Lecture 25: W 10/19 Lecture 26: F 10/21 Lecture 27: M 10/24. Reading:

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1 Lecture 25: W 10/19 Lecture 26: F 10/21 Lecture 27: M 10/24 Reading: Week 9: Lectures BLB Ch , 5.5, Homework: BLB 18: 9, 11, 15, 29, 69; Supp 18: 1 3; BLB 11:33, 37, 39, 31, 43, 47, 50; Supp 11:5-13 Reminder: Angel Quiz 8 due on Thur 10/20 ALEKS Objective 9 due on Tue 10/25 Jensen Office Hour: 501 Chemistry Building Tuesdays & Thursdays, 10:30 11:30 am TA office hours, evening tutor sessions, guided study group info posted on Angel -> Help Available Exam 3: Monday Nov 7 6:30 7:45 pm Real gases deviate from ideal behavior The 5 key postulates of KMT! Straight-line motion in random directions! Molecules are small & have no volume compared to the total volume! No intermolecular forces! Elastic collisions! Mean kinetic energy! " T (in K) For a non-ideal gas (real gas) this is not true for two types of conditions: high P low T Reasons: Molecules (or atoms) have a finite size molecules occupy space Molecules have attractive forces these forces become stronger when they are close together Jensen Chem 110 Chap 10 Page: 2

2 Real Gases Pressure For 1 mol of any ideal gas: PV/(RT) = 1 For 1 mol of different gases at same T & different P: Real Gases Temperature For 1 mol of any ideal gas: PV/(RT) = 1 For 1 mol of N 2 gas at different T & different P Low P PV/(RT) ~ 1 High P: PV/(RT) < 1 Very high P PV/(RT) > 1 At low P, the ideal-gas equation is valid: PV/(RT) = 1 In general, deviation from ideal behavior increases as P increases: At high P, attractive forces between molecules lead to the appearance of a smaller n PV/(RT) < 1 Example: CO 2 at 200 atm At very high P, finite molecular volume leads to repulsion and the appearance of a larger n PV/(RT) > 1 Example: CO 2 at 800 atm Jensen Chem 110 Chap 10 Page: 3 At low T, molecules stick together: PV/(RT)! 1 As temperature increases, molecules move faster, the behavior of real gases becomes more ideal. Jensen Chem 110 Chap 10 Page: 4

3 Summary High P PV/RT < 1 Very high P PV/RT > 1 Low P PV/RT! 1 As T #, real gas behavior becomes more ideal Correcting the Ideal Gas Law Attractive forces lead to the appearance of a smaller n Finite molecular volume leads to repulsion and the appearance of a larger n Attractive forces and finite molecular volume have minimal impact At high T, Kinetic energy overcomes attractive forces Example: Under which conditions is Ar(g) most likely to approach ideal behavior? A. B. C. D. E. 10 atm and 100 C 1.0 atm and -200 C 0.10 atm and 100 C 20.0 atm and 100 C 0.10 atm and 200 C Jensen Chem 110 Chap 10 For any gas, we can measure P, V, T At higher P, measured P is too small due to attractive forces between molecules The amount of missing P is proportional to 1. size of attractive forces (a): 2. frequency of collisions: (n2/v2) To compensate, use: P+ At higher P, measured V is too large due to finite molecular volume or excluded volume per mole (b) Actual volume: Vactual = Vmeasured Vexcluded To compensate, use: Page: 5 Jensen V nb Chem 110 Chap 10 Page: 6

4 Real Gases: the van der Waals Equation The Equation of state for IDEAL gases: PV = nrt The Equation of state for REAL gases: correction for intermolecular forces between gas molecules NOTE: a & b are specific for the particular gas we are talking about (BLB Table 10.3). a (L 2 -atm/mol 2 ) b (L/mol) He Xe H Cl CH CCl correction for excluded volume (size) of gas molecules Outline: Chapter 18: Atmospheric Chemistry Composition of Atmosphere Regions of Atmosphere Photochemical reactions in the upper atmosphere $ Chemistry in upper atmosphere protects us from UV radiation Photoionization Photodissociation(natural ozone cycle) $ Human activity has impact on the chemistry of the upper atmosphere Ozone depletion by CFCs Photochemical reactions in the lower atmosphere $ Water Vapor, CO 2 and the Climate $ Sulfur Compounds and Acid Rain $ Nitrogen Oxides and photochemical Smog Jensen Chem 110 Chap 10 Page: 7

