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Chem 111 Notes Part 2

by: Caroline Hurlbut

Chem 111 Notes Part 2 Chem 111

Caroline Hurlbut
GPA 3.7

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This is the second set of notes from Chem 111 that cover the material for the rest of the semester, continuing from my upload of the first part of notes last week.
General Chemistry I
Dr. Kerry MacFarland
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This 27 page Bundle was uploaded by Caroline Hurlbut on Friday January 29, 2016. The Bundle belongs to Chem 111 at Colorado State University taught by Dr. Kerry MacFarland in Fall 2015. Since its upload, it has received 75 views. For similar materials see General Chemistry I in Chemistry at Colorado State University.

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Date Created: 01/29/16
Molecule Polarity Bond Dipoles • how do you know if a diatomic molecule is polar or non polar? —EN difference between atoms • how do you know if a triatomic molecule is polar or non polar? —EN difference and shape • bond dipole - partial separation of charges that occurs when polar bonds form • molecular dipole (dipole moment) - representation of bond dipoles calculated by adding them together • bond dipoles vs molecular dipoles —bond dipoles add to give molecular dipole (unless they cancel each other) • larger EN difference—>larger dipole Octet Exceptions Incomplete Octets • molecules with boron, beryllium, hydrogen, or helium as central atom can have less than an octet —ex. BH3 —ex. BF3 Odd-Electron Octets • odd-electron molecules are called free radicals (very reactive) • every atom can’t have a full octet, so include an incomplete octet • atom with higher EN gets full octet to ensure best structure —ex. NO: oxygen gets full octet and nitrogen is one electron short —ex. NO2 Expanded Octets • nonmetals from period 3 and beyond can expand valence shells to hold more than an octet because d orbitals are included —ex. PF5: phosphorus has expanded valence shell with 10 electrons —ex. SF6: sulfur has expanded valence shell with 12 valence electrons VSEPR: Octet Exceptions • incomplete octets —most common is when B is central atom —electron geometry usually trigonal planar • odd number of electrons —one orbital with one electron instead of lone pair —if the electron is on central atom, treat it like lone pair • expanded octets —SN=5 (trigonal bipyramidal electron geometry, sp3d hybridization), no lone pairs A. trigonal bipyramidal molecular geometry B. 90˚ and 120˚ bond angles C. axial and equatorial positions —SN=5, 1 lone pair A. seesaw molecular geometry B. 90˚ and 120˚ bond angles C. lone pair takes equatorial position (because it’s larger than a bonding pair) and shrinks opposing bond angles a little less than ideal —SN=5, 2 lone pairs A. T-shaped molecular geometry B. 90˚ bond angles —SN=5, 3 lone pairs A. linear molecular geometry B. 90˚ and 120˚ bond angles —SN=6 (octahedral electron geometry, 90˚ bond angles, sp3d2 hybridization), no lone pairs A. octahedral molecular geometry —SN=6, 1 lone pair A. square pyramidal molecular geometry —SN=6, 2 lone pairs A. square planar molecular geometry • expanded octets and resonance —ex. SO4 2- —formal charges not ok (left structure) —give expanded octet if: A. central atom is nonmetal in period 3 or beyond B. composed of strongly electronegative elements (F,O,Cl) C. results in smaller formal charges (right structure) Intermolecular Forces London Forces • intermolecular forces - attractions between all molecules • all atoms and molecules are attracted by London dispersion forces, which create partial