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CHE 110 Dr. Nicholas Bizier Office DS 337b email: [email protected] What is General Chemistry about? Chemistry = Study of properties and changes in matter Matter= anything that takes up space Learn to describe states of matter, both in words and mathematically You will learn to predict properties of matter You will learn how to describe and predict chemical reactions that take place. You will learn about the energy associated with chemical reactions You will be introduced to the periodic table. How is Science Done? Hypothesis- a proposed explanation The Hypothesis is then tested. From this type of testing you get two types of data: Qualitative Data- nonnumerical data Quantitative Data-obtained from measurements to produce numeric data. Theory-A Hypothesis that has not been disproven Substance Substance= matter that has a fixed composition and distinct properties These properties are divided into two categories: 1. Physical and Chemical Physical properties- are properties that can be measured without changing the basic identity of the substance. Examples of this: color, m.p. b.p. density Let’s first look at physical properties because they are things we can visualize and measure Geometric shape area and volume formula’s you must memorize Area of rectangle= l x w, Volume of a cube= l x w x h Cylinder = area of base = πr2, volume= area of base x h Volume of a Sphere= 4/3 πr3 Greek Prefixes Modifying the units of measurement are Greek Prefixes The ones you will have to be comfortable with are: mega 106 kilo 103 centi 10 2 milli 10 3 micro 10 6 Density Density is how much matter (expressed in mass) is found in a given volume The mathematical expression for this: Typically this value is expressed as Density mass / volumne D g / ml or D g / cm3 Units of Measurement A closer look at the last equations D g / ml or D g / cm3 Weights: typically given in grams (scientific measurement) Volume in liters While not in the last equation: Temp in degree’s Celsius Dimensional Analysis Using a conversion factor-fraction whose numerator and denominator express the same quantity given in different units. Ex: Butane has a density of 0.579 g/mL or 1mL/0.579g What is the volume of 5.25g of butane. 1 ml 5.25 g 9.07 ml 0.579 g This type of calculation is an example of a physical property because we can measure both volume and mass without changing the substance. Chemical Properties Chemical Properties describe the ways a substance reacts to form other substances. Ex. Flammability (oxygen combines with what you are burning to give off heat and gas) During a chemical reaction the substance(s) which are called the (reactants) are transformed into different substances(s) (products). Mixture Combination of 2 or more substances. Two types: (1) heterogeneous- compositions in different pats of the sample. (2) homogeneous-uniform compositions throughout the sample One use various means to separate mixtures into pure substances. Ex. Filtration or chromatography Significant Figures Rules: (1) All nonzero numbers are significant: 376.8 (4 sig. figs), 921 (3 sig. figs) (2) Zeros between two nonzero numbers are significant: 1004 (4 sig. fig), 20.025 (5 sig. fig). (3)Zeros to the right of a decimal point are significant. (4) Zeros at the end of number and to the left of a decimal point, change to scientific notation. 200 2.00 x 102 (3 sig. figs ) 2.0 x 102 (2 sig. figs ) Preforming Math with Sig. Figs (1) Addition and Subtraction-the final answer should have as may digits beyond the decimal point as the number with the fewest digits beyond the decimal. 100.09 20.1 75.989 (196.179) 196.2 Multiplication and Division-the final answer should have as many sig. figs as the number with the fewest sig. fig. 4 459 92.91 (42645.69) 4.26 x 10 3.58 x 104 6.84 5231 Examples using the principles covered thus far How many liters are there in 10.0 gal of water? 4 qt 0.946 L 10.0 gal 37.8L 1 gal 1 qt One edge of a cube is measured and found to be 13 cm. What is the volume of the cube in m3? 1m 3 3 3 13 cm (0.13 m ) 0.002197 2.2 x 10 m 100 cm Policy on Sig Figs in Class vs lab You will follow instructions on quizzes and test. All answer will be 3 sig figs. (You will be reminded of this on each test and quiz at the top). You will follow your lab instructor directions regarding what you turn in for lab. Any question regarding any assignment in lab will be directed directly to the lab instructor. Treat this as two separate courses. Theory in class, practicle application in lab. Examples continued. Alcohol has a density of 0.76 g/ml. How many grams of alcohol would it take to fill a 2 fluid ounce shot glass? 32 oz 1 qt 1.057 qt 1 L mL L qt oz 1L 1.057qt 32 oz 2 1mL x x x 3.4 x 10 oz 1000 mL 1l 1 qt 0.78 g x 2 45.9 g 46 g 2 3.4 x oz Percent by Mass This is a type of measurement which tells you what the weight of substance out the total mass present. weight x precent by mass x 100 total weight Ex, what is percent by mass of sucrose in a water solution add 15 g of sucrose to 200 grams of water. 15 g ( sucrose) percent by weight sucrose x100 7.0% 200 g ( water ) 15 g ( sucrose) Make up off Atoms Atoms are made up of three different types subatomic particles: Protons which have positive charge and are in the nucleus Neutrons which have no charge and are also in the nucleus Electrons which are negatively charged, which are found in diffuse orbitals outside of the nucleolus. Make up of atoms cont. The number of protons in an atom is the atomic number and this determines the identity of the atom. The number of protons and neutrons is equal to the atomic mass. Atoms made of same element must have the same number of protons, but can have different numbers of neutrons Two atoms with same number of protons, but different number of neutrons are called isotopes of one another. Elemental symbol The elemental symbols used in the periodic table can have different amounts of information on them: Ex: 12 C 6 protons, 6 electrons, and 6 neutrons 14 6 C 6 protons, 6 electrons, and 8 neutrons These two types of carbon atoms are isotopes of one another Average Atomic weight: When you look at the periodic table You will find decimal numbers for atomic weights. These represent a weighted average of the naturally occurring isotopes. total wt. sample of x atoms AWaverage total number of atoms x in sample no. atoms isotope %isotope total no. atoms % #1 %#2 AWaverage xAW isotope #1 xAWisotope #2 .... 