G. L. * Effusion of Gases * Real Gases * van der Waals equation * Your extra credit question will have to do with this topic! * Atmospheric Chemistry * 3 types of pollution-very, very basic question Chapter 6 – Thermochemistry * Nature of Energy * System versus Surroundings * Definition of Energy, internal energy, law of conservation of energy * 1st Law of Thermodynamics * ? E = q + w * Sign convention, (is it positive or negative) * Heat and work * pV work * m Cs ? T heat transfer * conservation of thermal energy * Calorimetry * Constant volume calorimetry * only heat contributes to ? E * Enthalpy * Definition, equation Calculation using constant pressure calorimetry * Exothermic versus Endothermic reactions (sign of ? H) * Hess's Law * Enthalpy of reactions manipulations * This is a hard topic, please, please, please review this after Wednesday! Examination #2 - Chapters 4, 5, and 6 Study Guide Chapter 4 - Chemical Quantities and Aqueous Reactions * Reactions Stoichiometry * mole-mole conversions * Needs a balanced chemical equation * **Again no clear examples. Let me know if you can find any** * mass-mass conversions * **No clear examples. Let me know if you can find any** * Limiting Reactants * What is the Limiting Reagent The limiting reagent is also known as the limiting reactant. It is the reactant that limits the amount of product in a chemical reaction. Notice that the limiting reactant is the reactant that makes the least amount of product. * How do we find the L. R. * Example: * How many grams of N2 (g) can be produced from 9. 05 g of NH3 (g) reacting with 45. 2 g of CuO (s)? Create and Balance a Chemical Equation: 2NH3 (g) + 3CuO (g) N2 (g) + 3Cu (S) + 3H2O (l) 9. 05 g NH3 x 1 mol NH3 x 1 mol N2 x 28. 02 N2 = (7. 44 g N2) 17. 04 g NH3 2 mol NH3 1 mol N2 45. 2 g CuO x 1 mol CuO x 1 mol N2 x 28. 2 N2 = (5. 31 g N2 Less = LR Cuo is the Limiting Reactant! * Solutions * Morality - definition and how to calculate * Definition: * Amount of solute (in moles) per amount of solution (in Liters) * Molarity (M) = Amount of Solute (in moles) Amount of Solution (in L) * **Side Note** * Homogenous Mixture = solutions (Salt Water) * Solvent (a component in a solution) : Majority component, what something is dissolved in. (Water) * Solute (another component in a solution) : Minority component, what is being dissolved (salt) * Example: What is the molarity of a solution containing 3. 4 g of NH3 (l) in 200. 00 mL of solution? Given: 3. 4 g of NH3M = moles of solute (NH3) 200. 00 mL L of Solution (200. 00 mL) Convert: 3. 4 g NH3 X 1 mol NH3 = (0. 20 mols NH3) 17. 04 G nh3 200 mL X 1 L = (0. 2L) 1000 mL M = 0. 20 mols NH3 / 0. 2 L = 1. 0 M NH3 **More examples in Notes! ** * Dilutions Calculations (M1V1 = M2V2, careful with M2) * Diluting a solution is a common practice and the number of moles of solute will not change! [ (M1)(V1) = (M2)(V2) ] * Examples: What is the concentration of a solution prepared by diluting 45. mL of 8. 25 M HNO3 to 135. 0 mL? M1V1 = M2V2 8. 25 M HNO3 X 0. 045 L = M2 X 0. 135 L 0. 135 L 0. 135 L M2 = 275 M HNO3 * Solution Stoichiometry * volume-volume conversions * When using morality, you can easily extract moles! * With a balance chemical equation, you can convert between amounts of substances. * Exampes: Look at notes OR page 145 TB * volume-mass conversions * Examples: Look at notes OR page 145 TB * **This wasn’t clear and If you know what this means, let me know. Or else I will ask Donavan on Saturday (Because there wasn’t a specific section for the two bullet points) Molecular interpretation of solubility * solubility rules – be familiar with the chart/table that Prof. Donavan gave out 2 interactive forces that affect solubility: 1. solute-solute interaction 2. solute-solvent interaction if solute-solvent interactions are strong