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This is an attempt to state how scientists do science. It is necessarily artificial. Here are MY five steps: 1 • Introduction The Scientific Method (1 of 20) • Make observations the leaves on my plant are turning yellow • State a Problem to be solved how can I get my plants healthy (non-yellow) • Form a hypothesis maybe they need more water • Conduct a controlled experiment water plants TWICE a week instead of once a week • Evaluate results if it works, good... if not, new hypothesis (sunlight?) Step 1 of the Scientific Method is Make Observations. These can be of general physical properties (color, smell, hardness, etc.) which are called qualitative observations. 1 • Introduction Observations and Measurements Qualitative, Quantitative, Inferences (2 of 20) These can be measurements which are called quantitative observations. There are also statements that we commonly make based on observations. “This beaker contains water” is an example. You infer (probably correctly) it is water because it is a clear, colorless liquid that came from the tap. The observations are that it is clear, it is colorless, it is a liquid, and it came from the tap. Consider: 16.82394 cm In a measurement or a calculation, it is important to know which digits of the reported number are significant. 1 • Introduction Significant Digits I What do they mean? (3 of 20) That means… if the same measurement were repeated again and again, some of the numbers would be consistent and some would simply be artifacts. All of the digits that you are absolutely certain of plus one more that is a judgment are significant. If all the digits are significant above, everyone who measures the object will determine that it is 16.8239 cm, but some will say …94 cm while others might say …95 cm. a 1 • Introduction Significant Digits II Some examples with rulers. (4 of 20) b 1 c 2 (A composite ruler) a- No one should argue that the measurement is between 0.3 and 0.4. Is it exactly halfway between (.35 cm)… or a little to the left (.34 cm)? The last digit is the judgment of the person making the measurement. The measurement has 2 significant digits. b- The same ruler… so the measurement still goes to the hundredths place… 1.00 cm (3 significant digits). c- A ruler with fewer marks reads 1.6 cm (2 sig digits). 1 • Introduction Significant Digits III Rules for Recognizing Sig. Digits (5 of 20) In a number written with the correct number of sig. digits... • All non-zero digits are significant. 523 grams (3) • 0’s in the MIDDLE of a number are ALWAYS significant. 5082 meters (4) 0.002008 L (4) • 0’s in the FRONT of a number are NEVER significant. 0.0032 kg (2) 0.00000751 m (3) • 0’s at the END of a number are SOMETIMES significant. • Decimal point is PRESENT, 0’s ARE significant 2.000 Liters (4) 0.000500 grams (3) • Decimal point is ABSENT, 0’s are NOT significant 2000 Liters (1) 550 m (2) NOTE: textbook values are assumed to have all sig. digits Scientific notation uses a number between 1 and 9.99 x 10 to some power. It’s use stems from the use of slide rules. 1 • Introduction Scientific Notation Useful for showing Significant Digits (6 of 20) Know how to put numbers into scientific notation: 5392 = 5.392 x 103 0.000328 = 3.28 x 10 –4 1.03 = 1.03 550 = 5.5 x 102 Some 0’s in numbers are placeholders and are not a significant part of the measurement so they disappear when written in sci. notation. Ex: 0.000328 above. In scientific notation, only the three sig. digits (3.28) are written. Scientific Notation can be used to show more sig. digits. Values like 550 ( 2 sig. digits) can be written 5.50 x 102 (3) When you perform a calculation using measurements, often the calculator gives you an incorrect number of significant digits. Here are the rules to follow to report your answers: 1 • Introduction Significant Digits IV Significant Digits in Calculations (7 of 20) x and ÷: The answer has the same # of sig. digits as the number in the problem with the least number of sig. digits. example: 3.7 cm x 8.1 cm = 29.97 ≈ 30. cm2 (2 sig. digits) + and –: The last sig. digit in the answer is the largest uncertain digit in the values used in the problem. example: 3.7 cm + 8.1 cm = 11.8 cm (3 sig. digits) Know how to ilustrate why these rules work. 1 • Introduction Accuracy vs. Precision (8 of 20) Accuracy refers to how close a measurement is to some accepted or true value (a standard). Ex: an experimental value of the density of Al° is 2.69 g/mL. The accepted value is 2.70 g/mL. Your value is accurate to within 0.37% % error is used to express accuracy. Precision refers to the reliability, repeatability, or consistency of a measurement. Ex: A value of 2.69 g/mL means that if you repeat the measurement, you will get values that agree to the tenths place (2.68, 2.70, 2.71, etc.) ± and sig. digits are used to express precision. We generally use three types of measurements: volume Liters (mL) length meters (km, cm and mm) mass grams (kg and mg) 1 • Introduction Metric System (9 of 20) We commonly use the prefixes: 1/ centi100 th 1 milli/1000th kilo1000 Occasionally you will encounter micro(µ), nano, pico, mega, and giga. You should know where to find these in chapter 1. Know that 2.54 cm = 1 inch and 2.20 lb = 1 kg 1 • Introduction % and ppm (10 of 20) Percentage is a mathematical tool to help compare values. Two fractions, 3/17 and 5/31 are difficult to compare: If we set up ratios so we can have a common denominator: 3 x 17.65 5 x 16.13 = = = = 17 100 100 31 100 100 3 5 so… we can see that > . 17 31 There are 17.65 parts per 100 (Latin: parts per centum) or 17.65 percent (17.65 %)… the % is a “1 0 0” ppm (parts per million) is the same idea, (use 1,000,000 3 x instead of 100) = = 176,470 ppm 17 1 000 000 Consider the metric/English math fact: 2.54 cm = 1 inch This can be used as the “conversion factor”: 2.54 cm 1 inch or 1 inch 2.54 cm 1 • Introduction Unit Analysis Converting between English and Metric Units (11 of 20) You can convert 25.5 inches to cm in the following way: Given: 25.5 in 2.54 cm Desired: ? cm 25.5 in x = 64.77 cm ≈ 64.8 cm 1 in This is the required way to show your work. You have two jobs in this class, to be able to perform the conversions and to be able to prove that you know why the answer is correct. The important idea is that temperature is really a measure of something, the average motion (kinetic energy, KE) of the molecules. 1 • Introduction Temperature Scales (12 of 20) Does 0°C really mean 0 KE? nope... it simply means the freezing point of water, a convenient standard. We have to cool things down to –273.15°C before we reach 0 KE. This is called 0 Kelvin (0 K, note: NO ° symbol.) For phenomena that are proportional to the KE of the particles (pressure of a gas, etc.) you must use temperatures in K. K = °C + 273 °C = K – 273 mass is the amount of something... weight is how much gravity is pulling on the mass. (Weight will be proportional to the mass at a given spot.) 1 • Introduction Mass vs. Weight Theory, Measuring, Conversions (13 of 20) Mass is what we REALLY want to use... measured in grams. You use a balance to measure mass... you compare your object with objects of known mass. Weight is measured with a scale (like your bathroom scale or the scale at the grocery store). If there is no gravity, it doesn’t work. Note: electronic balances are really scales! You convert mass / weight using: 1 kg 2.205 lbs or 2.205 lbs 1 kg 1 You can calculate the KE of an object: KE = mv2 2 m = mass, v = velocity [Note units: 1 J = 1 kg·m 2 ·s–2] Temperature is a measure of the average kinetic energy. 1 • Introduction Potential Energy (PE) and Kinetic Energy (KE) (14 of 20) PE = the potential to do work which is due to an object’s position in a field. For example, if I hold a book 0.5 m above a student’s head it can do some damage... 1.0 m above her/his head, more work can be done. Important ideas: Objects tend to change from high PE to low PE (downhill). High PE is less stable than low PE. Extensive properties depend on the amount of substance. We measure these properties frequently... (mass & volume... mostly). 1 • Introduction Mass, Volume, and Density Intensive vs. Extensive Properties (15 of 20) Intensive properties are independent of the size of the sample. These are useful for identifying substances... (melting point, boiling point, density, etc.) mass volume is the ratio of two extensive properties... the size of the sample sort of “cancels out.” Be able to do density problems (3 variables) and know the usefulness of specific gravity. It is interesting that an intensive property, density = Heat is the total KE while temperature is the average KE. 1 • Introduction Calorimetry (16 of 20) A way to measure heat is to measure the temperature change of a substance... often water. It takes 1 calorie of heat energy (or 4.184 J) to heat 1 gram of H 2 O by 1 °C. cal J The specific heat of water = 1 = 4.184 g·°C g·°C heat = specific heat x mass H2 O x ∆T You can heat other substances as well, you just need to know their specific heats. Notice that this is simply heating or cooling a substance, not changing its phase. Equations to symbolize changes: reactants → products 1 • Introduction Physical and Chemical Properties Physical and Chemical Changes (17 of 20) Physical Properties can be measured from a sample of the substance alone... (density, MP, BP, color, etc.) Chemical Properties are measured when a sample is mixed with another chemical (reaction with acid, how does it burn in O2 ) Physical Changes imply that no new substances are being formed (melting, boiling, dissolving, etc.) Chemical Changes imply the substance is forming new substances. This change is accompanied by heat, light, gas formation, color changes, etc. Matter Pure Substances Elements 1 • Introduction Pure Substances, Elements, & Compounds Homogeneous & Heterogeneous Mixtures (18 of 20) Compounds Energy Mixtures Homogeneous Heterogeneous This chart should help you sort out these similar terms. Be able to use chemical symbols to represent elements and compounds. For example... CuSO4 •5H2 O, a hydrate, contains 21 atoms & 4 elements. Memorize the 7 elements that exist in diatomic molecules: HONClBrIF or BrINClHOF or “H and the 6 that make a 7 starting with element #7” 1 • Introduction Separating Mixtures by Filtration, Distillation, and Chromatography (19 of 20) Mixtures are substances the are NOT chemically combined... so if you want to separate them, you need to exploit differences in their PHYSICAL properties. Filtration: some components of the mixture dissolve and some do not. The filtrate is what passes through the filter. Distillation: some components vaporize at different temperatures or one component may not vaporize at all (e.g.: salt+water) complete separation may not be possible. Chromatography: differences in solubility vs. adhesion to the substrate. Substaratemay be filter paper (paper chromatography), or other substances, GLC, TLC, HPLC, column, etc. Definite Composition: samples of the same substance from various sources (e.g. water) can be broken down to give the same %’s of elements. Calculation: percent composition 1 • Introduction Early Laws: the Law of Definite Composition & the Law of Simple Multiple Proportions (20 of 20) Multiple Proportions: samples of 2 substances made of the same 2 elements... (e.g. CO2 & CO or H2 O and H2 O2 or CH4 and C3 H8 ) if you break down each to give equal masses of one element, the masses of the other element will be in a simple, whole-number ratio. Calculation: proportions to get equal amounts of one element and then simple ratios. Superscripts used to show the charges on ions Mg2+ the 2 means a 2+ charge (lost 2 electrons) 2•Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) Subscripts used to show numbers of atoms in a formula unit H2 SO4 two H’s, one S, and 4 O’s Coefficients used to show the number of formula units 2Br– the 2 means two individual bromide ions Hydrates CuSO4 • 5 H2 O some compounds have water molecules included stoichiometry atomic mass 2•Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24) formula mass molecular mass study of the quantitative relationships in chemical formulas and equations. weighted average mass of an atom, found on the periodic table sum of the atomic masses of the atoms in a formula sum of the atomic masses of the atoms in a molecular formula gram molecular mass molecular mass written in grams molar mass same as gram molecular mass empirical formula formula reduced to lowest terms Formula or molecular mass is found by simply summing the atomic masses (on the periodic table) of each atom in a formula. 