or white glow = radiation within the visible region – warmth = radiation within the infrared region speed of light, c=3.00 x 108 m/s, depends on frequency and wavelength c " ! # " distance distance wave " # time wave time Know how to convert between wavelength and frequency using the speed of light! Ex. 1 The wavelength for the electromagnetic radiation responsible for a blue sky is about 473 nm. What is the frequency of this radiation in Hz? CHM 151: Chapter 7 Notes page 1 of 6 7.2 Quantum Effects and Photons Classical Descriptions of Matter John Dalton (1803) – atoms are hard, indivisible, billiard-like particles – atoms have distinct masses (what distinguishes on type of atom from another) – all atoms of same element are the same JJ Thomson (1890s) – discovered charge-to-mass ratios of electrons! ! atoms are divisible because the electrons are one part of atom Ernest Rutherford (1910) – shot positive alpha particles at a thin foil of gold! ! discovery of the atomic nucleus James Maxwell (1873) – visible light consists of electromagnetic waves Transition between Classical and Quantum Theory Max Planck (1900); Blackbody Radiation – heated solids to red or white heat – noted matter did not emit energy in continuous bursts, but in whole-number multiples of certain well-defined quantities ! matter absorbs/emits energy in bundles = "quanta" (single bundle of energy= "quantum") Albert Einstein (1905); Photoelectric Effect – Photoelectric Effect: Light shining on a clean metal ! ! emission of electrons – Einstein applied Planck's quantum theory to light ! !!light exists as a stream of "particles" called photons Energy is proportional to the frequency (") and wavelength (#) of radiation, and the proportionality constant (h) is now called Planck's constant E " h" = CHM 151: Chapter 7 Notes hc # where h = 6.626#10–34 J·s page 2 of 6 Ex. 1. Excited mercury atoms emit light strongly at a wavelength of 436 nm. a. What is the energy (in J) for one photon of this light? b. What is the energy (in kJ/mol) for a mole of photons of this light? Ex. 2. Certain elements emit light of a specific color when they are burned. When an potassium solution is burned in a flame test, the energy of the light emitted is 4.909!10-19 J. Calculate the wavelength for this light, and use the visible spectrum (Fig. 7.5) to determine the color of the light. 7.4 Quantum Mechanics Louis de Broglie (1924); Dual Nature of the Electron – If light can behave like a wave and a particle ! " matter (like an electron) can behave like waves – if electron behaves like a standing wave ! an electron can only have specific wavelengths " ! an electron can only have specific frequencies and thus, energies " for wave: E = mc2 for matter: E = h" Combining the two equations, we can solve for the wavelength for any matter. ! de Broglie relation# h #"$" mv CHM 151: Chapter 7 Notes page 3 of 6 Ex. 1 a. A baseball with mass 0.143 kg is thrown towards a batter at a velocity of 42.5 m/s, calculate the wavelength (in m) associated with the baseball’s motion. b. How does the !! compare in size to the baseball (diameter"0.08 m)? The baseball’s !!"""""""! baseball’s size. (Circle one) " >> << Ex. 2 a. Calculate the wavelength (in m) associated with an electron traveling at the same velocity, 42.5 m/s. (The mass of the electron is 9.1095 #10–31 kg.) b. How does the !! compare in size to the electron (diameter"1#10–10 m)? The electron ‘s !!"""""""! electron’s size. (Circle one) " >> << Thus, although all matter can have wave properties, such properties are only significant for submicroscopic particles. 7.3 Bohr Theory of the Hydrogen Atom Bohr Postulates: Bohr Model of the Atom 1. Energy-level Postulate – Electrons move in discrete (quantized), circular orbits around the nucleus – "tennis ball and stairs" analogy for electrons and energy levels – a ball can bounce up to or drop from one stair to another, but it can never be halfway between two levels – Each orbit has a specific energy associated with it, indicated as n=1, 2,... – ground state or ground level (n = 1): lowest energy state for atom –!when the electron is in most stable orbit – excited state: when the electron is in a higher energy orbit (n = 2,3,4,...) CHM 151: Chapter 7 Notes page 4 of 6 2. Transitions Between Energy Levels – When the atom absorbs energy, an electron can jump from a lower energy level to a higher energy level. – When an electron drops from a higher energy level to a lower energy level, the atom releases energy, sometimes in the form of visible light. Atomic Line Spectra Emission Spectra: continuous or line spectra of radiation emitted by substances – a heated solid (e.g. the filament in an incandescent light bulb) emits light that can be spread out to give a continuous spectrum = spectrum containing all wavelengths of light, like a rainbow – an atom in the gas phase emits light only at specific wavelengths = line spectrum – each element has a unique line spectrum ! !!can be used to identify unknown atoms in chemical analysis – why line spectrum for each element called "atomic fingerprint" Limitations of the Bohr Model ! Quantum Mechanical Model Unfortunately, the Bohr Model failed for all other elements that had more than one proton and one electron. (The multiple electron-nuclear attractions, electronelectron repulsions, and nuclear repulsions make other atoms much more complicated than hydrogen.) In 1920s, a new discipline, quantum mechanics, was developed to describe the motion of submicroscopic particles confined to tiny regions of space. – Quantum mechanics makes no attempt to specify the position of a small particle at a given instant or how the electron got there – It only gives the probability of finding small particles – Just like taking snapshot of a location and estimating where greatest number of people are likely to be ! Instead we take a snapshot of the atom at different times and “see” where the electrons are usually found (See. Figs. 7.24, 7.26) Erwin Schrödinger (1926) – developed a differential equation that allows us to find the electron's wave " ), which ultimately allows us to determine the probability of finding function (" the electron in a given place – probability density for an electron is called the "electron cloud" ! !“shape” Quantum Numbers, Energy Levels, and Orbitals – 4 quantum numbers describe distribution and behavior of electrons in atoms – Each wave function corresponds to a set of 3 quantum numbers and is referred to as an atomic orbital CHM 151: Chapter 7 Notes page 5 of 6 We will only discuss the first 2 Quantum Numbers: First (or Principal) Quantum Number (n): n=1,2,3,... – relates the average distance of electron from nucleus – higher n means electron is further from nucleus and higher energy (less stable) orbital Second (or Angular Momentum) Quantum Number (l); ! Sublevels (s, p, d, f) – gives the "shape" of the electron clouds associated with each orbital – The limitations on n and l ! ! for n=1, only 1s sublevel ! ! for n=2, only 2s and 2p sublevels ! ! for n=3, there are 3s, 3p, and 3d sublevels ! ! for n=4, there are 4s, 4p, 4d, and 4f sublevels Atomic Orbital Shapes: s orbital: spherical (see Fig. 7.25) p orbitals: dumbbell-shaped (see Fig. 7.26) – only for n=2 or greater – 3 types: px, py, pz (where x, y, and z give axis on which orbital aligns) d orbitals: various shapes (see Fig. 7.27) – only for n=3 or greater – 5 types: dxy, dxz, dyz, d x 2 " y 2 , dz2 f orbitals: Don’t worry about these. The interesting part... – Consider Fig. 7.26 showing the electron distribution for the 2px orbital. – Note that the electron is never along the y-axis. There is a “node” in that region, indicating the electron is never found there. ! !How can the electron be on either side of the y-axis without going through it? CHM 151: Chapter 7 Notes page 6 of 6 Chapter 8: Electron Configurations and Periodicity Problems: 7-8, 12, 15, 19, 23, 33-56, 61-66, 71-72 8.2 Building-Up Principle and the Periodic Table Electrons are distributed in orbitals of increasing energy levels, where the lowest energy orbitals are filled first. Once an orbital has the maximum number of electrons it can hold, it is considered “filled.” Remaining electrons must then be placed into the next highest energy orbital, and so on. – Parking garage analogy Orbitals in order of increasing energy: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 5d < 6p electron configuration: – Shorthand description of the arrangement of electrons by sublevel according to increasing energy REMEMBER! – s orbitals can hold 2 electrons – a set of p orbitals can hold 6 electrons – a set of d orbitals can hold 10 electrons – a set of f orbitals can hold 14 electrons Ex. 1 Li!! atomic number=3! ! 3 eelectron configuration for Li: Ex. 2 F ! ! _____ e– electron configuration for F: Ex. 3 Fe ! _____ e– electron configuration for Fe: CHEM 151: Chapter 8 Notes page 1 of 8 Exceptions to the Building-Up Principle Atoms gain extra stability when their d subshells are half-filled or completely filled. ! If we can fill or half-fill a d subshell by promoting an electron from an s orbital to a d orbital, we do so to gain the extra stability. Ex. 1 Cr!! _____ e– electron configuration for Cr: actual electron configuration for Cr: Ex. 2 Ag!!