enough, solute will dissolve (solute-solvent interaction ; solute-solute interaction) * Precipitation Reactions * Determining reaction products General Form: AX (aq) + BY (aq) > AY (aq) + BX (s) Example: 2KI (aq) + Pb (NO3)2 (aq) >2KNO3 (aq) + PbI2 (s) * Following Solubility rules Molecular Formula, Total ionic formula, net ionic formula Examples: Molecular Formula: 2KOH (aq) + Mg(NO3)2 (aq) > 2KNO3 (aq) + Mg(OH)2 (s) Total ionic formula: * 2K+ (aq) + 2(OH)– (aq) + Mg2+ (aq) + 2(NO3)– (aq) > 2K+ (aq) + 2(NO3)- (aq) + Mg(OH)2(s) Net Ionic formula: (remove all spectator ions : ions that are aqueous as reactants and stay aqueous when they turn into products) Mg2+(aq) + 2(OH)-(aq) > Mg(OH)2(s) * Acid-Base Reactions General Form: HA (aq) + BOH (aq) > H2O (l) + BA (aq) Example: HCl (aq) + NaOH (aq) > H2O (l) + NaCl (aq) * Oxidation-Reduction reactions Oxidation is the loss of electrons * Reduction is the gain of electrons * Oxidation states: charges that allow us to keep track of electrons in chemical reactions * Identify oxidation states 1. Charge states of neutral compounds are zero 2. Charge of atoms in polyatomic ions need to add up to the total charge of the polyatomic 3. Keep Alkali metals as +1 alkali earth metals as +2 4. Keep F (fluorine’s) as -1 H as +1 O as -2 * Identify which species was oxidized and reduced * Look in last section of Chapter 4 Notes Chapter 5 – Gases * Pressure – definition Pressure: The force per unit area * Pressure comes from the constant interaction with a container * Standard Pressure = Normal Atmospheric Pressure * 760. 0 mm Hg = 1 atm * 760. 0 torr = 1 atm * 1. 000 atm * 101, 325 pa (pascals) = 1 atm * 14. 7 psi (lbs per square inch) = 1 atm * Example: * (45. 0 psi) x (101, 325 pa) x (1 k pa) _____________________________ = 310. kPa (14. 7 psi) x (1000 pa) * Simple Gas Laws * Boyle's Law – pV * The volume of a gas inversely proportional to its pressure, provided the temperature and quantity of gas don’t change. * V= k/p Actual Equation: pV= K * Example: A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0. 500 atm. If the volume of the balloon is now 2. 78 x 10^3 mL, what was it originally? V1 = 782 torr x 1. 000 atm/760 torr = 1. 03 atm (1. 03 atm)(V1) = (. 500 atms)(2. 78 x 10^3 mL) After Rearranging the equation: V1= 1350 mL or 1. 35 x 10^3 mL * Charles's Law - P/T * The volume of a gas is diretly proportional to its temperature, provided the pressure and quantity of the gas that don’t change. (V= KT) **Temp in Kelvin Only** * For changes in Volume (involving temperature): * V1/T1 = V2/T2 For Changes in Pressure: * P/T (initial) = P/T (final) * Example: (LOOK IN NOTES ) * Avogadro's Law – nT * The volume of a gas is directly proportional to the quantity of gas, provided the pressure and temperature of the gas don’t change. (V=Kn) * For changes in volume (involving moles) * V1/n1 = V2/n2 * Example: (LOOK IN NOTES ) * Ideal Gas Laws * pV=nRT * NEED TO KNOW THIS FORMULA! * P = pressure (atm) * V = volume (L) * n = quantity (moles) * T = temperature (K) * R = Universal Gas Constant * (0. 08206 Latm/molK) OR * (8. 314 J/molK) * Example: (look in notes ) Density calculations * Density of a gas @ STP: * For an Ideal gas @ STP, the molar volume = 22. 7 L * Density = mass/volume = mass/1mole = molar mass/molar volum * volume/1mole * Density for a gas NOT @ STP: * If gas isn’t at stp * Then D = P(MM)/ RT or D = m/v * Molar Mass calculations * From the equations: pV = mRT/MM You get: MM = mRT/ pV * Example (Look in notes ) * Molar Volume * At STP, all ideal gases take up the same volume. * Molar Volume = # of L of gas 1 mole of gas This also works: V/n = RT/P * Partial Pressures Dalton's Law of Partial Pressures * The total pressure of a mixture of gases is the sum of the pressures by each gas. * The pressure of a gas would exert if it were alone in a container. * You can calculate the Partial Pressure from Ideal gas Law * If 2 gases , A and B are mixed together * P(A) = (nA)(R)(T)/ (V) and P(B) = (nB)(R)(T)/ (V) * Since R, T, and V are all constant for a mixture * P(total) = P(A) + P(B) = (nTotal)(R)(T)/ (V) * nTotal = sum of nA + nB * Example: (Look in notes ) Eudiometer calculations * An Eudiometer is a gas collecting Tube * Example: 2Zn (s) + 6HCl (aq) 3H2 (g) + 2ZnCl3 (aq) H20 (l) H2O (g) P(total) = P(H2) + P(H20) (value may be looked up at table 5. 