5 Composition of the Atmosphere Composition of Dry Air Near Sea Level Component Content (mole fraction) Molar Mass Nitrogen Oxygen Xenon Mole fraction: x i = Express mole fraction in parts per million (ppm): 1 ppm = x i! 10 6 moles of i total moles Partial Pressures of Components Example: The mole fraction of NO on a smoggy day is measured at 20 ppm. If the barometric pressure is 732 torr, what is the partial pressure of NO in the atmosphere? A torr B torr C torr D torr E torr Example: Neon x Ne = , what is the mole fraction of Ne in ppm? Jensen Chem 110 Chap 18 Page: 2 Jensen Chem 110 Chap 18 Page: 3

6 Regions of the Atmosphere Based on temperature profile 1. Troposphere: T % as altitude # 2. Stratosphere: T # as altitude # 3. Mesosphere: T % as altitude # Pressure Profile Pressure at any altitude depends on the weight of gas above it. Pressure decreases exponentially with altitude. At high altitude, pressure is low... How does low pressure affect density? How does low pressure affect molecular collisions? 4. Thermosphere: T # as altitude # How does low collision frequency affect the frequency of chemical reactions? Jensen Chem 110 Chap 18 Page: 4 Jensen Chem 110 Chap 18 Page: 5

7 Important Chemistry in the Atmosphere Photochemical Reactions: reactions that involve the absorption or emission of a photon. Example: What is the maximum wavelength of a photon needed to photoionize O 2 in the upper atmosphere? Photoexcitation: electronic excitation NO 2 + h& ' NO 2 * (* = excited state) Photodissociation: bond broken by absorption of a photon (light) O 2 + h& ' O + O Requires energy " the bond energy O 2 + h& ' O e A nm B. 165 nm C. 274 nm D nm E nm IE = 1205 kj/mol Photoionization: removal of a valence e from a molecule by absorption of a photon N 2 + h& ' N e Requires energy " the ionization potential Jensen Chem 110 Chap 18 Page: 6 Jensen Chem 110 Chap 18 Page: 7

8 Importance of Atmospheric Ozone (O 3 ) The Natural Ozone Cycle Formation of O 3 O 2 (g) + h& ' 2 O (g) O (g) + O 2 (g) 'O 3 (g) Destruction of O 3 O 3 (g) + O (g) '2 O 2 (g) Protection of Earth#s surface from UV-radiation: EDG =, MG =, bond angle = 117, bond length = 1.28Å Light blue gas Pungent odor (smell near electrical discharges) (H f = 142kJ/mol (reactive less stable than O 2 ) Atmospheric Concentration: In the troposphere: O 3 is an irritant (smog) in the stratosphere: O 3 is essential; peak concentration at~25 km; [O 3 ] ~ 10 ppm The (small) amount of O 3 in the stratosphere reflects the delicate balance between creation and destruction of O 3. O 3 (g) + h& ' O (g) + O 2 (g) Ozone Depletion Chlorofluorocarbons (CFCs), CFCl 3, CF 2 Cl 2, CF 3 Cl, destroy ozone cycle CF 2 Cl 2 + h& ' CF 2 Cl + Cl () < 240 nm) 2Cl + 2O 3 ' 2ClO + 2O 2 2ClO + h& ' 2Cl + O 2 NET: 2 O 3 + h& ' 3 O 2 Cl atom acts as a catalyst to destroy O 3 1 Cl atom can destroy > 100,000 O 3 molecules!?!! Jensen Chem 110 Chap 18 Page: 8 Jensen Chem 110 Chap 18 Page: 9