_____ e– electron configuration for Ag: actual electron configuration for Ag: 8.3 Writing Electron Configurations Using the Periodic Table Electron Configuration from the Periodic Table – The Periodic Table's shape actually corresponds to the filling of energy sublevels. – See Fig. 8.12 (p. 325), to see how electrons for each element are distributed into the energy sublevels. Electron configurations of atoms with many electrons can become cumbersome. ! Abbreviated electron configurations (“noble-gas core” notation): – Since noble gases are at the end of each row in the Periodic Table, all of their electrons are in filled orbitals. – Such electrons are called “core” electrons since they are more stable (less reactive) when they belong to completely filled orbitals. valence electrons: electrons that are in the outermost shell (unfilled orbitals) Noble gas electron configurations can be used to abbreviate the “core” electrons of all elements. [He] [Ne] [Ar] [Kr] [Xe] = 1s2 = 1s2 2s2 2p6 = 1s2 2s2 2p6 3s2 3p6 = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 CHEM 151: Chapter 8 Notes page 2 of 8 [Fe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 = [Cd] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 = [Ni] = [Y] = [I] = Remember again that for some transition metals, extra stability with filled and half-filled d orbitals, so electrons are excited from the s orbitals to d orbitals: [Cu] = [Mo] = 8.1 Electron Spin and the Pauli Exclusion Principle Each electron in an atom can have one of two possible orientations, spin ! or spin ". electron configuration: arrangement of electrons in an atom among sublevels CHEM 151: Chapter 8 Notes page 3 of 8 orbital diagram: diagram showing how electrons exist within an atom’s orbitals Pauli Exclusion Principle: no 2 e-s in an atom can have same four quantum #s ! Two electrons in the same orbital must have opposite spins – For example, with the helium atom, there are three ways to represent two electrons in 1s orbital (where spin is represented with the electron pointing up or down): for He: "" ## "# (a) (b) (c) – but the Pauli exclusion principle rules out (a) and (b) since these show two electrons in the same orbital with the same spin. 8.4 Orbital Diagrams of Atoms; Hund’s Rule Hund's Rule: the most stable arrangement of electrons in subshells has the greatest number of parallel spins – i.e. distribute electrons with same spin (up or down) and do not pair electrons until all subshells have an electron For example, if carbon’s electron configuration is: 1s2 2s2 2p2 ! carbon’s orbital diagram can be shown in the following ways: (a) "# "# "# "# "# " # "# "# " " 1s (b) 1s (c) 1s 2s 2s 2s 2p 2p 2p – but using Hund's rule, we know (c) would be the most stable. General Rules for Assigning Electrons in Atomic Orbital Diagrams 1. First, determine the electron configuration. 2. There is only one s orbital for each level: one 1s, one 2s, one 3s, etc. – There are 3 p orbitals for each p sublevel. – There are 5 d orbitals for each d sublevel. 3. Each orbital orbital can only hold 2 electrons ! Each s orbital can hold 2 electrons, each p sublevel can hold 6, d’s can hold 10. 4. Electrons in the same orbital must have opposite spins. 5. To fill sublevels, put one electron in each orbital (with same spin) before pairing. CHEM 151: Chapter 8 Notes page 4 of 8 Ex. 1 Draw the atomic orbital diagram for oxygen. Ex. 2 Draw the atomic orbital diagram for phosphorus. (Use full notation) Ex. 3 Draw the atomic orbital diagram for the valence (outermost shell) electrons in cobalt. (Use core notation.) Magnetic Properties of Atoms paramagnetic: substance that contains unpaired electrons ! weakly attracted to magnetic field diamagnetic: substance that contains only paired electrons ! slightly repelled by magnetic field To determine if a substance is paramagnetic or diamagnetic, you must draw its orbital diagram and determine if it has an unpaired electrons. Ex. Is cobalt paramagnetic or diamagnetic? CHEM 151: Chapter 8 Notes ____________________ page 5 of 8 8.3 Mendeleev’s Predictions from the Periodic Table Dimitri Mendeleev proposed that elements display recurring properties according to increasing atomic mass ! Periodic Table originally arranged with elements in order of increasing atomic mass Henry G.J. Moseley’s high-energy x-ray radiation experiments of atomic nuclei ! Repeating properties of elements more clearly reflected by the arrangement of elements according to increasing atomic number ! Periodic Table’s arrangement today Trends for increasing atomic mass are identical with those forincreasing atomic number, except for Ni & Co, Ar & K, Te & I. Neils Bohr’s introduction of electron energy levels ! Periodic Table’s shape – Indicates filling of electron orbitals and element’s electron configuration periodic law: When elements are arranged in terms of increasing atomic number, the trends within a row or column form patterns that allow us to predict the physical and chemical properties of elements 8.6 Some Periodic Properties Atomic radius: distance from nucleus to outermost electrons – Increases down a group: More p+, n, and e– ! bigger radius – Decreases from left to right along a period. effective nuclear charge: – positive charge an electron experiences because of charge from protons in the nucleus minus any shielding due to electrons closer to the nucleus Consider the atoms of Ti and Ni: Thus, the higher the effective nuclear charge, the smaller the atomic radius! CHEM 151: Chapter 8 Notes page 6 of 8 Atomic radius trend: – Trend from top to bottom ! like a snowman – Trend from left to right ! like a snowman that fell to the right First Ionization Energy: Energy necessary to remove first electron from a neutral atom in gaseous state to form negatively charged ion. First Ionization Energy TRENDS – Decreases down a group: Bigger the atom, the further away e–s are from +vely charged nucleus ! e–s held less tightly and are more easily removed – Increases from left to right along a period: – Elements with fewer (1–3) valence e–s can more easily give up e–s to gain noble gas configuration (stability) – Elements with more (4–7) valence e–s can more easily gain e–s togain noble gas configuration (stability) Trend from top to bottom ! like an upside-down snowman Trend from left to right ! like a upside-down snowman that fell to the right Variations in Successive Ionization Energies – Recognize that it becomes more difficult to remove electrons from stable ions. – See Table 8.3 on p. 337. – For example, the ionization energy to remove the first electron from Li is much smaller than the any successive ionization energy since electrons are now removed from a stable ion (with a positive charge AND sometimes a noble gas electron configuration). CHEM 151: Chapter 8 Notes page 7 of 8 We can indicate first and successive ionization energies in the following way: First ionization energy = IE1 Second ionization energy = IE2 Third ionization energy = IE3 Ex. 1: Between which two ionization energies (IE1 and IE2, IE2 and IE3, etc.) would you expect there to be the largest jump for the following elements? Explain. a. Mg: Between _____ and _____ Explain why: b. K: Between _____ and _____ Explain why: c. Al: Between _____ and _____ Explain why: Ex. 2: 8.7 This 3rd period element has a large jump between IE5 and IE6. Explain how you can identify the element. Periodicity in the Main-Group Elements Metallic character: – Decreases from left to right along a period: Metals concentrated on left-hand side of P.T., nonmetals on right-hand side – Increases down a group: Looking at groups IVA and VA, go from nonmetals (C & N) to semimetals (Si & As) to metals (Sn & Bi) ! Same snowman trends as for atomic radius! CHEM 151: Chapter 8 Notes page 8 of 8 Chapter 6: Thermochemistry Problems: 1, 6, 8-11, 13, 18-22, 25-28, 33-60, 63-78, 85-94, 97-102 heat: form of energy that is transferred from a body at a higher temperature to one at a lower temperature – "heat flow" means heat transfer thermodynamics: study of heat and its transformations thermochemistry: study of heat flow that accompanies chemical reactions 6.1 ENERGY AND ITS UNITS energy: potential or capacity to move matter kinetic energy (KE): energy associated with an object’s motion – e.g. a car moving at 75 mph has much greater KE than the same car moving at 15 mph Greater damage if the car crashes at 75 mph than at 15 mph KE = 1 mv2 2 potential energy (PE): energy associated with an object’s position or its chemical bonds – A 10-lb bowling ball has higher PE when it’s 10 feet off the ground as opposed to 10 inches off the ground Greater damage on your foot if it hit your foot after falling 10 feet than if it hit after falling only 10 inches – In terms of chemical bonds, the stronger the bond more energy required to break the bond higher the potential energy of the bond joule (J): 1J= 1 kg m2 s2 – SI (i.e. standard) unit of energy – To appreciate the size of a joule, note that 1 watt = 1 So a 100-watt bulb uses 100 J every second. J s – Heat can also be reported in kilojoules (kJ), where 1 kJ = 1000 J CHM 151: Chapter 6 Thermochemistry page 1 of 12 Example: Calculate the kinetic energy (in joules) Randy Johnson fastball if the baseball has a mass of 143 g and travels at 95 miles per hour. (1 mph = 0.4469 m/s approximately) calorie (cal): unit of energy used most often in the U.S. – amount of energy required to raise the temperature of 1 g of water by 1˚C 1 cal 4.184 J (Note: This is EXACT!) Example: What is the kinetic energy in calories of the Randy Johnson fastball described above? law of conservation of energy: 6.2 energy is neither created nor destroyed but converted from one form to another HEAT OF REACTION system: that part of the universe being studied surroundings: the rest of the universe outside the system Direction and Sign of Heat Flow Let q = heat flow, q is + when heat flows into the system from the surroundings q is – when heat flows out of the system into the surroundings CHM 151: Chapter 6 Thermochemistry page 2 of 12 heat of reaction (qreaction): heat associated with a chemical reaction Where Does the Heat of Reaction Come From? Bond Energy – energy required to break a particular bond in 1 mol of gaseous molecules – always positive since breaking a bond always requires energy – a quantitative measure of the strength of a bond (i.e. stability of molecule) Breaking and Forming Bonds – energy is absorbed by reactants when their bonds are broken, and energy is released by products when their bonds are formed If energy released energy required > when products' to break bonds are formed reactants' bonds endothermic reaction If energy released energy required < when products' to break bonds are formed reactants' bonds exothermic reaction Endothermic reaction: qreaction = + – Heat flows from surroundings to reaction system surroundings feel cooler – e.g. water evaporating is endothermic H2O (l) H2O (g) qreaction = +ve Exothermic reaction: qreaction = – – Heat flows from reaction system to surroundings: surroundings feel hotter – e.g. propane burning is exothermic C2H5OH (l) + 3 O2 (g) 3 H2O (g) + 2 CO2 (g) qreaction = –ve 6.6 MEASURING HEATS OF REACTION Heat Capacity and Specific Heat specific heat (s): amount of heat necessary