100 100 Osmium (Os) as an example of Average Atomic weight Osmium has atomic number 76 (# of protons) It has 6 stable isotopes Isotope %Abundance So the equation for the atomic weight of Osmium would be: 186Os 1.6 187Os 1.6 188Os 189Os 13.4 16.2 190Os 192Os 26.4 41.1 1.6 1.6 13.4 16.2 26.4 41.1 193.2 amu x 186 amu x 187 amu x 188 amu x 189 amu x 190 amu x 192 amu 100 100 100 100 100 100 This comcept can also be integrated with mass: 100 g of naturally occuring Osmiun would cotain 41.1 grams of 192Os Finding an unknown isotope Chlorine consists of a mixture of 35Cl (34.97 amu) and 37Cl (36.97 amu). Given the average atomic weight for Cl on the periodic table (35.45 amu), what is the relative % abundance of each isotope in an average sample? 𝑥 100 − 𝑥 35 𝑎𝑚𝑢 + 37 𝑎𝑚𝑢 100 100 3545 𝑎𝑚𝑢 = 35 𝑥 𝑎𝑚𝑢 + 3700 − 37𝑥 𝑎𝑚𝑢 𝑥 = 37.78% 𝑤ℎ𝑖𝑙𝑒 100 − 𝑥 = 62.22 % 35.45 𝑎𝑚𝑢 = Weights of subatomic particles 1 amu (atomic mass unit)= 1.66054 x 10-24 g 1 g =6.02 x 1023 amu Proton = 1.0073 amu Neutron = 1.0087 amu Electron = 5.486-4 amu To find the atomic mass of an element = add the number of protons and neutrons together. Finding the Formula weight/Molecular weight The molecular formula is a list of the identity and number of atoms that make up molecule: Ex. Ethanol: CH3CH2OH 12.01 amu 16.00 amu 1.01 amu 2C x 6 H 1 O 78.01amu 1O 1H 1H Formula weight is similar to molecular weight, however it used when referring to ionic compounds Ex. FeCl3 Note we use the term term formula weight, because ions are discreet molecules, however the formula weight is how the ions add together to form a neutral speices. 55.85 amu 35.5 amu 1 Fe x 3 Cl 162.4amu 1 Fe 1Cl Concept of Moles A mole is the amount of matter that contains as many objects as the number of atoms in 12 g of 12C. It is just a collection of 6.02 x 1023 objects just as a dozen is a collection of 12 objects. Avogadro’s number 6.02x 1023. A mole is the connection between nanoscale and macroscale The Molar Mass (g)= mass of 1 mol of substance. Ex. FeCl3 = 162.4 amu, on mole of FeCl3 weighs 162.4 grams FeCl3= 162.4 grams/mole Using the concept of moles How many moles of COCl2 (Phosgene) are in 352g? 12.01g 16.00 g 35.5 g 98.91 g 1C 1 O 2 Cl COCl2 mol mol mol mol 1 mol 352 g x 3.56 mol COCl2 98.91 g How many grams of PBr3 are in 0.789 mol? 30.97 g 79.90 g 270.7 g 1P PBr3 3Br mol mol mol 270.7 g 0.789 mol 217 g PBr3 mol Mass→ Moles→ Molecules How many molecules of CH4(methane) are in 48.2 g? 12.01g 1.008 g 16.04 g 1C 4 H mol mol mol 1 mol 48.2 g 3.00 mol CH 4 16.04 g 6.02 x1023 molecules 24 3.00 mol CH 4 x 1.806 x 10 molecules of CH 4 mol Molecules and Ions There are two types of bonding, molecular and ionic Molecular bonding involves sharing of electrons between two nonmetals Molecular formula-indicates the actual numbers and types of atoms in a molecule: Ex. H2O2, C2H4 Empirical formula-a relative number of atoms in a molecule Ex. HO, CH2 Structural formula-a pictorial representation of a molecular formula Organic compounds that contain carbon and hydrogen and usally oxygen, nitrogen, and sulfur Naming (non-ionic) Inorganic compounds For Binary compounds (1) Element having more positive character is named first. (2) Element having more negative character is named second, and ide is added at the end of the element. (3) Use prefixes (mon (1),di (2), tri (3), tetra (4), pent (5), hexa (6), hepta (7), octa (8), nona (9), deca (10) to denoted how many elements there are. E.x N2O5= dinitrogen pentoxide, CO= carbon monoxide, CO2carbon dioxide Strong Acids There are six inorganic molecules you need to know by site and name. These are strong acids (something we will expand upon later). HCl= Hydrochloric acid HBr= Hydrobromic acid HI = Hydroiodic acid H2SO4= Sulfuric acid HNO3= Nitric acid HClO4= Perchloric acid One weak acid There is one weak organic acid you will need to know by site and name: CH3CO2H = acetic acid Inorganic Compounds that don’t contain metals or charged species These are molecules that have covalent bonds but contain no carbon-hydrogen bonds Ex. N2O5, CO,CO2, HCl Ionic Compounds These compounds are made of a combination of cations (positively charged species which have lost electrons) and anions (negatively changed species which have gained electrons. Most of the time they are made up of a positively charged metal and a negatively charged non-metal, or group of non-metals which are called poly atomic anions. Simple Ionic compound ex: NaCl, MgBr2 Metal and Polyatomic ion examples: NaNO3 (made of Na+ and NO3- ) CaSO4 (made of Ca+2 and SO4-2) Na2SO3 (made of 2 Na+ and SO3-2) What are metals, non-metals, and metalloids. You will be expected to know what each one is by sight For ionic compounds: Metals=form cations (+) Nonmetals=form anions (-) Metalloids form both (you have to look at what it is bonded to and make a judgement call) Charge states of certain groups Example to try Give an element that will react with magnesium to form an ionic compound with general formula (Mg+2)3 (X+3)2 X= N or P Naming inorganic compounds Very important to do this stepwise and pay attention to what is the cation and what is the anion Step 1: Cation A. Monatomic cation: Na+ = Sodium, Ca+2 = calcium B. This Elements that can form more than one positive ion-transition metals- use Roman numeral C. Cu+- copper(1) ion written as Cu(I), Cu+2- copper(2) written as Cu(II) D. Name the anion with -ide at the end: Cl-=chloride, Br-= bromide N-3= nitride Naming inorganic compounds Put the two ions together: NaCl = sodium chloride, Copper (II) Bromide Cu(II)Br2 Note figuring out what the roman numeral on is for one of the transition metals can be accomplished by noting the charge on the anion, and the number. The charge on the anion multiplied by the number of atoms tells you what the roman numeral is on the transition metal cation. List of poly atomic ions: NO3- = nitrate H3O+ = hydronium NO2- = nitrite NH4+ = ammonium SO42- = sulfate CrO42- = chromate SO32- = sulfite Cr2O72- = dichromate PO43- = phosphate MnO4- = permanganate CO32- = carbonates BO3- = borate HCO3- = hydrogen carbonate SiO4- = silicate Cl- = chloride OH- = hydroxide All of polyatomic ions in white will be in the place of the non-metal of a simple binary compound, because they are anions. The ones in red are cations, and will take the place of the metal in a simple binary compound. Naming with Polyatomic cations Name the simple cation or anion as you would a typical binary cation, and then use the name of polyatomic ion in its place. Barium hydroxide= Ba+2 + 2OH- → Ba(OH)2. Ammonium Iodide= NH4+ + I- → NH4I Copper(II) Sulfate= Cu+2 + SO42- → Cu(ii)SO4 Which of the following are ionic substances and which are molecular? (a) CaO (b) SiCl4 (c) Mg(NO3)2 (d) NOCl (e) B2H6 (f) CH3OH (g) Ag2SO4 Percent Composition This calculation tells what percent the weight of a type of atom in a compound or molecule is of the whole (in terms of weight) (# of atoms ) x ( AW ) % x 100 FW What is the % composition of each type of atom in glucose (C6H12O6)? Step 1 calculate MW 