positive/negative charges on the attracted atoms and are the weakest intermolecular force • temporary dipoles are induced when electrons spontaneously exist in close proximity, unevenly distributing the electrons and making one side of the atom more negative than the other (even in non polar molecules) • larger electron cloud (as atomic size increases)—>more polarizability, stronger London forces, higher boiling point • shape vs. dispersion forces —increased surface area—>stronger dispersion forces —longer shape molecules can have more contact with each other than rounder shape molecules • viscosity - resistance to flow —increases as intermolecular forces increase Polar Interactions • dipole interactions generally stronger than dispersion forces • ion-dipole interactions —between ions and polar molecules (ex. NaCl and H2O) —partial charges of ion attract opposite partial charges on ends of polar molecule and disperse intermolecular forces — hydration if polar molecule is water or solvation if not water • dipole-dipole interactions —between polar molecules (same or different) —oppositely charged ends attract (permanent dipoles) • dipole-induced dipole interactions —between polar and non polar molecules —permanent dipole and temporary dipole • ion-induced dipole interactions —between ions and non polar molecules Hydrogen Bonds • special type of dipole-dipole interaction • strongest intermolecular force • between molecules that have hydrogen directly bonded to oxygen, nitrogen, or fluorine —because O, N, F are small atoms with high EN and a lone pair on each • can occur between identical or nonidentical molecules (ex. water can hydrogen bond with water) • tends to increase boiling point Solubility Forces in Mixtures • solution - homogeneous mixture uniformly mixed • solvent - what is doing the dissolving (larger quantity present—mole proportion) • solute - what is being dissolved • solubility - measure of how much solute can dissolve in given volume of solvent • miscible - 2 liquids that can mix in any proportion (unlimited solubility) immiscible - 2 liquids that are not soluble in each other • • when a solute dissolves in a solvent: —solute-solute attractions + solvent-solvent attractions—>solute-solvent attractions —solute-solvent attractions must be stronger than replaced attractions for substance to dissolve —*like dissolves like* • hydrophilic interactions —ion-dipole —dipole-dipole —hydrogen bonding • hydrophobic interactions —mostly dispersion —ion-induced dipole —dipole-induced dipole (allows for slight solubility of O2 in water) Chemical Reactions Conservation of Mass • reactants - substances that go into a reaction • products - substances that come out of a reaction • chemical equation written like this: —reactant (physical state)—>product (physical state) —include physical state for all reactants and products (s, l, g, or aq) • elements in their standard states are mostly solids, but there are some exceptions —mercury is a liquid at room temp. —noble gases —diatomic elements are gases except Br2 (liquid) and I2 (solid) • law of conservation of mass - total mass of reactants=total mass of products and # of atoms of each element doesn’t change • ex.A+ B—>C + D —reactants=A+ B —products=C + D Writing/Balancing Chemical Equations • ex. hydrogen gas and fluorine gas combine to make hydrogen fluoride gas —hydrogen and fluorine are diatomic elements —H2(g) + F2(g)—>HF(g) —2 H on left and 1 H on right, 2 F on left and 1 F on right —balanced equation: H2(g) + F2(g)—>2HF(g) • guidelines —change coefficients, not the subscript —use smallest whole numbers possible as coefficients (don’t write “1”) —start balancing atoms in more complex substances first, save free elements for last —check everything at the end • ex. combustion of methane gas (combustion reactions always include O2 as a reactant and CO2 + H2O as products)) —write reactants and products with states: CH4(g) + O2(g)—>CO2(g) + H2O(g) —write out #’s of each atom A. 1 C—>1 C B. 4 H—>2 H C. 2 O—>3 O —balance: CH4(g) + 2O2(g)—>CO2(g) + 2H2O(g) A. 1 C—>1 C4 H—>4 H B. 4 O—>4 O • ex. combustion of octane (C8H18) —C8H18(g) + O2(g)—>CO2(g) + H2O(g) —8:1 C, 18:2 H, 2:3 O — 2C8H18(g) + 25O2(g)—>16CO2(g) + 18H2O(g) Stoichiometry Part 1 Mole Ratios & Conversions • # of things/6.022 x 10^23 things=1 mole of that thing • g/molar mass=moles mole ratios can be used to convert moles of X to moles of Y • • ex. How many moles of H2 are needed to react with 2 moles of N2? —start with balanced equation: N2(g) + 3H2(g)—>2NH3(g) — 2 mol N2 x 3 mol H2 = 6 mol H2 1 mol N2 • convert grams of X to moles of Y —use molar mass to get g of X—>mol of X —use mole ratios to get moles of X—>mol of Y • ex. How many moles of ammonia can be made from 63 g N2? —N2(g) + 3H2(g)—>2NH3(g) — 63 g N2 = 2.248 mol N2 28.02 g — 2.248 mol N2 x 2 mol NH3 = 4.5 mol NH3 1 mol N2 • convert grams of X to grams of Y —g of X—>mol of X —mol of X—>mol of Y —mol of Y—>g of Y • ex. How many grams of ammonia are produced from 82.0 g of H2? — 82.0 g H2 = 40.683 mol H2 2.0156 g H2 — 40.683 mol H2 x 2 mol NH3 = 27.122 mol NH3 3 mol H2 — 27.122 mol NH3 x 17.03 g NH3 = 462 g NH3 1 mol NH3 Stoichiometry Part 2 Percent Composition • find % composition from chemical formula —ex. Find mass % of Cu in Cu2O — mass % of X = (mass of X/total mass) x 100 —start with 1 mole of Cu2O —2 mol Cu x 63.55 g/mol = 127.1 g —1 mol O x 16.00 g/mol = 16.00 g O —% Cu = (127.1 g/ 143.1 g) x 100 = 88.82 % Empirical Formulas • empirical formula shows smallest whole number ratio of elements • using mass % to find empirical formula —%—>g—>moles—>ratio—>simplified ratio • ex.Acompound contains 73.9% Hg and 26.1% Cl by mass. —start with 100 g of compound —73.9 g Hg and 26.1 g Cl — 73.9 g Hg = 0.368 mol Hg 200.6 g Hg — 26.1 g Cl = 0.736 mol Cl 35.45 g Cl —divide by smallest # A. Hg: 0.368/0.368 = 1 B. Cl: 0.736/0.368 = 2 —empirical formula = HgCl2 • ex.Ascorbic acid is 40.92% C, 4.58% H, and 54.50% O by mass. — 40.92 g C = 3.407 mol C 12.01 g C — 4.58 g H = 4.544 mol H 1.008 g H — 54.50 g O = 3.406 mol O 16.00 g O —C: 3.407/3.406 = 1 mol —H: 4.544/3.406 = 1.33 mol (must be x3 to be whole #, also multiply other elements x3) —O: 3.406/3.406 = 1 mol —C: 1 x 3 = 3 —H: 1.33 x 3 = 4 —O: 1 x 3 = 3 —empirical formula = C3H4O3 Molecular Formulas • molecular formula shows whole number ratio of elements and the number of atoms in the compound, may be the same as or a multiple of the empirical formula • empirical formula is always the same as molecular formula for ionic compounds, but not necessarily molecular compounds • steps to find molecular formula —find empirical formula —use molar mass to find n value and multiply subscripts in empirical formula by n • ex. Empirical formula of naphthalene is C5H4 and molar mass is 128.16 g/mol. —mass of C5H4 = 64.082 g/mol —n = 128.16 g/mol = 2 64.082 g/mol —multiply subscripts by 2 A. C: 5 x 2 = 10 B. H: 4 x 2 = 8 —molecular formula = C10H8 Stoichiometry Part 3 Limiting Reactants and Yields • 2 ways to find limiting reactant (what runs out first) —find theoretical yield for each reactant and choose smallest one (easier) —compare ratio of reactants to stoichiometric ratio of reactants—>find theoretical yield • find remainder with ICE table —I=initial, C=change, E=end —subtract what’s