2•Stoichiometry: Chemical Arithmetic Calculating Formula Mass (3 of 24) H2 SO4 1.01 + 1.01 + 32.06 + 16.0 + 16.0 + 16.0 + 16.0 = 98.08 u 2(1.01) + 32.06 + 4(16.0) = 98.06 u or 98.06 g/mole Generally, round off your answers to the hundredths or tenths place. Don’t round off too much (98.06 g/mol or 98.1 g/mol is OK, but don’t round off to 98 g/mol) Units Use u or amu if you are referring to one atom or molecule A mole (abbreviated mol) is a certain number of things. It is sometimes called the chemist’s dozen. A dozen is 12 things, a mole is 6.02 x 1023 things. 2•Stoichiometry: Chemical Arithmetic Mole Facts (4 of 24) Avogadro’s Number 1 mole of any substance contains 6.02 x 1023 molecules Molar Volume (measured at P = 760 mmHg and T = 0 °C) 1 mole of any gas has a volume of 22.4 Liters Molar Mass (see gram formula mass) 1 mole 6.02 x 10 23 molecules 1 mole 22.4 L 1 mole molar mass A Line Equation is the preferred way to show conversions between quantities (amount, mass, volume, and number) by canceling units (moles, grams, liters, and molecules) 2•Stoichiometry: Chemical Arithmetic Line Equations (5 of 24) The line equation consists of the Given Value, the Desired Unit, and the line equation itself. Example: What is the mass of 135 Liters of CH4 (at STP)? Given: 135 L CH4 Desired: ? g CH 4 1 mol CH4 16.0 g CH4 135 L CH4 x x = 96.43 g CH 4 22.4 L CH4 1 mol CH4 The “Mole Map” shows the structure of mole problems Mass 2•Stoichiometry: Chemical Arithmetic Mole Relationships (6 of 24) Mass ➀ Volume at STP ➀ ➂ 1 mol molar mass 2) Volume at STP ➂ number of atoms or molecules 1) ➁ moles ➁ number of atoms or molecules 1 mol 22.4 L 3) 1 mol 6.02 x 10 23 molecules Percentage Composition quantifies what portion (by mass) of a substance is made up of each element. 2•Stoichiometry: Chemical Arithmetic Percentage Composition (by mass) (7 of 24) mass of element mass of molecule mass of element Change to percentage: 100 x mass of molecule Set up a fraction: Generally, round off your answers to the tenth’s place. The percentage compositions of each element should add up to 100% (or very close, like 99.9% or 100.1%) 2•Stoichiometry: Chemical Arithmetic Formula from % Composition (8 of 24) Given the Percentage Composition of a formula, you can calculate the empirical formula of the substance. Step 1 assume you have 100 g of substance so the percentages become grams Step 2 change grams of each element to moles of atoms of that element Step 3 set up a formula with the moles example: C2.4 H4.8 Step 4 simplify the formula by dividing moles 2.4 4.8 by the smallest value C 2.4 H2.4 = CH2 Step 5 If ratio becomes… 1:1.5 multiply by 2 1:1.33 or 1:1.66 multiply by 3 equation condensed statement of facts about a chemical reaction. substances that exist before a chemical rxn. Written left of arrow. substances that come into existence as a result of the reaction. Written to the right of the arrow. an equation describing a chemical change using the names of the reactants and products. a number preceding atoms, ions, or molecules in balanced chemical equationns that showing relative #’s. reactants → 2•Stoichiometry: Chemical Arithmetic Equation Terms (9 of 24) → products word equation coefficients The gas density is often converted to molar mass: 2•Stoichiometry: Chemical Arithmetic Other Mole Problems and Conversions (10 of 24) Example : The gas density of a gas is 3.165 g/Liter (at STP). What is the molar mass of the gas? Knowing that 22.4 L is 1 mole, you can set up the ratio: 3.165 g molar mass = 1 Liter 22.4 L Other metric conversions you should know: 1000 mL 1 kg 1 Liter 1000 grams Example: Write the formula equation of... sodium metal + water → sodium hydroxide + hydrogen gas Na° + H 2 O → NaOH + H2 2•Stoichiometry: Chemical Arithmetic Writing Formula Equations Things To Remember (11 of 24) • metals often are written with the ° symbol to emphasize that the metal is in the neutral elemental state, not an ion. • some compounds have common names that you should just know... water, H 2 O; ammonia, NH3 ; methane, CH4 • remember the seven diatomic elements so they can be written as diatomic molecules when they appear in their elemental form. Other elemental substances are written as single atoms (e.g. sodium metal or helium gas, He) Since every gas takes up the same amount of room (22.4 L for a mole of a gas at STP), the coefficients in an equation tell you about the volumes of gas involved. 2•Stoichiometry: Chemical Arithmetic Coefficients and Relative Volumes of Gases (12 of 24) Example: N2 (g) + 3 H2 (g) → 2 NH 3 (g) + The “heart of the problem” conversion factor relates the Given and the Desired compounds using the coefficients from the balanced equation. 2•Stoichiometry: Chemical Arithmetic Heart of the Problem (13 of 24) Example: N2 + 3 H2 → 2 NH3 3 moles H ♥ could be 2 moles NH2 3 …which means that every time 2 moles of NH 3 is formed, 3 moles of H2 must react. The format is always, moles of Desired moles of Given Mass-Mass problems are probably the most common type of problem. The Given and Desired are both masses (grams or kg). 2•Stoichiometry: Chemical Arithmetic Mass-Mass Problems Mass-Volume Problems (14 of 24) 2•Stoichiometry: Chemical Arithmetic Mass-Volume-Particle Problems (15 of 24) The pattern is: molar mass Given x of Given x mass ♥ x molar of Desired In Mass-Volume problems, one of the molar masses is 22.4 L replaced with depending on whether the Given or 1 mole the Desired is Liters. If the Given or Desired is molecules, then the Avogadro’s 6.02 x 10 23 molecules Number conversion factor, is used 1 mole and the problem is a Mass-Particle or Volume-Particle problem. The units of the Given and Desired will guide you as to which conversion factor to use: Mass Volume Particles grams or kg Liters or mL molecules or atoms In a problem with two Given values, one of the Given’s will limit how much product you can make. This is called the limiting reactant. The other reactant is said to be in excess. 2•Stoichiometry: Chemical Arithmetic Limiting Reactant Problems (16 of 24) Solve the problem twice using each Given… the reactant that results in the smaller amount of product is the limiting reactant and the smaller answer is the true answer. Example: N2 + 3 H2 → 2 NH3 When 28.0 grams of N2 reacts with 8.00 grams of H 2 , what mass of NH3 is produced? (in this case, the N2 is the limiting reactant) 2•Stoichiometry: Chemical Arithmetic How Much Excess Reactant is Left Over (17 of 24) Example: N2 + 3 H2 → 2 NH3 When 28.0 grams of N2 reacts with 8.00 grams of H 2 , what mass of NH3 is produced? (in this case, the N2 is the limiting reactant) To find out how much H2 is left over, do another line equation: Given: 28.0 g N2 Desired: ? g H2 subtract the answer of this problem from 8.00 g H2 It is difficult to simply guess which reactant is the limiting reactant because it depends on two things: (1) the molar mass of the reactant and (2) the coefficients in the balanced equation 2•Stoichiometry: Chemical Arithmetic Limiting Reactants (18 of 24) The smaller mass is not always the limiting reactant. Example: N2 + 3 H2 → 2 NH3 1 mole (28 g N 2 ) will just react with 3 moles (6.06 g H2 ) so, if we react 28.0 g N2 with 8.0 g H2 , only 6.06 g H2 will be used up and 1.94 g of H2 will be left over. In this case, N2 is the L.R. and H2 is in X.S. The answer you calculate from a stoichiometry problem can be called the Theoretical Yield. Theoretically, you should get this amount of product. 2•Stoichiometry: Chemical Arithmetic Theoretical Yield and Percentage Yield (19 of 24) In reality , you often get less than the theoretical amount due to products turning back to reactants or side reactions. The amount you actually get is called the Actual Yield. Percentage Yield = 2•Stoichiometry: Chemical Arithmetic Balancing Chemical Equations (20 of 24) Actual Yield x 100 Theoretical Yield The balanced equation represents what actually occurs during a chemical reaction. Since atoms are not created or destroyed during a normal chemical reaction, the number and kinds of atoms must agree on the left and right sides of the arrows. __Na 2 CO3 + __HCl → __ NaCl + __H2 O + __CO2 To balance the equation, you are only allowed to change the coefficients in front of the substances... not change the formulas of the substances themselves. Reduce the coefficients to the lowest terms. Fractions may be used in front of diatomic elements. The burning of fuels made of C, H, and O is called combustion. You need to memorize O2 , CO2 and H 2 O Example: The combustion of propane, C3 H8 , is written: 2•Stoichiometry: Chemical Arithmetic Combustion Equations (21 of 24) C 3 H8 + 5O2 → 3CO2 + 4H2 O Be careful when writing equations for alcohols, such as butanol, C 4 H9 OH • don’t forget to add the H’s (a total of 10 of them) • don’t forget to take account of the O atom in the alcohol C 4 H9 OH + 6O2 → 4CO2 + 5H2 O 2•Stoichiometry: Chemical Arithmetic Solutions -- Molar Concentration (22 of 24) Many reactions are carried out in solution. Solutions are convenient and speed up many reactions. Concentration is often expressed as moles of solute Molarity (M) = Liters of solution You can calculate the molarity of a solution when given moles (or grams) of a substance and its volume. You can use the molarity of a solution as a conversion factor 0.150 moles HCl 1 Liter HCl 0.150 M HCl ≈ or 1 Liter HCl 0.150 moles HCl to convert moles to Liters and vice versa. Volumetric flasks are used to make solutions. You can calculate the moles of a solute using the volume and molarity of the substance. Since diluting a solution adds water and no solute, the moles of solute before and after the dilution remains constant. So... 2•Stoichiometry: Chemical Arithmetic Dilution Problems (23 of 24) Vi · Mi = Vf · Mf where “I” means “initial” and “f” means “final” The units of volume or concentration do not really matter as long as they match on the two sides of the equation. Acids form the H+ ion. Bases form the OH– ion. Acids + bases mix to form H 2 O (HOH) and a salt. The moles of H+ = the moles of OH– in a neutralization. 2•Stoichiometry: Chemical Arithmetic Acid-Base Titrations (24 of 24) An acid-base titration is the technique of carefully neutralizing an acid with a base and measuring the volumes used. An indicator (we used phenolphthalein) allows us to observe when the endpoint is reached. If a monoprotic acid is neutralized with a base that only has one OH– ion per formula unit, the simple formula: Va · Ma = Vb · Mb allows you to determine the molarity of the unknown. 3 • The Periodic Table & Makeup of Atoms The Subatomic Particles (1 of 12) Name Symbol Mass Charge protons p+ 1u 1+ neutron n° 1u 0 1 – electron e u 1– 1837 Location part of the nucleus part of the nucleus normally at large distances from the nucleus J.J. Thompson is given credit for discovering electrons using a Crookes tube and testing many different gases. Cathode rays were found to be beams of electrons. Chadwick is given credit for the discovery of the neutron. 3 • The Periodic Table & Makeup of Atoms Terms I-- Atomic Structure (2 of 12) atoms the smallest particle of an element . It consists of a central nucleus and electron clouds outside the nucleus. nucleus the dense central portion of an atom. subatomic smaller than an atom. The proton, neutron, and electron are subatomic. net charge the difference in the positive charge due to protons and the negative charge due to electrons in an atom. nucleons the particles that make up the nucleus. atomic number the number of protons in an atom. This # determines the identity of an element. 3 • The Periodic Table & Makeup of Atoms Terms II-- Atomic Structure (3 of 12) mass number the number of protons + neutrons isotopes atoms with the same number of protons, but different numbers of neutrons. Atoms with the same atomic number, but different mass numbers. isotopic notation shorthand notation for a nucleus that shows the mass #, atomic # and the 238 symbol. U-238 would be 92 U Any real sample of an element contains more than one naturally occuring isotope. For instance, boron isotope 3 • The Periodic Table & Makeup of Atoms Calculating Atomic Mass (4 of 12) 10 boron-10 5 B 11 boron-11 5 B abundance mass # isotopic mass 19.78% 10 mass = 10.013 u 80.22% 11 mass = 11.009 u The atomic mass is the weighted average of the isotopes. (19.78%)(10.013u) + (80.22%)(11.0009u) or 100 at. mass = (0.1978)(10.013u) + (0.8022)(11.0009u) = 10.81 u at. mass = 33 Consider the following symbol: 16S2– The 16 is the atomic number which is the number of protons. 3 • The Periodic Table & Makeup of Atoms Determining Numbers of Protons, Neutrons, and Electrons from the Isotopic Notation (5 of 12) The 33 is the mass number which is the mass of one of the isotopes. This mass is due to the protons and neutrons. The number of neutrons is the mass number - the atomic number. 