to raise the temperature of 1 gram of any substance by 1°C; has units of J/g°C – water has relatively high specific heat (4.184 J/g°·C) – because the earth is covered by so much water why temperatures on earth do not vary by large degree CHM 151: Chapter 6 Thermochemistry page 3 of 12 heat capacity (C): amount of heat necessary to raise the temperature of a given amount of any substance by 1°C; in units of J/°C molar heat capacity: heat capacity per mole of a substance (in J/mol·°C) Use the following equations to solve problems: q = heat capacity T or q = (specific heat) (mass) T where T=change in temperature Ex. 1. How much heat is released by a 150.0-g sample of copper that cools from 100.0°C to 25.0°C? (The molar heat capacity for copper is 24.6 J/mol·°C) Ex. 2 To raise the temperature from 23.5°C to 100.0 °C for a 15.5-g sample of silver, 279.9 J is required. What is the specific heat of silver? Ex. 3 A beaker with 250.1 g of water is heated from 25.0°C to its boiling point. If the specific heat of water is 4.184 J/g·°C, how much heat is required to heat the water? CHM 151: Chapter 6 Thermochemistry page 4 of 12 Ex. 4 When drinking an ice-cold beverage, a person must raise the temperature of the beverage to 37.0°C (normal body temperature). One argument for losing weight is to drink ice-cold beverages since the body must expend about 1 calorie per gram of water per degree Celsius—i.e. the specific heat of water = 1.00 cal/g·°C —to consume the drink. a. Calculate the amount of energy expended (in cal) to consume a 12oz beer (about 355 mL) if the beer is originally at 4.0°C. Assume the drink is mostly water and its density is 1.01 g/mL. b. If the label indicates 103 Calories (where 1 nutritional Calorie (Cal) equal 1 kcal), what is the net calorie gain or loss when a person consumes this beer? Is this a viable weight loss alternative? Measurement of Heat of Reaction calorimeter: – an instrument that measures heat changes for physical and chemical processes – insulated, so the only heat flow is between reaction system and calorimeter Coffee-Cup Calorimeter – also called constant-pressure calorimeter since under atmospheric pressure – polystyrene cup partially filled with water – since polystyrene is good insulator, very little heat lost through cup walls – heat evolved by a reaction is absorbed by water, and the heat capacity of calorimeter is the heat capacity of the water heat of reaction: CHM 151: Chapter 6 Thermochemistry qreaction = – mwater 4.184 J T g ˚C page 5 of 12 Ex. 1 A 28.2 g sample of nickel is heated to 99.8°C and placed in a coffee cup calorimeter containing 150.0 g of water at 23.5°C. After the metal cools, the final temperature of metal and water is 25.0°C. Calculate the specific heat of the metal assuming no heat is lost to the surroundings or the calorimeter. 6.2 Enthalpy and Enthalpy Change Enthalpy (H): refers to heat flow into and out of a system under constant pressure Under constant pressure (usually the case since under atmospheric pressure), qreaction = H = Hproducts – Hreactants 6.4 For an endothermic reaction: H = +ve For an exothermic reaction: H = –ve Thermochemical Equations: thermochemical equation: shows both mass and enthalpy relationships Consider: Ice melting to form water at 0°C and 1 atm — requires energy (6.01 kJ/mol) We can represent the melting of 1 mol of ice as an equation: H2O (s) H2O (l) H = +6.01 kJ Note: H is +ve since ice must absorb heat to form water CHM 151: Chapter 6 Thermochemistry page 6 of 12 Consider: The formation of water releases heat: 2 H2 (g) + O2 (g) 2 H2O (g) H = –571.6 kJ Note: H is –ve since heat is lost to the surroundings Guidelines for Thermochemical Equations 1. Sign of H indicates if reaction is exothermic (H<0) or endothermic (H>0). 2. The coefficients in the chemical equation represent the numbers of moles of reactants and products for the H given. – e.g. H is –185 kJ for 1 mol H2 + 1 mol Cl2 to form 2 mols of HCl 3. The physical states must be indicated for each reactant and product. – e.g. For water, the solid and the liquid states vary by 6.01 kJ 4. H generally reported for reactants and products at 25°C. 6.5 APPLYING STOICHIOMETRY TO HEATS OF REACTION Rule 1. The magnitude of H is directly proportional to the amount of reactant or product. Consider heat as you would a reactant or product in a mole-tomole ratio, where H is the heat released or absorbed for the moles of reactants and products indicated in the equation Example: Consider H2 (g) + Cl2 (g) 2 HCl (g) H = –185 kJ a. What is H for when 5.00 mol of hydrogen gas reacts completely? b. What is H for the formation of 1.00 g of hydrogen chloride gas? CHM 151: Chapter 6 Thermochemistry page 7 of 12 Rule 2. For a reverse rxn, H is equal in magnitude but opposite in sign. If H2O (s) H2O (l) then H2O (l) H fusion): heat of fusion ( H2O (s) H = +6.01 kJ H = –6.01 kJ heat associated with a solid melting, in kJ/mol; i.e. solid liquid HHfusion H vapor): heat of vaporization ( heat associated with a liquid vaporizing, in kJ/mol; i.e. liquid gas HHvapor Example: If a 24.6-g ice cube requires 8.19 kJ of heat to melt completely, calculate Hfusion for ice (in kJ/mol)? Rule 3: If all the coefficients in a chemical equation are multiplied by a factor n H is multiplied by factor n: If H2O (s) H2O (l) H = +6.01 kJ 2 [H2O (s) H2O (l)] = 2 H2O (s) 2 H2O (l) H = 2(+6.01 kJ) = 12.0 kJ 1 1 1 [H2O (s) H2O (s)] = H2O (s) H2O (l) 2 2 2 H = 6.7 1 (+6.01 kJ) =3.01kJ 2 Hess’s Law Hess’s Law of heat summation: The value of H for a reaction is the same whether it occurs in one step or in a series of steps H = H1 + H2 + H3 + ... CHM 151: Chapter 6 Thermochemistry page 8 of 12 Example 1: Cgraphite (s) + 2 H2 (g) CH4 (g) Here are the reactions involved in the formation of CH4: (a) Cgraphite (s) + O2 (g) CO2 (g) H = –393.5 kJ (b) 2 H2 (g) + O2 (g) 2 H2O (l) H = –571.6 kJ (c) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) H = –890.4 kJ Rearranging the data allows us to calculate H for the reaction: (a)Cgraphite (s) + O2 (g) (b)2 H2 (g) + O2(g) (c) CO2 (g) H = –393.5 kJ 2 H2O (l) H = –571.6 kJ C O2 (g) + 2 H2O (l) Cgraphite (s) + 2 H2 (g) CH4 (g) + 2 O2 (g) CH4 (g) H = 890.4 kJ H = –74.7 kJ If the coefficients in an equation can be simplified, multiply by the necessary factors to cancel all intermediate compounds and get the correct coefficients for your final equation. Ex. 2 Rearrange the following data: (a) Cgraphite (s) + O2 (g) CO2 (g) H = –393.5 kJ (b) 2 CO (g) + O2 (g) H = 566.0 kJ 2 CO2 (g) to calculate the enthalpy change for the reaction: 2 Cgraphite (s) + O2 (g) 2 CO (g) CHM 151: Chapter 6 Thermochemistry page 9 of 12 Ex. 3 From the following data: (a)N2 (g) + 3 H2 (g) 2 NH3 (g) H = –92.6 kJ (b)N2 (g) + 2 O2 (g) 2 NO2 (g) H = (c) H = –571.6 kJ 2 H2 (g) + O2 (g) 2 H2O (l) 67.70 kJ Calculate the enthalpy change for the reaction: 4 NH3 (g) + 7 O2 (g) 4 NO2 (g) + 6 H2O (l) 6.6 Standard Enthalpies of Formation ( Hf ) Definitions of standard state (or reference form) 1. A gas at 1 atm 2. An aqueous solution with a concentration of 1 M at a pressure of 1 atm 3. Pure liquids and solids 4. The most stable form of elements at 1 atm and 25°C allotrope: one or 2 or more forms of an element in the same physical state – e.g. diamond and graphite are allotropes of carbon; O 2 (g) and ozone, O3 (g) are allotropes of oxygen standard enthalpy of formation ( Hf ) – enthalpy change for formation of 1 mole of an element or compound from its elements in their standard states (i.e. naturally occurring form at 1 atm and 25˚C) Note: Hf =0 for any element in its naturally occurring (most stable) form CHM 151: Chapter 6 Thermochemistry page 10 of 12 Examples of Hf are as follows: 1 Ag (s) + Cl2 (g) AgCl (s) 2 1 N2 (g) + O2 (g) NO2 (g) 2 H = –127.1 kJ Hf (AgCl, s) = -127.1 kJ H = +33.2 kJ Hf (NO2, g) = +33.2 kJ Example: Explain for each of the following if H˚ is a heat of formation. a. N2 (g) + 3 H2 (g) 2 NH3 (g) b. H2O (l) H2O (g) c. Cgraphite (s) + 2 H2 (g) d. 2 CO (g) + O2 (g) e. CO2 (s) CO2 (g) CHM 151: Chapter 6 Thermochemistry CH4 (g) 2 CO2 (g) page 11 of 12 Calculation of H° – superscript ° denotes taken under 1 atm and 25°C For the reaction: aA + bB cC + dD where a,b,c,d = stoichiometric coefficients = n Hf (products) – m Hf (reactants) Hrxn = [c Hf (C) + d Hf (D)] – [a Hf (A) + b Hf (B)] Use Table 6.2 to solve for Hrxn for each of the following: Ex. 1: C graphite (s) + O2 (g) CO2 (g) = (1mol) Hf (CO2, g) – [(1mol) Hf (Cgraphite, s) + (1mol) Hf (O2, g)] Hrxn = Ex. 2: 2 CH3OH (l) + 5 O2 (g) 2 CO2 (g) + 4 H2O (g) = Hrxn Ex. 3 Consider the reaction for the combustion of glucose: C6H12O 6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (g) Hrxn = –2803 kJ Calculate the enthalpy of formation for glucose using Table 6.2. CHM 151: Chapter 6 Thermochemistry page 12 of 12 Chapter 7: Quantum Theory of the Atom Problems: 1, 8-11, 15, 18, 25, 27-40, 49-54, 63-66, 71-72, 83-84 7.1 The Wave Nature of Light – Light travels through space as a wave – Waves have the following characteristics (see Fig. 7.4) =Greek “lambda”): distance between successive peaks wavelength ( = distance ; generally in units of m, cm, nm wave =Greek “nu”): number of waves passing by a given point in 1 s frequency ( = wave 1 cycle ; generally in hertz (Hz) = or time s s Electromagnetic (EM) spectrum: continuum of radiant energy (Fig. 7.5) visible region: portion of the EM spectrum that we can perceive as color For example, a "red-hot" or "white-hot" iron bar freshly removed from a hightemperature source has forms of energy in different parts of the EM spectrum – red or white glow = radiation within the visible region – warmth = radiation within the infrared region speed of light, c=3.00 x 108 m/s, depends on frequency and wavelength c distance distance wave time wave time Know how to convert between wavelength and frequency using the speed of light! Ex. 1 The wavelength for the electromagnetic radiation responsible for a blue sky is about 473 nm. What is the frequency of this radiation in Hz? CHM 151: Chapter 7 Notes page 1 of 6 7.2 Quantum Effects and Photons Classical Descriptions of Matter John Dalton (1803) – atoms are hard, indivisible, billiard-like particles – atoms have distinct masses (what distinguishes on type of atom from another) – all