6 x C (12.01amu ) 72.06amu 12 x H (1.008amu ) 12.12amu 96.00amu 6 x O (16.00amu ) 180.19amu % comp Glucose continued (6)(12.01 amu ) %C x 100 40.00% 180.2 amu (12)(1.01 amu ) %H x 100 6.73% 180.2 amu 6(16.00 amu ) %O x 100 53.28% 180.2 amu Calculation of Empirical and Molecular Formulas Molecular formula-indicates the actual numbers and types of atoms in a molecule Empirical formula-indicates the ratio of the types of atoms to one another Glucose Molecular C6H12O6 Empirical CH2O Combustion Analysis Combustion analysis is a process that determines the empirical formula of molecules by burning them in the presence of excess oxygen. The process involves placing a sample of know weight, in the analyzer and recording the weight of both CO2 and H2O which is emitted from the machine. From this information you can calculate the amount of carbon and hydrogen in the sample. However since oxygen is in excess you must find oxygen through indirect means (the mass comes from what is not accounted for by carbon and hydrogen, in a sample that only contains CHO). The information you get from this analysis provides you with the empirical formula of the molecule. If there are other elements present in the molecule, such nitrogen or chlorine, you make get information by decomposition analysis (same process as described for combustion, however a different term is used when it is not carbon or hydrogen. Combustion Analysis of Vitamin C If 0.200 g of Vitamin C are subject to combustion analysis, the mass of CO2 and water found is 0.2889 g and 0.0819 g respectively. What is the Empirical formula. Step 1: First determine the mass of carbon in 0.2998g of CO2, by converting to mol C. Then to grams of C. 1 mol CO2 1 molC 0.2998 mol CO2 x x 0.006812 mol C 44.01 g CO2 1 mol CO2 12.011 g C 0.006812 g C x 0.0812 g C 1 mol C Determining H in Vitamin C Similar fashion for finding H2O. 1 mol H 2O 2 mol H 0.819 g x H 2O x x 0.00909 mol H 18.02 g H 2O 1 mol H 2O 1.008 g H 0.00909 mol H x 0.00916 g H 1 mol H Determining O in Vitamin C and percent composition of elements Weight of original sample minus the weight of carbon and hydrogen (0.200g –(0.08182 g C + 0.00916 g H) = 0.1090 g O 0.08182 g C x 100 % 40.91% C 0.2000 g sample 0.00916 g H %H x 100% 4.58% H 0.2000 g sample 0.1090 g O %O x 100 % 54.50 % O 0.2000 g sample 1 mol O ? mol O 0.1009 g O x 0.006813 mol O 15.999 g O 1 C0.006812 H 0.0909O0.006813 x CH1.33O C3 H 4O3 0.006812 %C Combustion and Decomposition While combustion analysis only gives information about carbon hydrogen and oxygen indirectly, you can get information about other elements from decomposition analysis. In these types of problems you will have combustion analysis data from one sample with a specific weight, and decomposition data from another sample with a different weight. Lysine Example Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 2.175 g of lysine was combusted to produce 3.94 g of CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine was burned to produce 0.436 g of NH2. The molar mass of lysine is 150 g/mol. Determine the empirical and molecular formula of lysine. Step One: determine the mass of each element present. Carbon: 3.94 g x (12.011 / 44.0098) = 1.0753 g Hydrogen: 1.89 g x (2.016 / 18.0152) = 0.2115 g Nitrogen: 0.436 g x (14.007 / 16.023) = 0.38114 g Oxygen: cannot yet be done Why can't the oxygen be determined yet? It is because our C, H, and N data come from TWO different sources. Lysine continued Step Two: Convert mass of each element to percentages. Carbon: 1.0753 g ÷ 2.175 g = 49.44 % Hydrogen: 0.2115 g ÷ 2.175 g = 9.72 % Nitrogen: 0.38114 g ÷ 1.873 g = 19.17 % Oxygen: 100 - (49.44 + 9.7 + 19.17) = 21.67 % Step Three: Determine the moles of each element present in 100 g of lysine. Carbon: 49.44 g ÷ 12.011 g/mol = 4.116 mol Hydrogen: 9.72 g ÷ 1.008 g/mol = 9.643 mol Nitrogen: 19.17 g ÷ 14.007 g/mol = 1.3686 mol Oxygen: 21.67 g ÷ 15.9994 g/mol = 1.3544 mol Lysine continued Step Four: find the ratio of molar amounts, expressed in smallest, whole numbers. Carbon: 4.115 mol ÷ 1.3544 mol = 3.04 Hydrogen: 9.643 mol ÷ 1.3544 mol = 7.12 Nitrogen: 1.3686 mol ÷ 1.3544 mol = 1.01 Oxygen: 1.3544 mol ÷ 1.3544 mol = 1 The empirical formula is C3H7NO. In order to determine the molecular formula, we need to know the "empirical formula weight." This value is 73.1. We see that the approximate molecular weight is just about double this value, leading to the molecular formula of C6H14N2O2 Molarity Unit of concentration. M is used to show this value. Molarity = moles of solute/volume of soln in liters(l) This unit of measurement is useful when you are given a volume and concentration of substance in a solvent (this solvent will be assumed to be water unless specified). Example calculations Calculate the molarity of the solution prepared by dissolving 62.3 g of sucrose C12H22O11 in enough water to form 0.500l of solution. Grams → moles → molarity 1 mol sucrose 1 62.3 g sucrose x x 0.364M sucrose 324.34 g sucorse 0.500 L How would you prepare, in grams, 250ml of 0.600M aq. KBr solution? 1l 0.600 mol KBr 119.00 g KBr 250 ml x x x 18.0 g KBr 1l 1000 ml 1 mol KBr Examples continued. Calculate the molarity of chloride ions in 7.68g of CaCl2 which was dissolved in water to make a 250ml solution. CaCl2(aq) → Ca2+(aq) + 2Cl-(aq) G CaCl2 → moles CaCl2 → moles Cl- → molarity of Cl- 1 mole CaCl2 2 moles Cl 1 1 7.68 g CaCl2 x x x 0.554 M Cl 110.98 g CaCl2 1 mole CaCl2 0.250 l Dilution Taking a concentrated solution to a diluted solution: M1V1= M2V2 How much 5.00M HCl would be need to make 250ml of 0.350M HCl? Vconc HCl (0.350 M HCl ) x0.250 ml 18.0 ml HCl 5.00 M HCl Stoichiometry Stoichiometry-calculation of the quantifies of elements or compounds involved in a chemical reaction Atoms are neither created nor destroyed during a chemical reaction Chemical reaction H2(g)+ O2(g)→ H2O(l) States g=gas, l=liquid, s-solid, aq=aqueous substance dissolved in water Balanced equation 2H2(g)+ O2(g)→ 2H2O(l) Rules for balancing chemical equations (1) The chemical formulas of the reactants and products are fixed and can’t be altered by changing the subscripts. Only change the coefficients in front of the formulas. (2) Balance elements contained in only one substance first (3) Balance H and O atoms last. Ex. H2SO4(aq) + KOH(aq)→ K2SO4(aq) + H2O(l) H2SO4(aq) + 2KOH(aq)→ K2SO4(aq) + 2H2O(l) Ex. AgClO3(aq)+ CaBr2(aq) → AgBr(s) + Ca(ClO3)2(aq) 2AgClO3(aq)+ CaBr2(aq) → 2AgBr(s) + Ca(ClO3)2(aq) Molecular reactions are balanced the same way Ex. C3H8(g) + O2(g)→ CO(g)+ H2O(g) 2C3H8(g) + 7O2(g)→ 6CO(g)+ 8H2O(g) Types of chemical reactions (1)Combination reactions=two or more substances react to form one product. Ex. 2H2(g) + O2(g)→ 2H2O(l) (2) Decomposition reactions= substance react to form two or more different products. Ex. BaCO3(s) → BaO(s) + CO2(g) (3)Displacement reactions= Single element reacts with a compound to form a new product and new element. Ex. 