CHEM210 PSU Oxidation Chemical Reaction Concept Research Paper

Anonymous
timer Asked: Feb 22nd, 2019
account_balance_wallet $10

Question Description

Please read the following article:

http://www.bbc.co.uk/news/technology-22191650

and write a response in 1-2 pages (minimum of 300 words, maximum of 600 words, not including the exam question) using the 3-2-1 format described below:



3: Find 3 concepts from within the article and relate them to 3 concepts within CHEM 210 we have discussed in class and cite 3 textbook references using the chapter and page number.



2: Find 2 concepts from within the article that you want to know more about (i.e. muddy points, have questions about, did not quite understand).



1: Write an exam question with the answer about 1 concept discussed from within the article. The exam question must be well thought out and appropriate to the subject matter.

I also have 3 questions nada to be answer but I couldn't upload it now. I will send them we you are going to do the assignment

Chapter 2. Elements and Compounds 2.1 The Structure of the Atom 2.2 Elements and the Periodic Table 2.3 Covalent Compounds 2.4 Ions and Ionic Compounds In this chapter, we discover that there are actually a number of different variations, or isotopes, of the atoms associated with each element. We explore the structure of the atom in further detail and learn about the composition of isotopes. We also discuss the different ways that molecules and compounds are represented and named. 2.1 The Structure of the Atom 2 Atomic Structure: Protons and Neutrons Mass of the atom is primarily in the nucleus. Charge of the proton is opposite in sign but equal to that of the electron. Scanning Tunneling Microscopy (STM) Scanning Tunneling Microscopy •Fe atoms arranged on Cu. Atom in Chinese Fe atoms arranged on Cu. Quantum Corral Ag atoms arranged on Si. Atomic Masses and the Mole The mass of 1 atom of carbon-12 is defined to be 12 amu. vb = #eZ = #p+ also vb A = #p+ + #n 2.1 Atomic Number, Mass Number, and Isotopes Most elements have two or more isotopes, -atoms that have the same atomic number (Z) but different mass numbers (A). 7 SAMPLE PROBLEM Determine the numbers of protons, neutrons, and electrons in each of the following species: Number of protons = Z number of neutrons = A – Z number of electrons = number of protons. Solution Atomic number is Z = 17 17 protons. Mass number is A = 35, neutrons is 35 – 17 = 18. Number of electrons = number of protons = 17 electrons. 8 2.1 The Atomic Mass Scale and Average Atomic Mass Atomic mass unit (amu) is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. 9 Percent Abundance • For each element’s isotopes, the percent abundance is found by: number of atoms of an individual isotope Percent abundance = total number of atoms of ×100 all isotopes of that element 10 Atomic Weight (Mass) • Atomic weight: Equals the average mass of all naturally occurring isotopes of the element – Accounts for relative abundance of isotopes average atomic weight =  ( exact mass )( fractional abundance ) all isotopes atomic mass = ( %abundance of isotope 1) × mass of isotope 1 100 %abundance of isotope 2 ) ( + × mass of isotope 2... 100 11 Isotope Abundance Element Symbol At. Wt. Mass number Isotope mass Natural abundance (%) Hydrogen H 1.00794 1 2 3 1.0078 2.1041 3.0161 99.85 0.015 0 Boron B 10.811 10 11 10.0129 11.0093 19.91 80.09 Magnesium Mg 24.305 24 25 26 23.985042 24.985837 25.982593 78.99 10.00 11.01 Gallium 69.723 69 71 68.926 70.925 60.40 39.60 Ga 1 SAMPLE PROBLEM 2.3 Oxygen is the most abundant element in both Earth’s crust and the human body. The atomic masses of its three stable isotopes, 168O (99.757 percent), 178O (0.038 percent), and 188O (0.205 percent), are 15.9949, 16.9991, and 17.9992 amu, respectively. Calculate the average atomic mass of oxygen using the relative abundances given in parentheses. Report the result to four significant figures. What should the answer be? What can you check to estimate the answer? 13 SAMPLE PROBLEM 2.3 Solution Atomic mass = Σ(fractional abundance)(isotope mass) 16 8O 17 (99.757 %) 18 8O 8O (0.038 %) (0.205 %) Check your answer, what is the value on the periodic table? 14 2.2 The Periodic Table Horizontal rows - periods Vertical columns - groups (families). 15 2.3 Molecules and Molecular Compounds Molecule is a combination of at least two non-metal atoms in a specific arrangement and ratio held together by electrostatic forces known as covalent chemical bonds. Homonuclear diatomic molecules, both atoms in each molecule are of the same element. A diatomic molecule can also contain atoms of different elements (Heteronuclear diatomic molecules). 7 elements exist as diatomic molecules are hydrogen (H2), nitrogen (N2), oxygen (O2), fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2). LEARN the 7 Diatomics! 16 2.3 Molecules and Molecular Compounds Chemical Formula denotes the composition of the substance, metal-nonmetal, non-metal-non-metal, or metal-metal. Molecular Formula shows the exact number of atoms of each element in a molecule of non-metal atoms. Allotrope is one of two or more distinct forms of an element. Two of the allotropic forms of the element carbon—diamond and graphite—have dramatically different properties. 17 2.3 Molecules and Molecular Compounds Condensed structural formula Bonding formula Structural formula Ball and Stick model Space-Filling model 18 2.3 Molecules and Molecular Compounds Molecular substances can also be represented using empirical formulas. empirical means “from experience” in the context of chemical formulas, “from experiment.” Empirical formula tells what elements are present in a molecule and in smallest whole-number ratio 19 2.3 Molecules and Molecular Compounds 20 2.3 Molecules and Molecular Compounds Binary compounds consist of just two different elements. 1st name the element that appears first in the formula (usually the metal atom). Name the second element (usually the nonmetal), changing the ending of its name to –ide. metal non-metal HCl hydrogen chloride NO nitrogen oxide SiC silicon carbide 21 Prefixes used in Naming Binary Nonmetal Compounds (Molecular Compounds) ONLY Number Prefix 1 Name mono Formula 2 di Name Formula Sulfur dioxide tri H2O2 4 Hydrogen peroxidetetra SO2 Sulfur trioxide SO3 5 Ammonia Carbon monoxide CO Carbon dioxide CO2 Chlorine monoxide ClO Water 3 6 7 Hydrazine 8 Nitric oxide H2O penta NH3 hexa N2H4 hepta octaNO 9 Nitrogen dioxide nona NO2 10 deca 12 Disulfur decafluoride S2F10 dodeca Prefix mono– is generally omitted for the first element except carbon monoxide (CO). 22 2.3 Oxoanions Two forms exist: –ate and –ite suffix endings are used. Increase # O ▪ More oxygen = “-ate” ▪ Less oxygen = “-ite” SO42- sulfate ion NO3- nitrate ion SO32- sulfite ion NO2- nitrite ion If they contain H, add a prefix “hydrogen” HSO4- hydrogen sulfate ion (common name=bisulfate ion) HCO3- hydrogen carbonate ion (common name=bicarbonate ion) 2.3 Oxoanions When four forms exist, add prefix Increase O –Add “per_____ate” and “hypo____ite” names FO4FO3FO2FO- perfluorate fluorate fluorite hypofluorite ClO4ClO3ClO2ClO- perchlorate chlorate chlorite hypochlorite 2.3 Ions and Ionic Compounds Ions that consist of a combination of two or more atoms are called polyatomic ions. LEARN THEM! Quiz in Lab! 25 2.3 Molecules and Molecular Compounds Acid is a substance that produces hydrogen ions (H+) when dissolved in water. Rules for naming simple acids of this type are as follows: 1. remove the –gen ending from hydrogen (leaving hydro–) 2. change the –ide ending on the second element to –ic 3. combine the two words 4. add the word acid. 26 2.3 Molecules and Molecular Compounds Binary compounds consist of just two different elements. 1st name the element that appears first in the formula (usually the metal atom). Name the second element (usually the nonmetal), changing the ending of its name to –ide. metal non-metal HCl hydrogen chloride SiC silicon carbide NaBr sodium bromide 27 Interactive Figure 2.4.2 - Explore Monoatomic ion Formulas Transition Metal have different oxidation states (charges) Charge of cation is determined by the non-metal anion paired with 28 2.4 Ions and Ionic Compounds Cations that can have more than 1 charge (oxidation state) are use Roman numerals. Cations with only two possible charges – use ending suffix – ous to the cation with the smaller positive charge – ic to the cation with the greater positive charge: Cu1+ : Cuprous Cu2+ : Cupric ▪ exceptions: lead (Pb2+, Pb4+), tin (Sn2+, Sn4+)… ▪ exceptions: silver (Ag+), zinc (Zn2+), cadmium (Cd2+) 29 2.4 Naming Ionic Compounds Ionic Compound: A neutral compound in which the total number of positive charges must equal the total number of negative charges. Binary Ionic Compounds aluminum sulfide: Al3+ +6 Al23+ S2+ -6 S32- =0 =0 Al2S3 2.4 Ions and Ionic Compounds Formulas of Ionic Compounds Aluminum Oxide Calcium Phosphate Check to make sure charges cancel out and equal 0! 31 2.4 Ions and Ionic Compounds Naming Ionic Compounds 32 2.4 Naming Ionic Compounds Metal of fixed charge with a polyatomic ion Cation Anion Formula Name K+ OH− KOH potassium hydroxide Ca2+ OH− Ca(OH)2 calcium hydroxide Al3+ SO24- Al2(SO4)3 aluminum sulfate Pb(SO4)2 lead(IV) sulfate Fe(NO2)2 iron(II) nitrite ferrus nitrite 24 Pb4+ SO Fe2+ NO2- 33 2.4 Ions and Ionic Compounds Familiar Inorganic Compounds 34