atoms of same element are the same JJ Thomson (1890s) – discovered charge-to-mass ratios of electrons atoms are divisible because the electrons are one part of atom Ernest Rutherford (1910) – shot positive alpha particles at a thin foil of gold discovery of the atomic nucleus James Maxwell (1873) – visible light consists of electromagnetic waves Transition between Classical and Quantum Theory Max Planck (1900); Blackbody Radiation – heated solids to red or white heat – noted matter did not emit energy in continuous bursts, but in whole-number multiples of certain well-defined quantities matter absorbs/emits energy in bundles = "quanta" (single bundle of energy= "quantum") Albert Einstein (1905); Photoelectric Effect – Photoelectric Effect: Light shining on a clean metal emission of electrons – Einstein applied Planck's quantum theory to light light exists as a stream of "particles" called photons Energy is proportional to the frequency () and wavelength () of radiation, and the proportionality constant (h) is now called Planck's constant E h = CHM 151: Chapter 7 Notes hc where h = 6.62610–34 J·s page 2 of 6 Ex. 1. Excited mercury atoms emit light strongly at a wavelength of 436 nm. a. What is the energy (in J) for one photon of this light? b. What is the energy (in kJ/mol) for a mole of photons of this light? Ex. 2. Certain elements emit light of a specific color when they are burned. When an potassium solution is burned in a flame test, the energy of the light emitted is 4.90910-19 J. Calculate the wavelength for this light, and use the visible spectrum (Fig. 7.5) to determine the color of the light. 7.4 Quantum Mechanics Louis de Broglie (1924); Dual Nature of the Electron – If light can behave like a wave and a particle matter (like an electron) can behave like waves – if electron behaves like a standing wave an electron can only have specific wavelengths an electron can only have specific frequencies and thus, energies for wave: E = mc2 for matter: E = h Combining the two equations, we can solve for the wavelength for any matter. de Broglie relation h mv CHM 151: Chapter 7 Notes page 3 of 6 Ex. 1 a. A baseball with mass 0.143 kg is thrown towards a batter at a velocity of 42.5 m/s, calculate the wavelength (in m) associated with the baseball’s motion. b. How does the compare in size to the baseball (diameter0.08 m)? The baseball’s baseball’s size. (Circle one) >> << Ex. 2 a. Calculate the wavelength (in m) associated with an electron traveling at the same velocity, 42.5 m/s. (The mass of the electron is 9.1095 10–31 kg.) b. How does the compare in size to the electron (diameter110–10 m)? The electron ‘s electron’s size. (Circle one) >> << Thus, although all matter can have wave properties, such properties are only significant for submicroscopic particles. 7.3 Bohr Theory of the Hydrogen Atom Bohr Postulates: Bohr Model of the Atom 1. Energy-level Postulate – Electrons move in discrete (quantized), circular orbits around the nucleus – "tennis ball and stairs" analogy for electrons and energy levels – a ball can bounce up to or drop from one stair to another, but it can never be halfway between two levels – Each orbit has a specific energy associated with it, indicated as n=1, 2,... – ground state or ground level (n = 1): lowest energy state for atom –when the electron is in most stable orbit – excited state: when the electron is in a higher energy orbit (n = 2,3,4,...) CHM 151: Chapter 7 Notes page 4 of 6 2. Transitions Between Energy Levels – When the atom absorbs energy, an electron can jump from a lower energy level to a higher energy level. – When an electron drops from a higher energy level to a lower energy level, the atom releases energy, sometimes in the form of visible light. Atomic Line Spectra Emission Spectra: continuous or line spectra of radiation emitted by substances – a heated solid (e.g. the filament in an incandescent light bulb) emits light that can be spread out to give a continuous spectrum = spectrum containing all wavelengths of light, like a rainbow – an atom in the gas phase emits light only at specific wavelengths = line spectrum – each element has a unique line spectrum can be used to identify unknown atoms in chemical analysis – why line spectrum for each element called "atomic fingerprint" Limitations of the Bohr Model Quantum Mechanical Model Unfortunately, the Bohr Model failed for all other elements that had more than one proton and one electron. (The multiple electron-nuclear attractions, electronelectron repulsions, and nuclear repulsions make other atoms much more complicated than hydrogen.) In 1920s, a new discipline, quantum mechanics, was developed to describe the motion of submicroscopic particles confined to tiny regions of space. – Quantum mechanics makes no attempt to specify the position of a small particle at a given instant or how the electron got there – It only gives the probability of finding small particles – Just like taking snapshot of a location and estimating where greatest number of people are likely to be Instead we take a snapshot of the atom at different times and “see” where the electrons are usually found (See. Figs. 7.24, 7.26) Erwin Schrödinger (1926) – developed a differential equation that allows us to find the electron's wave ), which ultimately allows us to determine the probability of finding function ( the electron in a given place – probability density for an electron is called the "electron cloud" “shape” Quantum Numbers, Energy Levels, and Orbitals – 4 quantum numbers describe distribution and behavior of electrons in atoms – Each wave function corresponds to a set of 3 quantum numbers and is referred to as an atomic orbital CHM 151: Chapter 7 Notes page 5 of 6 We will only discuss the first 2 Quantum Numbers: First (or Principal) Quantum Number (n): n=1,2,3,... – relates the average distance of electron from nucleus – higher n means electron is further from nucleus and higher energy (less stable) orbital Second (or Angular Momentum) Quantum Number (l); Sublevels (s, p, d, f) – gives the "shape" of the electron clouds associated with each orbital – The limitations on n and l for n=1, only 1s sublevel for n=2, only 2s and 2p sublevels for n=3, there are 3s, 3p, and 3d sublevels for n=4, there are 4s, 4p, 4d, and 4f sublevels Atomic Orbital Shapes: s orbital: spherical (see Fig. 7.25) p orbitals: dumbbell-shaped (see Fig. 7.26) – only for n=2 or greater – 3 types: px, py, pz (where x, y, and z give axis on which orbital aligns) d orbitals: various shapes (see Fig. 7.27) – only for n=3 or greater – 5 types: dxy, dxz, dyz, d x 2 y 2 , dz2 f orbitals: Don’t worry about these. The interesting part... – Consider Fig. 7.26 showing the electron distribution for the 2px orbital. – Note that the electron is never along the y-axis. There is a “node” in that region, indicating the electron is never found there. How can the electron be on either side of the y-axis without going through it? CHM 151: Chapter 7 Notes page 6 of 6 Chapter 8: Electron Configurations and Periodicity Problems: 7-8, 12, 15, 19, 23, 33-56, 61-66, 71-72 8.2 Building-Up Principle and the Periodic Table Electrons are distributed in orbitals of increasing energy levels, where the lowest energy orbitals are filled first. Once an orbital has the maximum number of electrons it can hold, it is considered “filled.” Remaining electrons must then be placed into the next highest energy orbital, and so on. – Parking garage analogy Orbitals in order of increasing energy: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 5d < 6p electron configuration: – Shorthand description of the arrangement of electrons by sublevel according to increasing energy REMEMBER! – s orbitals can hold 2 electrons – a set of p orbitals can hold 6 electrons – a set of d orbitals can hold 10 electrons – a set of f orbitals can hold 14 electrons Ex. 1 Li atomic number=3 3 eelectron configuration for Li: Ex. 2 F _____ e– electron configuration for F: Ex. 3 Fe _____ e– electron configuration for Fe: CHEM 151: Chapter 8 Notes page 1 of 8 Exceptions to the Building-Up Principle Atoms gain extra stability when their d subshells are half-filled or completely filled. If we can fill or half-fill a d subshell by promoting an electron from an s orbital to a d orbital, we do so to gain the extra stability. Ex. 1 Cr _____ e– electron configuration for Cr: actual electron configuration for Cr: Ex. 2 Ag_____ e– electron configuration for Ag: actual electron configuration for Ag: 8.3 Writing Electron Configurations Using the Periodic Table Electron Configuration from the Periodic Table – The Periodic Table's shape actually corresponds to the filling of energy sublevels. – See Fig. 8.12 (p. 325), to see how electrons for each element are distributed into the energy sublevels. Electron configurations of atoms with many electrons can become cumbersome. Abbreviated electron configurations (“noble-gas core” notation): – Since noble gases are at the end of each row in the Periodic Table, all of their electrons are in filled orbitals. – Such electrons are called “core” electrons since they are more stable (less reactive) when they belong to completely filled orbitals. valence electrons: electrons that are in the outermost shell (unfilled orbitals) Noble gas electron configurations can be used to abbreviate the “core” electrons of all elements. [He] [Ne] [Ar] [Kr] [Xe] = 1s2 = 1s2 2s2 2p6 = 1s2 2s2 2p6 3s2 3p6 = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 CHEM 151: Chapter 8 Notes page 2 of 8 [Fe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 = [Cd] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 = [Ni] = [Y] = [I] = Remember again that for some transition metals, extra stability with filled and half-filled d orbitals, so electrons are excited from the s orbitals to d orbitals: [Cu] = [Mo] = 8.1 Electron Spin and the Pauli Exclusion Principle Each electron in an atom can have one of two possible orientations, spin or spin . electron configuration: arrangement of electrons in an atom among sublevels CHEM 151: Chapter 8 Notes page 3 of 8 orbital diagram: diagram showing how electrons exist within an atom’s orbitals Pauli Exclusion Principle: no 2 e-s in an atom can have same four quantum #s Two electrons in the same orbital must have opposite spins – For example, with the helium atom, there are three ways to represent two electrons in 1s orbital (where spin is represented with the electron pointing up or down): for He: (a) (b) (c) – but the Pauli exclusion principle rules out (a) and (b) since these show two electrons in the same orbital with the same spin. 8.4 Orbital Diagrams of Atoms; Hund’s Rule Hund's Rule: the most stable arrangement of electrons in subshells has the greatest number of parallel spins – i.e. distribute electrons with same spin (up or down) and do not pair electrons until all subshells have an electron For example, if carbon’s electron configuration is: 1s2 2s2 2p2 carbon’s orbital diagram can be shown in the following ways: (a) 1s (b) 1s (c) 1s 2s 2s 2s 2p 2p 2p – but using Hund's rule, we know (c) would be the most stable. General Rules for Assigning Electrons in Atomic Orbital Diagrams 1. First, determine the electron configuration. 