33 - 16 = 17 neutrons. Since the charge is 2-, there are 2 more electrons than protons. In this case, there are 18 electrons. 3 • The Periodic Table & Makeup of Atoms Important People in the Development of the Atomic Theory (6 of 12) Democritus [atomos] philosopher who argued that matter was discontinuous John Dalton [billiard-ball model] experimented with gases… different substances are different combinations of atoms J.J. Thomson [plum-pudding model] experimented with gas-discharge tubes… atoms have + and – parts… the negative e– ’s are the same for any atom Ernest Rutherford [nuclear model/solar system model] most of the mass of the atom is concentrated in a tiny, positively-charged nucleus Niels Bohr [quantized electron energy levels] the electrons have only certain allowed energy levels The elements can be classified as metals, nonmetals, and metalloids. Memorize the elements classified as metalloids (also called semi-metals or semiconductors). 3 • The Periodic Table & Makeup of Atoms Metals, Nonmetals, and Metalloids (7 of 12) Properties of metals include: ductility - the ability to pull a substance into a wire sectility - the ability to cut with a knife malleability - the ability to pound substance into a sheet conductivity - the ability to carry an electrical current Gold is the most malleable of all the metals. Ernest Rutherford’s classic gold foil experiment led to the nuclear model of the atom. 3 • The Periodic Table & Makeup of Atoms Rutherford’s Gold Foil Experiment (8 of 12) α a few bounced back at a large angle gold foil most alpha's α came straight through here • the nucleus is tiny - because most of the alpha’s missed the nucleus and went straight through the foil • the nucleus is positively charged - because the (+) charged alpha was repelled by the (+) charged nucleus • the nucleus is incredibly dense - because the nucleus was able to bounce back at a very large angle 3 • The Periodic Table & Makeup of Atoms The First Periodic Table (9 of 12) Meyer and Mendeleev are given credit for developing the first version of the periodic table. Mendelleev’s true claim to fame was that he actually predicted the existence of several element that had not been discovered. He found gaps in the table when he tried to organize the atoms and left spaces for those elements (ekasilicon = "like silicon", etc.) He predicted Ga, Ge, and Sc. He also arranged elements in order of atomic number rather than the previous idea of atomic mass. Several of the elements change order... (like Te and I). Horizontal rows of the table are called periods. Vertical columns are called groups or families. 3 • The Periodic Table & Makeup of Atoms Families of the Periodic Table (10 of 12) Memorize the names of some groups: IA - the alkali metals IIA - the alkaline earth metals VIIA - the halogens 0 - the noble gases Also know the transition metals, the inner transition metals (composed of the lanthanide series and the actinide series... the lanthanides are also called the rare earth metals) Families IA - VIIIA are called the representative elements. Radioactivity was discovered by Henri Becquerel (but named by Marie Curie). 3 • The Periodic Table & Makeup of Atoms Radioactivity Basics (11 of 12) “Becquerel rays” were found to consist of 3 types or radiation: alpha particles (α) a helium nucleus - 2 protons + 2 neutrons …easily stopped by paper or skin beta particles (β) a high energy electron …stopped by Al foil (several thicknesses of foil) gamma radiation (γ) a very high energy form of light (EMR) …the most penetrating and dangerous of the rays. 3 • The Periodic Table & Makeup of Atoms JJ Thomson & cathode ray tubes (12 of 12) Know the design of a cathode rays tubes. Realize that cathode rays are really beams of electrons. The cathode rays are the same for any substance, but the canal rays (the positive ions left after ionizing the gases) are different for each gas. Know how the bending of cathode rays can tell you the charge-to-mass ratio (e/m) (but not the mass or the charge of the electron). Millikan’s oil drop experiment gave evidence for the charge of the electron. Knowing this and the e/m ratio, you can calculate the mass of the electron. 4 • Electronic Structure & the Per. Table Wave Ideas You Should Know (1 of 16) EMR electromagnetic radiation … oscillating electric & magnetic fields at right angles wavelength (λ) the distance from crest to crest or trough to trough. amplitude the distance from the equilibrium point to the crest or trough. frequency (ν) the number of waves that pass a point per second. (Hz, s–1, 1/s) continuous spectrum a “normal” rainbow that contains all of the colors (ROYGBV). line spectrum a spectrum that only contains certain bright lines that result from electron transitions within an atom. Exchanging wavelength, frequency, & energy of light: λ·ν=c c = 3.00 x 10 8 m/s [speed of light] 4 • Electronic Structure & the Per. Table Wave Calculations (2 of 16) 4 • Electronic Structure & the Per. Table The Balmer Series (3 of 16) E = hν h = 6.626 x 10–34 J·s [Planck’s constant] hc E= λ Energy of Level “n” in the Hydrogen atom: A En = – 2 A = 2.18 x 10–18 J [Arrhenius constant?] n NOTE: You can learn the Balmer equation or the Rydberg equation listed on pp. 108 & 109. I suggest you use the above equations to calculate the wavelengths or frequencies of light emitted when electrons change energy levels. When the electron “drops” to energy level n=2, visible light is emitted. The bight line spectrum observed is called the Balmer Series of lines. [Memorize this info...] 3→2 red 4→2 blue-green 5→2 blue-violet 6→2 violet All the lines result from an electron transition to level n=1 are too high energy to be visible… UV [Lyman Series] All the lines that result from an electron transition to levels n=3 [Paschen], n=4 [Brackett] and n=5 [Pfund] are too LOW energy to be visible. 4 • Electronic Structure & the Per. Table The De Broglie Wavelength of Electrons (4 of 16) Two equations can be combined into one to allow you to calculate the wavelength of a particle: hc E = mc2 E= λ h c mc2 = λ h h mc = or more generally, mv = λ λ h λ= mv h = Planck’s constant, m = mass of particle (in kg), v = velocity of particle (in m/s). Standing waves are something that waves do… resulting from having repeating waves that always cancel at the nodes and always add up at the antinodes. 4 • Electronic Structure & the Per. Table Standing Waves (5 of 16) We have seen standing waves of strings: of sound: of drumheads: + + – + – We hypothesized the standing waves of electrons: (4-dimensional vibrating wavicles) the orbitals (s, p, d, f… probability waves) These are variables in some unseen equation: The Rules: n = 1, 2, 3, ... determines the energy of the e– 4 • Electronic Structure & the Per. Table Quantum Numbers (n, l, m, s) (6 of 16) l = 0 → (n–1) the type of orbital (subshell) 0 ≈ s, 1 ≈ p, 2 ≈ d, 3 ≈ f m = –l → +l which orientation of the orbital (x, y, z… for p orbitals) 1 1 s = + or – 2 2 the “spin” of the electron NOTE: m is also called ml and s is ms The Pauli Exclusion Principle states that no two electrons in an atom may have the same four quantum numbers. This translates to the idea that an orbital may contain no more than two electrons. 4 • Electronic Structure & the Per. Table Three Rules for Filling Orbitals (7 of 16) The Aufbau Principle states that electrons occupy the lowest energy available orbital. You must memorize the orbital chart on page 120. [Aufbau = “building up”] Hund’s Rule states that when you have several orbitals of the same energy (e.g. three p orbitals or five d orbitals) place one electron in each orbital before doubling them up. NOTE: Remember the Aufbau hotel analogy... This is a shorthand notation for the arrangement of electron in the orbitals. 1s2 means 2 electrons in the 1s orbital 4 • Electronic Structure & the Per. Table Electron Configurations (8 of 16) Consider Arsenic, As This is the order in which the electrons fill… As 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 however, we write the orbitals according to the distance from the nucleus (n=3’s then n=4’s, etc.) As 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 long form: (shown above) show each subshell short form: (show the last filled NRG level...noble gas core) As [Ar] 3d10 4s2 4p3 4 • Electronic Structure & the Per. Table The s-block, p-block... of the Periodic Table Exceptions to the Filling Rules (9 of 16) The shape of the Periodic Table comes from the way the electrons fill the orbitals. s-block Families I and II p-block Families III, IV, V, VI, VII, and VIII d-block The Transition Elements f-block The Inner Transition Elements Six elements do not follow the rules. They are in the dblock of the periodic table… the transition elements. Cu, Ag, and Au... instead of a full s-orbital and almost filled d-orbital, they have a filled d and half-filled s-orbital Cr, Mo, W... instead of a full s-orbital and an almost halffilled d-orbital, they have d5 and s1 Know the general shapes of the s (spherical) orbitals, 4 • Electronic Structure & the Per. Table Shapes of the orbitals (10 of 16) 4 • Electronic Structure & the Per. Table Predicting the Atomic Size (radius) Trends in the Periodic Table (11 of 16) p (perpendicular) orbitals, and d (diagonal) orbitals. What: As you move ACROSS a period, the size of the atom DECREASES Why: As you move ACROSS a period, the number of protons in the nucleus increases... so the protonelectron attraction increases... the size decreases. NOTE: The increased e– -e– repulsion that one might expect is not important, because the outer electrons do not SHIELD the electrons from the nucleus.. What: As you move DOWN a family, the size of the atom INCREASES Why: As the value of “n” increases, the average distance of the electron from the nucleus increases. The size of the atom IS the electron cloud... so the size increases The change in size of ions (compared to the neutral element) depends on the electron-electron repulsion. 4 • Electronic Structure & the Per. Table Explaining The Sizes of Ions The Lanthanide Contraction (12 of 16) If an atom gains electrons, the increased repulsion increases the size of the electron cloud. So... negative ions are larger than the neutral atom. Positive ions form by losing electrons (less repulsion) and get smaller. The Lanthanide contraction makes the third row of the transition elements about the same size as the second row... This causes these elements to be much more dense. Following the decrease in size, the ionization energy GENERALLY increases as you move across a period. If the atom is smaller, the electron being removed is closer to the nucleus and therefore feels a stronger attraction. 4 • Electronic Structure & the Per. Table Ionization Energy Trends Across a Period (13 of 16) The Jags: (why are some electrons EASIER to remove) In Family III, the electron being removed comes from the porbital rather than the s-orbital. The p-orbital electron is at a higher energy and requires LESS energy to ionize. In Family VI, the electron being removed comes from an orbital with TWO electrons. The e– - e– repulsion felt by the electron allows it to be ionized with less added energy. 4 • Electronic Structure & the Per. Table Ionization Energy & Reactivity Trends Down a Family (14 of 16) 4 • Electronic Structure & the Per. Table Ionization Energy Trends in Successive Ionizations (15 of 16) What: Why: Down a family, Ionization energy DECREASES The trend here goes right with the size of the atom. Since “n” increases, the distance of an electron from the nucleus increases and is easier to remove. What: Why: Down family I and II, the reactivity INCREASES. These families lose electrons to become + ions. The easier it is (lower ionization energy) the more reactive they are. What: Why: Down family VII, the reactivity DECREASES. These elements GAIN electrons to become – ions. Smaller atoms mean the attracted electrons will be closer to the nucleus... more effective attractions. What: You can tell the family of an element by observing when the ionization energies get very large. For example: family III elements Al + energy = Al+ + e– (3p electron) Al+ + energy! = Al2+ + e– (3s electron) Al2+ + energy!! = Al3+ + e– (3s electron) Al3+ + ENERGY = Al4+ + e– (2p electron!) Why: The “easy” electrons to remove are in an orbital with a higher value of n. When n decreases, the average distance of the electron from the nucleus decreases and the attractions between the proton and electrons increase. Electron affinity is the energy involved when an atom gains an electron to become a negative ion. F + e – → F– + energy 4 • Electronic Structure & the Per. Table Electron Affinity (16 of 16) Elements in the upper right corner of the periodic table have the greatest electron affinity (greatest attraction for electrons). The electron affinity may be + or –. Negative values mean energy is released and also counts as a greater electron affinity. Electron affinity data is not complete, but it gives SOME evidence for trends in the periodic table. 