CHAPTER 3.1 3.2 3.3 3.4 3.5 Chapter 3. Stoichiometry The Mole and Molar Mass Stoichiometry and Compound Formulas Stoichiometry and Chemical Reactions Stoichiometry and Limiting Reactants Chemical Analysis In this chapter we will explore the chemical counting unit that links the atomic and macroscopic scales, the mole. The mole will allow us to study in greater detail chemical formulas and chemical reactions. Specifically, we will investigate stoichiometry, the relationship between quantities of materials in chemical reactions. 3.1 The Mole and Molar Masses The Mole People use a variety of counting groups to conveniently indicate the number of objects in some set: 1 “pair” objects 2 objects 1 “dozen” objects 12 objects 1 “gross” objects 144 objects 1 “million” objects 1,000,000 objects 1 “trillion” objects 1012 objects 1 “mole” objects 6.023 × 1023 objects 2 3.1 The Mole and Molar Masses The Mole A “mole” is a counting group, defined as the number of atoms in exactly 12 g of carbon-12, number of atoms in 12 g of carbon-12 is known as Avogadro’s Number (NA): NA = 6.0221418 × 1023 objects 1 mol objects 6.022  1023 objects 6.022  10 objects 1 mol objects 23 If you order one dozen doughnut, you are asking for 12 doughnuts. If you order one mole of doughnuts, you are asking for 6.022 × 1023 doughnuts! Fun Fact: “Avogadro’s Number” is named for Amedeo Avogadro, 1776-1856 3 Interactive Figure 3.1.1 - Recognize How the Mole Connects Macroscopic and Atomic Scales One mole quantities of (from left to right) the elements copper, aluminum, sulfur and the compounds potassium dichromate, water and copper (II) chloride dihydrate 4 SAMPLE PROBLEM 3.5 A typical human body contains roughly 30 moles of calcium. Determine (a) the number of Ca atoms in 30.00 moles of calcium and (b) the number of moles of calcium in a sample containing 1.00 × 1020 Ca atoms. Setup 1 mol Ca atoms 6.022  1023 Ca atoms 6.022  10 Ca atoms 23 1 mol Ca atoms 5 3.1 Molar Mass • Molar mass (MM): Molecular weight in grams of one mole of a substance • Molecular weight – Formula weight for substances that exist as individual molecules • Formula weight – Sum of the atomic weights of the elements that make up a substance multiplied by the number of atoms of each element in the formula 6 3.1 The Mole and Molar Masses Interconverting Mass, Moles, and Numbers of Particles 7 3.1 The Mole and Molar Masses Formula MM, Molar mass (g/mol) m, Mass of sample (g) SO2 n, Moles of NA, Number of Atoms, Sample (mol) Molecules or Formula Units 4.25 2.65 x 1024 molecules PCl3 OCl2 0.0186 6.38 x 1022 molecules NaBr Calculate the mass of carbon contained in 3.45 x 1021 molecules of aspirin, C9H8O4. [MM C9H8O4 = 180.158] 21 3.45 x 10 molecules x 1 mol 9 mol C x 12.01 1g = 0.619 g C x 23 1 mol 1 mol C 6.022 x 10 molecules atoms 8 3.2 Element Composition • Stoichiometry: Study of the relationship between relative amounts of substances – Formula of a compound provides information on the relative amount of each element in either one molecule or one mole of the compound • One molecule of acetic acid, CH3CO2H, contains: – Two atoms of oxygen and one mole of acetic acid – 2 mol of oxygen atoms 9 3.2 Percent Composition of Compounds (2 * Composition 1.008 amu) Percent - Mass of an element present in exactly 100 g of a compound composition by mass. % element = (number of atoms of element in formula) (molar mass of element) x 100% mass of 1 mol of compound Formula mass is: (2 * 1.008 (2 * 1.008 amu) amu) + (2 * 16.00 amu) = (2 * 1.008 34.02 amu amu) (2 * 1.008 amu) (2 * 1.008 amu) 10 3.2 Empirical Formula • Whole-number ratio of elements present in the compound • Determined by converting mass of each element to an amount in moles • Reverse of the process that is used to determine percent composition 11 3.2 Molecular Formula (2 * 1.008 formula amu) • Molecular is a multiple scaled by a factor “n”, the molecular and empirical molar masses must scale by the same ratio  g molar formula mass  mol =n g empirical formula mass mol   empirical formula molecular formula n x C XH Y O Z = CnXHnYOnZ n = 2,3,4… 12 The molecular formula can be determined if the molecular mass is known and vice versa. Example: Vitamin C has the empirical formula C3H4O3 and molecular mass = 176.12 g/mol. Empirical mass: 3(12.01 g/mol) + 4(1.008 g/mol) + 3(15.99 g/mol) = 88.03 g/mol C3H4O3 Multiple repeat units = Molecular mass Empirical formula mass = 176.12 g/mol 88.03 g/mol Molecular formula = 2(C3H4O3) = C6H8O6 =2 3.2 Percent Composition & Empirical Formulas Check your WORK! What is Empirical Formula Mass? How many Repeat Units? What is the Molecular Formula Mass? Do the masses and the repeat units make sense? 14 Hydrated Compounds • Hydrated ionic compound: Ionic compound that has a well-defined amount of water trapped within the crystalline solid • Water of hydration: Water associated with the compound • Hydrated compound formula includes the term nH2O – Where n is the number of moles of water incorporated into the solid per mole of ionic compound 15 3.3 Chemical Equations Interpreting and Writing Chemical Equations Chemical Reaction, as described in the third hypothesis of Dalton’s atomic theory, is the rearrangement of atoms in a sample of matter. Examples include the rusting of iron and the explosive combination of hydrogen and oxygen gases to produce water. Chemical Equation uses chemical symbols to denote what occurs in a chemical reaction. 16 3.3 Chemical Equations Chemical Equations = Chemical Sentence “Ammonia and hydrogen chloride react to produce ammonium chloride.” “Calcium carbonate reacts to produce calcium oxide and carbon dioxide.” 17 3.3 Chemical Equations Each chemical species that appears to the left of the arrow is called a reactant. Reactants are those substances that are consumed in the course of a chemical reaction. Each species that appears to the right of the arrow is called a product. Products are those substances that are produced in the course of a chemical reaction Physical state of compounds are indicated using symbols 18 3.3 The Atomic Theory Dalton’s Atomic Theory 1. Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass, and chemical properties. The atoms of one element are different from the atoms of all other elements. 1. Compounds are composed of atoms of more than one element. In any given compound, the same types of atoms are always present in the same relative numbers. 2. Chemical Reaction rearranges atoms in chemical compounds; it does not create or destroy them. 19 3.3 Law of Conservation of Mass Hg(NO3)2(aq) + 2KI(aq) 3.25 g 3.32 g 3.25 g + 3.32 g = 6.57 g HgI2(s) + 2KNO3(aq) 4.55 g 2.02 g 4.55 g + 2.02 g = 6.57 g Balancing Chemical Equations • Relative amounts of the reactants and products involved in the reaction must be determined by balancing the equation 21 Balancing Chemical Equations 1. Write an unbalanced equation with correct formulas for all substances and states (). 2. Balance the atoms of one element. a. Start with the most complex molecule (keep polyatomics together) b. Change the coefficients in front of the molecules c. DO NOT alter the chemical formulas (i.e. do not change the subscripts!!!!) 3. Balance the remaining elements. 4. Check the atoms are all balanced. 3.3 Balancing Chemical Equations Balance : Al + Fe2O3 step 1 Al Al2O3 + Fe + Fe2O3 1 Al Al2O3 + Fe (2Fe + 3O) (2Al + 3O) 1Fe not balanced not balanced step 2 Al 1 Al + Fe2O3 Al2O3 + 2 Fe (2Fe + 3O) (2Al + 3O) balanced not balanced 2Fe step 3 2 Al + Fe2O3 2Al (2Fe + 3O) Al2O3 + 2 Fe (2Al + 3O) 2Fe step 4 – balanced 3.3 Balancing Chemical Equations Polyatomic ion on both sides of an equation? Balance Polyatomic ion as a collective species. NaNO3(s) + H2SO4(aq) Na + NO3 + 2H + SO4 Na2SO4(aq) + HNO3(aq) 2Na + SO4 + H + NO3 not balanced Balance Na in Na2SO4 2 NaNO3(s) + H2SO4(aq) 2Na + 2NO3 + 2H + SO4 2Na + SO4 + H + NO3 not balanced 2 NaNO3(s) + H2SO4(aq) 2Na + 2NO3 + 2H + SO4 Na2SO4(aq) + HNO3(aq) Na2SO4(aq) + 2 HNO3(aq) 2Na + SO4 + 2H + 2NO3 balanced 3.3 Chemical Arithmetic: Stoichiometry Stoichiometry: Relative proportions in which elements form compounds or in which substances react. aA + bB Grams of A Molar Mass of A Moles of A cC + dD Moles of B Mole Ratio Between A and B (Coefficients) Grams of B Molar Mass of B Stoichiometric Factor or Ratio 2A + 3B 2 As combine with 3Bs A2B3 3B 2A or Conversion factors 3B 2A 26 Stepwise method for solving, need to determine what is being asked and what you know. 1. Write correct formulas for reactants and products 2. BALANCE the chemical equation! 3. Decide what is known and what is unknown from the question • write it down under the chemical species 4. Map out a strategy for answering the question • What insights/hints are in the question • What are you solving for? 5. Convert everything to moles! 6. Use mole-mole stoichiometric ratio to convert from 1 chemical species to another chemical species 7. Check the answer! Make sure it is reasonable, think about the answer, does it Make Sense? Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas. How many grams of NaOH are needed to react with 25.0 g Cl2? 2 NaOH(aq) + Cl2(g) ____ g NaOCl(aq) + NaCl(aq) + H2O(l) 25.0 g Grams of Cl2 Molar Mass Moles of Cl2 Mole Ratio Moles of NaOH Grams of NaOH Molar Mass Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas. How many grams of NaOH are needed to react with 25.0 g Cl2? 2NaOH(aq) + Cl2(g) 25.0 g Cl2 x 1 mol Cl2 70.9 g Cl2 NaOCl(aq) + NaCl(aq) + H2O(l) 2 mol NaOH x 1 mol Cl2 = 28.2 g NaOH Significant figures!!! 40.0 g NaOH x 1 mol NaOH Sec 3.3 Exercise: Strategy Map Data Information: Mass of reactant Step 1: Write the balanced chemical equation Eq. gives mole ratios (stoichiometry) Step 2: Convert mass to moles amount of reactant in moles Step 3: Convert moles reactant to moles product amount of products in moles Step 4: Convert moles of products to mass mass of products in grams 30 3.4 Limiting Reactants The reactant used up first is called the limiting reactant, , the amount of this reactant limits the amount of product that can form. When all the limiting reactant has been consumed, NO MORE PRODUCT can be formed! Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant. Initial amount – amount Used = amount Remaining (excess) 31 3.4 Reactions with Reactant in Limited Supply Given 10 slices of cheese and 14 slices of bread and a sandwich is 1 cheese and 2 breads, how many sandwiches can you make? Balanced equation 1 cheese + 2 bread 1 sandwich 1 cheese ≡ 2 bread 1 cheese ≡ 1 sandwich 2 bread ≡ 1 sandwich Product Method- calculate the product from each starting material reactant, must do the calculation 2x’s. Reactant giving the smallest amount of product is the Limiting Reactant (Reagent)! 1 cheese + 2 bread  10 slices 14 slices 1 sandwich 1 cheese 10 cheese x 14 bread x 1 sandwich 2 bread 1 sandwich = 10 sandwiches = 7 sandwiches Correct answer Bread is the Limiting Reactant, used up first smallest product Reactant Method- calculate the amount of the OTHER reactant (2nd reactant) needed (from the 1st reactant) and check to see/ask,“Do I have enough?” Repeat 2x’s! 1 cheese + 2 bread  10 slices 14 slices 10 cheese x 14 bread x 2 bread 1 cheese 1 cheese 2 bread 1 sandwich = 20 bread Do you have enough bread? = 7 cheese Do you have enough cheese? 