2. There is only one s orbital for each level: one 1s, one 2s, one 3s, etc. – There are 3 p orbitals for each p sublevel. – There are 5 d orbitals for each d sublevel. 3. Each orbital orbital can only hold 2 electrons Each s orbital can hold 2 electrons, each p sublevel can hold 6, d’s can hold 10. 4. Electrons in the same orbital must have opposite spins. 5. To fill sublevels, put one electron in each orbital (with same spin) before pairing. CHEM 151: Chapter 8 Notes page 4 of 8 Ex. 1 Draw the atomic orbital diagram for oxygen. Ex. 2 Draw the atomic orbital diagram for phosphorus. (Use full notation) Ex. 3 Draw the atomic orbital diagram for the valence (outermost shell) electrons in cobalt. (Use core notation.) Magnetic Properties of Atoms paramagnetic: substance that contains unpaired electrons weakly attracted to magnetic field diamagnetic: substance that contains only paired electrons slightly repelled by magnetic field To determine if a substance is paramagnetic or diamagnetic, you must draw its orbital diagram and determine if it has an unpaired electrons. Ex. Is cobalt paramagnetic or diamagnetic? CHEM 151: Chapter 8 Notes ____________________ page 5 of 8 8.3 Mendeleev’s Predictions from the Periodic Table Dimitri Mendeleev proposed that elements display recurring properties according to increasing atomic mass Periodic Table originally arranged with elements in order of increasing atomic mass Henry G.J. Moseley’s high-energy x-ray radiation experiments of atomic nuclei Repeating properties of elements more clearly reflected by the arrangement of elements according to increasing atomic number Periodic Table’s arrangement today Trends for increasing atomic mass are identical with those forincreasing atomic number, except for Ni & Co, Ar & K, Te & I. Neils Bohr’s introduction of electron energy levels Periodic Table’s shape – Indicates filling of electron orbitals and element’s electron configuration periodic law: When elements are arranged in terms of increasing atomic number, the trends within a row or column form patterns that allow us to predict the physical and chemical properties of elements 8.6 Some Periodic Properties Atomic radius: distance from nucleus to outermost electrons – Increases down a group: More p+, n, and e– bigger radius – Decreases from left to right along a period. effective nuclear charge: – positive charge an electron experiences because of charge from protons in the nucleus minus any shielding due to electrons closer to the nucleus Consider the atoms of Ti and Ni: Thus, the higher the effective nuclear charge, the smaller the atomic radius! CHEM 151: Chapter 8 Notes page 6 of 8 Atomic radius trend: – Trend from top to bottom like a snowman – Trend from left to right like a snowman that fell to the right First Ionization Energy: Energy necessary to remove first electron from a neutral atom in gaseous state to form negatively charged ion. First Ionization Energy TRENDS – Decreases down a group: Bigger the atom, the further away e–s are from +vely charged nucleus e–s held less tightly and are more easily removed – Increases from left to right along a period: – Elements with fewer (1–3) valence e–s can more easily give up e–s to gain noble gas configuration (stability) – Elements with more (4–7) valence e–s can more easily gain e–s togain noble gas configuration (stability) Trend from top to bottom like an upside-down snowman Trend from left to right like a upside-down snowman that fell to the right Variations in Successive Ionization Energies – Recognize that it becomes more difficult to remove electrons from stable ions. – See Table 8.3 on p. 337. – For example, the ionization energy to remove the first electron from Li is much smaller than the any successive ionization energy since electrons are now removed from a stable ion (with a positive charge AND sometimes a noble gas electron configuration). CHEM 151: Chapter 8 Notes page 7 of 8 We can indicate first and successive ionization energies in the following way: First ionization energy = IE1 Second ionization energy = IE2 Third ionization energy = IE3 Ex. 1: Between which two ionization energies (IE1 and IE2, IE2 and IE3, etc.) would you expect there to be the largest jump for the following elements? Explain. a. Mg: Between _____ and _____ Explain why: b. K: Between _____ and _____ Explain why: c. Al: Between _____ and _____ Explain why: Ex. 2: 8.7 This 3rd period element has a large jump between IE5 and IE6. Explain how you can identify the element. Periodicity in the Main-Group Elements Metallic character: – Decreases from left to right along a period: Metals concentrated on left-hand side of P.T., nonmetals on right-hand side – Increases down a group: Looking at groups IVA and VA, go from nonmetals (C & N) to semimetals (Si & As) to metals (Sn & Bi) Same snowman trends as for atomic radius! CHEM 151: Chapter 8 Notes page 8 of 8 Chapter 9: Ionic and Covalent Bonding Problems: 13, 15-16, 18, 22-24, 26-28, 31-38, 47-70, 75-78, 81-84, 91 9.1 Describing Ionic Bonds A metal transfers valence electrons to a nonmetal metal cations and nonmetal anions – the charged ions are attracted to each other ionic bond: electrostatic attraction between positively charged cation and negatively charged anion core electrons: innermost electrons belonging to filled electron shells valence electrons: Electrons in the outermost shell – Since atoms want filled electron shells to be most stable, they’ll combine with other atoms with unfilled shells (gaining or losing e–s) to get stability. Valence e–s lead to chemical bonds and reactions between atoms For Main Group (A) elements, Group # # of valence e–s – Elements in Group IA: each have 1 valence e– – Elements in Group IIA: each have 2 valence e– – Elements in Group IIIA: each have 3 valence e– – Elements in Group IVA: each have 4 valence e– – Elements in Group VA: each have 5 valence e– – Elements in Group VIA: each have 6 valence e– – Elements in Group VIIA: each have 7 valence e– – Elements in Group VIIIA: each have 8 valence e– How many valence electrons do each of the following have? Mg: ____ val. e– N: ____ val. e– Br: ____ val. e– Al: ____ val. e– Rb: ____ val. e– Si: ____ val. e– Se: ____ val. e– Xe: ____ val. e– Lewis Electron-Dot Symbols – Shows the atom or ion of an element with its valence electrons 1. Element symbol representing the nucleus and core electrons 2. Dots representing the valence e– CHM 151: Chapter 9 Notes page 1 Rules for writing Electron Dot Symbol 1. Write down the element symbol 2. Determine the number of valence electrons using the group number 3. For ions, account for electrons gained or lost leading to the charge. 4. Assume the atom has four sides, and distribute electrons with one electron per side before pairing electrons. Ex. 1 Give the Lewis electron-dot symbol for the following elements: Si K Cl P Al Ex. 2 Give the Lewis electron-dot symbol for the following ions: calcium ion: potassium ion: sulfide ion: Why do ions form if formation requires energy (i.e. ionization energy)? – When ionic compounds form, energy is released due to so many bonds formed amount of energy released more than makes up for the ionization energy Properties of Ionic Substances: – An ionic compound is actually a network of ions, with each cation surrounded by anions, and vice versa. A 2-dimensional slice would look like this: Most ionic compounds are solid at room temperature and have very high melting points, since every bond between each ion must be broken to melt the substance. CHM 151: Chapter 9 Notes page 2 Coulomb's law: Relative strength of ionic bond given by the following: 1. Charges of ions: Higher the charge the stronger the bond – The bond in CaO (+2 and –2) is stronger than that in NaCl (+1 and –1) why melting point of CaO (2927°C) much higher than NaCl's (801°C) 2. Distance between two ions: Shorter distance stronger the bond – internuclear distance for NaCl (0.276 nm) is shorter than KBr's (0.328 nm). why NaCl melting point (801°C) is higher than KBr's (734°C). 