4) * 0. 12 moles of Hz is collected over H20 in a total 10. 0 L container at 323 K. Find the total pressure. P = nRT/V P(H2) = (0. 12 mol H2) (0. 08206 Latm/molK) (323 K)= 0. 3181 am (10. 0L) P(total) = P(H2) + P(H20) P(H2O) @ 50 degrees Celsius = 92. 6 mmHg P(total) = 240mmHg + 96. 6mmHg = 330mmHg * Gas Reaction Stoichiometry * General Concept plan on most problems: P, V, T of Gas A Amount A (in moles) Amount B (in moles) P, V, T of Gas B * Volume - moles conversions * Ex: Methanol CH3OH can be synthesized by the following reaction * CO2 (g) + 2H2(g) CH3OH(g) * What is the volume (in liters) of hydrogen gas @ a temperature of 355 K and pressure of 738 mmHG, is required to synthesize 35. 7 g of methanol * Given: 35. 7 g CH3OH temp: 355 K pressure: 738 mmHG * Find: V of H2 * 1. G of CH3OH mols * 35. 7g CH3OH x 1 mol CH3OH = 1. 1142 mol CH3OH 31. 04 g CH3OH * 2. Mol CH3OH mol H2 * 1. 11 mol CH3OH x 2 mols H2 = 2. 23 mols H2 1 mol CH3OH 3. N(mol H2), P, T VH2 * Convert your mmhg to ATM, and get . 971 atm * VH2= (2. 23 mol H2) (. 08206 l atm/ mol K) (355 K) = 66. 9 L .971 atm * VH2= 66. 9 L * Kinetic Molecular Theory * In this theory a gas is modeled as a collection of particles (either molecules or atoms depending on the gas ) in constant motion. * Ex, a single particle moves in a straight line until it collides with another particle (or with the walls of its container). * 4 components of the theory 1. Particles are infinitely small and have no volume 2. Average kinetic energy of a particle is proportional to the temperature (k). . Particles travel in two straight lines following Newtonian Laws 4. All collisions are elastic (no attractive or repulsive forces) * You DO NOT need to know the derivation of I. G. L. * Effusion of Gases * Effusion: the process by which a gas escapes from a container into a vacuum through a small hole. * The rate of effusion (the amount of gas that effuses in an amount of time) is also related to the root mean square velocity * Rate is ? 1M * Grahms law of effusion: * The ratio of effusion rates of two different gases. * For example (look in notes, end of chapter 5) Real Gases * van der Waals equation is an equation used to correct for the discrepancies from the Kinetic Molecular Theory that real gases undergo. Real gases attract each other, therefore, real pressure ; ideal pressure. Real gases also take up space, therefore, real volume ; ideal volume. [P + a (n/v)? ] x (V – nb) = nRT where: a – corrects for molecular interaction. It makes the real pressure larger so it equals the ideal pressure b – corrects for molecular size. It decreases the volume of the container. * Your extra credit question will have to do with this topic! * Atmospheric Chemistry 3 types of pollution-very, very basic question * 3 types of pollution-very, very basic question 1. Hydrocarbon combustion for automobiles 2C8H18 + 2SO2 > 16CO2 + 18 H2O At high temperature, nitrogen can also be combusted, which causes a problem. N2 + O2 > 2NO 2NO + O2 > 2NO2 (nitrogen dioxide) – photochemical smog (causes problem in the environment) 2. Combustion of coal from power plants (Ex. Electrical cars) C + O2 > CO2 (Coal contains a significant amount of sulfur and it further combusts) S8 + 8O2 > 2SO3 2SO2 + O2 > 2SO3 SO3 + H2O > H2SO4 (H2SO4 results to acidification)