3.4 Reactions with Limited Reactants Bread is limiting… …base all other calculations on the limiting reactant. Sandwiches made 14 bread (1 sandwich / 2 bread ) = 7 sandwiches Cheese remaining 14 bread (1 cheese / 2 bread ) = 7 cheese used. Started with 10 cheese. Cheese remaining Initial Amount – Used Amount = Remaining Amount 10 – 7 = 3 slices Stepwise method for solving, need to determine what is being asked and what you know. 1. Write correct formulas for reactants and products 2. Balance the chemical equation 3. Decide what is known and what is unknown from the question 4. Map out a strategy for answering the question • What insights/hints are in the question • What are you solving for? 5. Convert everything to moles, use mole-mole stoichiometric ratio to convert from 1 chemical species to another chemical species 6. Check the answer! Make sure it is reasonable, think about the answer, does it Make Sense? Reactions with Limiting Amounts of Reactants Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: Li2O(s) + H2O(g) 2LiOH(s) If 80.0 g of water are to be removed and 65.0 g of Li2O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced? Li2O(s) + H2O(g) 65.0 g 80.0 g 2LiOH(s) _____ g Which reactant is limiting? Amount of H2O that will react with 65.0 g Li2O: 65.0 g Li2O x 1 mol Li2O 29.9 g Li2O x 1 mol H2O 1 mol Li2O = 2.17 moles H2O Amount of H2O given: 80.0 g H2O x 1 mol H2O 18.0 g H2O = 4.44 moles H2O Li2O is limiting Li2O(s) + H2O(g) 65.0 g 80.0 g 2LiOH(s) _____ g How many grams of excess H2O remain? 2.17 mol H2O x 18.0 g H2O 1 mol H2O = 39.1 g H2O (consumed/used up) 80.0 g H2O - 39.1 g H2O = 40.9 g H2O initial consumed remaining How many grams of limiting reactant, Li2O remain? 0 g all used up! Reactions with Limiting Amounts of Reactants Li2O(s) + H2O(g) 65.0 g 80.0 g 2LiOH(s) _____ g How many grams of LiOH are produced? 2.17 mol H2O 2 mol LiOH x 1 mol H2O 23.9 g LiOH x 1 mol LiOH = 104 g LiOH 3.4 Percent Yield Theoretical yield Maximum amount of product predicted by stoichiometry calculations (Calculated on paper!) Experimental yield Quantity of desired product actually formed from the experiment, sometimes called experimental yield Percent yield Efficiently of the chemical reaction experimental yield percent yield = x 100% theoretical yield Percent Yield You heat 2.50 g of copper with an excess of sulfur and synthesize 2.53 g of copper(I) sulfide 16 Cu(s) + S8(s) 8 Cu2S(s) What was the percent yield for your reaction? 1 mol Cu = 0.03934 mol Cu nCu used: 2.50 g Cu x 63.55 g Cu 16 mol Cu used  8 mol Cu2S made Theoretical yield: 0.03934 molCu 8 mol Cu2S = 0.01967 mol Cu2S 16 mol Cu Percent Yield Heat 2.50 g of Cu with excess S8 and make 2.53 g of copper(I) sulfide: 16 Cu(s) + S8(s) → 8 Cu2S(s). What was the %-yield for your reaction? Theoretical yield = 0.01967 mol Cu2S = 0.01967 mol Cu2S 159.2 g Cu2S = 3.131 g Cu2S 1 mol Cu2S Actual yield = 2.53 g Cu2S (in problem) Percent yield = 2.53 g x 100% = 80.8% 3.131 g Sec 3.4 Exercise: Strategy Map Data Information: Masses of two reactants. Recognize as a limiting reactant problem Step 1: Write the balanced chemical equation Balanced Eq. gives mole ratios (stoichiometry) Step 2: Convert masses of each reactant to moles amount of each reactant in moles Step 3: Limiting reactant is known Step 4: Calculate the moles of product based on limiting reactant Amount of product based on the L.R Step 5: Calculate the mass of product based on moles This is the Theoretical Yield Decide which reactant limits by comparing mole amounts of product formed 44 3.5 Combination Reactions (Addition Reaction) + X Z XZ Element plus halogen or O2: 2 Mg(s) + O2(g)  2 MgO(s) I2(s) + Zn(s)  ZnI2(s) There are other types: 2 SO2(g) + O2(g)  2 SO3(g) 3.5 Decomposition Reactions (Dissociation Reaction) + XZ X Z Often initiated by heat: CaCO3(s) 2 KNO3(s) 800 - 1000°C heat CaO(s) + CO2(g) 2 KNO2(s) + O2(g) Occasionally by shock: 4 C3H5(NO3)3(l) 12 CO2(g) + 10 H2O(l) + 6 N2(g) + O2(g) nitroglycerin 3.5 Decomposition Reactions 3.5 Single Displacement Reactions + A + XZ AZ X Some metals displace another metal from its salt Fe(s) + CuSO4(aq) Zn(s) + 2 AgNO3(aq) FeSO4(aq) + Cu(s) Zn(NO3)2(aq) + 2 Ag(s) Some other examples: F2(g) + 2 LiCl(s) 2 Na(s) + 2 H2O(l) 2 LiF(s) + Cl2(g) 2 NaOH(aq) + H2(g) 3.5 Displacement Reactions 3.5 Double Replacement Reactions Exchange Reactions + AD + XZ AZ XD AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2 KNO3(aq) 3.5 Limiting Reactants Types of Chemical Reactions Combustion. A combustion reaction is one in which a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and water. By convention, we will consider the water produced in a combustion reaction to be liquid water. 51 SAMPLE PROBLEM 3.14 Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: Solution (a) combination (b) combustion (c) decomposition 52
Chapter 4. Chemical Reactions and Solution Chemistry 4.1 4.2 4.3 4.4 4.5 Types of Chemical Reactions Aqueous Solutions Reactions in Aqueous Solutions Oxidation-Reduction Reactions Stoichiometry of Reactions in Aqueous Solutions The science of chemistry is unique in that it involves the ability to create new forms of matter by combining the elements and existing compounds in new but controlled ways. The creation of new chemical compounds is the subject of this chapter. Here, we explore different types of chemical reactions, emphasizing the reactions that take place in aqueous solution. Reaction Types Gas-forming Acid base Reactions Types Precipitation Redox 2 Classifying Compounds • Salts (ionic compounds) - Composed of a metal and non metal element(s) • Acids – Arrhenius definition: Produce H+(aq) in water – HCl, HNO3, HC2H3O2 • Bases – Arrhenius definition: Produce OH(aq) in water – NaOH, Ba(OH)2, NH3 • Molecular compounds - Covalently bonded atoms, not acids, bases, or salts – Compounds like alcohols (C2H5OH) or table sugar (C6H12O6) – Never break up into ions 3 Classifying Compounds Compound Type Na2SO4 Ba(OH)2 H3PO4 CH4 P2O5 NH3 HCN Ionic-salt Base Acid Molecular Molecular Base Acid 4 4.2 General Properties of Aqueous Solutions A solution is a homogeneous mixture of two or more substances. Solutions may be gaseous (such as air), solid (such as brass), or liquid (such as saltwater). Usually, the substance present in the largest amount is referred to as the solvent and any substance present in a smaller amount is called the solute. 5 4.1 General Properties of Aqueous Solutions Electrolytes and Nonelectrolytes An electrolyte is a substance that dissolves in water to yield a solution that conducts electricity. nonelectrolyte is a substance that dissolves in water to yield a solution that does not conduct electricity. Difference between an aqueous solution that conducts electricity and one that does not is the presence or absence of ions. nonelectrolyte electrolyte 6 Aqueous Solubility of Compounds • Not all compounds dissolve in water. • Solubility varies from compound to compound. Soluble ionic compounds dissociate. Ions are solvated Most molecular compounds stay associated in water. 4.1 General Properties of Aqueous Solutions Strong Electrolytes and Weak Electrolytes © The McGraw-Hill Companies, Inc./Stephen Frisch, photographer 8 Electrolytes in Aqueous Solution Electrolytes: Substances which dissolve in water to produce conducting solutions of ions-Ionic Compounds, Strong Acids, and Strong Bases H 2O NaCl(s) Na1+(aq) + Cl1-(aq) Nonelectrolytes: Substances which do not produce ions in aqueous solutions- Weak Acids and Weak Bases H 2O C12H22O11(s) C12H22O11(aq) Chapter 7/9 Solute in Aqueous Solution Ionic Compound Metal-Nonmetal Strong Electrolyte IONS Molecular Compound Non-metals only Acids and Bases Molecular Compound Nonelectrolyte MOLECULES Strong Acids Strong Bases Strong Electrolyte IONS Weak Acids Weak Bases Weak Electrolyte MOLECULES & IONS 4.1 Classify each of the following compounds as a nonelectrolyte, a weak electrolyte, or a strong electrolyte: (a) (b) (c) (d) methanol (CH3OH) sodium hydroxide (NaOH) ethylamine (C2H5NH2) hydrofluoric acid (HF) A -Nonelectrolyte, molecular compound B -Strong electrolyte, strong base C -Nonelectrolyte, molecular compound D-Weak electrolyte, weak acid 11 4.2 Precipitation Reactions and Solubility Guidelines Solute – the material being dissolved in a solution Solvent – the medium in which a solute is being dissolved to form a solution, usually H2O aq Solubility: States how much of a compound will dissolve in a given amount of solvent at a given temperature. Precipitation Reaction – 2 aqueous solutions are mixed in a double exchange reaction and produces an insoluble, solid product Precipitate – insoluble product of a chemical reaction, solid state(s) Chapter 7/12 4.2 Precipitation Reactions © The McGraw-Hill Companies, Inc./Charles D. Winters, photographer 13 4.2 Classify each of the following compounds as soluble or insoluble in water: (a) AgNO3, (b) CaSO4, (c) K2CO3. AgNO3 CaSO4 K2CO3 soluble insoluble soluble 15 Chemical Reactions and Solution Stoichiometry 4.3 Reactions in Aqueous Solution 1 4.2 Precipitation Reactions Net Ionic Equations The steps necessary to determine the molecular, ionic, and net ionic equations for a precipitation reaction are as follows: 1. Write and balance the molecular equation, predicting the products by assuming that the cations trade anions. 2. Write the ionic equation by separating strong electrolytes into their constituent ions. 3. Write the net ionic equation by identifying and canceling spectator ions on both sides of the equation. 