5 • Chemical Bonding: Gen Concepts Some Properties of Ionic and Molecular Compounds (1 of 12) Compound Conducts as Solid Conducts as Liquid Conducts in Solution Conducts as Gas Molecular NO NO NO NO Ionic NO YES YES YES Hardness soft hard MP / BP low high Bonding covalent ionic Examples He, CH 4 , CO2 , C 6 H12O6 NaCl, KI, AgNO 3 Lewis symbols conisist of the atomic symbol surrounded by valnece electrons. The four sides represent the four valence orbitals. Atoms are usually shown in their excited states (II, III, IV) 5 • Chemical Bonding: Gen Concepts Lewis Symbols of Atoms and Ions (2 of 12) Li Be B C N O F Ne Ions include brackets. Positive ions show no valence electrons while negative ions usually have an octet. [Li]+ [Mg] 2+ [O] 2– Many ions can be explained because they have gained or lost electrons and attain a noble gas configuration. For example: P 3– S2– Cl– Ar K+ Ca 2+ all have the same electron arrangement: 1s2 2s2 2p 6 3s2 3p 6 5 • Chemical Bonding: Gen Concepts The Ionic Bond Noble and Pseudonoble Gas Configurations (3 of 12) A pseudonoble gas configuration is: 1s2 2s2 2p 6 3s2 3p 6 3d 10 This is found in Cu+ Zn 2+ Ga3+ and Ge4+ Similar configurations are found in the next two periods. The importance of this configuration is that there is more than one reason why ions form what they do. Many ions are not explained. Know the 5 steps that can be thought to occur when an ionic bond forms. Note whether each is exo- or endothermic... whether a larger energy helps or hinders the bond formation. 5 • Chemical Bonding: Gen Concepts Factors that Influence the Formation of Ionic Bonds (4 of 12) Overall: 1. 2. 3. 4. 5. Li(s) + 1/2F2 (g) → LiF(s) heat of vaporization heat of decomposition ionization energy electron affinity lattice energy Li(s) + NRG → Li(g) 1/2F2 (g) + NRG → F(g) Li(g) + NRG → Li+(g) + e– F(g) + e– → F– (g) Li+(g) + F– (g) → LiF(s) Large energy values for 1,2,3 hinder ionic bond formation. The covalent bond between two atoms depends on the balanc e of attractions between one atom’s + nucleus and the other atom’s – electrons and the proton-proton repulsions as well as electron-electron repulstions. 5 • Chemical Bonding: Gen Concepts The Covalent Bond Attractions and Repulsions (5 of 12) PE Distance between nuclei If two atoms have half-filled orbitals , the interactions balance at a small enough distanc eso the e – ’s can be close to both nucle i at the same time... this is a covalen tbond. Count up your valence electrons. 5 • Chemical Bonding: Gen Concepts Groves’ Electron Dot System Multiple & Extended Valence Bonds (6 of 12) Give every atom who “wants” and octet an octet. [the first 5 elements do not need octets... too small] [Family I, II, and III do not form octets] If you have drawn too man y electrons... “Take away a lone pair... take away a lone pair... make these two atoms share” If you have drawn too fe w electrons... place the extra electrons on the central atom (extended valence shell) 5 • Chemical Bonding: Gen Concepts Bond Order: Bond Length, Strength, & Vibrational Frequency (7 of 12) Bond orde r is the number of pairs of electrons bonding two atoms together. single bond bond order = 1 double bond bond order = 2 triple bond bond order = 3 single bonds have the longest bond length single bonds have the weakest bond strength single bonds have the lowest vibrational frequency (think of single bonds as soft, springy springs... triple bonds are tight springs...sproinnnnng) Bonds in resonance structure must be averaged... the S-O bond in SO 2 has a bond order of 1.5. C-O in CO3 2– is 1.33 When you draw a Lewis structure (SO2 , O3 , CO3 2–, etc.) in which you must make a choic e as to who gets a double bond, the structure is actually a blend of two or three structures. 5 • Chemical Bonding: Gen Concepts Resonance (8 of 12) We “say” that the structure “resonate s” or we say that the structure contains contributions from each of the resonance structures. Resonance occurs simply because the electron-dot model (while very useful) is too limited to show how the electrons are being shared between the atoms... wait for π bonding. Coordinate covalent bond: When a covalent bond is formed by sharing a pair of electrons BUT the electron pair belonged to only one of the atoms. 5 • Chemical Bonding: Gen Concepts Coordinate Covalent Bonds (Preview: Lewis Acids) (9 of 12) Classic Example: NH3 + BF3 → NH3 BF3 The bond between the N and the B is coordinate covalent. The lone pair donor is called a Lewis Base. (this atom has a lone pair of electrons) The lone pair acceptor is called a Lewis Acid. (this atom has an empty orbital) “Have Pair Will Share” --Lewis Base 5 • Chemical Bonding: Gen Concepts Electronegativity and Polar Bonds (10 of 12) 5 • Chemical Bonding: Gen Concepts Naming Ionic Compounds Traditional and Stock Names (11 of 12) You will be given a chart of electronegativity values. Memorize the most electronegative element s (F = 4.0) then oxygen (O = 3.5) and chlorine (Cl = 3.0). The noble gases have no electronegativity values… no bonds. Trend is large electronegativity in the upper right of the per. table and small in the lower left portion of the table. Classify the bond between any two atoms by subtracting their electronegativity values (∆e) Non-polar covalent 0 < ∆e < 0.5 Polar covalent 0.5 ≤ ∆e ≤ 1.7 Ionic ∆e > 1.7 The more electronegative atom is more negative. Polar covalent bonds have partial charges δ+ and δ– The Stock System of naming compounds is used… • when a positive ion has more than one possible charge Traditional: mercurous, Hg2 2+ mercuric, Hg2+ Stock: mercury(I) mercury(II) Traditional: cuprous, Cu+ cupric, Cu 2+ Stock: copper(I) copper(II) • for molecular compounds where the elements have many different oxidation states (i.e. N in NO 2 , NO, N 2 O, etc.) Stock Name: Traditional Name: NO2 nitrogen(IV) oxide nitrogen dioxide NO nitrogen (II) oxide nitrogen monoxide N2 O nitrogen(I) oxide dinitrogen monoxide Acids are ionic formulas in which the positive ion is H+. Use as many H+ ions as the charge on the negative ion. 5 • Chemical Bonding: Gen Concepts Naming Acids (12 of 12) Three rules for naming: if the anion ends with: –ite –ate –ide the acid is named: ********ous acid ********ic acid hydro********ic acid • Acids from sulfide, sulfite, and sulfate include a “ur” H2 S is hydrosulfuric acid, not hydrosulfic acid • Acids from phosphate and phosphite include a “or” H3 PO4 is phophoric acid, not phosphic acid 7 • Chemical Reactions & the Per. Table Solutions and Solubility (1 of 16) We learned the terms solute, solvent, and solution. Solubility (how MUCH solute will dissolve) is measured in g/100 mL of H2 O or sometimes in M. This information may be given numerically or graphically. A saturated solution is one in which any additional solute added will simply settle on the bottom of the container. An unsaturated solution is any amount less than saturated. Supersaturated implies that the solution was saturated at some higher temperature and then carefully cooled. This unstable situation can be changed with a “seed” crystal. Recall the supersaturated solution of NaC2 H3 O2 demo. The terms concentrated and dilute refer to the amount of solute and do not necessarily coincide with saturation. 7 • Chemical Reactions & the Per. Table Electrolytes: Weak and Strong (2 of 16) Solutions of acids, bases, and salts contain mobile ions and conduct electricity. These solutions are called electrolytes. Salts are ionic compounds that dissociate in water. Acids are actually molecular compounds (covalently bonded) the become ions when dissolved in water. Only 8 acids are strong electrolytes and completely dissociate when dissolved. All others dissolve completely, but only partially dissociate into ions. Only 8 bases are strong electrolytes because they dissolve completely. All others have low solubility and remain solids rather than dissolve. One common exception is the weak base NH4 OH . It dissolves, but partially dissociates. 7 • Chemical Reactions & the Per. Table Ionic Reactions (3 of 16) A common class of chemical reactions occurs when two ionic solutions are mixed. The double replacement or metathesis reaction involves the formation of two new combinations of ions. AgNO 3 + NaCl → AgCl + NaNO3 (molecular equation) The new combinations may be more stable than the original due to low solubility (precipitate forms), weak electrolyte, gas formation, or complex ion formation. The reaction is written above as though the substances exist as molecules. This is the easiest time to balance. The ionic equation shows strong electrolytes as separate ions. The net equation eliminates “spectator ions”. 7 • Chemical Reactions & the Per. Table Arrhenius Acids and Bases (4 of 16) Acid: a substance that increases the [H+] in solution. Base: a substance that increases the [OH–] in solution. Diprotic acids have more than one removable H. (H 2 SO4 ) However, only the first H is ever easily dissociated. Oxides of nonmetals (SO2 ) are acid anhydrides. Oxides of metals (Na2 O)are basic anhydrides. Just add water... to get the acid or base. SO2 + H2 O → H2 SO3 Na 2 O + H2 O → 2 NaOH Acids and bases neutralize each other because the H+ and OH– ions form the very weak electrolyte... H2 O (and a salt). Acid salts are the partially neutralized polyprotic acids. NaH2 PO4 or Na2 HPO4 or NaHSO4 , etc... solid acids. 7 • Chemical Reactions & the Per. Table Brønsted-Lowry Acids and Bases (5 of 16) Acid: a proton (H+) donor. Base: a proton acceptor. This is a more general definition of acids and bases because it does not require the substance to be dissolved in water. Consider the following equations: The species that accepted the proton (base) can be considered a donor (conjugate acid) in the reverse reaction. HF + H 2 O ⇔ H3 O+ + F– (acid) (base) (acid) (base) NH3 + H2 O ⇔ NH4 + + OH– (base) (acid) (acid) (base) Strong base≈weak conjugate acid, (good acceptor≈lousy donor). Conjugates differ by only a H+ (e.g. HF and F – ) 7 • Chemical Reactions & the Per. Table Ions in Water Some Metal Ions Make Acidic Solutions (6 of 16) Since water molecules are polar, they surround ions in solution (called hydration). When we write Na+(aq) and Cl– (aq) we are implying this more complex situation. Some highly charged ions (Al3+ , Cr3+ , Fe3+ ) tend to tightly bind the water molecules. We can write them as complex ions: Al(H2 O) 6 3+ The electron clouds are drawn toward the central ion and away from the oxygen and therefore the O-H bond. This extra-polar O-H bond results in the H atom more readily joining with a passing water molecule... making the solution acidic. [Note: this is a conjugate acid-base situation.] Al(H 2 O) 6 3+ + H2 O ⇔ H3 O+ + Al(H2 O) 5 OH2+ 7 • Chemical Reactions & the Per. Table Trend in Strengths of Acids and Bases Three cases to explain (7 of 16) 7 • Chemical Reactions & the Per. Table Lewis Acids and Bases (8 of 16) Case 1: The more oxygens on an oxoacid, the stronger the acid. H2 SO4 > H2 SO3 .. HClO4 > HClO 3 > HClO 2 > HClO Why?: The electronegative oxygens draw electron density away from the central atom and therefore from the H-O bond... making it more polar. H leaves more easily. Case 2: The more electronegative the central atom, the stronger the oxoacid. H 3 PO4 < H2 SO4 < HClO 4 Why? Same as above... the more electronegative atom in the center makes the O-H bond more polar. Case 3: Binary acid strength depends on the SIZE of the atom... HF < HCl < HBr < HI... not the electronegativity. Why?: The greater distance means a weaker attraction. Consider the formation of a coordinate covalent bond: Acid: electron pair acceptor [Note: proton donor] Base: electron pair donor [Note: proton acceptor] This definition is more general than the other two because this covers cases that don’t even involve hydrogen (protons). Classic case: HN3 + BF3 → NH3 BF3 Other important cases: oxide of metal + oxide of nonmetal → salt (base anhydride + acid anhydride) CaO + SO2 → CaSO3 (s) and the oft-confusing reactivity of CO2 OH– + CO 2 → HCO3 – 7 • Chemical Reactions & the Per. Table Oxidation Numbers (9 of 16) 7 • Chemical Reactions & the Per. Table Balancing Redox Equations Oxidation Number Change Method Step-By-Step (10 of 16) 7 • Chemical Reactions & the Per. Table Balancing Redox Equations Ion-Electron Method (Half-reaction Method) Step-By-Step (11 of 16) Definition: Oxidation numbers are the apparent charges atoms have if shared e– ’s are assigned to the more electronegative atom. You can assign ox. #’s by studying a Lewis diagram or... The Rules: …ox. # of neutral atoms is 0 In compounds… …simple ions have ox. #’s equal to their charge. …F (-1), Family I (+1), II (+2), Al (+3) …O is usually (-2) except peroxide and OF 2 …H is usually (+1) except hydrides of Fam. I, II, Al …the sum of the ox. #’s of individual atoms equals the charge on the entire species. 