9.2 Electron Configuration of Ions Ions of the Main-Group Elements – Representative elements generally form ions—ie. gain or lose electrons—to achieve a noble gas electron configuration Ions from representative metals are usually isoelectronic with—i.e. have the same electron configuration as—one of the noble gases! For IONS, one must account for the loss or gain of electrons: # electrons = atomic # – (charge = change in # of valence electrons) Or you can simply use the Periodic Table – Find out with which element the ion is isoelectronic — Move to the left for electrons lost — Move to the right for electrons gained write the electron configuration for that element Fill in the table below: Ion Isoelectronic with this element Electron Configuration (using core notation) P–3 Na+ Al+3 CHM 151: Chapter 9 Notes page 3 Transition-Metal Ions – Transition metals lose their s electrons first when forming ions Ion Electron Configuration for element (use core notation) Electron Configuration for ion (use core notation) Zn2+ Ag+ Fe+3 9.3 IONIC RADII: cation radius < neutral atom radius < anion radius (loses e-s) (gains e-s) Cl atom Na+ ion Cl – ion Na atom 9.4 Describing Covalent Bonds covalent bond: sharing of a pair of electrons between 2 nonmetal atoms – the type of bond that holds atoms together in a molecule H. + • H H—H bonding electrons: electrons involved in covalent bond formation nonbonding electrons (or lone pairs): belong solely to an individual atom bonding electron pair : : : : + : : : : :F . :F : nonbonding electrons or lone pairs : F—F : Coordinate covalent bond: – When one atom donates both electrons to make the bond :F : – : : CHM 151: Chapter 9 Notes : : H+ + H—F : page 4 9.5 Polar Covalent Bonds; Electronegativity Electronegativity (EN): Ability of an atom in a bond to draw e–s to itself – F is the most electronegative (EN) element Elements are less electronegative the further away from F EN increases Periodic Table EN increases Except for H, which has EN between B and C. In some covalent bonds, one of the two atoms holds bonding e–s more tightly polar covalent bond results between two atoms Because the electrons spend more time around F, they must spend less time around H F gets a partial –ve charge (–) and H gets partial +ve charge (+): : : H—F : In other covalent bonds, both atoms have equal EN values share the bonding electrons equally nonpolar covalent bond 9.6 Writing Lewis Electron-Dot Formulas 1. Count the total number of valence electrons present – from atoms – from charge if polyatomic ion (add or subtract e–s for charge) 2. Write the skeleton structure of the compound – Put least EN atom as central atom and surround with other atoms – Note: H and F atoms will always be outer atoms 3. Connect all atoms by drawing single bonds between all atoms, then distribute the remaining valence electrons as lone pairs around outer atoms so each has 8 e–s octet rule: atoms form bonds such that all atoms get eight electrons – BUT hydrogen (H) can only have 2 electrons (to fill 1s orbital) CHM 151: Chapter 9 Notes page 5 4. If there are not enough electrons for each atom to have an octet, make double and/or triple bonds between central atom and surrounding atoms – BUT fluorine can only form a single bond – Note that double bonds are shorter than single bonds, and triple bonds are shorter than double bonds 5. For polyatomic ions, square brackets are drawn around the Lewis structure, and the charge is put in the upper right-hand corner Give the Lewis formula for the following: a. H2O b. O 2 c. NF3 d. HCN e. NH4+ f. NO2– g. SO4–2 h. CO32– CHM 151: Chapter 9 Notes page 6 Lewis Electron-Dot Formulas for Ternary Oxyacids (e.g. HNO3, H2SO4, etc.) – Ternary oxyacids contain hydrogen, oxygen, and one other element. – In ternary oxyacids, the central atom is the “other element” and hydrogen atoms are bonded directly to oxygen atoms. more than one central atom in the molecule Example: Draw the electron-dot formula for each of the following: a. HNO3 (aq) 9.9 b. H2SO4 (aq) Formal Charge and Lewis Formulas Example: Give two different Lewis formulas for CO2 that satisfy the octet rule. Because there are many ways to draw Lewis structures, we need a way to determine the most plausible structure formal charge: hypothetical charge an atom would have if bonding electrons are shared equally and lone pairs belong solely to a single atom total number of total number of formal charge = valence electrons — nonbonding — total number of bonds in free atom electrons 1. For neutral molecules, sum of formal charges must equal zero. 2. For ions, the sum of the formal charges must equal charge. CHM 151: Chapter 9 Notes page 7 Guidelines for formal charges and plausible structures: – Lewis formulas with no formal charges is preferable to one with formal charges. – Lewis formula with large formal charges (+2, +3, and/or –2, –3, etc.) is less plausible than one with lower formal charges (+1 and –1, etc.) – When Lewis formulas have similar formal charge distributions, most plausible formula has negative formal charge on more electronegative atom Ex. 1 Determine the most plausible formula for CO2 above using formal charges. Ex. 2 Determine the most likely Lewis structure for formaldehyde, CH2O, starting the with following skeletal structures and by determining formal charges: H H C O H C O H 9.7 Delocalized Bonding; Resonance Given the Lewis structure for ozone, we expect either of the following structures: : : O : : : : : O : O : O : : O : O so one bond (O—O) bond should be longer than the other (O=O). BUT experimental evidence indicates that both oxygen-oxygen bonds in ozone are identical, so neither of the structures accurately represents the molecule. page 8 : O : O : CHM 151: Chapter 9 Notes O : These types of electrons are called delocalized electrons because they are spread between more than two atoms. : The actual structure is a cross between the two structures, where the electron pair is actually spread over all three atoms: To correctly represent such delocalized electrons using Lewis formulas we show all the Lewis formulas with a double-arrow between each: : : O : : : O : : : : O : O : O : O where each of these structures is called a resonance structure. resonance structure:one of two or more Lewis structures representing a single molecule that cannot be described fully with only one Lewis structure resonance: the use of two or more Lewis structures to represent a molecule – a real ozone molecule doesn't oscillate between the resonance structures above but is a unique, stable structure we cannot adequately represent with one Lewis structure Ex. 1 Give the resonance structures for NO2–: O Ex. 2 9.8 N O O N O Give the resonance structures for the carbonate ion, CO32–: Exceptions to the Octet Rule The Incomplete Octet: – Some atoms are satisfied with less than an octet – Be (1s22s2) is stable with only four valence electrons :F B : : : : – Boron (1s2 2s2 2p1) also tends to form compounds with less than eight electrons H—Be—H F: :F : : CHM 151: Chapter 9 Notes page 9 : : : : The Expanded Octet: – Atoms in and beyond the 3rd period can have more than eight electrons when in a compound :F :F : : : : : : . : : : Odd-Electron Molecules: – Some molecules have an odd number of electrons can't satisfy octet rule; usually N has the odd number N=O :F : F: S :F : F: Examples: Draw the Lewis structure for the following molecules: BeCl2: BH3: PF5: XeF4: NO2: triiodide ion, I3–: CHM 151: Chapter 9 Notes page 10 9.10 Bond Length and Bond Order Bond Length: distance between 2 nuclei of two bonded atoms – shorter the bond length, the stronger the bond – triple bonds are the shortest, so they are the strongest bonds – double bonds are the next shortest, so they are next in strength – single bonds are longest, so they are weaker than double and triple bonds Bond order: number of pairs of electrons in a bond – single bond: bond order=1 – double bond: bond order=2 – triple bond: bond order=3 9.11 Bond Energy bond energy: energy required to break a particular bond in the gas phase – always positive since it takes energy to break a bond – a quantitative measure of a bond’s strength (i.e. stability of molecule) – higher the bond energy stronger the bond Use of Bond Energies in Thermochemistry – We can estimate the enthalpy change (H˚) for any reaction using bond energies: H° = BE (reactants) – BE (products) = total energy input – total energy released Note: if if BE (reactants) BE (reactants) > < BE (products) BE (products) endothermic reaction exothermic reaction Ex. 1 Use Table 9.5 to calculate the enthalpy of reaction for the following: H2 (g) + F2 (g) 2 HF (g) CHM 151: Chapter 9 Notes page 11 Ex. 2 Use Table 9.5 to calculate the enthalpy of reaction for the following: H :NN: (g) CHM 151: Chapter 9 Notes + 2 H2 (g) H H—N—N—H (g) page 12 Chapter 10: Molecular Geometry and Chemical Bonding Theory Problems: 2-4, 7-9, 17-18, 23-32, 35-40, 43-44, 47-48, 55-60 10.1 The Valence-Shell Electron-Pair Repulsion (VSEPR) model – refers to three-dimensional arrangement of atoms in molecule – responsible for many physical and chemical properties of molecules Repulsion between electrons causes them to be as far apart as possible valence-shell electron-pair repulsion (VSEPR) model – accounts for geometric arrangements of electron pairs around central atom in terms of repulsion between electron pairs General Rules 1. Consider double and triple bonds like single bonds (an approximation since extra electrons occupy space but sufficient for qualitative purposes) 2. If two or more resonance structures can be drawn for a molecule, VSEPR model can be applied to any one of them 3. Don't show formal charges If there are only two atoms, the molecule must be linear. Ideal Geometries with Two to Six Electron Pairs on the Central Atom (where central atom has no lone pairs) — consider a molecule composed of only two types of atoms, A and X A=central atom X=outer atoms For three or more atoms in a molecule, general formula: AX# (where #=2 to 6) CHM 151: Chapter 10 Notes page 1 of 13 Molecules Where Central Atom Has No Lone Pairs # of BONDS General Formula 2 AX 2 3 AX 3 4 AX 4 MOLECULAR GEOMETRY 180˚ 120˚ 109.5˚ NAME Examples linear BeCl2 trigonal planar BF3 tetrahedral CH4, NH4+ trigonal bipyramidal PCl5 octahedral SF6 90˚ 5 AX 5 6 AX 6 120˚ all angles are 90˚ CHM 151: Chapter 10 Notes page 2 of 16 AX 2: linear – the two outer atoms are 180° from each other – e.g. HCN AX 3: trigonal planar – three outer atoms at the corners of an equilateral triangle – each outer atom is 120° from the other two outer atoms – e.g. CH2O AX 4: tetrahedral (tetra = four) since four-sided, or four faces – maximum distance between electrons requires 3D structure with – each outer atom is 109.5° from the other outer atoms – e.g. CH4 AX 5: trigonal bipyramidal – trigonal = three outer atoms form planar triangle around central atom – bipyramidal = two outer atom directly above and below central atom, connecting outer atom forms two 3-sided pyramids – equatorial positions: ends of planar triangle – 3 of outer atoms are at equatorial positions, 120° from each other – axial positions: above and below central atom – 2 atoms are at axial positions, 90° from equatorial atoms – e.g. PF5 AX 6: octahedral (octa=eight) connecting six atoms eight faces – all outer atoms are 90° away from each other – the terms "axial" and "equatorial" do not apply because all six positions are identical – e.g. SF6 Molecules Where Central Atom Has One Or More Lone Pairs A central atom with lone pairs