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) (4) Exchange reactions= exchanging the partners between two compounds. The driving force is the removal of a solid, molecule or gas from solution. Ex Na2SO4(aq) + PbCl2(aq)→ 2NaCl(aq) + PbSO4(s) Coefficients Coefficients in a chemical eq. can be interpreted as numbers of molecules at a nanoscale level. Ex. 3H2(g) + N2(g)→ 2NH3(g) 3 molecules of H2 reacts with 1 molecule of N2 to produce 2 molecules of NH3 When then can use Avagadro’s number to bring it to macroscale: 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3 Use of mass (in grams) of reactants and chemical equations If you are given grams of substance you can use moles to predict the moles of product and ultimately the grams of the product Grams of reactant → moles of reactant →use coefficient multipliers → moles of product → grams of products Consider the following equation: 3H2(g) + N2(g) → 2NH3(g) How many moles of NH3(g) are produced from 10.0 moles of N2(g)? 2 mol NH 3 10.0 mol N 2 x 20.0 mol NH 3 1 mol N 2 Stoichiometry cont. 3H2(g) + N2(g) → 2NH3(g) (2) How many grams of NH3 are produced from 280 g of N2? Grams of reactant → moles of reactant →use coefficient multipliers → moles of product → grams of products 1 mol N 2 2 mol NH 3 17.04 g NH 3 280 g N 2 x 340 g NH 3 x x 28.02 g N 2 1 mol N 2 1 mol NH 3 Stoichiometry cont. Consider the combustion of propane in oxygen Chemical equation is: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) How many grams of O2 are required to burn 75.0 g of C3H8? g C3H8(g) → moles C3H8(g) → moles O2(g) → grams O2(g) 1 mol C3 H 8 5 mol O2 32.00 g O2 75.0 g x x x 272 g O2 44.11 g C3 H 8 1 mol C3 H 8 1 mol O2 Stoichiometry cont. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) How many grams of H2O(l) and CO2(g) are produced? 4molH 2O 18.02 gH 2O 1.70molC3 H 8 x x 122 gH 2O 1molC3 H 8 1molH 2O 3molCO2 44.01gCO2 1.70molC3 H 8 x x 224 gCO2 1molC3 H 8 1molCO2 Limiting reagents The limiting reagent is the reactant which is consumed completely and therefore limits the amount of product formed. (other reactants are in excess taking into account stoichiometry) How many moles of magnesium oxide are produced by the reaction 3.82g of magnesium nitride with 7.73g of water? Mg3N2(s) + 3 H2O(l) → 2NH3(g) + 3MgO(s) Step 1: find moles of each reactant 1 mol Mg3 N 2 3.82 g Mg3 N 2 x 0.0378 mol Mg 3 N 3 100.9 g Mg3 N 2 1 mol H 2O 7.73g H 2O x 0.429 mol H 2O 18.02 g H 2O Limiting reagent Step 2: Determine # of moles of product produced from each. 3 mols MgO 0.0378 mol Mg3 N 2 x 0.113 mol MgO 1 mol Mg3 N 2 3 mols MgO 0.429 mol H 2O x 0.429 mol MgO 3 mols H 2O Step 3: Which ever reactant produces the smallest # of moles of product is the limiting reagent. 40.3g MgO 0.113 mol MgO x 4.55 g MgO 1 mol MgO Limiting reagent (what’s left over) How much excess reactant is left in the reaction from the previous slide? Step 1: Find moles of excess reagent used: 3 mols H 2O 0.133 mol MgO x 0.113 mols H 2O 2 mols MgO 0.429 moles H 2O( started with) 0.113 mols H 2O (used ) 0.316 mols H 2O (leftover ) More examples A 2.00g sample of Fe3O4(s) is reacted with 7.50 g of O2(g) to produce Fe2O3(s). Calculate the number of grams of Fe2O3(s) produced. 4Fe3O4(s) + O2(g)→ 6Fe2O3(s) 1 mols Fe3O4 3 2.00 g Fe3O4 x 8.64 x10 mols Fe3O4 231.54 g Fe3O4 1 mols O2 7.50 g O2 x 0.234 mols O2 32.00 g O 2 6 mols Fe2O3 8.64 x103 mols Fe3O4 x 0.0130 mols Fe2O3 4 mols Fe O 3 4 6 mols Fe2O3 0.234 mols O2 x 1.40 moles O2 1 mols O 2 159.7 g Fe2O3 0.0130 mols Fe2O3 x 2.08 g Fe2O3 1 mol Fe O 2 3 Iron example cont. 4Fe3O4(s) + O2(g)→ 6Fe2O3(s) How much excess reagent is left over? 1 mols O2 3 0.0130 mols Fe2O3 x 2.17 x 10 mols O2 used 6 mols Fe2O3 0.234 mols O2 ( started with) 2.17 x103 mols O2 (used ) 0.23 mols O2 excess More examples A mixture is prepared from 25.0 g of aluminum and 85.0 g of Fe2O3(s). How much iron is produced? Fe2O3(s) + 2Al(s)→ Al2O3(s) + 2Fe(l) How much iron is produced in the reaction in grams? 1 mol Al 25.0 g Al x 0.927 mols Al 26.98 g Al 1 mol Fe2O3 85.0 g Fe2O3 x 0.532 mol Fe2O3 159.7 g Fe O 2 3 2 mol Fe 0.927 mol Al x 0.927 mol Fe 2 mol Al 2 mol Fe 0.523 mol Fe2O3 x 1.06 mol Fe 1 mol Fe O 2 3 55.87 g Fe 0.927 mol Fe x 51.8 g Fe 1 mol Fe Al plus Fe cont. How much excess reagent is there? Fe2O3(s) + 2Al(s)→ Al2O3(s) + 2Fe(l) 1 mol Fe2O3 0.927 mol Fe x 0.463 mol Fe2O3 used 2 mol Fe 0.532 mol ( started with) 0.463 mol (used ) 0.069 excess Fe2O3 Theoretical Yield/Actual Yield Theoretical Yield= calculated amount of product to form when all of the limiting reagent reacts Actual Yield= amount of product that is actually obtained actual yield % yield x100 theoretical yield Sample Yield problem If the reaction of 3.50 g of magnesium nitride with 7.00 g of water produced 3.60g of magnesium oxide, what is the percent yield of the reaction? Mg3N2(s) + 3 H2O(l) → 2NH3(g) + 3MgO(s) 1 mol Mg3 N 2 3.50 g Mg3 N 2 x 0.0347 mol Mg3 N 2 100.92 g Mg3 N 2 1 mol H 2O 7.00 g H 2O x 0.388 mol H 2O 18.02 g H O 2 3 mol MgO 0.0347 mol Mg3 N 2 x 0.104 mol MgO 1 mol Mg3 N 2 3 mol MgO 0.388 mol H 2O x 0.388 mol MgO 3 mol H 2O Yield continued 40.30 g MgO 0.104 mol MgO x 4.19 g MgO theoretical yield 1 mol MgO actual 3.60 g MgO x 100 x 100 85.4% theoretical 4.19 g MgO Net Ionic equations A solution is homogenous mixture of 2 or more substances. An aqueous solution is a mixture where water is the solvent Solvent-component of a solution present in the greatest quantity Solute-all other components Electrolytes-solutes that exist as ions in solution and conduct electricity. NaCl(s)→Na+(aq) + Cl-(aq) Strong electrolytes-completely ionize in solution (salts). Ex. NaCl Weak electrolytes-only partially ionize in solution. HC2H3O2(aq) H+(aq) + C2H3O2- (aq) Non-Electrolytes Nonelectrolytes-molecular substances that do not form ions in solution. Ex. Sugar C12H22O11 Precipitation reactions Positive and negative ions exchange partners. Driving force is the removal of ions from solution. There are three general types of these reactions (1) Formation of a precipitate (solid) (2) Formation of a weak or nonelectrolyte (Acid-base reactions fall in this category (3) Formation of a gas Solubility rules Examples NaCl(aq) + AgNO3(aq) NaCl(aq) + AgNO3(aq)→ NaNO3(aq) + AqCl(s) Pb(NO3)2(aq)+ 2KBr(aq) Pb(NO3)2(aq)+ 2KBr(aq) → PbBr2(s) + 2KNO3(aq) Writing total ionic and Net ionic equations A total ionic equation shows all molecules and ions (drawn separately as cation and anions) on both sides of the equation A net ionic equation leaves out the spectator ions=ions that appear in the identical