1. Identify the ox. #’s of elements that change PER ATOM. One element will change UP (oxidation / lose e– ’s) one element changes DOWN (reduction / gain e– ’s). 2. Adjust coefficients and e – changes for situations where more than one atom MUST change: ex: 2HCl + K2 Cr 2 O7 → KCl + 2CrCl3 + Cl 2 + H2 O Cr: 2 x (3 e – per Cr) Cl: 2 x (1 e– per Cl) 3. Balance these changes in e– ’s (e– gained = e– lost). 4. Balance all elements except H and O. 5. Balance O’s (add H2 O’s as needed). 6. Balance H’s (add H+’s as needed). 7. If solution is basic, see card #12. 1. Identify the substances involved in the oxidation and reduction changes. Include substances so that all elements are represented (except H and O). For each half-reaction… 2. Balance all elements except H and O. 3. Balance O’s (add H2 O’s as needed). 4. Balance H’s (add H+’s as needed). 5. Balance charges (add e– ’s to the more positive side). 6. Balance e– ’s in the two half-reactions. 7. Combine the two half-reactions. Cancel substances that show up on both sides of the equation (e – ’s must cancel). 8. If solution is basic, see card #12. If the reaction occurs in a basic solution (usually stated clearly in the problem) then instead of H+’s and H2 O’s, you utilize OH– ’s and H2 O’s. 7 • Chemical Reactions & the Per. Table Balancing Redox Equations Reactions in Basic Solution (12 of 16) An easy method is: - balance as though the reaction were in an acid solution. - add OH– ’s to each side of the equation until all H+’s are turned into H2 O’s. - cancel H2 O’s that show on both sides of the equation. In each case, the metal changes to the + ion: M → M+ + e– Since the metal is oxidized, it is a reducing agent. Metals that most easily lose e– ’s (those with low ionization energy and low electronegativity) make the best reducers. 7 • Chemical Reactions & the Per. Table Metals as Reducing Agents (13 of 16) M E M O R I Z E T H I S: Some metals react with H2 O [to H 2 (g) and OH – ] Some react with non-oxidizing acids such as HCl, and cold, dilute H2 SO4 (the H+ is the reacting species) [to H 2 (g)] Some react only with oxidizing acids: • dilute HNO3 [to colorless NO(g) + H2 O], • conc. HNO 3 [to red-brown NO 2 (g) and H2 O], and • hot, conc. H2 SO4 [to SO2 (g) and H2 O] FOUR GROUPS OF METALS 7 • Chemical Reactions & the Per. Table The Activity Series of Metals (14 of 16) 1 - Most active - Families I and II - great reducing agents reduction half-reaction: 2H2 O + 2e– → H2 + 2OH– 2 - Most metals... Zn, Fe, Al, etc. reduction half-reaction: 2H+ + 2e– → H 2 3 - Ag, Cu, Hg ex. half-reaction: 2H+ + NO3 – + e – → NO2 + H2 O 4 - Noble Metals - Au, Pt, Ir only changed by “aqua regia” [HNO3 + HCl forms Cl 2 ] A more active metal can reduce or displace the ion of a less active metal. Ex. Zn° is more active than Cu°, so… Zn° + Cu2+ → Zn2+ + Cu° Cu° + Zn2+ → no reaction Elemental nonmetals (O, Cl, F, S, etc.) form negative ions by gaining electrons (reduction) and are oxidizing agents. 7 • Chemical Reactions & the Per. Table Nonmetals as Oxidizing Agents Oxygen as an Oxidizing Agent Combustion as a Redox Reaction (15 of 16) 7 • Chemical Reactions & the Per. Table Amphiprotic/Amphoteric & Leveling Effect (16 of 16) The strongest oxidizing agents are those to the right of each period (excluding noble gases) and those at the top of each family. So we can predict that F>Cl>Br>I and O > S. Example: Cl will displace Br – , but not F– Cl2 (g) + 2NaBr → Br2 (l) + 2NaCl • Cl2 + NaF → N.R. Oxygen, O2 , is a common and powerful oxidizing agent. Corrosion of metals, formation of oxides, and combustion are all examples of redox with O2 as the oxidizer. Anything with a lone pair can act as a proton acceptor (base) Anything with a H atom can act as a proton donor (acid). See water on card 5 act as either an acid or a base. See acetic acid on page 241 act as both acid and base. We say these are amphiprotic or amphoteric substances. Leveling effect... You cannot tell which strong acid is strongest in water, because donating a proton to water is not a good enough challenge. Strong acids completely dissociate in water because water is a pretty good acceptor of protons. We say that water has a leveling effect. Acetic acid is amphiprotic, but a poor proton acceptor and therefore is a great test for which strong acid is the strongest... 8 • Ionic Reactions in Solution Driving Forces for Metathesis Reactions (1 of 12) 8 • Ionic Reactions in Solution Precipitates as a Driving Force The Solubility Rules (2 of 12) 8 • Ionic Reactions in Solution Weak Electrolyte Formation as a Driving Force Weak Acids and other Weak Electrolytes Neutralization Reactions (3 of 12) 8 • Ionic Reactions in Solution Gas Formation as a Driving Force Gases That Commonly Form (4 of 12) During a double replacement or metathesis reaction, two new combinations of ions are produced. We identify four reasons why these NEW combinations are more stable than the original combos. • a precipitate forms memorize your solubility rules • a gas forms which leaves the system memorize the list of gases that form • a weak electrolyte forms memorize the strong acid list so you will recognize weak acids, also H2 O and NH4 OH • a complex ion forms learn the structure of complex ions and common ligands Always Always Soluble compounds with alkali metal ions (Li+, Na +, K+, Cs +, Rb +), NH4 +, NO3 – , C 2 H3 O2 – , ClO 3 – & ClO4 – Usually Soluble Cl– , Br– , I– [except “AP/H”... Ag+, Pb 2+ , Hg2 2+ ] SO4 2– [except “CBS”: Ca 2+ , Ba2+ , Sr2+ & “PBS”: Pb2+ ] Usually NOT Soluble O2–, OH– [except alkali and “CBS” Ca2+ , Ba2+ , Sr2+ ] Never Soluble CO3 2–, SO3 2–, S 2–, PO4 3– [except NH4 + & alkali] NOTE: some of these insoluble compounds WILL dissolve in acid solutions because of gas formation... useful idea! Weak electrolytes are substances that break up into ions only a LITTLE in solution... therefore, the two ions are MOSTLY in a combined state... not likely to re-form the reactants. H2 O, weak acids, NH4 OH Memorize the 8 strong acids so you can recognize a weak acid when you see one... HCl, HBr, HI, HNO3 , H2 SO4 , HClO3 , HClO4 , HIO4 Acids (forming H+ ions) and bases (forming OH – ions) combine to form a salt (an ionic compound) and H2 O... the very weak electrolyte. Neutralization of the acid and base occur because the H+ & OH– ions are “tied up” as H2 O. If you see the following substances formed during metathesis, realize that they will breakup into gases and leave the system (preventing re-formation of the reactants). Watch For… It Turns Into… H2 CO3 → CO2 (g) + H 2 O H2 SO3 → SO2 (g) + H 2 O H2 S → H 2 S(g) [rotten egg smell] NH4 OH → NH3 (g) + H 2 O 2 HNO2 → NO(g) + NO 2 (g) + H 2 O NOTE: these compounds are formed from acids with carbonates, sulfites, sulfides, nitrites, and bases with ammonium compounds. If you want to make the ionic compound, XY, you mix AY + XB to make XY + AB Either XY or AB need to drive the reaction (ppt, gas, etc.) 8 • Ionic Reactions in Solution Preparation of Salts (5 of 12) 8 • Ionic Reactions in Solution Comparing Driving Forces (6 of 12) 8 • Ionic Reactions in Solution More Concentration Units Weight Percent, ppm, and ppb (7 of 12) You may need to do this in two steps... make a carbonate (or sulfite, sulfide, hydroxide or oxide) of the cation (+ ion) you need and then react it with an acid that has the proper anion. The Practical Side: Keep in mind how you could recover the product you want... could you filter the product mixture? Do you want what is in the filter paper or what is in the filtrate? If you need the filtrate, you need to be careful not to have excess ions in it. Gas formation is a very strong driving force... even compounds that exist as insoluble solids will react (slowly) to form gases because gases leave the system and CANNOT re-form reactants. CaCO3 (s) + 2HCl → CO 2 (g) + H 2 O + Ca2+ + 2 Cl – The tendency to form H2 O is very strong. Insoluble oxides will react with acids. ZnO(s) + 2HCl → Zn2+ + 2Cl– + H2 O Sometimes, one insoluble solid can change into another even MORE insoluble solid... but you need more than the solubility rules to predict this (you need Ksp’s). AgCl(s) + Br– → AgBr(s) + Cl– Percent means “parts per 100” 96 96% means , that is, 96 out of every 100. 100 96g 96g Weight Percent (w/w) means … (w/v) means 100g 100mL ppm means “parts per million” 96 96 ppm means , 96 out of every 1 million. 1 000 000 ppb means “parts per billion” 96 96 ppb means , 96 out of every BILLION! 1 000 000 000 8 • Ionic Reactions in Solution Chemical Analyses Precipitations, Combustions, and Titrations (8 of 12) Real chemistry often deals with testing what is in a particular reaction mixture, environmental sample, etc. Stoichiometry is used to analyze the compositions. • You can precipitate an ion you are interested in, filter the precipitate and then determine from its mass the amount of compound in the original sample. • You can burn a sample and collect the combustion products (CO2 & H2 O) to determine the amount of C and H in the original sample. • You can carefully measure the volumes of solutions used during a titration. The endpoint must have some sort of indicator to allow you to recognize when the correct amounts of reactants have been added. 8 • Ionic Reactions in Solution Titration Terminology Acid-Base and Redox Titrations (9 of 12) A titration is a volumetric analysis because you carefully measure the volume of titrant, dispensing it from a buret. When you have added just enough titrant to completely react with the sample, you have reached the endpoint. This is usually apparent because of the color change of some indicator molecule (such as phenolphthalein). The endpoint can also be tracked because of changes in pH or changes in voltage due to the amount of some ion. Acid-Base & Redox titrations follow the formula: V·N=V·N where the N indicates [H+] or [OH– ] in acid-base titrations and [Oxidizer]· e– gained or [Reducer] · e– lost in redox titrations. 8 • Ionic Reactions in Solution Three Most Common Oxidizing Agents and what they turn into (10 of 12) purple permanganate ion acid solution MnO4 – + 8H+ 5e– → Mn2+ + 4H2 O Mn2+ ion is colorless neutral/basic MnO4 – + 2H2 O + 3e– → MnO2 (s) + 4OH – MnO2 (s) is a black solid yellow chromate / orange dichromate ion depends on [H+] → Cr2 O7 2– + H2 O 2CrO4 2– + 2H+ ← 2– acid solution Cr2 O7 + 14H+ + 6e– → 2Cr3+ + 7H2 O slightly basic CrO4 2– + 4 H2 O + 3e– → Cr(OH)3 + 5OH– Cr(OH)3 is a solid very basic CrO4 2– + 2H2 O + 3e– → CrO2 – + 4OH– 8 • Ionic Reactions in Solution Common Reducing Agents and what they turn into (11 of 12) Tin(II) (a gentle reducing agent) Sn 2+ → Sn4+ + 2 e – Sulfites and Bisulfites acidic solution: HSO3– + H2 O → SO4 2– + 3H+ + 2e– basic solution: SO3 2– + 2OH– → SO 4 2– + H2 O + 2e– Thiosulfate ion (also called “hypo” in photography) strong oxidizer: S2 O3 2– + 5H2 O → 2SO4 2– + 10H+ + 8e– half-reaction w/I2 : 2S 2 O3 2– → S4 O6 2–+ 2e – complete: I2 + 2S2 O3 2– → 2I– + S4 O6 2– – excess I : I2 + I– → I3 – → starch•I2 complex (blue-black) I2 + starch ← 8 • Ionic Reactions in Solution Equivalents, Equivalent Weights, and Normality (an old-fashioned, but useful idea) (12 of 12) equivalents = H+, OH– , or electrons gained or lost n = # of equivalents in the balanced chemical equation example: I2 + 2S2 O3 2– → 2I– + S4 O6 2– n = 2 equivalent weight = molar mass ÷ n …mass of a chemical that provides 1 mole of equivalents. moles equivalents Normality, N, = n · M, = Liter solution This idea is useful in acid-base and redox titrations because it takes into account the differences of acids and bases or oxidizers and reducers. This concept allows the use of the simple formula: V·N = V·N 9 • Properties of Gases Boyle’s Law (P and V) (1 of 12) General: Formula: When P↑, V↓ (inversely proportional) P·V = constant or P1 V1 = P2 V2 Restrictions: P 1 and P2 must be in the same units V1 and V 2 must be in the same units Convert pressures using conversion factors using the fact that 1 atm = 760 mmHg = 760 torr = 101.3 kPa = 14.7 psi lb psi = 2 in 101.3 kPa Example: 730 mmHg x = 97.3 kPa 760 mmHg Graphically: V 1/V 9 • Properties of Gases Boyle’s Law Lab (2 of 12) P P In our lab, we had to add the atmospheric pressure to our measurements because tire gauges only measure the pressure ABOVE atmospheric pressure. 