has three types of repulsive forces lone - pair vs. lone - pair repulsion > lone - pair vs. bonding - pair repulsion > bonding - pair vs. bonding - pair repulsion – bonding-pair: takes up less space than lone pairs since held by attractive – lone pairs: take up more space than bonding-pair A=central atom CHM 151: Chapter 10 Notes X=outer atoms E=lone pairs page 3 of 13 Molecules Where Central Atom Has Lone Pairs Original Shape General Formula Molecular Geometry Name trigonal planar (AX3) <120˚ 120˚ tetrahedral (AX4) AX2E bent or angular AX3E trigonal pyramidal < 109.5˚ 109.5˚ trigonal pyramidal AX3E < 109.5˚ AX2E: bent – start with AX3 molecule (trigonal planar) and replace an X atom w/ lone pair – bond angles are now less than 120° – e.g. SO2 AX3E: trigonal pyramidal (central atom + 3 outer atoms make a pyramid) – start with AX4 molecule (tetrahedral) and replace a X atom w/ lone pair – bond angles are now less than 109.5° – e.g. NH3 AX2E2: bent – start with AX4 molecule (tetrahedral) and replace 2 X atoms with 2 lone pairs – bond angles are now less than 109.5° –eg. H2O CHM 151: Chapter 10 Notes page 4 of 16 Molecules Where Central Atom Has Lone Pairs (Cont’d) Original Shape General Formula Molecular Geometry Name < 180˚ AX4E See-saw < 90˚ trigonal bipyramidal (AX5) 90˚ < 120˚ < 180˚ AX3E2 T-shaped < 90˚ 120˚ AX2E3 linear 180˚ AX4E: seesaw – start with AX5 molecule and replace one X atom with one lone pair – X atom can be taken from an axial or an equatorial position –from axial: lone pair is 90° from three equatorial X atoms and – from equatorial: lone pair is 90° from two axial and 120° from two other equatorial X atoms – taking X atom from equatorial position maximizes space between lone pair and X atoms – bond angles are now less than 90° and less than 120° – eg. SF4 CHM 151: Chapter 10 Notes page 5 of 13 AX3E2:T-shaped – start with AX5 molecule and replace two X atoms with two lone pairs –both X atoms taken from equatorial positions to maximize distance between the lone pairs bond angles for remaining atoms are now less than 90° –eg. ClF3 AB2E3:linear – start with AX5 molecule and replace three X atoms with three lone pairs –first two X atoms taken from equatorial positions; third X atom taken: – from axial: third lone pair would be 90° from other lone pairs – from equatorial: third lone pair would be ~120° from other lone pairs taking third X atom from equatorial position maximizes space – bond angle is now 180° – eg. XeF2 Molecules Where Central Atom Has Lone Pairs (Cont’d) Original Molecular Geometry General Formula Molecular Geometry Name octahedral (AX6) square pyramidal AX5E all angles are < 90˚ square planar AX4E2 all angles are 90˚ all angles are 90˚ CHM 151: Chapter 10 Notes page 6 of 16 AX 5E: square pyramidal (central atom + 5 X atoms make 4-faced pyramid) – start with AX6 (octahedral) and replace one X atom with one lone pair – since all six outer positions are identical, doesn't matter which one you take – bond angles are now less than 90° – e.g. BrF5 AX4E2: square planar (central atom + 4 X atoms form square all in 1 plane) – start with AX6 (octahedral) and replace 2 X atoms with 2 lone pairs – doesn't matter which X atom taken first, second X atom taken 180° away from it to maximize space between lone pairs square planar shape – bond angles are now exactly 90° since lone pairs balance each other – e.g. XeF4 Guidelines for Applying the VSEPR Model 1. Draw Lewis formula 2. Count number of atoms and lone pairs around central atom, treating double and triple bonds as single bonds – Consider molecule's shape if lone pairs around central atom were actually bonded pairs, then remove number of outer atoms equal to number of lone pairs. – Remember that lone pair(s) occupy more space than bonded pairs of electrons, so the bond angles are compressed when the central atom has lone pairs. Example: For each of the following, i. Draw the Lewis formula ii. Indicate the molecular geometry (or shape). iii. Indicate the bond angles. PCl3: CHM 151: Chapter 10 Notes IBr4–: page 7 of 13 NH2–: SCN–: SeF4: SnCl5–: Geometry of Molecules with More than One Central Atom – Overall geometry of entire molecule cannot be described instead describe shape around each central atom in molecule – e.g. CH3OH Lewis structure H H : : C H—C—O—H H H H O tetrahedral around C and H bent around O What is the geometry around each central atom in CH3COOH? O : : H : : H—C—C—O—H H CHM 151: Chapter 10 Notes page 8 of 16 10.2 Dipole Moment and Molecular Geometry Polar Covalent Bonds – Electrons concentrate around the more electronegative atom in a molecule Atom gains a partial –ve charge – Since atoms spend less time around the other atom Other atom gains a partial +ve charge – molecules that have separated +ve and –ve ends are polar molecules – overall dipole indicated with an arrow pointing to more electronegative end – : H—F Nonpolar covalent bond: Covalent bond between 2 atoms w/ same EN – Best example is found between two identical atoms: – H2, O2, N2, Cl2, F2, I2, Br2 – Nonpolar covalent bonds can also occur between different atoms which have identical EN values: (See table of EN values in Table 9.15 on p. 367) – eg. N—Cl, C–S, S—I, etc. Example: Indicate the type of bond (ionic, polar covalent, nonpolar covalent) for each of the following bonds: a. S—O in SO2 b. C—C in C2H4 c. Cl—Ba in BaCl2 ionic ionic ionic polar covalent polar covalent polar covalent nonpolar covalent nonpolar covalent nonpolar covalent For diatomic molecules: – nonpolar molecules: when the 2 atoms have equal EN values – polar molecules: when the 2 atoms have different EN values –have dipole (+ve and –ve ends) For molecules of three of more atoms: — polarity depend on individual bonds and geometry around central atom Example: Determine whether the following have dipole moments: BeCl2: H2O: Cl —Be—Cl CHM 151: Chapter 10 Notes O H H page 9 of 13 CCl4 versus CHCl3: H CCl4 CHCl3 10.3 Valence Bond (VB) Theory Basic Theory – a covalent bond consists of a pair of electrons in atomic orbitals H atom has the atomic orbital diagram: 1s Each H in a H2 (H—H) molecule 1s Hybridization of Atomic Orbitals (explaining polyatomic molecules) hybridization: mixing of atomic orbitals in an atom (usually central atom) to generate a new set of atomic orbitals, called hybrid orbitals sp Hybridization: mixing one s orbital and one p orbital – Consider BeCl2: – the ground state orbital diagram for Be should be 2s 2p – BUT this indicates that Be does not form covalent bonds with Cl since Be's electrons are already paired one of Be's electrons must be promoted to 2p: 2s CHM 151: Chapter 10 Notes 2p page 10 of 16 2s 2p – Now there are two Be orbitals for bonding, but this indicates that the two Be–Cl bonds are different since one forms from a 2s orbital and other from a 2p – BUT VSEPR predicts BeCl2 is linear and experiment indicates that the two Be–Cl bonds are equivalent hybridize a 2s and a 2p orbital to get 2 equivalent sp hybrid orbitals: 2s unhybridized p orbitals sp 2p 2p linear shape (180˚ bond angle) corresponds to sp orbitals sp2 Hybridization: mixing one s orbital and two p orbitals – Consider BF3: – ground state orbital diagram for B should be 2s 2p and promoting one of Be's 2s electrons to 2p gives: 2s 2p – Now there are three B orbitals for bonding, but this indicates that two of the B–F bonds (from 2s) should be the same but one (from 2s) should be different. But VSEPR predicts BF3 is trigonal planar and experiment indicates that the three B–F bonds are equivalent hybridize a 2s and two 2p orbitals to get 3 equivalent sp2 hybrid orbitals: 2s 2p sp 2 last unhybridized p orbital 2p 2 trigonal planar shape (120˚ bond angle) corresponds to sp orbitals CHM 151: Chapter 10 Notes page 11 of 13 sp3 Hybridization: mixing one s orbital and three p orbitals — Consider CH4: — ground state orbital diagram for C should be 2s 2p — form 4 bonds, C’s orbital diagram should be: 2s 2p But VSEPR predicts CH4 is tetrahedral and experiment indicates that the four C–H bonds are equivalent hybridize a 2s and three 2p orbitals to get 4 equivalent sp3 orbitals: 2s sp 3 2p tetrahedral shape (109.5˚ bond angle) corresponds to sp 3 orbitals sp3 hybridization also applies to ammonia (NH3), where N's e–s are arranged as sp 3 – The lone pair on the N takes up more space than the e–s in N–H bonds, so the H–N–H angles are all less than 109.5°. sp3d Hybridization: mixing one s orbital, three p orbitals, and one d orbital — Consider PCl5: 3p — ground state orbital diagram for P should be 3s — the necessary configuration for P to form 5 bonds should be: 3s , 3d 3p – But VSEPR predicts PCl5 is trigonal bipyramid and experiment indicates that the five P–Cl bonds are equivalent hybridize the 3s, all three 3p, and one 3d orbital to get 5 equal sp3d orbitals: sp 3d 3d 3 trigonal bipyramidal shape corresponds to sp d orbitals CHM 151: Chapter 10 Notes page 12 of 16 sp3d2 Hybridization: mixing one s orbital, three p orbitals, two d orbitals – Consider SF6: 3p – ground state orbital diagram for S should be 3s – the necessary configuration for S to form 6 bonds should be: 3s , 3d 3p – Now VSEPR predicts SF6 is octahedron and experiment indicates that the six S–F bonds are equivalent hybridize the 3s, all three 3p, and two 3d’s to get 6 equal sp3d2 orbitals: sp 3d 2 3d octahedral shape corresponds to sp 3d 2 orbitals Predicting what hybrid orbitals form: 1. Draw Lewis structure to determine total number of bonds on central atom 2. Given the shape of molecule or bond angle from the VSEPR model, we can determine what hybrid orbitals must be involved: Molecular Geometry (Shape) Hybrid Orbitals linear (e.g. HCN, BeCl2, CO2) sp trigonal planar (120˚), bent (<120˚) sp 2 tetrahedral (109.5˚) or shapes with <109.5˚ angles sp 3 trigonal bipyramidal, see-saw, T-shape, linear with expanded octet on central atom sp 3d octahedral (90˚) or shapes with all 90˚ or <90˚ angles sp3d2 Example: What are the hybrid orbitals for the central atom in each of the following molecules from pages 7-8 of your notes? PCl3: IBr4–: NH2–: SCN–: SeF4: SnCl5–: CHM 151: Chapter 10 Notes page 13 of 13 Chapter 11: States of Matter; Liquids and Solids Problems: 1, 5-6, 9, 10, 12-13, 27, 29-32, 35-36, 45-46, 49-50, 53-64 11.1 Comparison of