used (change) from what you started with (initial) to get what remains (end) • ex. How many grams of ammonia could be made from 25.0 g N2 and 15.0 g H2? —equation: N2 + 3H2—>2NH3 —find theoretical yield for each reactant A. 25.0 g N2 = 0.892 mol N2 x 2 mol NH3 = 1.784 mol NH3 x 17.03 g = 30.4 g 28.02 g N2 1 mol N2 1 mol NH3 B. 15.0 g H2 = 7.44 mol H2 x 2 mol NH3 = 4.96 mol NH3 x 17.03 g = 84.5 g 2.016 g H2 3 mol H2 1 mol NH3 C. 30.4 is smaller, so N2 is limiting reactant • ex. Find theoretical yield of ammonia from 82.4 g NO and 18.7 g H2 —equation: 2NO + 5H2—>2NH3 + 2H2O — 82.4 g NO = 2.745 mol NO x 2 mol NH3 = 2.745 mol NH3 x 17.03 g = 46.8 g 30.01 g NO 2 mol NO 1 mol NH3 — 18.7 g H2 = 9.275 mol H2 x 2 mol NH3 = 3.71 mol NH3 x 17.03 g = 63.2 g 2.016 g H2 5 mol H2 1 mol NH3 —46.8 g is smaller, so that is the theoretical yield of ammonia for this reaction and the limiting reactant is NO • using ICE table to find remainder — N2 + 3H2 —> 2NH3 I 0.892 mol 7.44 mol 1.78 mol C -0.892 mol -2.68 mol 0 mol E 0 mol 4.76 mol 1.78 mol —get amount used up by using moles of limiting reactant with mole ratio of reactants A. 0.892 mol N2 x 3 mol H2 = 2.68 mol H2 used 1 mol N2 —convert mol to g: 4.76 mol H2 x 2.016 g = 2.36 g H2 remains 1 mol H2 • find percent yield — actual yield x 100 theoretical yield —ex. theoretical yield: 46.8 g, actual yield 31.9 g —% yield = (31.9/46.8) x 100 = 68.2 % Solutions and Dilutions Solutions/Concentrations • concentration usually expressed in some form of amount of solute amount of total solution • concentration can be used as conversion factor • ex. 1.5 % (m/m) NaCl solution —m/m = mass of solute/mass of solution —1.5 g NaCl per 100 g solution (conversion factor) • 63 mg/L glucose solution —63 mg glucose per L solution • parts per million = mg solute kg solution • molarity (n/V) = moles solute L solution • ex. 0.50 mol NaCl dissolved in water, total volume is 2.0 L. —molarity = 0.50 mol NaCl = 0.25 mol/L or 0.25 M NaCl 2.0 L • molarity can have standard SI prefixes —ex. 1M = 1000 mM • ex. How many mol of NaCl are in 0.25 L of 0.10 M NaCl solution? —0.10 M = x mol NaCl = 0.025 mol NaCl 0.25 L Dilutions • concentrated stock solution—>diluted solution • any units can be used for concentration and volume as long as they are the same before and after • M1xV1=M2xV2 —M1 = concentration before —V1 = volume before —M2 = concentration after —V2 = volume after —mol before = mol after • ex. If you dilute 11 mL of 0.25 M HCl solution to final volume of 75 mL, what is the concentration of diluted solution? —0.25 M x 11 mL = M2 x 75 mL — 0.25 M x 11 mL = 0.037 M HCl 75 mL Electrolytes and Neutralization Reactions Electrolytes • electrolytes - solutions that conduct electricity, particularly aqueous ionic compounds with free ions in solution • strong electrolytes - solutions with ions that completely dissociate (all ions dissolve) in water, include strong acids and bases —ex. NaCl (aq)—>Na+ (aq) + Cl- (aq) —ex. Ba(OH)2 (aq)—>Ba2+ (aq) + 2OH- (aq) A. 0.02 M Ba(OH)2 solution—>0.04 M Ba2+ and 0.02 M OH- • weak electrolytes - solutions with ions that partially dissociate in water, include weak acids and bases —ex. acetic acid CH3COOH • nonelectrolytes - solutions that do not ionize or conduct electricity, include most molecular compounds Acids and Bases • acids tend to increase concentration of hydronium (H3O+) ions in solution, have more than one acidic H+ (proton) which when released in solution forms H3O+ and decreases pH of solution (H+ and H3O+ used interchangeably) —most acids are molecular compounds that start with H —mostly weak acids —also identify acids by COOH (ex. acetic acid