1 Consistent (“ good”) data form a straight line (P vs. ). V K = °C + 273 °C = K – 273 Examples: 0 °C + 273 = 237 K 25 °C + 273 = 298 K 100 °C + 273 = 373 K 9 • Properties of Gases Kelvin Temperature Scale (3 of 12) 300 K – 273 = –27 °C The Kelvin scale is used in gas law problems because the pressure and volume of a gas depend on the kinetic energy or motion of the particles. The Kelvin scale is proportional to the KE of the particles… that is, 0 K (absolute zero) means 0 kinetic energy. 0 °C is simply the freezing point of water. Charles’ Law General: 9 • Properties of Gases Charles’ Law (V and T) Gay-Lussac’s Law (P and T) (4 of 12) When T↑, V↑ (directly proportional) V V V Formula: = constant or 1 = 2 T T1 T2 Restrictions: T must be in Kelvins V1 and V 2 must be in the same units Gay-Lussac’s Law General: When T↑, P↑ (directly proportional) P P P Formula: = constant or 1 = 2 T T1 T2 Restrictions: T must be in Kelvins P 1 and P2 must be in the same units Formula: Restrictions: 9 • Properties of Gases The Combined Gas Law (5 of 12) P ·V P ·V P·V = constant or 1 1 = 2 2 T T1 T2 T must be in Kelvins V1 and V 2 must be in the same units P 1 and P2 must be in the same units STP (“standard temperature and pressure”) is often used as one of the two conditions T = 0 °C = 273 K P = 1 atm = 760 mmHg = 101.3 kPa Each of the three gas laws is really a special case of this law. Example: If T1 = T2 , the law becomes P1 V1 = P2 V2 Formula: where 9 • Properties of Gases The Ideal Gas Law (6 of 12) 9 • Properties of Gases Dalton’s Law of Partial Pressure (7 of 12) P·V = n·R·T or PV = nRT P = pressure V = volume n = number of moles R = the ideal gas constant T = temperature (in Kelvins) The value of R depends on the P and V units used. PV R= so you can use the molar volume info to calculate R nT (101.3 kPa)(22.4 L) L·kPa R= = 8.31 (1 mole)(273 K) mol·K L·mmHg L·atm R = 62.4 = 0.0821 mol·K mol·K When you have a mixture of gases, you can determine the pressure exerted by each gas separately. This is called the partial pressure of each gas. Since each gas has the same power to cause pressure (see card #8) the partial pressure of a gas depends on how much of the mixture is composed of each gas (in moles) Example: Consider air, a mixture of mostly O2 and N 2 P O2 P N2 moles O 2 moles N 2 = = moles total P total moles total P total Also: P total = P O2 + P N2 This idea is used when a gas is collected over water P atm = P gas + P H2 O P H2 O is found on a chart The gas laws work (to 3 significant digits) for all gases… that is, all gases have the same power to cause pressure. 9 • Properties of Gases Why Do All Gases Cause the Same Pressure? (8 of 12) At the same temperature, the KE of each gas is the same. KE = 1 /2 mass·velocity2 … if two particles have different masses, their velocities are also different. So… SMALL particles move FAST LARGE particles move SLOWLY v2 m m v2 We can use this idea with numbers as well: (Graham’s Law) KEA = KEB mAvA2 = mBvB2 [another version of this formula is on the next card] mAvA2 = mBvB2 can also be used as the equation… rate of effusion of A rate of effusion of B = 9 • Properties of Gases Graham’s Law of Effusion (9 of 12) MB MA Notice that the A is in the numerator in the ratio of the rates and in the denominator in the radical. “Effusion” is similar to diffusion. It means to escape through a small opening. The ratio of the rates (or velocities) of CH4 (mass=16 u) to SO2 (mass=64 u) is 64 = 16 4 = 2 Ideal gases have no volume & no attractions for each other. Luckily, real gases act pretty much like ideal gases at room temperature and pressure. The most ideal of real gases is He. 9 • Properties of Gases The Real Gas Law (10 of 12) 9 • Properties of Gases Kinetic Molecular Theory (11 of 12) The REAL GAS Law is: n2a (P + 2 )(V - nb) = nRT V where: a corresponds to the attractions between real gas particles b corresponds to the size of the real gas particle Explaining the behavior of gases involves the kinetic molecular theory. Here are the main ideas: • all particles are in constant, random motion • temperature is a measure of the average kinetic energy • pressure is due to collisions of gas particles with the walls of the container • increased temperature causes more collisons as well as harder collisions • some particles are moving fast, some are moving slowly F P=A Pressure is proportional to the force pushing and inversely proportional to the area over which that force pushes. 9 • Properties of Gases Pressure = Force ÷ Area (12 of 12) P=F A P=F A Properties of matter depend on the model that gas particles are spread out; liquids are close together, but random; and solids are close together and arranged in a crystal lattice. 10 • States of Matter & IMFs Comparing Gases, Liquids, and Solids (1 of 16) gas liquid solid Volume and shape, compressibility, and the ability of substances to diffuse depend on these models. Gases have no set shape or volume. Liquids have a constant volume, but no set shape. Solids have constant volume and shape. Surface tension- a measure of the amount of energy needed to expand the surface area of a liquid. 10 • States of Matter & IMFs Surface Tension (2 of 16) An interior molecule is surrounded by molecules to which it is attracted… no net attraction. A surface molecule feels a net attraction toward the interior. To move a molecule to the surface (i.e. increase the surface area), energy must be used, work must be done. The potential energy of the liquid is increased. Substances tend toward the lowest potential energy so liquids tend toward the minimum surface area. A sphere is the smallest surface area for given volume. 10 • States of Matter & IMFs KE Distributions and Evaporation (3 of 16) In any sample of liquid, the distribution of KE varies. Particles to the right of the line (the “threshold energy” have enough KE to escape the IMF’s holding them in the liquid. Increasing the temperature (average KE) of the liquid moves the curve to the right. The line depends on the IMF of the liquid. Only particles at the surface of the liquid may escape (evaporate.) 10 • States of Matter & IMFs Molecular Crystals and IMFs (4 of 16) Substances that exist as molecules (as opposed to ionic, metallic, or covalent network crystals) are in three groups: nonpolar polar molecules polar with H-O-, molecules Dipole-dipole H-N-, or H-F London Forces attractions H-bonding Weak IMF. Due + end of one Strong dipole to polarizable emolecule because of high clouds & temp. attracting - end electroneg. /small attraction of other molecule size of O, N, & F In London Forces--larger atoms and larger molecules have stronger London forces due to more sites or more polarizable electron clouds . Heat of Vaporization, ∆ Hvap, can be thought of as the energy needed to vaporize a mole of a liquid. It can be used as a conversion factor in a calculation of heat during a phase change. [Calorimetry is used for temperature changes between the phase changes.] 10 • States of Matter & IMFs Hvap and IMFs (5 of 16) Ex: 111g H 2 O × 1 mol H 2 O 40.6 kJ × = 250. kJ 18.0 g H 2 O 1 mol H 2 O It can also serve as a indicator of the strength of the IMF (intermolecular forces of attraction) in the liquid. Ex: CH4 (9.20 kJ/mol) vs. C3H8 (18.1 kJ/mol) Larger molecule… greater IMF… greater Hvap In a closed container, the number of particles changing from liquid → vapor will eventually equal the number of particles changing from vapor → liquid. 10 • States of Matter & IMFs Vapor Pressures of Liquids (6 of 16) The amount of vapor when this balance is reached depends on the IMF and the KE of the liquid (& not on the volume of the container). The pressure exerted by this vapor is called the equilibrium vapor pressure (VP) of the liquid. As temperature increases, the VP increases. (This is important for why/when a liquid boils.) VP is another indicator of the strength of the IMF. The stronger the IMF, the smaller the VP. Boiling occurs when the vapor pressure (VP) of a substance = the air pressure above the liquid. You can boil a liquid by increasing the VP of the liquid (heating) or by lowering the pressure above the liquid. 10 • States of Matter & IMFs Boiling Point and IMFs (7 of 16) The temperature at which a liquid reaches 760 mmHg is called the “normal boiling point” of the liquid. ---Again, BP is an indicator of IMF.--↑ boiling points (BP) ~ ↑ IMF’s ~ ↓ vapor pressures. Altitude (low air pressure) lowers the boiling temperature of water in an open container (increases cooking time). Pressure cookers ↑ BP by ↑ the pressure above the liquid. Freezing Point and Melting Point are the same temperature, just opposite directions. That is, a substance will freeze and melt at the same temperature. 10 • States of Matter & IMFs Freezing Point, Melting Point, ∆ Hfusion (8 of 16) When you fuse two metals, you MELT them… thus the term fusion means melting. ∆ Hfusion is the energy needed to melt a mole of solid into a mole of liquid. As with ∆Hvap, ∆Hfus can be used as a conversion factor as well as an indicator of IMF strength. Note: Freezing is more complicated than vaporization because the process of forming a crystal causes some subtle considerations which we will not deal with in this course. 10 • States of Matter & IMFs Common Crystal Structures and Unit Cells (9 of 16) Three common crystal structures are Simple Cubic; BodyCentered Cubic (which means there is an atom, “a body,” in the center of the cubic structure; and Face-Centered Cubic (with an atom in the center of each side or “face”). A unit cell is the theoretical arrangement of atoms that, if repeated, will recreate the crystal. This topic, although interesting, is no longer on the AP curriculum and will not be dealt with here. One resource is a page by OSU From Dr. John Gelder’s Solid State Chemistry Page chemist, Dr. John Gelder. (See Card 16) Crystal lattice points: IMF’s 10 • States of Matter & IMFs Four Types of Crystals -- Summary (10 of 16) Props. Ex Molecular molecules or atoms London, dipole, H-bonding soft, low MP, nonconduct Ionic + and - ions Covalent atoms Metallic positive ions attraction between + & - ions hard, brittle, high MP, (l) (aq) conduct covalent bonds I2 H 2O HI NaBr attr. between + ions & “sea of e-‘s” high luster, conductor, variable MP, soft/hard Na° Fe° Cu° v. hard, high MP, nonconduct (graphite) C SiC WC Students are often confused between molecular crystals in which covalent bonds hold the molecules together (but the IMF = London forces, dipole-dipole attractions or hydrogen bonding) and covalent crystals in which covalent bonds hold the crystal together (the IMF = covalent bonds). 10 • States of Matter & IMFs Crystal Types -- Further Notes (11 of 16) Substances can conduct electricity for two reasons: freely moving ions or delocalized electrons. Ionic compounds have freely ions in the liquid state and when dissolved in water. Metals have delocalized electrons -- the “sea of electrons.” Graphite has a chicken-wire shaped π-bond above & below each sheet of sp 2-hybridized C atoms, allowing it to conduct. (g) (l) 10 • States of Matter & IMFs Heating and Cooling Curves (12 of 16) (s) & (l) (s) PE KE (l) & (g) PE KE KE KE = kinetic energy changes which are times when the heat energy speeds up the molecules. Time (min) PE = potential energy changes which are times when the heat energy separates the molecules from solid to liquid or liquid to gas. The phase diagram shows the phases of a substance at all temperatures and pressures. 10 • States of Matter & IMFs Phase Diagrams -I (13 of 16) Moving across the diagram gives you the MP and then BP of a substance. There is a point above which it is no longer possible to liquefy a substance (the critical point, C). Moving vertically you can see the effect of pressure on the phase of the substance. This is a diagram for a substance like CO2, in which the liquid can be compressed into the solid. (Unlike H2O.) Water’s phase diagram is unique because the liquid phase is less dense than the solid phase. To maximize hydrogen bonding, the solid must expand. 