Gases, Liquids, and Solids phase (=physical state): solid, liquid, or gas Solids have the lowest kinetic energy (KE)—i.e. do not move very much – Highest attraction between particles particles are stuck in specific sites = very confined Liquids have slightly higher KE—i.e. particles moving more than in solid – Particles are still attracted and maintain contact with one another but can move past one another particles are less confined Gases have greatest KE—i.e. particles move quickly and randomly – Attractive forces almost (if not) completely overcome, so particles can fly particles are far away from each other = unrestricted 11.5 Intermolecular Forces; Explaining Liquid Properties intermolecular forces: Attraction between 2 or more molecules – e.g. between 2 water molecules or between a water and a CO2 molecule London (Dispersion) Forces – Attraction between temporary or induced dipoles in adjacent molecules – Electrons are constantly shifting and can sometimes concentrate in one region instantaneous (temporary) dipole that goes away once electrons shift – Chocolate-chip cookie dough analogy – Most common type and weakest of intermolecular forces, found in all molecules – Only type of intermolecular force between nonpolar molecules – The strength of theses forces depends on – Number of electrons in a molecule – How easily the electrons are dispersed or (polarized) to form temporary dipoles – In general, larger molecules have more electrons and are more “polarizable” as molar mass increases, dispersion/London forces become stronger boiling point of nonpolar molecules increases with molar mass CHEM 151: Chapter 11 page 1 of 8 Dipole-Dipole Forces: Attraction between polar molecules – strong because attraction is due to permanent electrostatic charges Note: Van der Waals Forces refer to intermolecular forces due to either London forces or dipole-dipole forces. Hydrogen Bonds: – Exist between molecules with following bonds: H–F, H–O, H–N – Special type of dipole-dipole force caused by small radii and large electronegativity differences between H and O, N, and F atoms – strongest type of intermolecular bond – Responsible for relatively high boiling points of molecules like H2O, NH3, and HF – Responsible for many of water's unusual properties, including why the density of ice (d=0.917 g/cm3) is less than the density of liquid water (d=1.000 g/cm3) – Arrangement of water molecules in ice crystal results in "holes" or empty space, but in water, the molecules arrange to fill in the holes Note: Although hydrogen bonds are the strongest type of intermolecular bonds, ionic and covalent bonds will always be stronger than hydrogen bonds because ionic and covalent bonds hold the ions or atoms together in a compound. How to determine type of intermolecular forces involved: yes polar Is the molecule polar or nonpolar hydrogen bonding Are there H–F, H–O, or H–N bonds no nonpolar CHEM 151: Chapter 11 dispersion (London) forces dipole-dipole forces page 2 of 8 Example Identify the type of bond holding atoms together in the molecules and the intermolecular forces between the molecules for the two examples below: O O H H H H H H H H H H Cl Cl H H O H H O H H O H H O H H O H H H Cl H H H H H H H Cl H H H Cl H H Cl Cl Ex. 2: Circle the type of bond or attraction described for each below: a. The C–C bond in C2H6. dispersion forces ionic bond dipole-dipole forces polar covalent bond hydrogen bond nonpolar covalent bond b. The Na–Cl bond in NaCl. dispersion forces ionic bond CHEM 151: Chapter 11 dipole-dipole forces polar covalent bond hydrogen bond nonpolar covalent bond page 3 of 8 c. What attracts HF molecules to each other in a sample. dispersion forces ionic bond d. dipole-dipole forces polar covalent bond hydrogen bond nonpolar covalent bond The H–O bond in a H2O molecule. dispersion forces ionic bond e. dipole-dipole forces polar covalent bond hydrogen bond nonpolar covalent bond What holds atoms together in a molecule of NH3. dispersion forces ionic bond Ex. 2: dipole-dipole forces polar covalent bond hydrogen bond nonpolar covalent bond The stronger the intermolecular forces, the more energy is required to break the attraction between molecules in liquids and solids higher melting and boiling points Explain each the following: a. Why O2's boiling point is -183°C while NO's boiling point is -151°C. b. Why N2's boiling point is -196°C while Br2's boiling point is 59°C. c. Why H2S’s boiling point is –61°C and H2O’s boiling point is 100˚C. CHEM 151: Chapter 11 page 4 of 8 Physical Properties and Intermolecular Forces: Vapor Pressure: partial pressure exerted by gas molecules above the liquid in equilibrium with the liquid – varies for different liquids, varies for different T’s – more gas molecules higher vapor pressure – weaker the intermolecular forces more molecules can go from liquid to vapor higher vapor pressure H vap): molar heat of vaporization ( – energy required to vaporize one mole of a liquid – stronger intermolecular forces higher molar heat of vaporization Practical example of molar heat of vaporization: – rubbing alcohol on your hands – heat from your hands increases the KE of alcohol molecules evaporates – this is an endothermic process – perspiration: body heat vaporizes water of perspiration from body's surface Boiling Point: temperature where vapor pressure of liquid is equal to external pressure (usually atmospheric pressure) – normal boiling point is boiling point at 1 atm – e.g. water boils at 100°C at 1 atm, but it boils at ~95°C in Albuquerque where the atmospheric pressure is ~0.85 atm – both Hvap and boiling point depend on intermolecular forces Surface Tension: attraction between surface molecules in a liquid – stronger intermolecular forces = stronger attraction surface molecules are held together more strongly higher surface tension 11.2 Phase Transition: change from one physical state to another sublimation SOLID fusion freezing LIQUID vaporization condensation GAS deposition CHEM 151: Chapter 11 page 5 of 8 The Equilibrium Nature of Phase Changes dynamic equilibrium: rate of forward process is exactly equal to the rate of reverse process Liquid-Gas Equilibrium: liquid vapor evaporation (or vaporization): liquid vapor condensation: vapor liquid – the process of a gas liquefying – can result from two techniques: 1. cooling sample of gas lower KE, and molecules start to aggregate to form small drops of liquid 2. applying pressure to gas minimize space between molecules, so molecules are attracted to each other Liquid-Solid Equilibrium freezing: liquid solid melting (or fusion): solid liquid melting point: temperature at which solid and liquid phases coexist in equilibrium – normal melting point is melting point at 1 atm Consider solid-liquid equilibrium of water and ice (at 0°C and 1 atm): ice water When ice cubes are placed into a glass of water, the ice cubes begin to melt, but some water between the ice cubes freezes, causing ice cubes to fuse H fus): energy required to melt one mole of solid molar heat of fusion ( supercooling: a liquid can be temporarily cooled to below its freezing point – results when a liquid is cooled so rapidly that molecules don't have time to arrange themselves properly – unstable condition; gentle stirring or adding "seed" crystal will cause solidification – Practical example: Sometimes happens in your refrigerator if a bottle of carbonated soda supercools. – You notice ice crystals upon opening. CHEM 151: Chapter 11 page 6 of 8 Solid-Gas Equilibrium: Consider the dynamic equilibrium: solid vapor Dry ice, CO2 (s), sublimes at room temperature, completely skipping a liquid phase. sublimation:solid gas (with no liquid phase) deposition: gas solid (with no liquid phase) H sub): Hsub = Hfus + Hvap molar heat of sublimation ( – energy required to sublime one mole of solid – given by sum of heats of fusion and vaporization – another illustration of Hess's Law Heat of Phase Transition heating-cooling curve: Shows the phase changes that occur when heat is added or removed from a sample Heating Curve Temperature (˚C) Heat added Draw a heating curve indicating the following: 1. Regions for solid only, liquid only, gas only, solid-liquid, liquid-gas 2. The relationship between melting point and phases present 3. The relationship between boiling point and phases present 4. Where the curve is flat, where the slope is positive CHEM 151: Chapter 11 page 7 of 8 11.3 Phase Diagrams – summarize the conditions at which a substance exists as solid, liquid, or gas – allow us to determine melting and boiling points at different external pressures Water: phase diagram shown in Fig. 11.11 – graph divided into three regions corresponding to each phase – lines separating two regions indicate conditions when both phases exist – triple point: point at which all three curves meet – when all three phases can be in equilibrium with each other – for water, at 0.01°C and about 4.56 mm Hg Phase Diagram for Water versus that for CO2: – Note that line separating solid and liquid phases has positive slope for CO2 but negative slope for H2O – for CO2, the solid is more dense than the liquid, so increasing pressure converts the liquid to a solid – for H2O, the solid is less dense than the liquid, so increasing pressure converts the solid to a liquid Critical Temperature and Pressure: critical temperature (Tc): above which its gas form cannot be made to liquefy, no matter how great the applied pressure – highest temperature at which a substance can exist as a liquid – intermolecular attraction is a finite quantity for given substance – below Tc, molecules are moving slowly enough to maintain contact – above Tc, molecular motion so energetic that molecules will always break away from attraction critical pressure (Pc): minimum pressure that must be applied to liquefy sample at the critical temperature Given a Phase Diagram, be able to do the following: – Determine what phase(s) is/are present at a given temperature and pressure – Indicate the boiling point or melting point at a given pressure – Describe what phase change occurs when temperature is changed at a constant pressure – Describe what phase change occurs when pressure is changed at a constant temperature. 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