CH3COOH or HC2H3O2) —different way to classify acids by # of acidic hydrogens A. monoprotic: 1 acidic H (ex. HCl) B. diprotic: 2 acidic H (ex. H2SO4) C. triprotic: 3 acidic H (ex. H3PO4) • bases tend to increase concentration of hydroxide (OH-) ions in solution, increase pH of solution —most bases are ionic compounds with OH —include alkali and alkaline earth metal hydroxides —mostly strong bases Neutralization • neutralization reactions involve an acid reacting with a base that result in water and a salt • ex. HCl (aq) + NaOH (aq)—>H2O (aq) +NaCl (aq) • ex. H2SO4 (aq) + 2LiOH (aq)—>2H2O (aq) + Li2SO4 (aq) titrations - neutralizations that involve stoichiometry and solution chemistry • — equivalence point: when 100% of the reactants have reacted (seen from indicator) • ex.A30.0 mL sample of LiOH is titrated with 0.250 M H2SO4 and 43.2 mL of the H2SO4 solution is required to reach equivalent point. What is the concentration of LiOH solution? —balanced equation: 2LiOH + H2SO4—>Li2SO4 + 2H2O —find moles of acid: 0.250 M = x mol H2SO4 = 0.0108 mol H2SO4 0.0432 L —convert mol H2SO4 to mol LiOH: 0.0108 mol H2SO4 x 2 LiOH = 0.0216 mol LiOH 1 mol H2SO4 —find molarity of LiOH: 0.0216 mol LiOH = 0.720 M LiOH 0.0300 L Solubility Rules Rules Precipitation Reactions • precipitation reaction - 2 soluble ionic compounds react to form insoluble or slightly soluble solid product that precipitates out of solution • ex. 2KI (aq) + PB(NO3)2 (aq)—> PbI2 (s) + 2KNO3 (aq) —using solubility rules, we find that PbI2 is not soluble in water, so that is the precipitate —this is an example of a molecular equation (full reaction written as compounds) • total ionic equation —write aqueous ionic compounds as separate ions —2K+ (aq) + 2I- (aq) + Pb2+ (aq) + 2NO3- (aq)—>PbI2 (s) + 2K+ (aq) + 2NO3- (aq) — spectator ions don’t participate in reaction and are on both sides of equation • net ionic equation —excludes spectator ions —2I- (aq) + Pb2+ (aq)—>PbI2 (s) Thermochemistry Part 2 Heating Curves & Heat Capacity • 1 calorie - amount of energy to raise 1 g of water by 1˚C • 1 Calorie (1 kcal) - dietary calorie • 1 Cal = 4,184 J • heating curves show phase changes and melting/boiling points of a substance with heat added compared to temp. • total heat capacity - amount of heat needed to raise temp of something, q/∆T • specific heat capacity - amount of heat needed to raise 1 g of something by 1˚C (or K) • q = mc∆T —q = heat —m = mass —c = specific heat —∆T = change in temp • ex.Achemist has 0.704 g of a pure substance. The temp increases from 8.4˚C to 18.6˚C, which requires 1.69 J of heat. Calculate the specific heat capacity of the substance. —c = q m∆T —c = 1.69 J 0.704 g x 10.2˚C —c = 0.235 J/g˚C molar heat capacity - same as specific heat capacity but in J/mol˚C • Calorimetry, Stoichiometry, & Hess’s Law • q(system) = -q(surroundings) • enthalpy/heat of reaction (∆H) = q(reaction) = -q(calorimeter) • ex.Areaction set up in a simple calorimeter with a water bath containing 243 g of water which cools from 25.3˚C to 22.1˚C. What is ∆H(reaction)? —q(reaction) = -q(calorimeter) — -q(calorimeter) = mc∆T — -q(calorimeter) = 243(4.184)(-3.2) = -(-3253 J) —∆H(reaction) = +3.3 kJ stoichiometry of thermochemical equations • —ex. 2H2O(l)—>2H2(g) + O2(g) ∆H=572 kJ —reverse eq: 2H2(g) + O2(g)—>2H2O(l) ∆H=-572 kJ —cannot have more than 1 mol of product —new eq: H2(g) + 1/2O2(g)—>H2O(l) —∆H proportional to amounts of substance (fraction coefficients acceptable) • Hess’s law - overall ∆H is the sum of the ∆H’s of all steps • ex. CH4 combustion (seen in 2 steps) —step 1: CH4(g) + 2O2(g)—>CO2(g) + 2H2O(g) ∆H=-802 kJ —step 2: 2H2O(g)—>2H2O(l) ∆H=-88 