17 Aqueous Reactions and Net Ionic Equations Molecular Equation = Balanced Chemical Equation: All substances in the chemical equation are written using their complete formulas as if they were molecules and include state (aq, l, s, g). Pb(NO3)2(aq) + 2KI(aq) strong electrolytes 2KNO3(aq) + PbI2(s) precipitate Aqueous Reactions and Net Ionic Equations Ionic Equation (Total Ionic Equation): All of the strong electrolytes (all chemical compounds in the aq state only) are written as ions. Pb(NO3)2(aq) 2KI(aq) Pb2+(aq) + 2NO31- (aq) + 2K1+(aq) + 2I1-(aq) 2K1+(aq) + 2NO31- (aq) + PbI2(s) 2KNO3(aq) + PbI2(s) Aqueous Reactions and Net Ionic Equations Spectator Ions: Ions that undergo no change (have no effect) during the reaction and appear on both sides of the reaction arrow. Pb2+(aq) + 2NO31- (aq) + 2K1+(aq) + 2I1-(aq) 2K1+(aq) + 2NO31- (aq) + PbI2(s) Aqueous Reactions and Net Ionic Equations Net Ionic Equation: Only the ions undergoing change are shown, must include charge and state, overall equation. Pb2+(aq) + 2I1-(aq) PbI2(s) 1. Balanced Chemical Equation Pb(NO3)2(aq) + 2KI(aq) 2KNO3(aq) + PbI2(s) 2. Total Ionic Equation (must have correct # of species, charges, & states), and then Cancel out Spectator Ions Pb2+(aq) + 2NO31- (aq) + 2K1+(aq) + 2I1-(aq) 2K1+(aq) + 2NO31- (aq) + PbI2(s) 3. Net Ionic Equation Pb2+(aq) + 2I1-(aq) PbI2(s) Write the molecular, ionic, and net ionic equations for the 4.3 reaction that occurs when aqueous solutions of lead acetate [Pb(C2H3O2)2], and calcium chloride (CaCl2), are combined. Molecular equation: Ionic equation: Net ionic equation: 23 4.3 Acid-Base Reactions Acid as a substance that ionizes in water to produce H+ ions Base is a substance that ionizes (or dissociates, in the case of an ionic base) in water to produce OH– ions LEARN THESE! 24 4.3 Acid-Base Reactions Brønsted Acids and Bases Base Acid Acid Base 25 4.3 Acid-Base Reactions Brønsted Acids and Bases H2SO4 diprotic acid (2 H’s removable) -is strong only in its first ionization in water. For all other polyprotic acids (many H’s removable), each ionization is incomplete 26 Acids, Bases, and Neutralization Reactions Acid-Base neutralization reactions are double-replacement reactions just like the precipitation reactions: HA + MOH MA + HOH or HA + MOH Acid Base MA + H2O Salt Water Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous HBr and aqueous Ba(OH)2. 1. Write the chemical formulas of the products (use proper ionic rules for the salt). HBr(aq) + Ba(OH)2(aq) Acid Base H2O + BaBr2 Water Salt Acids, Bases, and Neutralization Reactions 2. Molecular Equation: Balance the equation and predict the solubility of the salt in the products. 2HBr(aq) + Ba(OH)2(aq) Strong Acid Strong Base 2H2O(l) + BaBr2(aq) Use the solubility rules. Acids, Bases, and Neutralization Reactions 3. Ionic Equation: Dissociate a strong acid and strong bases and the soluble ionic compounds (i.e. aq states). 2HBr(aq) Ba(OH)2(aq) 2H1+(aq) + 2Br1-(aq) + Ba2+(aq) + 2OH1-(aq) 2H2O(l) + Ba2+(aq) + 2Br1-(aq) BaBr2(aq) Acids, Bases, and Neutralization Reactions 4. Net Ionic Equation: Eliminate the spectator ions from the ionic equation. 2H1+(aq) + 2Br1-(aq) + Ba2+(aq) + 2OH1-(aq) 2H2O(l) + Ba2+(aq) + 2Br1-(aq) 2H1+(aq) + 2OH1-(aq) 2H2O(l) or H1+(aq) + OH1-(aq) H2O(l) Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous NaOH and aqueous HF. 1. Write the chemical formulas of the products (use proper ionic rules for the salt). HF(aq) + NaOH(aq) Acid Base H2O Water + NaF Salt Acids, Bases, and Neutralization Reactions 2. Molecular Equation: Balance the equation and predict the solubility of the salt in the products. HF(aq) + NaOH(aq) Weak Acid H2O(l) + NaF(aq) Use the solubility rules. Strong Base Acids, Bases, and Neutralization Reactions 3. Ionic Equation: Dissociate a strong acid and the soluble ionic compounds but HF is a weak acid NaOH(aq) HF(aq) + Na1+(aq) + OH1-(aq) H2O(l) + Na1+(aq) + F1-(aq) NaF(aq) Weak Acid Acids, Bases, and Neutralization Reactions 4. Net Ionic Equation: Eliminate the spectator ions from the ionic equation. HF(aq) + Na1+(aq) + OH1-(aq) HF(aq) + OH1-(aq) H2O(l) + Na1+(aq) + F1-(aq) H2O(l) + F1-(aq) Gas-Forming Reactions • Special type of acid–base reaction involving less obvious bases • Metal carbonate is involved, one product of the reaction is CO2(g) • Ionic compound (not a metal carbonate) is involved, it is not always easy to predict the products of the reaction NaHCO3  s   H3C6H5O7 (aq)  NaH2C6H5O7 (aq)  H2CO3 (aq) H2CO3 (aq)  CO2 (g)  H2O( ) https://www.youtube.com/watch?v=xRfPvDEs2gM 36 Gas-Forming Reactions Summary of Gas-Forming Reactions Metal carbonate (MCO3) + acid solution → metal salt + CO2(g) + H2O(l) Metal hydrogen carbonate (MHCO3) + acid solution → metal salt + CO2(g) + H2O(l) Metal sulfite (MSO3) + acid → metal salt + SO2(g) + H2O(l) Metal in acid solution → metal salt + H2(g) Group I metal in water → metal hydroxide + H2(g) Ammonium salt + basic solution → metal salt + NH3(g) + H2O(l) 37 Chemical Reactions and Solution Stoichiometry 4.4 Oxidation-Reduction Reactions 3 4.4 Oxidation-Reduction Reactions Oxidation Numbers The following guidelines will help you assign oxidation numbers. There are essentially two rules: 1. Oxidation number of any element, in its elemental form, is zero. 2. Oxidation numbers in any chemical species must sum to the overall charge on the species. -must sum to zero for any molecule -must sum to the charge on any polyatomic ion. -Oxidation number of a monatomic ion is equal to the charge on the ion. 39 Oxidation Numbers Most common oxidation numbers 40 SAMPLE PROBLEM 4.5 Determine the oxidation number of each atom in the following compounds and ion: (a) (c) (b) (d) 41 4.4 Oxidation-Reduction Reactions Oxidation of Metals in Aqueous Solutions displacement reaction Zinc displaces, or replaces, copper in the dissolved salt by being oxidized from Zn to Zn2+. Copper is displaced from the salt (and removed from solution) by being reduced from Cu2+ to Cu. Chloride (Cl–), which is neither oxidized nor reduced, is a spectator ion in this reaction. 42 4.4 Redox Reactions and Electron Transfer Loss of electrons is oxidation Gain of electrons is reduction Leo says Ger Oxidation is loss Reduction is gain Oil Rig 4.4 Oxidation-Reduction (Redox) Reactions • OXIDATION— loss of electron(s) by a species; increase in oxidation number (more positive). • REDUCTION—gain of electron(s); decrease in oxidation number (more negative). • OXIDIZING AGENT— electron acceptor, causes oxidation; species itself is reduced. • REDUCING AGENT— electron donor, causes reduction; species itself is oxidized. 4.4 Oxidation-Reduction Reactions Oxidation-Reduction Reaction, or Redox Reaction is a chemical reaction in which electrons are transferred from one reactant to another. Oxidation is the loss of electrons, and the gain of electrons is called reduction. • Zn is called the reducing agent donates electrons, causing Cu2+ to be reduced. • Cu2+ is called the oxidizing agent accepts electrons, causing Zn to be oxidized. 45 4.4 Oxidation-Reduction Reactions Oxidation of Metals in Aqueous Solutions Higher up in the table; Product species will be in the oxidized state (reducing agents) Any element higher in the activity series will reduce any element lower in the activity series. Table of who wins to be the oxidized product. 46 Activity Series of the Elements Which one of these reactions will occur? And why? Cu(s) + 2Ag1+(g) Cu2+(aq) + 2Ag(s) Yes, Cu is higher than Ag on the Activity Series Scale 2Ag(s) + Cu2+(g) 2Ag1+(aq) + Cu(s) No, Ag is lower than Cu on the Activity Series Scale Chapter 7/47 Chemical Reactions and Solution Stoichiometry 4.5 Stoichiometry of Reactions in Aqueous Solution 4 4.5 Concentration of Solutions © The McGraw-Hill Companies, Inc./Charles D. Winters, photographer 80 Molarity and Reactions in Aqueous Solution nA = [ A ] x V [conc] = nconc/ (total volume) Moles, Liters, Molarity If you know molarity and volume, you know moles Molarity  Volume = moles mols  L = mols L If you know mols and molarity, you know volume 1 mols  = Volume (L)  mol  M   L  L mol   L mol 51 4.5 Concentration of Solutions Solution Stoichiometry A solution that is 0.35 M in Na2SO4 has what concentration of Na+ and SO42–ions? 0.35 M Na2SO4 x 0.35 M Na2SO4 x 2 mol Na+ 1 mol Na2SO4 = 0.70 mol Na+ 1 mol SO42– 1 mol Na2SO4 = 0.35 mol SO42– 52 Example: Determine Grams of Solute Given: 35.0 mL of a 1.51 M Na3PO4 solution Find: grams of sodium phosphate Roadmap: mL solution  L  mols Na3PO4  g Na3PO4  use M as a conversion factor  molar mass Solve: 1.51 mol Na 3PO 4 163.94 g Na 3PO 4 L 35.0 mL × 3 × × L 1 mol Na 3PO 4 10 mL Molarity term = 8.66 g Na 3PO 4 53 4.5 Concentration of Solutions Dilution is the process of preparing a less concentrated solution from a more concentrated one. initial final Number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent. Can ONLY USE this Eqn’ when you see the words; Dilute, Diluted, and/or Stock Solution. 54 4.9 What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare 250.0 mL of 0.125 M HCl? Strategy Use Mc × mLc = Md × mLd to determine the volume of 12.0 M HCl required for the dilution. Setup Mc = 12.0 M, Md = 0.125 M, mLd = 250.0 mL Solution 55 Interactive Figure 4.5.6 - Explore Stoichiometric Relationships Involving Aqueous Solutions 56 Solution Stoichiometry What volume of 0.250 M H2SO4 is needed to react with 50.0 mL of 0.100 M NaOH? H2SO4(aq) + 2NaOH(aq) Volume of Solution of H2SO4 Molarity of H2SO4 Moles of H2SO4 Na2SO4(aq) + 2H2O(l) Moles of NaOH Mole Ratio Between H2SO4 and NaOH Volume of Solution of NaOH Molarity of NaOH Solution Stoichiometry H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) Moles of NaOH available: 50.0 mL NaOH 0.100 mol x 1L 1L x 1000 mL = 0.00500 mol NaOH Volume of H2SO4 needed: 0.00500 mol NaOH 1 mol H2SO4 1 L solution 1000 mL x x x 2 mol NaOH 0.250 mol H2SO4 1L 10.0 mL solution (0.250 M H2SO4) Aqueous Solution Titrations Titrant: Base of known concentration Buret = volumetric glassware used for titrations. Slowly add standard solution. End point: indicator changes color. Determine Vtitrant added. Unknown acid + phenolphthalein (colorless in acid)… …turns pink in base Titration 48.6 mL of a 0.100 M NaOH solution is needed to react with 20.0 mL of an unknown HCl concentration. What is the concentration of the HCl solution? HCl(aq) + NaOH(aq) Volume of Solution of NaOH Molarity of NaOH Moles of NaOH NaCl(aq) + H2O(l) Moles of HCl Mole Ratio Between NaOH and HCl Volume of Solution of HCl Molarity of HCl Chapter 6/60 Titration HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Moles of NaOH available: 48.6 mL NaOH 0.100 mol 1L x x = 0.00486 mol NaOH 1L 1000 mL Moles of HCl reacted: 0.00486 mol NaOH 1 mol HCl = 0.00486 mol HCl x 1 mol NaOH Concentration of HCl solution: 0.00486 mol HCl 1000 mL = 0.243 M HCl x 20.0 mL solution 1L