forms on both side of the equation. HCN(aq)+ KOH(aq)→KCN(aq) +H2O(l) Total: HCN(aq)+ K+(aq) + OH-(aq) → K+(aq) + CN-(aq) + H2O(l) Net Ionic: HCN(aq)+ OH-(aq) → CN-(aq) + H2O(l) Gas Formation H2CO3(aq) → H2O(l) +CO2(g) FeCO3(s) + 2HCl(aq) → H2CO3(aq) + FeCl2(aq), (H2O(l) + CO2(g) from the breakdown of H2CO3(aq)) Total: FeCO3(s) + 2H+(aq) +2Cl-(aq)→ H2O(l) +CO2(g) + Fe2+(aq) + 2Cl-(aq) Net: FeCO3(s) + 2H+(aq) → H2O(l) +CO2(g) + Fe2+(aq) Oxidation/Reduction Reactions Oxidation=loss of e-1 by a substance. Charge on the atom increase Reduction=grain of e-1 by a substance. Charge on the atom decreases. Oxidation Numbers: looks at the charge on an atom if it wasn’t in a compound. Rules for assigning oxidation numbers: (1) The oxidation of an atom of a pure element is zero. (2) The oxidation number of a monoatomic ion equals its charge. (3) Some elements have the same oxidation numbers in all of their compounds: (a) hydrogen is +1 except when it’s combined with a metal, then it’s -1. (b) fluorine is -1. (c) oxygen is -2 expect when it’s in a peroxide like H2O2 when it is -1. Oxidation numbers cont. The sum of the oxidation numbers in a neutral compound is 0. The sum of the charges in a polyatomic ion equals the charge on the ion. 2H2(g) + O2(g) → 2H2O(g) 2H2(g) → 0 to +1 losing electrons-oxidized (reducing agent) O2 → 0 to -2 gaining electrons-reduced (oxidizing agent) 0 0 2(+1) -2 Assigning oxidation states cont. Fe2S3 + 12HNO3 → 2Fe(NO3)3 + S3 + 6NO2 + 6H2O 2(+3) 3(-2) (+1),(+5),3(-2) (+3),(+5),3(-2)X3 0 +4,2(-2) 2(+1), -2 Balancing redox reaction in a step wise manner using half reactions (1) Identify elements changing oxidation state (2) separate oxidation from reduction into two half reactions (a) balance the non-H,O atoms (b) Determine the electrons lost or gained per atom in each halfreaction (3) Determine total electron change for element by Multiplying no. atoms changing times electron change per atoms, then write total half reaction (4) Add half reactions together and use whole number multipliers as need to make sure total gain of electrons equals total loss. Redox steps cont. (5) After addition the electrons cancel, add up total charge on reactant side and compare to total charge on product side. If it is not equal: (a) If acidic add appropriate number of H+ ions to whichever side need to balance the charges. (b) If basic, add appropriate number of OH- ions to whichever side needed to balance the charges. (6) After charges balanced, add H2O as needed to balance H’s. Once the H’s are balanced, the O’s should also be balanced. If not, you made a mistake in steps (1-3). Balancing redox reactions examples Cr2O7-2 + CH3OH → HCO2H + Cr+3 (acidic) 2(+6),7(-2) 4(+1), (-2),(-2) 2(+1),+2,2(-2) +3 Reduction half reaction: 6e- +Cr2O7-2 → 2Cr+3 Oxidation half reaction: CH3OH → HCO2H + 4e- Addition of reaction: 2Cr2O7-2 + 3CH3OH → 3HCO2H + 2Cr+3 Charge on reactant side is = -4 Charge on product side is = +6 X2 X3 Final Steps example 1 redox Add H+ to balance charge 16H+ + 2Cr2O7-2 + 3CH3OH → 3HCO2H + 2Cr+3 Hydrogens on reactant: 28H Hydrogen on Product: 6H Add H2O: 16H+ + 2Cr2O7-2 + 3CH3OH → 3HCO2H + 2Cr+3 + 11H2O Example two redox HClO2 + MnO2 → Cl2 + MnO4- (acidic) (+1),(+3),2(-2) (+4),2(-2) 0 (+7),4(-2) Reduction: 6e- + 2HClO2 → Cl2 Oxidation MnO2 → MnO4- + 3e- Addition of reaction: 2HClO2 + 2 MnO2 → Cl2 + 2 MnO4- x1 x2 Example 2 cont. Add H+ to balance charge 2HClO2 + 2 MnO2 → Cl2 + 2 MnO4- +2H Hydrogens are already equal at this point, so no water needed! Basic example CrO2- + ClO- → CrO4-2 +Cl2 (basic) +3,2(-2) (+1),(-2) (+6),4(-2) 0 Reduction: 2e-1 + 2ClO- → Cl2 Oxidation: CrO2- → CrO4-2 + 3e-1 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 X3 X2 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 Balance charge with OH- 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 + 4OH- Balance H with water: 2H2O + 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 + 4OH- Additional Notes about redox Once you have the final equation you can use this equation just like any chemical equation you have before. However, you need to be ready for spectator ions again: Take the previous equation we looked at: 2H2O + 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 + 4OH- A question might ask how many grams of NaClO would it take to produce 3 grams of Cl2. In this case the Na really isn’t part of the calculations other then final weight of reactant. (it is part of question because you can’t physically add a anion alone, it has to have a cation with it). Gases Gases can be monoatomic, diatomic (two atoms which the same), or polyatomic Gases expand to fill the container they are in If you apply pressure to a gas, the volume of a gas decreases If you release pressure it increases the volume. If you have two or more gases they form homogeneous mixtures Gases molecules are far aprat Pressure Pressure = Force/area F=mass x accerlation F=Kgm/s2= 1 newton (N) A=m2 P= N/m2 = 1 Pascal (Pa) There are number of ways pressure is can be reported: 1 atm = 760mmHg= 760torr=1.01335x105 Pa Pressure example A lead cube with edges 25.0 cm in length rest on a flat surface. The density of lead is 11.34 g/cm3. Calculate the pressure in Pascals exerted by the cube on the surface. F P A F ma 9.807m / s 2 25.0cm 11.34 g 1kg 9.807m 3 2 F x x 1.74 x 10 Kgms x 3 2 1 cm 1000kg 1sec 3 2 1m 2 Areaofsurface 25.00cmx 0.625 m 100cm 1.74 x103 Kgms 2 1 2 3 P 2784 Kgm s 2.78 x 10 Pa 2 0.625m Simple Gas laws Boyles Law (pressure and volume, temp and amount of gas kept constant PV=constant P1V1= P2V2 Charles law-volume varies directly temp (when pressure and the number of moles is kept constant V cons T V1 V2 T1 T2 Simple Gas Laws cont Avogradro’s Law-volume varies directly with the number of moles when pressure and temperature remain constant. V cons n V1 V2 n1 n2 Ideal Gas Law PV=nRT P=pressure (usually in atm’s) V=volume (L) T= temp (always in Kelvin) n= # of moles R= gas constant (0.8206 L atm mol-1 K-1) Standard tempt= 0⁰ C, (273 K) Standard pressure 1 atm Ideal gas examples The pressure of oxygen in inhaled air is 157 torr. The total volume of the average adult lung when expanded is about 6.00 L, and body temperature is 37⁰ C. Calculate the mass of O2. 1 atm P 157 torr x 0.207 atm 760 torr V 6.00 L T 37 o C 273 310 K n? atm L R 0.8206 mol K 0.207 atm x 6.00 L 32.00 g O 2 x n 0.156 g O2 atm L 0.8206 x310 K 1 mol O2 mol K Ideal gas examples How many additional moles of gas would have to be added to a flask containing 2.00 moles of gas at 25⁰ C and 1 atom in order to increase the pressure to 1.60 atm under conditions of constant temp and volume? Initail Final P = 1.00 atm 1.60 atm V = Constant Constant n = 2 moles T= Constant ? Constant PV nRT P RT n V P1 P2 n1 n2 1.00atm 1.60atm 3.2molesgas 2.00moles N2 2.00moles 1.2molesgas Gas Laws examples cont. If all of the propane in a tank (C3H8) at 20 C and a pressure of 300 atm is burned in a combustion reaction, where the O2 is also at 20 C, but 756 mmHg. How many liters of O2 would be required to completely burn the propane? 