kJ —write net reaction: CH4(g) + 2O2(g)—>2CO2(g) + 2H2O(l) —sum ∆H’s to find total ∆H: -802 - 88 = -890 kJ Standard Heat of Reaction • ∆Hrxn˚ = heat of reaction under standard conditions (constant pressure of 1 atm and usually 25˚C) • standard state of element is most stable physical state under standard conditions • calculate ∆Hrxn˚ from standard heat of formation (∆Hf˚) of 1 mol of substance (get from a table) • ex. 2H2(g) + O2(g)—>2H2O(l) becomes H2(g) + (1/2)O2(g)—>H2O(l) ∆Hf˚=-285.8 kJ • ∆Hf˚ of element in standard state is 0 Bond Strength and Reaction Enthalpy • bond formation is exothermic and bond destruction is endothermic • triple bonds are the strongest and shortest bonds, while single bonds are the weakest and longest bonds • bonds with smaller atoms are usually shorter and stronger than bonds with larger atoms (ex. H-F stronger than H-Cl) • ways to find ∆Hrxn —calorimetry —Hess’s law —use ∆Hf˚ —estimate from bond energies Thermochemistry Part 1 Intro • thermochemistry - study of energy changes that accompany chemical reactions • internal energy (E) - sum of an object’s kinetic energy and its potential energy of composition • potential energy of composition - chemical energy due to bonding of atoms • ∆E = energy of products - energy of reactants • ex. combustion of hydrogen, energy of products lower than reactants ∆E 2H2 + O2 —> reactants 2H2O products Progress of reaction • law of conservation of energy - energy can neither be created nor destroyed, but transformed (energy of universe cannot change) • energy of universe = energy of system + energy of surroundings = 0 • how is energy transferred? —∆E = q + w —q=heat, w=work —energy flow due to temperature difference — work: exerting a force (F) through a distance (d), w=F x d • pressure-volume (P-V) work —involves gases —system does work—>expansion of gases —work done on system—>compression of gases —w=-p∆V • for a system: —∆E>0: system gains energy —q>0: system gains heat from surroundings —w>0: work done on system by surroundings —∆E<0: system loses energy —q<0: system loses heat which goes to surroundings —w<0: work done by system on surrounding • heating a system (q>0) or doing work on it (w>0) increases internal energy (∆E>0) • internal energy is a state function (depends on state of system, but not how it got there) —ex. altitude: going up a mountain ∆E is the same no matter what path you take —∆E depends on initial and final energy levels, not on path between them Systems/Surroundings • 3 general types of systems —isolated —closed —open • isolated systems don’t exchange energy or matter • closed systems exchange energy but not matter open systems exchange energy and matter • • ex. common system —reaction in container placed in water bath —piston in container and thermometer in water bath —container=system, water bath=surroundings • ex. System absorbs 469 kJ heat from water bath and does 327 kJ work on piston. What is ∆E? —∆E = q + w —∆E = +469 kJ - 327 kJ = +142 kJ Heat: Exothermic/Endothermic Properties • exothermic reaction - releases heat to surroundings, feels warm (q<0) • endothermic reaction - absorbs heat from surroundings, feels cold (q>0) • phase changes —solid—>liquid—>gas is endothermic —gas—>liquid—>solid is exothermic • reactions often occur at constant (atmospheric) pressure • enthalpy (H) - measure of total energy of system —H = E + pV —∆H = ∆E + p∆V —∆H = q at constant pressure —q sub p = heat at constant pressure • ∆H>0—>endothermic • ∆H<0—>exothermic • ∆H fusion = enthalpy/heat of fusion, energy required for 1 mole of a solid to become a liquid • ∆H vaporization = enthalpy/heat of vaporization, energy