Chapter 5. Thermochemistry 5.1 5.2 5.3 5.4 5.5 5.6 Energy Enthalpy Energy, Temperature Changes, and Changes of State Enthalpy Changes and Chemical Reactions Hess’s Law Standard Heats of Reaction This chapter begins an exploration of thermochemistry, the study of the role that energy in the form of heat plays in chemical processes. We will investigate the energy changes that take place during phase changes and the chemical reactions you have studied in previous chapters and learn why some chemical reactions occur while others do not. 1 5.1 Energy and Reactions Thermochemistry: Study of how energy in the form of heat is involved in chemical change Exothermic process is a process that gives off thermal energy as heat (energy is a Product), system feels hot to the touch. Endothermic process is a process that absorbs thermal energy as heat (energy is a Reactant), system feels cold to the touch 2 5.1 The Nature of Energy Energy: Capacity to supply heat (q = smDT) or do work (w = DPV). Kinetic energy (Ek) - Energy of motion Ek = ½mv2 m = mass kg v = velocity of object m/s ▪ macroscale = mechanical energy ▪ random nanoscale = thermal energy-vibrations, rotations, translation ▪ periodic nanoscale = acoustic energy Potential energy (Ep) – Energy of position, stored energy Ep = m g h m =mass (kg) g= gravity (g =9.8 m/s2) ▪ charges held apart -bond energy h = height (m) 5.1 Energy Units joule (J) - SI unit 1 J = 1 kg m2/s2 1 J is a relatively small amount of energy. 1 kJ (1000 J) is more common in chemical problems. calorie (cal) 1 cal = 4.184 J (exactly) Dietary Calorie (Cal) - the “big C” calorie Used on food products 1 Cal = 1000 cal = 1 kcal Problem 5.1.1 - Use and Interconvert Energy Units 42.0 gallons of oil = 4.5 x 1010 J Calculate how much of energy this represents in Kcal. Given: Find: Roadmap: Factors: J  cal  kcal 1 cal 1 kcal and 4.184 J 1000 cal Solve: 1 cal 1 kcal 4.50  10 J   =1.08 107 kcal 4.184 J 1000 cal 10 Check: Joule is a smaller unit than cal; therefore the answer makes sense 5 5.1 Energy and Its Conservation Thermal Energy: Kinetic energy of molecular motion, measured by finding the temperature of an object. Heat (q): Amount of thermal energy transferred from one object to another as the result of a temperature difference between the two. Thermochemistry is the study of heat (the transfer of thermal energy) in chemical reactions Temperature measures heat (energy), Temperature is NOT energy! First Law of Thermodynamics: Energy cannot be created or destroyed; it can only be converted from one form to another. Table 5.1.1 - Types of Energy 7 5.1 Energy and Energy Changes Energy is interconvertible, BUT the total amount of energy in the universe is constant! If the Energy of one form disappears, then the same amount of energy must appear in another form or forms -Law of Conservation of energy = 1st Law of Thermodynamics. DEuniverse  0 system is defined as the specific part of the universe that is of interest to us. BE THE SYSTEM!! The rest of the universe outside the system constitutes the surroundings. 8 5.1 Introduction to Thermodynamics States and State Functions Thermodynamics, which is the scientific study of the interconversion of heat (q) and other kinds of energy. Open system can exchange matter and energy with its surroundings. Closed system allows the transfer of energy but not matter, can be passed to or from the surroundings Isolated system does not exchange either matter or energy can be passed to or from surroundings 9 State Function: A function or property whose value depends only on the initial and final, not on the path used to arrive at. State of a System - defined by the values of all relevant macroscopic properties, composition, energy, temperature, pressure, and volume. Internal Energy (E) is a state function ∆E = Efinal - Einitial whose value depends only on the current state of the system, not on the DE = Eproducts - Ereactants path taken to arrive at that state. 5.1 Internal Energy DEsystem  q  w heat lost or gained by the system change in = internal energy ΔE Heat (q) + work done by or on the system Work (w) System gains energy system gains heat energy (+) system gains energy from work (+) System loses energy system loses heat energy system loses energy by (–) doing work (–) Neither q nor w is a state function-both are path dependent. The sum, DE, does not depend on the path between initial and final states, E is a state function -path independent. 11 Figure 5.1.2 - Define System and Surroundings Heat (q) and work (w) sign conventions in thermodynamics ΔE positive: internal energy increases, nonspontaneous ΔE > 0 ΔE negative: internal energy decreases, spontaneous ΔE < 0 12 5.1 Introduction to Thermodynamics Work = Force x Distance = F x d w = P*A x d Force = Pressure x Area F = P x A w = -P(DV) DEsystem  q Heat Released Exothermic q<0 Expansion work done BY system w = - (PDV) V2 > V1 then w < 0 Heat Absorbed Endothermic q>0  w Compression work done ON system w = - (PDV) V2 < V1 then w > 0 13 5.1 w=Fxd Expansion Work w = -P x DV w = -P x (Vf-Vi) Perspective is from the System to determine DE!!!!!! • Energy flows OUT of the system, DE is Negative • Expansion, increase in volume (DV > 0), molecules inside the system do work to move the piston • Work < 0, work is negative: w = -P(4-3) = -P(1) = -w • System has lost energy, Efinal < Einitial • Energy flows INTO of the system, DE is Positive • Contraction, decrease in volume (DV < 0), molecules outside the system do work to move the piston • Work > 0, work is positive: w = -P(3-4) = -P(-1) = +w • System has gained energy, Efinal > Einitial SAMPLE PROBLEM 5.2 Calculate the overall change in internal energy, DU, (in joules) for a system that absorbs 188 J of heat and does 141 J of work on its surroundings. Setup The system absorbs heat, so q is positive. The system does work on the surroundings, so w is negative. Solution 15 Thermochemistry 5.2 Enthalpy 1 5.2 For a system that can do only PV work: ∆E = q - PDV DE = q + w If volume is constant: q = ∆E + PDV ∆V = Vf - Vi = 0 q = ∆E + P (0) qv = ∆E qv = heat at constant volume If pressure is constant: P = 1 atm q = ∆E + PDV qP = ∆E + P∆V = ∆ H qP = ∆H = Enthalpy = heat at constant pressure 17 5.2 DH Enthalpy Diagram • Energy of reactants higher than final energy of the products, downhill, spontaneous reaction • DH < 0, exothermic • Creates stronger bonds in product species • Temperature system decreases • Energy of reactants lower than final energy of the products, uphill, nonspontaneous reaction • DH > 0, endothermic • Creates weaker bonds in product species • Temperature system increases 18 Thermochemistry 5.3 Energy, Temperature Changes, and Changes of State 19 5.3 Temperature and Heat Important aspects of thermal energy & temperature – Heat is NOT the same as temperature – Greater thermal energy, greater motions of atoms – Total thermal energy is the sum of all individual energies of the atoms, molecules, and ions of an object • When an object gains thermal kinetic energy, the energy raises the object’s temperature • Three factors control temperature change – Amount of heat added – Mass of the object – Material the object is made of 20 5.3 Temperature and Heat Example • A cup of scalding hot coffee can be sipped safely, but if you chug it, or spill it on yourself, get at a serious burn – Why? All the coffee is at the same temperature! – Think about this in macroscopic terms • Temperature is a measure of heat per unit mass • Heat is the total amount of 'heat energy' present • A sip is a small mass and the chug is a much greater mass 21 5.3 22 Exploring Heat • There are three identical cubes made of the same material and same mass • Heat cube A to 100C • Heat cube B to 150C • Heat cube C to 200C • Place each cube into their own cup of water that are all at room temperature • Which cup of water will get the warmest? A B or C A B C The greatest initial temperature 22 5.3 Exploring Heat • You have three cubes made of the same material • The cubes have different masses • Heat all cubes to the same temperature • Place each cube in a cup of water that is at room temperature. • Which cup of water will get the warmest? A B or C A B C The greatest mass 23 5.3 Exploring Heat • You have three cubes made of different material • The cubes have the same mass • Heat all three cubes to the same temperature of 100C • Place each cube in a cup of water that is at room temperature • Which cup of water will get the warmest? A B or C A B C No real way to tell because we do not know what the metals are 24 5.3 Calorimetry and Heat Capacity Heat Capacity (C): The amount of heat required to raise the temperature of an object or substance a given amount. q = C x ∆T Molar Heat Capacity (Cm): The amount of heat required to raise the temperature of 1 mol of a substance by 1 °C. q = (moles of substance) x (Cm) x ∆T q = m x c x ∆T Specific Heat (s): The amount of heat required to raise the temperature of 1 g of a substance by 1 °C. q = (specific heat) x (mass of substance) x ∆T q = s x m x ∆T Table 5.3.1 - Specific Heat Capacity Values of Common Substances 26 A 200. g block of Cu at 500.°C is plunged into 1000. g of water of T = 23.4 °C in an insulated container. What will be the final equilibrium T of water and Cu? sCu = 0.385 J g-1 °C-1 sH2O = 4.184 J g-1 °C-1 Cu cools (−q) and water heats (+q); q = s m ΔT Tfinal must be between Thot and Tcold based on s-values, Tfinal should be closer to H2O or Cu? qCu = -s m ΔT Heat lost by Cu: qH2O = +s m ΔT -qCu = +qH2O qcu = − (0.385 J g-1 °C-1) (200. g) (Tfinal− 500°C) Heat gained by H2O: qH2O = +(4.184 J g-1 °C-1) (1000. g) (Tfinal− 23.4°C) A 200. g block of Cu at 500.°C is plunged into 1000. g of water of T = 23.4 °C in an insulated container. What will be the final equilibrium T of water and Cu? sCu = 0.385 J g-1 °C-1 sH2O = 4.184 J g-1 °C-1 -qCu = +qH2O −77.0 J/oC(Tfinal – 500oC) = +4184J/oC (Tfinal – 23.4oC) (4184 J/oC + 77.0 J/oC)Tfinal = 38500 J + 97906 J (4261 J/oC)Tfinal = 136406 J Tfinal = 32.0°C Note: Tfinal must be between Thot and Tcold based on svalues, Tfinal should be closer to H2O or Cu? 5.3 Phase Changes ∆Hfusion = - ∆Hfreezing Phase Change (State Change): A change in physical form but not the chemical identity of a substance. NO bonds are broken! 