10 • States of Matter & IMFs Phase Diagrams -II (14 of 16) 10 • States of Matter & IMFs Name of the Phase Changes (15 of 16) The B-D boundary of the phase diagram has a negative slope. The “triple point” is the temperature and pressure in which the solid, liquid, and vapor phases of a substance can coexist. I visualize this as a boiling glass of ice water. By increasing the pressure, dry ice can melt. By decreasing the pressure, solid water can sublime. NRG is REQUIRED solid → liquid melting or fusion NRG is RELEASED liquid → solid freezing liquid → gas vaporization evaporation or boiling gas → liquid condensation solid → gas sublimation gas → solid solidification The energy involved in the phase change is calculated using heat of fusion (solid → liquid or liquid → solid) heat of vaporization (liquid → gas or gas → liquid) Searching the Internet, I found an interesting set of topic reviews. These are from Purdue University (Indiana) I was looking at the topic, LIQUIDS, but there are many topics to choose from. 10 • States of Matter & IMFs More Internet Resources (16 of 16) chemed.chem.purdue.edu/genchem/topicreview/ The unit cell is a frame from an online movie file on Dr. John Gelder’s Solid State Chemistry page. (OK State Univ.) www.okstate.edu/jgelder/solstate.html thermodynamics system 12 • Chemical Thermodynamics Commonly Used Terms (1 of 12) surroundings adiabatic isothermal state functions Study of NRG changes & flow of NRG. Tells whether a reaction is possible. That portion of the universe on which we focus. Everything outside the system. A change without heat transfer between the system and its surroundings. A change that occurs at constant temperature. Properties that depend on the initial and final state, not on how the change was made. ex: ∆H, temp, but not work Energy changes can be measured using calorimetry. Often this involves heating water under controlled conditions (a bomb calorimeter). 12 • Chemical Thermodynamics Heat Capacity and Specific Heat (2 of 12) Three closely related terms are: heat capacity is the amount of heat needed to change a system by 1°C. molar heat capacity is the amount of heat needed to change a mole of a substance by 1°C. specific heat is the amount of heat needed to change 1 gram of a substance by 1°C. (Water: 1 cal/g°C = 4.184 J/g°C) Note that heat capacity is an extensive property whereas the other two are intensive. The first law: “if a system undergoes some series of changes that ultimately brings it back to its original state, the net energy change is zero.” ∆E = Efinal - Einitial 12 • Chemical Thermodynamics First Law of Thermodynamics (3 of 12) ∆E = 0 when Efinal = Einitial The usefulness of this idea is that the internal energy, ∆E, depends only on the initial and final state, not on how you get there. Every path you take from Einitial to Efinal takes the same amount of energy. (no perpetual motion) There are two ways for a system to exchange energy with the surroundings, heat and work. ∆E = q - w [q = heat absorbed by system, w = work done by system] From physics, work = force × distance = F × d. In chemistry there is electrical work (e-‘s through a wire) and P∆ ∆ V work as a gas expands. 12 • Chemical Thermodynamics Work and PV Work (4 of 12) Use pressure = force/area and area × distance = volume to derive work = P∆ ∆ V from work = F × d. Note that units of work are units of energy. Energy and work are two forms of energy. Doing work on a system increases the potential energy of the system. P∆V work is in L⋅⋅ atm 1 L⋅atm = 24.2 cal = 101.3 J PV = nRT… work can be calculated as work = ∆ nRT for chemical reactions where the # of moles of gas change, ∆n. 12 • Chemical Thermodynamics Work is NOT a State Function (It depends on HOW you do the work.) (5 of 12) Q: Gas in a piston at a pressure of 10 atm is allowed to expand from 1 L to 10 L. How much work done? A: It depends on the resisting force. If it expands against 0 atm pressure, P∆V = 0 atm × 9 L = 0 L⋅⋅ atm Expanding against 1 atm pressure, Vfinal is 10 L P∆V = 1 atm × 9 L = 9 L⋅⋅ atm Expanding in two steps, first, against 2 atm, Vfinal = 5 L P∆V = 2 × 4 L = 8 L⋅atm then in a second step against 1 atm, Vfinal = 10 L P∆V = 1 × 5 L = 5 L⋅atm… Total = 13 L⋅⋅ atm From the P∆V work example, you can see that the more steps (and smaller increments) you use to get the work from the system, the more work you can get. 12 • Chemical Thermodynamics Reversible Processes (6 of 12) There is an upper limit to how much work can be derived from any system. That maximum work is called the Gibb’s Free Energy, ∆G. The theoretical maximum work can be achieved if the steps are so small that they can go either way, if they are reversible. If a chemical reaction occurs in a closed container ∆ V = 0 and so P∆ ∆ V work = 0. ∆ E = q - w becomes ∆ E = q. 12 • Chemical Thermodynamics ∆ H = ∆ E + P∆ ∆V (7 of 12) In a system at constant pressure (a common situation) the energy change involves both heat (q) & work (P∆V). ∆ E = q - P∆ ∆ V [P∆V = work done BY the system] or q = ∆ E + P∆ ∆ V q at constant pressure is called ∆ H. ∆ H = ∆ E + P∆ ∆ V or ∆ E = ∆ H - P∆ ∆V The work (P∆V) is generally insignificant unless the # of moles of gas (∆n = n final - n initial) is changing. P∆ ∆ V = ∆ nRT Recall that if ∆n is negative (# moles decreasing) work is negative (work is being done ON the system). If several reactions add up to give an overall reaction, the ∆ H’s of the reactions will add up to the overall ∆ H. 12 • Chemical Thermodynamics Hess’s Law of Heat Summation (8 of 12) Standard Heats of Formation, ∆ Hf° , are useful for this purpose. This is the energy involved in making a mole of a substance from its elements at 25°C and 1 atm pressure. This law is often written as: ∆ H°° reaction = Σ ∆ Hf° products - Σ ∆ Hf° reactants Note: ∆Hf° for elements is 0. If you are NOT using heats of formation, you need to write out the equations to see how they combine. Bond energy is the amount of energy needed to BREAK a certain bond. 12 • Chemical Thermodynamics Bond Energies (9 of 12) You can determine the approximate energy change in a chemical reaction by summing the bond energies of the reactants and subtracting the bond energies of the products. Note: this is opposite to the Hess’s Law (products reactants) because bond energy involves breaking bonds whereas ∆H°f involve forming bonds. Things in the world tend toward lowest energy (-∆H) and also tend toward greatest disorder (+∆S). 12 • Chemical Thermodynamics Two Big Driving Forces of the Universe Enthalpy (∆ ∆ H) and Entropy (∆ ∆ S) (10 of 12) More disorder can be recognized as: • greater # of moles of gas formed • gas > liquid > solid and (aq) > (s) • greater volume formed • mixed molecules (HI) formed from diatomic molecules example: H2 (g) + I2 (g) → 2HI(g) +∆S Note: ∆S° can be calculated using Hess’s Law, but S° of elements is NOT 0. Also, S° is often reported in J/mol⋅K and cal/mol⋅K, not kJ and kcal… watch your units! The second law defines a value called Gibb’s Free Energy symbolized as ∆G. This represents the theoretical maximum work that can be done by a system. 12 • Chemical Thermodynamics The Second Law of Thermodynamics ∆ G = ∆ H - T∆ ∆S (11 of 12) A reaction is spontaneous when ∆G is negative (∆G<0). The reverse reaction is spontaneous when ∆G is positive. When ∆G = 0, the reaction is at equilibrium. ∆H + + ∆S + + Spontaneous… … at all temperatures … at LOW temperatures … at HIGH temperatures the reverse reaction is spontaneous We have been calculating ∆G° values, that is, ∆G under standard conditions. Once a reaction begins, the concentrations change and the value of ∆G is different. 12 • Chemical Thermodynamics ∆ G (thermodynamics) and Keq (equilibrium) (12 of 12) If ∆ G°° < 0, then the reaction will proceed in a forward direction. As it does, however, the value of ∆G will increase until the forward and reverse reactions are balanced and ∆ G = 0. This is called equilibrium. If ∆ G°° > 0, then the reverse reaction will proceed and the value of ∆G will decrease until equilibrium is reached. ∆G° gives information about where we are in relation to equilibrium. There is a formula that combines ∆ G and Keq. 22 • Nuclear Chemistry The People (1 of 16) 22 • Nuclear Chemistry Terms I-- Radioactivity (2 of 16) 22 • Nuclear Chemistry Terms II--Radioactivity (3 of 16) • Wilhelm Roentgen (1845-1923) discovered X-rays, a high energy form of light. (1895) • Henri Becquerel (1852-1909) found that uranium ores emit radiation that can pass through objects (like x-rays) and affect photographic plates. (1896) • Marie Sklodowska Curie (1867-1934) Marie and Pierre worked with Becquerel to understand radioactivity. The three shared a Nobel Prize in Physics in 1903. Marie won a second Nobel Prize in Chemistry in 1911 for her work with radium and its properties. • E. O. Lawrence invented the cyclotron which was used at UC Berkeley to make many of the transuranium elements. radioactivity the spontaneous breakdown of atomic nuclei, accompanied by the release of some form of radiation (also called radioactive decay) half-life time required for half of a radioactive sample to decay transmutation one element being converted into another by a nuclear change nuclides isotopes of elements that are identified by the number of their protons and neutrons decay series the sequence of nuclides that an element changes into until it forms a stable nucleus radioactive dating using half-life information to determine the age of objects. C-14/C-12 is common for organic artifacts. Uranium is common for rocks. nuclear fission large nucleus breaking down into pieces of about the same mass nuclear fusion two or more light nuclei blend to form one or more larger nuclei Alpha particles are the same as a helium nucleus, 4 2 He, with a mass of 4 amu. It travels about 1/10th the speed of light and is the most easily stopped of the three particles (a sheet of paper will stop them). It is the least dangerous. 22 • Nuclear Chemistry Types of Radiation (4 of 16) 0 Beta particles are high speed electrons, −1 e, with a mass of 0.00055 amu and travel at nearly the speed of light. They can be stopped by a sheet of aluminum. It is more penetrating and therefore more dangerous than alpha. Gamma rays are extremely high energy light, γ, with no mass, and are the most penetrating (several cm’s of lead are needed to stop them). They can cause severe damage. In each half-life problem there are basically four variables: • total time • half-life • starting amount • ending amount 64g 22 • Nuclear Chemistry Half-Life Problems (5 of 16) 32g 16g 8g 4g 2g 1g 0.5g 0.25g Question: If you have 0.25 g of a radioactive substance with a half life of 3 days, how long ago did you have 64 grams? Answer: Draw the chart to determine the number of halflives to get from the ending amount to the starting amount… each half-life is worth 3 days…24 days. Half-Life The time it takes for half of a radioactive substance to decay. The decay graph has a characteristic shape: # 22 • Nuclear Chemistry Half-Life (6 of 16) time The time it takes for the amount of substance or the activity of the substance to drop to half is the same WHEREVER you start on the graph. This is a first-order reaction. Half-lives can range from microseconds to thousands of years and is characteristic of each substance. Memorize the symbols for the important particles alpha beta positron neutron 4 2 22 • Nuclear Chemistry Nuclear Equations (7 of 16) He 0 −1 e 0 +1 1 0n e Decay means the particle is on the right side of the equation: example: alpha decay of U-238 238 4 92 U → 2 He + 234 90 Th The 234 and 90 are calculated… the Th is found on the periodic table (find the element with atomic # = 90). Several neutrons can be shown together and written as… 1 3 3( 0 n) and would be counted as 0 n in the equation. 22 • Nuclear Che mistry How Each Type of Decay Can Stabilize an Unstable Nucleus (8 of 16) Certain values of p +’s and n°’s in the nucleus are stable. A nucleus can be unstable (radioactive) for 3 reasons: • the nucleus has too many protons compared to neutrons solution: positron decay (change a proton into a neutron and a positive electron… …a positron) • the nucleus has too many neutrons compared to protons solution: beta decay (change a neutron into a proton and a negative beta particle) • the nucleus is too big (too many protons and neutrons) solution: alpha decay (lose 2 p + and 2 n°) 22 • Nuclear Chemistry Uses of Radioactivity (9 of 16) Radioactive Dating: In every living thing there is a constant ratio of normal C-12 and radioactive C-14. You can calculate the time needed to change from what is expected to what is actually found. Radioisotopes: Many substances can be radioactive and then followed as they move through the body. Fission Reactors: Current nuclear reactors use fission reactions to produce heat which is used to turn water into steam and drive turbine engines that produce electricity. The Sun and Stars are powered by nuclear fusion… this is related to the fact that the most abundant element in the universe is hydrogen… followed by helium. U-235 is “fissionable” which means it can be split when bombarded by neutrons. 