9 The Greenhouse Effect Energy coming to Earth is mostly solar $ Some is reflected by atmosphere (~30%) $ Some UV is absorbed by atmosphere $ Most hits Earth#s surface, where it is absorbed and causes warming Energy radiated by Earth#s surface is mainly infrared (IR) heat The atmosphere is transparent to UV and Visible light, but NOT to IR The H 2 O, CO 2, CH 4 and O 3 in atmosphere absorbs IR light Sulfur Compounds and Acid Rain Natural rain: ph ~ 6 (slightly acidic due to dissolved CO 2 ) Acid rain: ph of 4 to 4.5 (more acidic due to dissolved pollutants like H 2 SO 4, HNO 3 ) Sources: bacterial decay of organic matter volcanic gases and forest fires fossil fuels combustion and industrial processes Example: Coal burning (contaminated by sulfur): S 8 (s) + 8 O 2 (g) ' 8 SO 2 (g) SO 2 reacts in the atmosphere: 2 SO 2 (g) + O 2 (g) ' 2SO 3 (g) SO 3 (g) + H 2 O (g) ' H 2 SO 4 sulfuric acid dissolves in water and falls to earth as acid rain affects ph of soil and water corrodes metals (Fe) dissolves stone (marble, limestone) Thermal regulation by these gases (mostly CO 2 and H 2 O) in the atmosphere leads to the Greenhouse Effect CaCO 3 (s) + H 2 SO 4 (aq) ' CaSO 4 (aq) + CO 2 (g) + H 2 O(!) Jensen Chem 110 Chap 18 Page: 10 Jensen Chem 110 Chap 18 Page: 11

10 Acid Rain Natural rain: ph ~ 6 slightly acidic due to dissolved CO 2 Acid rain: ph of 4 to 4.5 more acidic due to dissolved pollutants like H 2 SO 4, HNO 3 Photochemical smog primary h& secondary pollutants pollutants NO, NO 2, CO O 3 hydrocarbons In auto engine: N 2 +O 2 ' 2NO In air: 1. 2NO + O 2 ' 2NO 2 NO 2 + h& ' 2NO + O (H =181 kj ) = *400 nm O + O 2 + M ' O 3 + M (M=another molecule) 2. NO 2 + H 2 O' HNO 3 (burns eyes) Ozone: good in the stratosphere, but not in the troposphere Diminishes respiratory capabilities Reacts with NO to form NO 2 and O 2 Photodissociates to form reactive oxygen radicals, which react with hydrocarbons, etc. Jensen Chem 110 Chap 18 Page: 12 Jensen Chem 110 Chap 18 Page: 13

11 Kinetic Molecular Description of Matter Gas: Kinetic energy >> intermolecular forces Liquid: Kinetic energy + intermolecular forces Solid: Kinetic energy << intermolecular forces Kinetic Energy " T: Functional group Structure Affects Function Structure MW g/mol BP C hydrocarbon aldehyde Heating: T #, KE # : Cooling: T %, KE % : solid ' liquid ' gas gas ' liquid ' solid ketone amine alcohol carboxylic acid Jensen Chem 110 Chap 11 Page: 1 Jensen Chem 110 Chap 11 Page: 2

12 Viscosity and Surface Tension Viscosity: resistance to flow; higher viscosity: slower flowing Vapor Pressure Vapor pressure (v.p.): pressure exerted by a vapor in equilibrium with its liquid or solid phase; $ stronger IM forces, more to flow $ IM forces increases, viscosity Surface tension: energy needed to increase surface area; $ Intermolecular interactions are favorable (heat is required to break them) $ Surface molecules have fewer interactions. $ IM forces increases, surface tension Example: Which of the following has the highest viscosity? A. B. C. Dynamic equilibrium: no NET change, but change is occurring on molecular level forward rate = backward rate evaporation = condensation Jensen Chem 110 Chap 11 Page: 3 Jensen Chem 110 Chap 11 Page: 4