29 5.3 Enthalpies of Physical and Chemical Change Enthalpy of Fusion (∆Hfusion opposite ∆Hfreezing): amount of heat necessary to melt a substance without changing its temperature, s  l or l  s ∆Hfusion = - ∆Hfreezing +6.02 kJ/mol -6.02 kJ/mol Enthalpy of Vaporization (∆Hvap opposite ∆Hcondensation): amount of heat required to vaporize a substance without changing its temperature, l  g or g l ∆Hvaporization = - ∆Hcondensation +40.7 kJ/mol -40.7 kJ/mol Enthalpy of Sublimation (∆Hsublimation opposite ∆Hdeposition): amount of heat required to convert a substance from a solid to a gas without going through a liquid phase, s  g or g s ∆Hsublimation = - ∆Hdeposition Chapter 8/30 5.3 Energy Change Calculations • Heating and cooling q heating or cooling = m × C × ΔT   • Heat absorbed or lost in a phase change q  phase change  =Δ  phase change  H × n  moles/grams  • When matter absorbs heat, temperature increases till undergoes a phase change where the temperature remains constant – Phase changes are “isothermal” processes 31 Heating Curve for Water Heating/Cooling to next Phase, Temperature changes: ΔH = qp = mcΔT ΔH5 = q = mcΔT ΔH4 = nΔHvapor Phase Change, Temperature constant: ΔH = nΔHphase ΔH3 = q = mcΔT ΔH1 = q = mcΔT ΔH2 = nΔHfusion Use the heating curve, convert 100 g of ice at -20°C into vapor at 120°C. Step 1: Heat the ice to 0°C. ΔH = mcΔT = (100g)(2.06 Jg-1°C-1)(0-[-20]°C) = 4.1 kJ Step 2: Convert the ice to water (no change in temp!) ΔH = nΔHfus = (100g/18.02g mol-1)(6.020 kJ/mol) = 33.4 kJ Step 3:Heat the water from 0°C to 100°C. ΔH = mcΔT = (100g)(4.184 Jg-1°C-1)(100 - 0°C) = 41.8 kJ Step 4: Convert water to steam (no change in temp!). ΔH = nΔHvap = (100g/18.02 g mol-1)(40.7 kJ/mol) = 225.9 kJ Step 5: Heat steam from 100 to 120°C. ΔH = mcΔT = (100g)(1.84 Jg-1°C-1)(120 - 100°C) = 3.7 kJ TOTAL heat required = 308.9 kJ Thermochemistry 5.4 Enthalpy Changes and Chemical Reactions 34 5.4 Enthalpy and Chemical Reactions Reactants  Products ΔH = Hfinal - Hinitial Energy Enthalpy of reaction  ΔH rxn  = H products - H reactants H (Products) H (Reactants) DHrxn > 0 DHrxn < 0 H (Reactants) H (Products) Endothermic Exothermic 35 5.4 Thermochemical Equations • Similar to regular chemical equations, but contains an energy term CH4 (g ) + 2O2 (g )  CO2 (g ) + 2H2O(g ) DHrxn = 802 kJ (Exothermic) CH4 (g ) + 2O2 (g )  CO2 (g ) + 2H2O(g ) + 802 kJ • Energy is a product just like CO2 or H2O – implies conversion factors +802 kJ of Energy Released 1 mol CH 4 (g) consumed +802 kJ of Energy Released 2 mol H 2O (g) produced 36 5.4 Example - Reaction Enthalpy How many kJ of energy are released when 128.5 g of methane, CH4 (g) is combusted? CH4 (g ) + 2O2 (g )  CO2 (g ) + 2H2O(g ) DHrxno =  802 kJ/mol g  mols  J  Molar mass 128.5 g CH 4 ×  Reaction enthalpy 1 mol CH 4 1 mol rxn - 802 kJ × × = - 6.43×103 kJ 16.04 g 1 mol CH 4 1 mol rxn 37 5.4 Enthalpy and Thermochemical Equations 1. Always specify the physical states (s, l, g) of all reactants and products for DH = Hproducts – Hreactants 2. If Multiply both sides by a factor n, then DH must also change by the same factor n since DH for 1 mole only. 3. If Reverse a chemical equation, flip the roles of reactants and products and the magnitude of DH for the equation remains the same, BUT the sign changes. 38 5.4 Measuring Enthalpy Changes Conservation of Energy, 1st Law of Thermo: qreaction + qbomb + qwater = 0 Bomb Calorimeter −qreaction = qbomb + qwater qreaction = - [qbomb + qwater ] Substitute in for each q: qbomb = ccalmcalΔT = CcalΔT qwater = CwatermwaterΔT qreaction = - [CcalΔT + CH2O mH2O ΔT] Coffee-cup calorimeter- Nested styrofoam cups prevent heat transfer with the surroundings. Constant P, ΔT measured. q = qp = ΔH Assume the cups do not absorb heat, Ccal = 0 qbomb = CcalΔT = 0 J°C-1 (T2-T1)= 0 0 = qreaction + qbomb + qwater −qreaction = qbomb + qwater −qreaction = + qwater ΔH = qreaction = - qwater = -(CwatermwaterΔT ) 5.4 Exercise - Calorimeter Calculation 5.44 g of ammonium nitrate was added to 150.0 mL of water in a coffee cup calorimeter, temperature changed from 18.6oC to 16.2oC. Calculate the enthalpy change (in kJ/mol) for dissolving NH4NO3(s) in water , assume the solution has a specific heat capacity of 4.18 J/g·K Data information: Mass of reactant, C, water, and temperature change Step 1: Calculate Moles of NH4NO3 qsolution Step 3: Equation gives mole ratios (stoichiometry) Step 2: Determine qsolution = mCDT qsolution + qrxn = 0 qrxn = - q[NH4NO3(aq)] Step 4: DHr=qrxn/mol NH4NO3 Enthalpy per mole of reactant 41 5.4 Exercise: Calorimeter Calculation Given: 5.44 g NH4NO3, 150.0 mL H2O, Ti = 18.6oC, Tf= 16.2 oC Find: Moles NH4NO3; DHrxn in kJ/mol Relationships: Factors: 5.44 g NH 4 NO 3 × Solve: q solution qsoln + qrxn =0; qsoln = - qrxn qsoln = m × C × ΔT dH2O  1.00 g mL Masssoln = 5.44 g + 150.0 g 1 mol NH 4 NO 3 80.04 g = 0.0680 moles NH 4 NO 3 J = 155.44 g × 4.18 × (16.2 °C -18.6 °C) = -1.56 ×103 J g × °C q rxn = -q solution = +1.56 ×103 J Δr H = q rxn mol rxn = +1.55 ×103 J × 1 kJ 3 0.0680 moles NH 4 NO3 10 J = +22.9 kJ mol 42 Thermochemistry 5.5 Hess’s Law 43 5.5 Hess’s Law Enthalpy (DH) is a state function, change in H occurs when reactants are converted to products is the same if the reaction takes place in one step or in a series of steps. This observation is called Hess’s law. Work out Hess’ Law Problems: – Inspect the net equation and identify DH want to calculate – Identify reactants & products species, find them in the individual step reaction equations – Set-up the equations where the reactants and products are on the correct side of eqn’ that may require reversing (flipping) – Get correct number of reactants and products on each side, may require to multiply by n (changes chemical coefficients) – Make sure ALL other substances cancel out that do not appear in the overall net chemical equation (self-check!) 5.5 Exercise: Hess’s Law Calculation • Determine the DHrxn for the following reaction: 3H 2  g  + N 2  g   2NH3  g  • Given: ∆H°rxn = ? 2H2 (g ) + N2 (g )  N2H4 (g ) ΔH1 = + 95.4 kJ N2H4 (g ) + H2 (g )  2NH3 (g ) ΔH2 = – 187.6 kJ Multiple-Step Process Intermediate species, product in step 1 then is a reactant in step 2 Step 1: 2H2(g) + N2(g) N2H4(g) ∆H°1 = +95.4 kJ Step 2: N2H4(g) + H2(g) 2NH3(g) ∆H°2 = -187.6 kJ 3H2(g) + N2(g) 2NH3(g) ∆H°1+2 = -92.2 kJ 45 5.5 Hess’s Law ∆H°1 + ∆H°2 = ∆H°1+2 ∆H°1 = ∆H°1+2 - ∆H°2 ∆H°1 = -92.2 kJ - (-187.6 kJ) = +95.4 kJ 5.5 Calculate ΔH for the reaction: C(graphite) + O2(g) 2 CO(g) + O2(g) 2C(graphite) + O2(g) → 2CO(g) CO2(g) 2 CO2(g) ΔH = −393.5 kJ ΔH = −566.0 kJ Rearrange steps and multiply to get correct # species: +2[C + O2 CO2] +2(−393.5) = −787.0 kJ −1[2 CO + O2 2 CO2] −1(−566.0) = +566.0 kJ Rewrite for clarity 2 C + 2 O2 2 CO2 2 CO2 2 CO + O2 ΔH° = −787.0 kJ ΔH° = +566.0 kJ Add steps and cancel like terms: 2 C + 2 O2 + 2 CO2 2 CO2 + 2 CO + O2 ΔH° = −221.0 Write Net Equation, compare to eqn’ given in problem- self check 2 C + O2 2 CO ΔH° = −221.0 kJ Thermochemistry 5.6 Standard Heats of Reaction 48 5.6 Standard Enthalpies of Formation Standard Enthalpy of Formation (DH°f), defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states at 25oC and 1 atm DHof = 0 for an element in standard state Units are kJ/mol, follow the units! Standard Enthalpy of Reaction (DH°rxn), defined as the enthalpy of a reaction carried out under standard conditions. 49 5.6 Standard Heat of Formation • Refers to the enthalpy change for the formation of one mole of a species from its constituent elements in their most stable form – enthalpies are always reported at standard state conditions, 1 atm and 25C (298 K) • Denoted by DHfo (kJ/mol) • Standard enthalpies of formation of the most stable form of any element is zero DHf˚ (element) = 0 50 Table 5.6.1 - Standard Heat of Formation Values 51 5.6 Standard Heats of Formation Using standard heats of formation provided (Table), calculate the standard enthalpy of reaction DHorxn for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O. 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆Hrxn° = ? H° = [1 ∆H°f (C6H12O6(s))+ 6 ∆H°f (O2(g))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))] ∆H° = [(1 mol)(-1273.3 kJ/mol) + (6 mol)(0 kJ/mol) ] - [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] ∆Hrxn° = 2802.5 kJ 5.6 Consider the following reaction and calculate the energy of this reaction using Hess’s Law & formation reactions C3H8  g  + 5O2  g   3CO2  g  + 4H 2O  g  C3H8 (g )  3C(graphite) + 4H 2  g  - ΔHof = +104.7 kJ / mol 3C graphite + 3O 2  g   3CO 2  g  3× ΔH of = 3  - 393.5  kJ / mol 4H 2  g  + 2O2  g   4H 2O  g  4× ΔHof = 4  - 241.8 kJ / mol C3H8  g  + 5O2  g   3CO2  g  + 4 H 2O  g  ΔHorxn = ΔH1o + ΔHo2 + ΔH3o ΔHorxn = 3(- 393 - 5) + 4(- 241.8) - (-104.7 kJ / mol) = - 2043 kJ / mol ΔHr ° =  ΔH°f  products  -  ΔH°f  reactants  ΔH o rxn = [3× ΔH o f(CO 2 ) + 4 × ΔH o f(H 2O) ] - [ΔH o f(CH 3OH) + ΔH o f(O 2 ) ] 53