235 1 92 U + 0 n 22 • Nuclear Chemistry Fission and Fusion Reactions (10 of 16) → 141 56 Ba 92 36 + 1 Kr + 3 0 n + energy The fact that each splitting nucleus can emit neutrons that can split other nuclei is the basis for the “chain reaction.” “Breeder reactors” use different isotopes. Fusion in the Sun involves several steps that can be 1 4 0 summed up as: 4( 1 H) → 2 He + 2 1 e + energy Thermonuclear devices use isotopes of hydrogen (deuterium and tritium): 2 1 3 H+ 1 H→ 4 2 1 He + 0 n + energy Einstein’s famous equation, E = mc2 , is the basis for explaining where the energy associated with nuclear changes comes from. 22 • Nuclear Chemistry Energy–Mass Conversions (11 of 16) When a nuclear change occurs, the mass of the products is slightly less than the mass of the reactants. This loss in mass is called the mass defect. E = the energy m = the mass defect c = the speed of light, 3.00 x 108 m/s 1 kg of mass converted into energy would be equivalent to burning 3 billion kg of coal! During beta decay, 1 neutron changes into 1 proton + 1 negative beta particle (The atomic # increases by one due to the new proton. The mass # is unchanged… a neutron is gone. To maintain electrical neutrality, a negative beta particle is also formed.) 22 • Nuclear Chemistry What Happens During Beta and Positron Decay (12 of 16) Example: 235 0 92 U → −1 e+ 235 93 Np During positron decay, 1 proton changes into 1 neutron + 1 positron particle (The atomic # decreases by one due to the loss of a proton. Since it changed into a neutron, the mass # is unchanged.) Example: 235 0 92 U → +1 e + 235 91 Pa When a problem involves whole numbers of half-lives, divide by 2 to determine the amounts involved. For other situations, the following equations are useful: ln 22 • Nuclear Chemistry Calculating Half-Lives (13 of 16) [A] 0 = kt and the special case for half-life, t ½, where by [A] t definition, [A]t = ½[A]0 ln 2 = 0.693= kt ½ [A] is the concentration (or activity) of the radioactive substance, t = time, k = the rate constant (the same that is in Rate Laws). Note: if you know the half-life, you can calculate the rate constant and vice-versa. Once a nucleus decays, the daughter isotope is often unstable as well. Many decays may occur before a stable nucleus is formed. 22 • Nuclear Chemistry Radioactive Decay Series (14 of 16) A classic example is U-238 that decays through 14 steps into stable Pb-206. Each step has a characteristic decay particle and half-life. This characteristic decay series is the method used to verify the identity of newly formed atoms. The fact that daughter products can be even more radioactive than the parent isotope adds to the problem of nuclear waste and its storage/disposal. An useful characteristic of decay particles are that they ionize the air they pass through by striking atoms and knocking off electrons. 22 • Nuclear Chemistry Geiger-Muller Tubes, Smoke Detectors, and Brushes for Cleaning Negatives (15 of 16) Geiger counters use this idea. As radioactive particles pass through a chamber with two electrodes, ionized particles migrate to the + and - electrodes and complete the circuit. Smoke Detectors use a tiny piece of radioactive Am to keep a circuit flowing due to ionized particles. Smoke particles attract ionized particles, break the circuit, & set off the alarm. Brushes are kept ionized by tiny bits of radioactive material to more easily attract tiny bits of dust. Uranium, Z=92, is the largest naturally-occurring element. Larger atoms were manufactured. Elements 93 and 94 were formed in atomic bomb tests and identified by Seaborg. Glenn Seaborg and Al Ghiorso at UC Berkeley were able to use E. O. Lawrence’s cyclotron to make larger atoms (95-103). 22 • Nuclear Chemistry Extending the Periodic Table (16 of 16) Some of these new elements have uses in the medical field as well as helping to further the understanding of the nucleus. For many of the larger elements, however, only a few atoms or even one atom formed. They were identified by their characteristic decay series. As of July 2000, 118 is the largest element. Chemicals from living things were thought to contain a “vital force” that could not be duplicated in the lab. This changed with Friedrich Wöhler who mixed cyanic acid (HCNO) with ammonium hydroxide making ammonium cyanate (NH4CNO). 23 • Organic Chemistry Historical Ideas (1 of 16) O C NH2 H2 N urea He usually allowed the salt solution to evaporate overnight, but tried heating it to hurry the process. The result was a crystal that he recognized as urea (H2NCONH2). The modern view of organic chemistry is the chemistry of carbon compounds. C is the key element. It can form four bonds and that are very strong bonds due to its small size. The alkanes (paraffins) follow the formula: CnH2n+2: These molecules contain ONLY single bonds. They are said to be “saturated” with hydrogens. 23 • Organic Chemistry Alkane Series -- Saturated Hydrocarbons (2 of 16) Memorize these prefixes also used with alkenes & alkynes. CH4 methane C6H14 hexane C2H6 ethane C7H16 heptane C3H8 propane C8H18 octane C4H10 butane C9H20 nonane C5H12 pentane C10H22 decane Given a formula, you can tell that it contains only single bonds because it fits the alkane formula. As the molecules increase in size, they tend to be liquids and 23 • Organic Chemistry Structural Formulas Can Be Misleading (3 of 16) CH4, can be drawn using a structural formula. This can be misleading. The molecule is not flat with bond angles of 90°. You must be aware of the 3-D structure and the 109.5 ° bond angles. Cl H C Cl H H C Cl Cl H H H C H H For example, there is only one isomer of dichloromethane, but you can draw it at least two ways. Building models of the molecules is an important part of strengthening this skill. H Alkenes contain 1 double bond. The formula is CnH2n. They are said to be “unsaturated” (like unsaturated fats). The double bond can be broken and more hydrogens added. 23 • Organic Chemistry Alkenes and cis- /trans- Isomerism (4 of 16) H C C H ethene Since double bonds cannot easily rotate (due to the pi bonding) cis- and trans- isomers can be formed. Example: 1,2-dichloroethene can be built two ways. Cl Cl C H Cl C H C H cis -1,2-dichloroethene (a polar molecule) H C Cl trans-1,2-dichloroethene (a nonpolar molecule) H 23 • Organic Chemistry Alkynes, Alkadienes, and Cyclic Hydrocarbons (5 of 16) H C C H Alkynes contain 1 triple bond (unsaturated). Formula: CnH2n-2. ethyne (acetylene) The triple bond is linear, so no cis/trans isomerism occurs. Alkadienes are molecules with two double bonds. They have the same formula as the alkynes, CnH2n-2. H Example: C4H6 is named 1,3-butadiene C CH2 because the double bonds start on H2C C carbons #1 and #3. H H2 C H2C Cyclic compounds contain rings having the same formula as the alkenes, CnH2n. Example: cyclopropane, C3H6. CH2 The basic idea is to name the molecule after the longest continuous chain of carbon atoms. Side groups are listed with #’s to indicate the C atom to which they are attached. 23 • Organic Chemistry Naming Organic Compounds (Organic Nomenclature Using IUPAC Rules) (6 of 16) 23 • Organic Chemistry Common Errors in Drawing/Naming Structures 1-methylsomething & 2-ethylsomething (7 of 16) Side Groups : -CH3 methyl -Cl chloro -C2H5 ethyl di- = 2 groups tri- = 3 groups 2,2,3-tribromobutane (not 2,3,3-) Note that we # the carbons from whichever end results in the smallest numbers. While drawing the isomers of pentane, C5H12, students draw this structure, naming it 2-ethylpropane. (a chain of 3 C’s with an ethyl group) The longest chain is four C’s, and should be named 2-methylbutane. tetra- = 4 groups H H Br Br H C C C H H Br H H H H C C C H CH2 H H C H H CH3 A similar error is to draw and name “1-methylsomething”. H H H C C H H O 5 bonds to C H2C H2 C H CH2 H3 C C 23 • Organic Chemistry Optical Isomers Chiral Compounds (8 of 16) -Br bromo -I iodo -C3H7 propyl, etc. CH3 Two more tips … double check that each C has four and only four bonds. Also, remember that N and O atoms have lone pairs of e-‘s although they are seldom drawn. (Impt. for steric #!) CH3 Some molecules have the ability to rotate polarized light. These molecules can be recognized by a C atom (the chiral carbon) bonded to four different groups . 3-methylhexane This carbon is bonded to H, methyl, ethyl, & propyl groups. You can build two versions of this molecule that are “nonsuperimposable mirror images of each other.” One will rotate light clockwise, one counterclockwise. In biology, these are called dextro- and levo- (D and L) forms. ethene is also called ethylene propene is also called propylene H3C 23 • Organic Chemistry Common Names You Should Know About (9 of 16) CH3 CH3 H3 C 2-methylbutane is also called isopentane. “Iso-“ means the same… the same two methyl groups come branch from C #2. 2-methylpentane is isohexane, etc. H2 C HC H3 C C CH3 CH3 2,2-dimethylpropane is called neopentane. These common names show up occasionally in names… such as in isopropyl alcohol. H C H C HC HC 23 • Organic Chemistry Aromatic Compounds Benzene and its Derivatives (10 of 16) C H CH HC CH HC Benzene, C6H6, is unique. It can be drawn as shown, but the actual structure involves a circular pi bond (sp 2 orbitals & delocalized e-‘s). CH C H CH two resonance structures Benzene is also shown with a circle as the pi bond. 1 The carbon #’s can be used for substituted 6 2 benzene. Example: dichlorobenzene 1,2- is known as the ortho- position 5 3 1,3- is known as the meta- position 4 1,4- is known as the para- position Paradichlorobenzene: the main ingredient in some moth balls. General formula: R-O-H [R ≈ Rest of molecule] C atoms are classified as primary (1), C secondary (2), or tertiary (3) by the number of C atoms it is bonded to. 1 C C 3 1C C A primary alcohol has the -OH group 2 bonded to a primary carbon, etc. This is not a base because the -OH is covalent, not ionic. Naming: group + “alcohol” (e.g. ethyl alcohol or ethanol) Ethers General formula: R-O-R’ [R’ can = R, but not H] Naming: two groups + “ether” H2 H2 C C diethyl ether was the 1st effective H3 C O CH3 surgical and dental anesthetic. Alcohols 1 23 • Organic Chemistry Functional Groups I Alcohols and Ethers (11 of 16) Aldehydes Ketones O O General formula: C C R 23 • Organic Chemistry Functional Groups II Aldehydes and Ketones (12 of 16) Naming: H R R' names end in “al” names end in or “aldehyde” “one” methanaldehyde propanone (formaldehyde) (acetone) Aldehydes and ketones both have a C=O group (carbonyl group). Aldehydes have it on an end carbon. Ketones have it on a middle carbon. Reactions: Primary alcohols can be oxidized into aldehydes. Secondary alcohols into ketones. Carboxylic Acids Esters O General formula: O C R 23 • Organic Chemistry Functional Groups III Carboxylic Acids and Esters (13 of 16) C OH R OR Naming: names end - “oic names end - “ate” acid” ethyl acetate ethanoic acid (acetic acid + (acetic acid) ethyl alcohol) Reactions: Acids can be made by oxidizing aldehydes. Esters are formed (“esterification”) from a carboxylic acid & an alcohol. Water is removed (a “condensation” reaction). Esters often have pleasant, agreeable odors (e.g. banana.) Amines H General formula: Amides H O N C R 23 • Organic Chemistry Functional Groups IV Amines & Amides (14 of 16) 23 • Organic Chemistry Polymers I Monomers & Addition Polymerization (15 of 16) R NH2 Naming: names contain names end in “amino” or end in “amide” “amine” acetamide aminomethane (methylamine) The N may have 1 or 2 or all 3 H atoms replaced with groups. The lone pair on the N atom makes these molecules basic. Your body needs certain amines “vital amines” ≈ “vitamins.” Monomer = one part Polymer = many parts H H One kind of polymer is made up of C C monomers that contain a double bond. The double bond can break and we can H H “ethylene” ADD to it… “Addition polymerization.” * H H C C H H H + * H C H → C * H H H H H C C H H H H C C * Different monomers form different polymers. This polymer would be called polyethylene. Replace on H on the monomer with Cl and you can make polyvinyl chloride, “PVC.” Another polymerization involves condensation reactions. O O C 23 • Organic Chemistry Polymers II Copolymers & Condensation Polymerization (16 of 16) HO R H H HO C C OH H OH C R H a di-alcohol (a glycol) a di-acid Esters form from an acid and an alcohol. Using a di-acid and a di-alcohol, you can make a continuous chain by removing water molecules. The resulting polymer is called a polyester. Soda bottles are made from a polyester, polyethylene terephthalate ester (PETE). Nylon (a polyamide) can be made from a di-amine & a di-acid.