13 Vapor Pressure and Temperature Vapor Pressure and Boiling Point $ At higher temperature, more molecules have energies sufficient to escape the liquid, so the vapor pressure is higher $ For different liquids, as IM forces increases, vapor pressure (at a given T) decreases Example: Which of the following has the highest vapor pressure? A. B. C. Boiling point (BP): T at which v. p. = P ext As P ext increases, BP Normal boiling point: T at which v. p. = Example: Why does it take longer to hard boil eggs at higher altitudes than at lower altitudes? Jensen Chem 110 Chap 11 Page: 5 Jensen Chem 110 Chap 11 Page: 6

14 Phase Diagrams Comparison of H2O and CO2 Phase Diagrams Plot of pressure vs. temperature of the system showing the boundaries between the phases. Differences: 1) Triple Point: Be able to identify the following; 2) Direction of the solid-liquid coexistence line: $ normal melting point CO2 slants to the as P increases: MP as P# $ normal boiling point $ critical point H2O slants to the as P increases: MP as P# $ triple point Example: Using the phase diagram for water, how many phases are encountered if the water sample is heated from -10ºC to 100 ºC? The pressure is kept at 780 torr through the heating process. $ coexistence curves Jensen Chem 110 Chap 11 Page: 7 Jensen Chem 110 Chap 11 Page: 8

15 Phase Changes Heating Curve and Calorimetry Two types of changes take place as heat is added: Endothermic Process: Requires energy to disrupt intermolecular forces. Melting (fusion) Vaporization Sublimation Exothermic Process: Energy is released when intermolecular interactions are formed Freezing Condensation Deposition 1. Heating within a single phase (in blue): q = m C p (T (p, constant pressure) T #, average KE #, total energy #,molecular separation #, molecular attractions %, molecular order %. 2. phase transition (in red): changes from one physical state to another q = n (H x (x = fusion, vaporization) T is constant, average kinetic energy is constant, but total energy #, separation between molecules #, molecular attractions %, molecular order %. Jensen Chem 110 Chap 11 Page: 9 Jensen Chem 110 Chap 11 Page: 10

16 Heating within a Single Phase: q = C m (T C = heat capacity (J/g-K or J/mol-K) m = amount of substance (g or mol) (T = T final - T initial Example: The specific heat of toluene (C 7 H 8 ) is 1.13 J/g-K. How many Joules of heat are needed to raise the temperature of 0.5 mole of toluene from 0 C to 25 C? Note: (T (in K) = (T (in C) Heat Capacity (C): Amount of heat required to raise an object#s temperature by 1K (or 1 C). Usually C is given for a specified amount of pure substance Example: liquid H 2 O specific heat, C = J/g-K molar heat capacity, C m = 75.2 J/mol-K $ Specific heat is different for each phase $ Specific heat is different per gram than per mole Jensen Chem 110 Chap 11 Page: 11 Jensen Chem 110 Chap 11 Page: 12

17 Practice Example: What is the specific heat of gold, if raising the temperature of 985 g of Au from 30 C to 45 C requires kj of energy? A J/g-K B J/g-K C J/g-K D J/g-K E J/g-K Heating during a Phase Transition: q = m (H x (x = fusion, vaporization) (H = heat (kj/g or kj/mol) given off (-) or absorbed (+) when a change occurs m = amount of substance (g or mol) (H fusion = heat needed to melt a mole of substance (H vap = heat needed to vaporize a mole of substance Jensen Chem 110 Chap 11 Page: 13 Jensen Chem 110 Chap 11 Page: 14

18 Calculation of Energy Change Measure the heat required for each segment of heating curve, then sum all individual steps. Scratch Paper Example: Ethanol (C 2 H 5 OH) melts at 114 C. The enthalpy of fusion is 5.02 kj/mol. The specific heats of solid and liquid ethanol are 0.97 J/g K and 2.3 J/g K, respectively. How much heat is needed to convert 25.0 g of solid ethanol at 135 C to liquid ethanol at 50 C? A kj B kj C kj D kj E kj Jensen Chem 110 Chap 11 Page: 15 Jensen Chem 110 Chap 11 Page: 16