Chapter 6. Electromagnetic Radiation and the Electronic Structure of the Atom 6.1 6.2 6.3 6.4 6.5 Electromagnetic Radiation Photons and Photon Energy Atomic Line Spectra and the Bohr Model of Atomic Structure Quantum Theory of Atomic Structure Quantum Numbers, Orbitals, and Nodes Understanding physical and chemical properties of chemical compounds relies on a detailed understanding of the arrangement of electrons in atoms and molecules. This chapter begins that exploration by examining what we know about atomic electronic structure and how we know it. In this chapter, we examine the ways we learn about the electronic structure of elements. This, for the most part, involves studying how electromagnetic radiation interacts with atoms. We therefore begin with the nature of electromagnetic radiation. 2 Electromagnetic Radiation and the Electronic Structure of the Atom 6.1 Electromagnetic Radiation © 2017 Cengage Learning. All Rights Reserved. 6.1 The Nature of Light Wavelength  (lambda) is the distance between identical points on successive waves (e.g., successive peaks or successive troughs). Units are meters Frequency  (nu) is the number of waves that pass through a particular point in 1 second. Units are hertz = s-1 Amplitude is the vertical distance from the midline of a wave to the top of the peak or the bottom of the trough. 3 4 6.1 Wavelength and Frequency λ(m) × ν(s ) = c (ms ) -1 -1 c c ν = or λ = λ ν • Wavelength and frequency are inversely proportional – Long wavelength = Low frequency – Short wavelength = High frequency • Speed of light: Speed at which all electromagnetic radiation travels in a vacuum – 2.99792458  108 m/s 6.1 The Nature of Light Properties of Waves © Jim Wehtje/Getty Images 5 SAMPLE PROBLEM 6.1 One type of laser used in the treatment of vascular skin lesions is a neodymium-doped yttrium aluminum garnet or Nd:YAG laser. The wavelength commonly used in these treatments is 532 nm. What is the frequency of this radiation? Setup Solving for frequency gives  = c/. 6 7 6.2 Electromagnetic Radiation and the Electronic Structure of the Atom 6.2 Photons and Photon Energy © 2017 Cengage Learning. All Rights Reserved. 8 6.2 Photoelectric Effect • Classical theory suggests that the energy of an ejected electron should increase with an increase in light intensity -Not experimentally observed • No ejected electrons were observed until light of a certain minimum energy is applied • Minimum energy is the ionization energy of the metal • Light with a particular wavelength is directed at a piece of metal in a vacuum – total energy and the wavelength are correct, electrons are ejected from that metal • energy travels in packets called photons • implies that light has particle-like properties – energy of a single packet is related to the wavelength of the light 6.2 Quantum Theory Quantization of Energy Planck gave the name quanta to the smallest quantity of energy that can be emitted or absorbed in discrete quantities, like small packages or bundles in the form of electromagnetic radiation. Energy E of a single quantum of energy is: E = h  = hc  c= c=  h is called Planck’s constant- the frequency of the radiation. Planck’s constant h = 6.63 × 10–34 J·s/mol. Follow your UNITS 9 h (Planck’s constant) = 6.626 x 10-34 J-s/mol 11 6.2 Exercise 1: Energy of Radiation  = 685 nm E (energy) in kJ/mol Given: Find: J kJ  E in photon mol 1m hc 6.022 x 1023 photons ; E = hυ = ; 9 10 nm λ 1 mol photons Roadmap: λ in nm  λ in m  E in Equations & conversions factors: Solve: J×s 8 m 6.626x10 x 3.00x10 hc J photon s E photon = = =2.902 x 10-19 1m λ photon 685nm x 9 10 nm -34 Answer: 2.902 × 10 -19 J photons  23 6.022 x 10 photons 1 mol photons  1 kJ 3 10 J = 175 kJ mole 12 Electromagnetic Radiation and the Electronic Structure of the Atom 6.3 Atomic Line Spectra and the Bohr Model of Atomic Structure © 2017 Cengage Learning. All Rights Reserved. 6.3 Particlelike Properties of Electromagnetic Energy Niels Bohr proposed in 1914 a model of the hydrogen atom as a nucleus with an electron circling around it. Energy levels of the orbits are quantized, only certain specific orbits corresponding to specific energies of electron are available. n = 1, 2, 3, and so on. n = 1, the most stable, lowest energy state ground state n > 1 excited state e- can only travel on the orbit line, space between is a void. The Bohr Model rn = n2ao ao = Bohr radius (53 pm) 6.3 Bohr Model of the H-Atom Bohr’s model exactly predicts the H-atom spectrum. E = h  = hc  400 500 600 wavelength (nm) 700 Radius of each circular orbit in Bohr’s model depends on n2 as n increases from 1 to 2 to 3, the orbit radius increases very rapidly e- can only travel on the orbit line, space between is a void. 15 6.3 Changes in Electronic State • When a hydrogen atom absorbs a photon of light – Ground state finds electron in the 1s orbital – Electron is promoted to a higher-energy orbital (n > 1) – Photon of light causes the electron to move farther away from the nucleus • The level the electron is promoted to depends on the wavelength of light absorbed • Used to depict orbital energies – Energy of each orbital depends only on its shell and not its subshell – hydrogen atom, energy of each shell is E n = - 2.179 × 10 -18 1 J( 2) n 6.3 Bohr’s Theory of the Hydrogen Atom Line Spectra are Atomic Finger Prints! 16 6.3 Bohr’s Theory of the Hydrogen Atom Johannes Rydberg developed an equation that could calculate all the wavelengths of hydrogen’s spectral lines: DE = En+1 - En E = h  = hc  Rydberg equation,  is the wavelength of a line in the spectrum; R∞ is the Rydberg constant (1.09737316 × 107 m–1); and n1 and n2 are positive integers. 17 18 Electromagnetic Radiation and the Electronic Structure of the Atom 6.4 Quantum Theory of Atomic Structure © 2017 Cengage Learning. All Rights Reserved. 6.4 Wave Properties of Matter The de Broglie Hypothesis e- can only travel on the orbit (circumference line). nodes, do not move at all; that is, the amplitude of the wave at these points is zero = to the void between the orbits in Borh Model. 19 20 6.4 Broglie Equation E = mc 2 and E = h ν so mc 2 = hν divide both sides of the equation by c : hν = mc = p (momentum) c ν 1 since = c λ h h = m or λ = λ m Conclusion: Light waves have mass and particles have wavelength!! 6.4 Wave Properties of Matter Diffraction: e- do indeed possess wavelike properties Thomson passed a beam of e- (particles) through a thin piece of gold foil; saw concentric rings White lines are e- and dark space are the nodes (voids), space between energy levels Diffraction pattern using X-rays (waves) were used saw concentric rings. 21 6.4 Quantum Mechanics Heisenberg Uncertainty Principle (HUP): impossible to know simultaneously both the momentum (p = mDu) and the position (x) of a particle with certainty. Stated mathematically, where Dx and Du are the uncertainties in measuring the position and velocity of the particle, respectively. If you measure the position x, it’s moved, if you measured the velocity u-it’s moved. Making measurement of the u of a particle more precise means that the x must become correspondingly less precise. If the x of the particle is known more precisely, its u measurement must become less precise. 22 23 Schrödinger Equation and Wave Functions • Erwin Schrödinger proposed that matter can be described as a wave • electron is treated as both a wave and a particle – An electron is described by a wave function • Wave function (): Equation that predicts the energy of an electron and the regions in space where an electron is most likely to be found • Probability Wave function (2): Square of the equation that gives boundaries to the predicted energy of an electron and the regions in space where an electron is most likely to be found 6.4 Quantum Mechanics 2 = probability of finding an e- at a point in space. An electron density (probability) map plots 2 for each point in space. Bigger value = darker shade = greater e- density Each  describes a different energy level known as an atomic orbital (AO) e- do not follow fixed orbits around the nucleus, kept in confined areas with boundaries (like a fence). 6.4 26 Electromagnetic Radiation and the Electronic Structure of the Atom 6.5 Quantum Numbers, Orbitals, and Nodes © 2017 Cengage Learning. All Rights Reserved. 6.5 Wave Functions and Quantum Numbers Wave function solve Wave or atomic equation orbital ( ) Probability of finding electron in a region of space ( 2) A wave function ( ), describes distribution of edensity is characterized by three parameters called- quantum numbers: n, l, ml An electron description is characterized by four parameters calledquantum numbers: n, l, ml, ms 6.5 Wave Functions and Quantum Numbers Principal Quantum Number (n) • Describes the size and energy level of the orbital (shell) • Positive integer (n = 1, 2, 3, 4, … ) Angular-Momentum Quantum Number (l) • Defines the 3-D shape of the orbital (subshell) • l = n -1 ….0 l = 0 s (sharp) l = 1 p (principal) l = 2 d (diffuse) l = 3 f (fundamental) Magnetic Quantum Number (ml ) • Defines the spatial orientation (x-, y-, z-axes) of the orbital • ml values are 2l + 1 ranging from -l …0…+l 29 6.5 Orbital Shapes • Each boundary surface contains the volume in space where an electron is most likely to be found – Surfaces contain regions of high electron density Atomic orbitals (boundary surfaces, n = 1–3) 30 6.5 • • • • Nodes Areas where there is no electron density The wave energy is canceled out at the node Crosses the axes Types – Planar or angular nodes – Radial or spherical nodes Radial nodes in the 2s and 3s orbitals 31 6.5 Determining the Type of Orbital • 1st decide on the shape to determine the ℓ value – example, shape of a dumbbell, it is a p orbital • 2nd Count up the total number of nodes (planar and spherical) and add one to determine the energy level – example, if there is one planar node and two spherical nodes, then 3+1 = 4 = 4p orbital 6.5 The Shape of S-Orbital S orbital (sphere) – probability of finding an s electron depends only distance from nucleus, not on direction s subshell in every shell for which n  1 1 s-orbital l=0 # e- = 2 2l + 1 ml values: 2(0)+ 1 = 1 ml = 0 6.5 The Shapes of P-Orbitals p subshell in every shell for which n  2 3 p-orbitals l=1 # e- = 6 px, py ,pz 2l + 1 ml values: 2(1)+ 1 = 3 ml = -1, 0, +1 6.5 ml = -2 The Shapes of d-Orbitals ml = -1 ml = 0 ml = +1 ml = +2 d subshell in every shell for which n  3 5 d-orbitals l=2 # e- = 10 2l + 1 ml values: 2(2)+ 1 = 5 ml = -2, -1, 0, +1, +2 34 6.5 35 6.5 37 6.5 Electrons in Orbitals • High energy electrons – Have a higher n value – Are farther from the nucleus – Have small negative energy (close to zero) • Low energy electrons – Have a lower n value – Are closer to the nucleus – Have large negative energy 38 6.5 Changes in Electronic State • When a hydrogen atom absorbs a photon of light – Ground state finds electron in the 1s orbital – Electron is promoted to a higher-energy orbital (n > 1) – Photon of light causes the electron to move farther away from the nucleus • The level the electron is promoted to depends on the wavelength of light absorbed • Used to depict orbital energies – Energy of each orbital depends only on its shell and not its subshell – hydrogen atom, energy of each shell is E n = - 2.179 × 10 -18 1 J( 2) n 39 6.5 Interactive Figure 6.5.5 - Add Energy to a Hydrogen Atom • Shows electron promoted from a 1s orbital to 3px orbital when a photon with wavelength of 102.6 nm is absorbed A hydrogen atom absorbs energy