5O2 + C3H8 → 3CO2 + 4H2O Gas law for tank → take into account Stoic → gas for oxygen Propane tank cont. Tank P 300atm V 30 L n? T 293K R 0.8206 atm L mol K 5O2 1 C H 3 8 Oxygen 756 mmHg P 0.995 atm 760 mmHg V ? n value from tank x 5 atm L R 0.8206 mol K T 293 atm Propane continued Propane tank → to moles of Oxygen PV RT 300 atm x 30.00 L 374 moles C3 H 8 n 0.8206 atm L x 293 K mol K 5O2 374 moles C3 H 8 x 1870 moles O2 1C3 H 8 n Volume of Oxygen nRT P atm L 1870 mol O x 0.8206 x 293 K 2 mol K 45200 L O2 V 0.995 atm V MW of a gas can be found using gas laws PV=nRT, n can also me expressed as grams/MW Acetaldehyde is a common liquid that vaporizes readily. If the pressure of acetaldehyde is observed to be 331 mmHg in a 125 mL flask a 0 C and the density of the gas is 0.0855g/L what is the molar mass of acetaldehyde? Denistry→grams, and Gas Law to get moles 1atm P 331mmHgx 0.436atm 760 mmHg V 0.125 L n ?( g / mw) atm L R 0.8206 mol K T 273K Dg/L 0.855 g g x 0.125L 0.107 g 1L Gas Law portion 1atm P 331mmHgx 0.436atm 760 mmHg V 0.125 L n ?( g / mw) atm L R 0.8206 mol K T 273K 0.436atmx0.125 L n 2.43x103 moles acetlaldehyde atm L 0.8206 x 273K mol K 0.107 grams MW 44.0 2.43 x103 moles Dalton’s law of partial Pressure Two or more gases will mix together to total occupy the volume they are contained in. Total pressure is equal to the sum of the individual pressures exerted by the various gases in a mixture PT= P1 + P2 + P3…… Total number of moles of gas in a mixture is equal to the sum of the number of moles of each gas in a mixture. nT= n1 + n2 + n3…… PTV=nTRT Example of Partial Pressure What is the total pressure exerted by a mixture of 3.00g of CH4 and 10.50g N2 at 301K in a 25.0L flask? PT PCH 4 PN 2 V 25.0 L 1 mol CH 4 1 mol N 2 nt nCH 4 nN2 3.00 g CH 4 x 10.50 g N 0.563 moles gas 2 16.0 g CH 28.0 g N 4 2 atm L R 0.08206 mol K T 301 K atm L 0.563 moles gas x0.08206 x 301K mol K PT 0.556 atm gas 25.0 L Collection of Gas over water We can use Dalton’s law of Partial Pressure to measure the amount of gas given off from a reaction This is done by subtracting the amount of water vapor that would normally be in flask at a given temperature. Collection of Gas over water picture. Collection of Gas over water problem A sample of KClO3 is partially decomposed and the gas collected over water. 2KClO3(s) → 2KCl(s) + 3O2(g) (a) How many moles of O2 are collected? (b) How many grams of KClO3 decomposed? © When dry, what volume would the collected O2 occupy at the same temp. and pressure? Part A Find partial pressure of O2 by subtracting the pressure of water from the total. PV nRT 1 atm PT 756 torr x 1.07 atmT 760 torr 1 atm PH 2O 23.8 mmHgx 0.0313 atmO2 760 mmHg PO2 107 atmT 0.0313 atmO2 1.04 atmo2 V 0.250 L n? R 0.08206 atm L mol K T 298 K 1.04atmo2 x0.250 L n 0.0106 moles O2 atm L 0.08206 x 298 K mol K Part B and C Part B 2 moles KClO3 122.55 g KClO3 0.0106 moles O2 x 0.866 g KClO3 x 3 moles O2 1 mole KClO3 Part C P 1.07 atm V ? n 0.0106 mole O2 R 0.08206 atm L mol K T 293K V 0.0106 mole O2 x0.08206 1.07 atm atm L x 293 K mol K 0.242 L O2 Energy and Chemical Reactions Thermodynamics=study of energy and its transformations C6H12O6(s)→6O2(g) → 6CO2(g) + 6H2O(g) + energy Energy-the capacity to do work or transfer heat Two types: (1)kinetic-energy of motion a. macroscale-moving ball-mechanical energy b. nanoscale-moving atoms, molecules,-etc. thermal energy C. Moving electrons-electrical energy Energy continued (2) Potential energy-stored energy-energy possessed by an object by virtue of its position relative to other objects. a. energy an object has in your hand due to the force of gravitygravitational energy b. attraction or repulsion of charge particles-electrostatic energy c. chemical potential energy-stored in foods and fuels Calculated in different ways depending on type. Energy continued Total energy= KE + PE Energy Units: Joule 2 2 1 2 1 m Kgm Ek mv ( Kg ) x 2 2 s2 s Calorie=amount of energy required to raise the temp of 1g of water by 1⁰ C. Work on Surroundings? ∆E=q+w Energy is a State function A State function= values which do not depend on the pathway, only on the initial and final states. Energy is a state function. So 100g of water going from 90⁰ C to 15 ⁰ C, will give off the same amount of energy, if you cooled the water down to 50 ⁰ C, and then heated to 75 ⁰ C then cooled it again. Heat and work are not state functions! When talking about chemical reactions, most of the energy change is in the form of heat. Work is negligible. Heat Capacity Heat Capacity=amount of energy required to increase the temp of that substance by one degree= Cp Cp=J/(Kor ⁰C) Specific heat=amount of energy needed to increase the temp of one gram of substance by one degree-like density, it’s characteristic of specific substance Cp/g=J/(Kor ⁰C)g Molar heat capacity= Cp/mole=J/(Kor ⁰C)mole Specific heat example The Specific heat capacity of copper metal is 0.380 J/⁰Cg. Assume you had a 75.0g cube of copper at 25.0g. What would be the final temperature of the copper if it absorbed 150 J of heat. q C p x mcopper x T T T2 T1 150 J 0.380 J x75.0 g x T o Cg T 5.26o C 5.26o C T2 25o C 30.3o C Molar heat capacity example If a 1.00 Kg block of Al (molar heat capacity of 24.2 J/⁰C mole) at 600 ⁰C is placed in contact with 1.00Kg. Block of copper (molar heat capacity of 24.2 J/⁰C mole), calculate the final temperature of the blocks. qal=qcu 1moleAl J J 1000 gAlx x 24.2 896 o o Cmole C 26.98 gAl 1moleCu J J 1000 gCux x 25.2 385 o o Cmole C 63.55 gCu C pAl C pCu J 896 o (600 oC T f ) 385(T f 100 oC ) C J J J J 5.38 x105 o 896 o T f 385 o T f 38500 o C C C C T f 450 oC Calorimetry Calorimetry-measuring the heat flow into or out of a substance, 1) if temp is increasing= Then heat is flowing from the reaction to the surrounding solution. 