But, people have found a way to eliminate the production of SO3 and that is by using “clean coal” and scrubbers. CaCO3 + SO2 > CaO + CO2 CaO + SO2 > CaSO3 (s) (calcium sulfite) 3. Stratospheric Ozone O3 + UV > O2 + O (oxygen radical) O2 + O > O3 + IR These two equations above just shows how ozone is used and how it is just regenerated again. But, in 1974, Sherwood Rowland discovered that CFCs from air conditioners, refrigerators, and spray cans destroy the atmospheric ozone. CF2Cl2 + UV > CF2Cl + Cl (chlorine radical) Cl + O3 + UV > O2 + ClO ClO + O > O2 + Cl ( 1 Cl radical can destroy a hundred thousands of ozone)
Practice test: answer keyChapter 6 – Thermochemistry * Nature of Energy * System versus Surroundings System – the part of the universe we want to focus on (like a chemical reaction inside a beaker) Surrounding – everything else in the universe (like the glass of the beaker and the air around it) * Definition of Energy, internal energy, law of conservation of energy Energy is classified into two types: a. heat (q) – energy transferred that causes a temperature change (due to a change in the random motion of molecules) b. work (w) – energy transferred that causes an object to move (due to a change in the concerted motion of the molecules in the object) c. nits of energy: I. Joule (J) – the amount of energy it take to move 1kg mass a distance of 1 meter (unit: kg*m2/s2) II. Calorie (cal) – the amount of energy needed to raise the temperature of 1 gram of water by 1 ? C 1 kcal = 1000 cal (food calories) 1 cal = 4. 184 J (exact measurement) Internal Energy – total energy of a system. (Esystem) Law of conservation of energy – energy is neither created or destroyed, only transferred. * 1st Law of Thermodynamics - The change in energy of a system is equal to heat that enters the system plus the work done on the system. * ? E = q + w a. ?E = change in the internal energy of a system E is (+) if the energy is absorbed by the system ?E is (-) if the energy is released by the system b. q = heat q is (+) if the heat is absorbed by the system q is (-) if the heat is released by the system c. w = work w is (+) if the work is done on the system w is (-) if the work is done by the system on the surrounding * Heat and work * pV work - is defined by the equation: w = -p? V * m Cs ? T heat transfer - q = m Cs ? T where: m = mass Cs = specific heat capacity (J/ g ? C) ?T = (Tfinal – Tinitial) - q = n Cm ? T where: n = number of moles Cm = molar heat capacity (J/ mol ? C) ?T = (Tfinal – Tinitial) conservation of thermal energy – the amount of energy that is given must be equal with opposite sign to that energy that is being taken. qsurr = - (qsys) msurr Cs(surr) ? T(surr) = -[msys Cs(sys) ? Tsys] * Calorimetry * Constant volume calorimetry * Constant volume calorimetry - “bomb” calorimetry, no pv work done, therefore only heat contributes to ? E qcal = Ccal ? T = -qrxn where: Ccal = calorimeter constant (KJ/ ? C) * * only heat contributes to ? E * Enthalpy * Definition, equation Enthalpy (? H) – the heat absorbed or released during a process taking place at a constant external pressure. ?H = qrxn = -qsurr ?H = -( m Cs ? T) Calculation using constant pressure calorimetry – refer to example in notes * Exothermic versus Endothermic reactions (sign of ? H) Endothermic reactions have (+) ? H because they are reactions that absorb heat. Exothermic reactions have (-) ? H because they are reactions that give off heat. * Hess's Law * Enthalpy of reactions manipulations 2 rules to remember: 1. If a reaction is reversed, the sign of ? H flips (from negative to positive or from positive to negative) 2. If you multiply coefficients by a number, ? H is also multiplied by that number. * This is a hard topic, please, please, please review this after Wednesday!