Tutor Answer

Marrie
School: Duke University

Attached.

INSERT SURNAME HERE 1
Article Writing
Institutional Affiliation
Date
What is Oxidation? In the chemical reactions chemistry concept, the two major kinds of
biochemical response of reactions are reduction and oxidation. Oxidation is indeed not
associated with oxygen. Oxidation is the occurrence where there is electrons loss by ion,
molecule or an atom at the time of a reaction. Oxidation takes place when a molecule, ion or
atom oxidation condition goes up. The adverse process is known as reduction. An ancient
definition of oxidation was during oxygen addition to a compound as oxygen gas was initially
known as an oxidation fac...

flag Report DMCA
Review

Anonymous
Thanks, good work

Similar Questions
Hot Questions
Related Tags
Study Guides

Brown University





1271 Tutors

California Institute of Technology




2131 Tutors

Carnegie Mellon University




982 Tutors

Columbia University





1256 Tutors

Dartmouth University





2113 Tutors

Emory University





2279 Tutors

Harvard University





599 Tutors

Massachusetts Institute of Technology



2319 Tutors

New York University





1645 Tutors

Notre Dam University





1911 Tutors

Oklahoma University





2122 Tutors

Pennsylvania State University





932 Tutors

Princeton University





1211 Tutors

Stanford University





983 Tutors

University of California





1282 Tutors

Oxford University





123 Tutors

Yale University





2325 Tutors