2)if temp decreasing=The heat is flowing from the solution into the reaction. There are two types of calorimetry: “coffee cup” and Bomb “Coffee Cup” vs. Bomb Law of Conservation of Energy In interactions between a system and its surrondings, the total energy remains constant-energy is neither created nor destroyed. Applied to the change of heat this means. qsystem + qsurrondings = 0 If this often conveint to rearrange this equation to: qsystem = - qsurrondings Example of calorimetry When 4.25g of solid ammonium nitrate dissolves in 60.0 g of water in a calorimetry, the temp drops from 22.0⁰C to 16.9 ⁰C. Calculate ∆H (in kJ/mol of NH4NO3) for the solution process. NH 4 NO3( s ) NH 4 ( aq ) NO3 ( aq ) Assume the specific heat of the solution is the same as that of pure water: qreaction qwater ( surrondings ) 0 qwater ( surrondings ) 4.184 J o 3 x (60.0 g 4.25 g ) x 5.1 C 1.37 x 10 kJ g ,o C qreaction (1.37 x 103 kJ ) 1.37 x103 kJ 1 mole NH 4 NO3( s ) 4.25 g NH 4 NO3( s ) x 0.531 moles NH 4 NO3( s ) 80.1 g NH NO 4 3( s ) 1 kJ 3 1.37 x10 kJ x 25.8 0.531 moles NH NO mol 4 3( s ) Enthalpy-heat content of the system (∆H) ∆H= q ∆H⁰=enthalpy under standard conditions: 1 atm, 25 ⁰C Enthalpy is a state function: ∆H=Hfinal-Hinitial H H final H initial (endothermic ) H H final H initial (exothermic ) Enthalpy of Reaction Enthalpy of reaction or heat of reaction-enthalpy change that accompanies a reaction ∆H=Hproducts-Hreactants Enthalpy of formation The Enthalpy of formation (Heat of formation ∆Hf)- the enthalpy for the reaction forming the substance from its pure elements Standard State-the most stable form of a substance at a particular temp (25 C) and standard atmospheric pressure (1 atm). Standard enthalpy of formation (∆H⁰f)= change in enthalpy that occurs with the formation of 1 mole of a substance from its elements in their standard state. (∆H⁰f)= 0 for C (graphite), H2(g), O2(g) Ex. C(s) + O2(g) → CO2(g) ∆H⁰f = -393.5kJ H2(g) + 1/2O2(g) → H2O(g) ∆H⁰f = -393.5kJ ∆H⁰rxn = ∑(x)∆H⁰f Products - ∑(y) ∆H⁰f reactants X and Y are coefficents Hess’s Law continued The enthalpy of a reaction is equal to the sum of the heats of formation of the product minus the heats of formation of the reactants. H o rxn H o f products H o f reac tan ts Ex.rxn : 2C2 H 2( g ) 5O2( g ) 4CO2( g ) 2 H 2O( l ) H o rxn (4H o f CO2 2H o f H 2O(l ) ) (2H o f C2 H 2( g ) 5H o f O2( g ) ) 4(395kJ ) 2(285.8kJ ) 2(226.7 kJ ) 5(0 KJ ) 2599.5kJ Calculate the ∆H⁰rxn for the dissolution of 0.8327 g of H3AsO4 in water give the following: 1mole H 3 AsO4( s ) 5.866 x103 mol H 3 AsO4( s ) 0.8327 g H 3 AsO4( s ) x 141.95 g H AsO 4( s ) 3 H 3 AsO4( s ) H o f 900.4kJ / mol H 3 AsO4( aq ) H o f 899.7kJ / mol H 3 AsO4( s ) H 3 AsO4( aq ) H o rxn H o fH3 AsO4 ( aq ) H o fH3 AsO4 ( s ) (899.7 kJ ) (900.4 kJ ) 0.7 kJ / mol 0.7 kJ / mol x 5.866 x103 mol 4.0 x103 kJ Application of Hess’s Law Indirect Determination of Hess’s Law ∆H is an Extensive Property (the amount matters) N 2( g ) O2( g ) 2 NO( g ) H o 180.5kJ 1 1 N 2( g ) O2( g ) NO( g ) H o 90.25kJ 2 2 ∆H Changes sign when a process is reversed. NO( g ) 1 1 N 2( g ) O2( g ) 90.25kJ 2 2 Hess’s Law of Constant Heat Summation If a process occurs in stages or steps (even if hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. Example: 1 1 N 2( g ) O2( g ) NO( g ) O2( g ) H o 90.25kJ 2 2 1 NO( g ) O2( g ) NO2( g ) H o 57.07kJ 2 1 N 2( g ) O2( g ) NO2( g ) H o 33.18kJ 2 A second example Use the following equation to determine ∆H⁰rxn for the following reaction. 3C( graphite ) 4 H 2( g ) C3 H 8( g ) H o ? 1)C3 H 8( g ) 5O2( g ) 3CO2( g ) 4 H 2O( l ) H o 2219.8 kJ 2)C( graphite ) O2( g ) CO2( g ) H o 393.5 kJ 1 3) H 2( g ) O2( g ) H 2O( l ) 2 H o 285 kJ 3CO2( g ) 4 H 2O( l ) C3 H 8( g ) 5O2( g ) H o 2219.8 kJ 3C( graphite ) 3O2( g ) 3CO2( g ) H o 1181kJ 4 H 2( g ) 2O2( g ) 4 H 2O( l ) H o 1143 kJ 3C( graphite ) 4 H 2( g ) C3 H 8( g ) H o 104kJ Electron Configuration Electron structure: arrangement of electrons Electromagnetic radiation carries energy through space (ex. visible light, ultraviolet, infrared, X-rays). All of these move through a vacuum at 3.00x108(speed of light). They all move as waves: Electromagnetic Spectrum Wavelength (λ)-distance between successive crests or troughs(units of length). Frequency (ν)-number of cycles that pass a given point in one second (cycles/s-Hertz(Hz)). λ ν=c (speed of light) High frequency = high energy (gamma rays) Low frequency = low energy (radio) X-ray Example X-rays are produced by bombarding a metal surface with a beam of energetic electrons. If copper is the metal used, the wavelength of the x-ray is 0.154mn. Calculate the frequency of this radiation. v c 8 18 3.00 x10 m 1.95 x v 9 0.154 x10 ms sec Max Plank (1900) He assumed that radiation could only be emitted in little packets (quanta). The energy associated with these quanta is proportional to the frequency of the reaction. E=hv E=energy of a quantum, h=Plank’s constant (6.63 x-34 J,s), v=frequency of radiation Energy is always emitted in whole numbers-multiples of hv (hv, 2hv, 3hv, etc) Example Cosmic rays are a penetrating electromagnetic radiation produced by certain catastrophic events. Calculate the energy associated with cosmic radiation on a per mole basis and compare your result with the energy of radio waves. The average wavelength of a radio wave is 500nm and that of a cosmic ray is 1x10-14m. Ecos mic hv Eradio hv v c cos mic 1x1014 m v c radio 500m 22 3.8 x108 m 1 3.00 x10 v x 14 sec sec 1x10 m 3.8 x108 m 1 6.00 x105 v x sec sec 500 m Ecos mic 3.00 x1022 6.02 x1023 1.2 x1013 J 34 x6.63x10 J sec x sec 1 mole mole Eradio 6.00 x105 6.02 x1023 2.4 x 104 J 34 x 6.63 x 10 J sec x sec 1 mole mole Photoelectric Effect-Albert Einstein If you shine light on a metal, electrons are emitted. There is a minimum energy of light that can be show where no electrons are emitted. One metal might require red light, whereas another might require yellow light. Radiation strikes the metal in a stream of tiny energy packets (massless). Photons - energy particles. Light can have particle-like properties as well as wavelengths. Duel nature of light. E=hv - energy of a photon. Monochromatic light- made up of a single wavelength Continuous spectrum (all wave lengths) Einstein Continued Apply high voltage to an element (gaseous)-atoms absorb energy and then emit energy as electromagnetic radiation. Ex. Neon-redorange light. Which wavelength of light is emitted is a characteristic of an element. Line spectrum - spectrum containing radiation of specific wavelengths. Niels Bohr - Bohr’s Model of the Hydrogen Atom 1 En RH 2 n En = energy of an electron in a hydrogen atom RH = Rydberg constant (2.18 x 10-18 J) n = principle quantum number- different allowed orbits for the electron n = 1 if electron is in first orbit, n= 2 if electron is in the second orbit, etc. All energies are negative- the more negative, the more stable an atom will be. Bohr Model Continued Ground state-lowest energy state; n = 1 Excited state- higher energy state; n = 2 n gets larger and larger until the electron completely breaks away from the nucleus. Zero-energy state-when an electron is removed from the nucleus. Highest energy state: n= infinity 1 18 1 E ( RH ) 2 2.18 x 10 J 2 0 E E f Ei Bohr Continued Energy is absorbed when it moves to a higher n. Energy is emitted when it moves to a lower n. 1 1 E E f Ei ( RH ) 2 ( RH ) 2 n ni f 1 1 RH 2 2 hv n ni f +∆E when nf › ni – radiant energy is absorbed -∆E when nf ‹ ni – radiant energy is emitted. Example How much energy is required for an electron in a hydrogen atom to make a transition from the n= 2 state to the n=3 state? 