required for 1 mole of a liquid to become a gas Gases Properties of Gases • dominated by kinetic energy rather than intermolecular forces • free flowing and fill containers • relatively low densities • volume changes significantly with pressure and temp. • form solution in any proportion (with other gases) Gas Pressure • gas pressure - result of collision between gas molecules, defined as force/area, measured with manometer • atmospheric pressure - atmosphere and gravity, measured with barometer (avg. 1 atm at sea level) • 1 atm=760 mmHg=760 torr • higher elevation—>lower atmospheric pressure Kinetic Molecular Theory of Gases • kinetic molecular theory states: —a gas consists of a large # of particles in constant motion —no attractions/repulsions between molecules —more volume taken up by space between particles than the particles themselves —speed of particles increases with temp. —collisions are elastic —total kinetic energy is constant —avg. kinetic energy (related to most probable speed) is proportional to absolute temp. (always in K) • most probable speed increases with temp. • kinetic energy related to mass and speed of particle —Ek=(1/2)mv^2 —at given temp. higher speed—>avg. lighter gases Gas Laws • components of gas laws —pressure (P) —temp. (T) in K —volume (V) —moles (n) • Boyle’s law relates volume and pressure, which are inversely proportional: PiVi = PfVf • Charles’s law relates volume and temp, which are directly proportional: Vi= Vf Ti Tf • pressure and temp. are directly proportional • the combined gas law relates pressure, volume, temp, and moles: PiVi = PfVf niTi nfTf • the ideal gas law relates properties of ideal gases: PV=nRT —R is the universal gas constant=0.08206 L x atm mol x K • ideal gas law + stoichiometry ex. What volume of Cl2 gas, measured at 631 torr and 25˚C, is required to form 25 g NaCl? 2Na(s) + Cl2(g)—>2NaCl(s) —convert g NaCl to g Cl2: 25 g NaCl = 0.428 mol NaCl x 1 mol Cl2 = 0.214 mol Cl2 58.44 g 2 mol NaCl —convert torr to atm: 631 torr = 0.830 atm 760 torr —PV=nRT: 6.3 L Cl2 Mixtures of Gases • each gas in a mixture behaves as if it were the only one present • partial pressure - pressure exerted by each gas in a mixture, proportional to mole fraction (mole fraction x total pressure) • sum of partial pressures is the total pressure • mole fraction - # moles component # total moles • ex. Dry air is 78.08% nitrogen (by volume) and 20.95% oxygen. At total pressure of 1 atm, calculate partial pressure of each gas. —mole fraction N2=0.7808 mol —mole fraction O2=0.2095 mol —partial pressure N2=0.7808 mol x 1 atm = 0.78 atm —partial pressure O2=0.2095 mol x 1 atm = 0.21 atm Solubility and Diffusion of Gases • gases dissolving in water relates to dipole-induced dipole forces • solubility depends on: —which gas —which solvent —temp. —partial pressure of gas surrounding liquid • Henry’s law: C(gas)=kH x P(gas) —C(gas)=max concentration of gas —kH=Henry’s law constant (for a certain gas) —P(gas)=partial pressure of gas • as temp. increases, gas solubility decreases • as partial pressure increases, gas solubility increases • gas diffusion - spread of one gas through another, faster moving particles—>faster diffusion • avg. kinetic energy depends on temp. —kinetic energy=(1/2)(mass)(root mean speed)^2 • avg. speed of molecules depends on kinetic energy and mass of particles • at given temp, larger mass—>slower avg. speed • higher temp and/or smaller mass—>faster avg. speed • most probable speed at certain temp. is highest point on graph, avg. speed is slightly faster Molecular Speed Distribution Graph


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