1 1 E RH 2 2 n n f f 1 18 1 E 2.18 x 10 J 2 2 2.18 x 1018 J x 0.139 3.03x1019 J 3 2 de Broglie If light can have both wave-like properties and particle-like properties, then why can’t matter (like electrons) have both? Both radiation (light) and matter obey the following equation. h mv λ= de Broglie wavelength (m) m= mass of the particle (kg) v= speed (m/s) h=Plank’s constant Can Look at Electrons, Baseballs, Molecules with the Equation. Calculate the de Brogile wavelength of an electron traveling at 1.00% the speed of light h 6.62 x1034 J s m 9.11x1031 kg v 0.01(3.00 x108 m / s ) 3.00 x106 m / s h mv 6.62 x1034 J s Js 10 2.43 x 10 31 6 m 9.11 x 10 kg x 3.00 x 10 m / s kg s Particle Wave Duality Since electron exhibit both wave and particle like properties, one can tell exactly how fast an electron is moving around an atom at a given time. Heisenberg Uncertainty Principle- It is impossible for us to know both the exact location of an electron and its speed at the same time. Only one or the other can be determined. Quantum Mechanics - Schrodinger - We can’t predict the exact speed and location of an electron, but we can determine a probability. Quantum Mechanics Wave functions (ψ)- ψ2 – probability density- probability of an electron’s location in an allowed state. Higher probability of finding an electron in an area of high density. Orbitals - not the same things as an orbit - In quantum mechanics n measure the most probable distance of an electron from the nucleus-not the radius of the a defined orbit. Quantum numbers: (1)n- principle quantum number-shell- a collection of orbitals with the same quantum number n. Specifies the size and extent of the orbital. Determines the energy of an electron in a hydrogen atom. 1 En 2.18 x1018 J x 2 n (2)azimuthal quantum number-specifies the shape of the orbital. Restricted to 0,1….n-1. Designated by letters: s,p,d,f which correspond to numbers Azimuthal quantum number continued l = 0 –s-sharp l = 1 –p-principle l = 2 –d- diffuse l = 3- f-fundamental Defines subshells n 1, n 2, n 3, n 4, l 0, s l 0,1, s and p l 0,1, 2 s, p, d l 0,1, 2,3, 4 s, p, d , f 1s 2 s, 2 p 3s3 p,3d 4 s , 4 p, 4 d , 4 f Magnetic Quantum Number Determines the spatial orientation of an orbital. Has values between l and –l including zero. Defines the # of orbitals in each subshell. n 1 l 0 (1s ) ml 0 1 orbital in subshell n 2 l 0 (2 s ) ml 0 1 orbital in subshell n 2 l 0 (2 p) ml 1, 0,1 3 orbitals in subshell n 3 l 0 (3s ) ml 0 1 orbital in subshell n 3 l 0 (3 p) ml 1, 0,1 3 orbitals in subshell n 3 l 0 (3d ) ml 2, 1, 0,1, 2 5 orbitals in subshell Spin Quantum Number (4) ms-electron spin quantum number-electrons spin in 2 opposite directions, -1/2 or + ½ Each orbital can hold a maximum of two electrons No two electrons in an atom can have the same two quantum numbers- Pauli Exclusion Principle Electron #1 n 1 l 0 ml 0 Electron # 2 n 1 l 0 ml 0 ms 1/ 2 ms 1/ 2 Using Arrows to Show the Pauli Exclusion Principle 1 electron 1s 2 electrons 1s can not be: or 1s 1s How Many Electrons are in the n=4 Shell? l ml ms 0 0 1/ 2 or 1/ 2 1 1, 0,1 1/ 2 or 1/ 2 2 2, 1, 0,1, 2 1/ 2 or 1/ 2 3 3, 2, 1, 0,1, 2 1/ 2 or 1/ 2 # orbitals 1 3 5 7 # electrons 2 6 10 14 32 e- Shapes of Orbitals Electron Configurations Arrangement of electrons in the orbitals. Orbitals fill up in order of increasing energy He 2 electrons 1s 2 1s B 5electrons 1s 2s 2p 1s 2 2 s 2 2 p Li 1s 2 1s C 6 electrons 1s 2s 2p 1s 2 2 s 2 2 p 2 Hund’s Rule The ground state configuration is obtained by placing the electrons in different orbitals with parallel spins. No orbital in the subshell contains two electrons until each on contains one electrons. O 7 electrons 1s 2s 2 2 3 1s 2 s 2 p 2p N 8 electrons 1s 2s 2 2 4 1s 2 s 2 p 2p Noble Gas Configurations Fr 87 electrons Valance electrons-outer shell electrons Core electrons-inner shell electrons Representing the inner shell electrons as a noble gas [ Rn]7 s1 K 19 electrons [ Ar ]4 s1 Na 11electrons [ Ne]3s1 Mn 25 electrons [ Ar ]4 s 2 3d 5 Pt 78 electrons [ Ar ]6 s 2 4 f 14 5d 8 Removing Electrons to Form Ions Do not remove electrons in the order in which they are filled. Always remove from valence shell (outermost energy level). Zn [ Ar ]4s 2 3d 10 Zn 2 [ Ar ]3d 10 Se [ Ar ]4 s 2 3d 10 4 p 4 Se 3 2 10 1 [ Ar ]4 s 3d 4 p Two cases when electrons shift from their ground state configuration. 1 chromium(Cr)-24 electrons [ Ar ]4 s 2 3d 4 but really [ Ar ]4s1 3d 5 4s 3d 2 copper(Cu)-29 electrons [ Ar ]4 s 2 3d 9 4s but really [ Ar ]4s1 3d 10 3d Periodic Trends Atomic radius-electrons distribute themselves about a nucleus in a diffuse, cloud-like manner-no sharp boundaries. (1)atomic radius increases going down the PT-adding a whole new shell. (2)atomic radius decreases going across the PT- increased positive charge in the nucleus. Therefore, pull on the electrons is greater, making the radius smaller. Ionic Radius (1) Positive or negative ions of elements in the same group increase in size moving down the PT. (2) Radius of cation is always smaller then the neutral atom. (3) Radius of anion is always larger than the neutral atom. Ionization Energy Amount of energy it takes to remove an electron completely from an atom (making a cation) First ionization energy (I1) =amount of energy associated with removing one electron from a neutral atom, forming a positive ion, A+. A g A ( g ) e Second ionization energy (I2) = amount of energy associated with removing one electron from the A+ ion to produce A2+ ion A g A 2 (g) e The higher the ionization energy, the harder it is to remove an electron. Ionization Energy Cont. Na 11electrons 1s 2 2 s 2 2 p 6 3s1 1s 2 2s 2 2 p 6 I1 496kJ / mole Na 10electrons 1s 2 2 s 2 2 p 6 1s 2 2 s 2 2 p 5 I 2 4560kJ / mole Generally, (1)ionization energy increases moving across the PT. (2)ionization energy increases going down the PT. Electron Affinity The energy change when an atom gains an electron in the gaseous state. Generally, for neutral and positively charged species, energy is released when an electron is added. Generally, for an anion, energy must be added for an electron to be added. A e A Cl e Cl EA 349kJ / mole Generally, the trend is moving from left to right, and up the periodic table (excluding the noble gases). Electronegativity This is a value only used when talking about atoms bound in covalent bonds. This represents the pull that the single bond places on the shared electrons. Trend increases going across and going up, making Fluorine the most electronegative . Periodic Trends