Chemistry: The Molecular Science Moore, Stanitski and Jurs

1. Chemistry: The Molecular Science Moore, Stanitski and Jurs Chapter6: Energy and Chemical Reactions

2. The Nature of Energy Energy (E) = the capacity to do work. Work (w) occurs when an object moves against a resisting force: w = −(resisting force) x (distance traveled) w = −F d All energy is either KineticorPotential energy.

3. The Nature of Energy Potential energy (Ep)– Energy of position. Stored E. It may arise from: • gravity: Ep = m g h (mass x gravity x height). • charges held apart. • bond energy. Kinetic energy (Ek) - Energy of motion • macroscale = mechanical energy • random nanoscale = thermal energy • periodic nanoscale = acoustic energy Ek = ½mv2 (m = mass, v = velocity of object)

4. Energy Units 2.0 kg mass moving at 1.0 m/s (~2 mph): Ek = ½ mv2 = ½ (2.0 kg)(1.0 m/s)2 = 1.0 kg m2 s-2 = 1.0 J 1 J is a relatively small amount of energy. 1 kJ (1000 J) is more common in chemical problems. joule (J) - SI unit (1 J = 1 kg m2s-2)

5. Energy Units calorie (cal) Originally: “The energy needed to heat of 1g of water from 14.5 to 15.5 °C.” Now: 1 cal = 4.184 J (exactly) Dietary Calorie (Cal) - the “big C” calorie Used on food products. 1 Cal = 1000 cal = 1 kcal

6. Conservation of Energy “Energy can neither be created nor destroyed” E can only change form. Total E of the universe is constant. Also called the1st Law of Thermodynamics

7. Conservation of Energy A diver: • Has Ep due to macroscale position. • Converts Ep to macroscale Ek. • Converts Ek,macroscale to Ek,nanoscale (motion of water, heat)

8. Energy and Working If an object moves against a force, work is done. • Lift a book • you do work against gravity. The book’s Ep increases. • Drop the book: • Ep converts into Ek • The book does work pushing the air aside. • The book hits the floor • no work is done on the floor (it does not move). • Ek converts to a sound wave and T of the book and floor increase (Ek converts to heat).

9. Energy and Working In a chemical process, work occurs whenever something expands or contracts. Expansion pushes back the surrounding air. Imagine the gas inside a balloon • heat the gas. • the gas expands and the balloon grows. • the gas does work pushing back the rubber and the air outside it.

10. Energy, Temperature and Heating Temperatureis a measure of the thermal energy of a sample. Thermal energy • E of motion of atoms, molecules, and ions. • Atoms of all materials are always in motion. • Higher T = faster motion.

11. Energy, Temperature and Heating Consider a thermometer. As T increases: • Atoms move faster, and on average get farther apart. • V of the material increases. • Length of liquid column increases.

12. Energy, Temperature and Heating Heat • Thermal E transfer caused by a T difference. • Heat flows from hotter to cooler objects until they reach thermal equilibrium (have equal T ).

13. Systems, Surroundings & Internal Energy System = the part of the universe under study • chemicals in a flask. • the coffee in your coffee cup. • my textbook. Surroundings= rest of the universe (or as much as needed…) • the flask. • perhaps the flask and this classroom. • perhaps the flask and all of the building, etc. Universe= System + Surroundings

14. Systems, Surroundings & Internal Energy Internal energy= E within the system because of nanoscale position or motion Einternal= sum of all nanoscaleEk and Ep • nanoscaleEk = thermal energy • nanoscale Ep • ion/ion attraction or repulsion • nucleus/electron attraction • proton/proton repulsion …..

15. Systems, Surroundings & Internal Energy Internal energy depends on • Temperature • higher T = larger Ek for the nanoscale particles. • Type of material • nanoscale Ek depends upon the particle mass. • nanoscale Ep depends upon the type(s) of particle. • Amount of material • number of particles. • double sample size, double Einternal, etc.

16. SURROUNDINGS SURROUNDINGS SYSTEM SYSTEM Efinal Einitial E out ΔE > 0 ΔE < 0 E in Einitial Efinal Calculating Thermodynamic Changes Energy change of system = final E – initial E ΔE = Efinal – Einitial A system can gain or lose E ΔE positive: internal energy increases ΔE negative: internal energy decreases

17. work heat SURROUNDINGS SYSTEM Heat transfer out q < 0 Heat transfer in q > 0 ΔE = q + w Work transfer out w < 0 Work transfer in w > 0 Calculating Thermodynamic Changes • No subscript? Refers to the system: E = Esystem • E is transferred by heat or by work. • Conservation of energy becomes: ΔE = q + w Note the same sign convention for q and w

18. Heat Capacity Heat capacity = E required to raise the T of an object by 1°C. Varies from material to material. Specific heat capacity (c) • E needed to heat 1 g of substance by 1°C. Molar heat capacity (cm) • E needed to heat 1 mole of substance by 1°C.

19. Heat Capacity E required to change the T of an object is: Heat required = mass x specific heat x ΔT q = mcΔT or… Heat required = moles x molar heat capacity x ΔT q = ncmΔT

20. Substance c (J g-1 °C-1) cm (J mol-1 °C-1) Elements Al(s) 0.902 24.3 C (graphite) 0.720 8.65 Fe(s) 0.451 25.1 Cu(s) 0.385 24.4 Au(s) 0.129 25.4 Compounds NH3(l) 4.70 80.1 H2O(l) 4.184 75.3 C2H5OH(l) 2.46 112. (CH2OH)2(l) 2.42 149. H2O(s) 2.06 37.1 CCl4(l) 0.861 132. CCl2F2(l) 0.598 72.3 Common solids wood 1.76 concrete 0.88 glass 0.84 granite 0.79 Heat Capacities

21. Heat Capacity Example How much energy will be used to heat 500.0 g of iron from 22°C to 55°C? cFe = 0.451 J g-1 °C-1. Heat required = mass x specific heat x ΔT q = mcΔT q = 500.0 g (0.451 J g-1 °C-1)(55−22)°C q = 7442 J = +7.4 kJ + sign, E added to the system (the iron)

22. Heat Capacity Example 24.1 kJ of energy is lost by a 250. g aluminum block. If the block is initially at 125.0°C what will be its final temperature? (cAl = 0.902 J g-1 °C-1) q = m cΔT ΔT = q / (m c) heat is lost, q is negative ΔT = −24.1 x 103 J /(250. g x 0.902 J g-1 °C-1) ΔT = Tfinal – Tinital = −107 °C Thus Tfinal = ΔT+ Tinital = −107 + 125°C = 18°C

23. Heat Capacity A 200. g block of Cu at 500.°C is plunged into 1000. g of water (T = 23.4 °C) in an insulated container. What will be the final equilibrium T of water and Cu? (cCu = 0.385 J g-1 °C-1) Cu cools (−q); water heats (+q); q = m c ΔT Heat lost by Cu = −(200. g)(0.385 J g-1 °C-1)(Tfinal− 500) Heat gained by H2O = +(1000. g)(4.184 J g-1 °C-1)(Tfinal− 23.4) So: −77.0(Tfinal – 500) = 4184(Tfinal – 23.4) (4184 + 77.0)Tfinal = 38500 + 97906 Tfinal = 32.0°C (Note: Tfinal must be between Thot and Tcold)

24. Conservation of Energy and Changes of State Added to a system • q is positive • the change is endothermic Removed from a system • q is negative • the change is exothermic. When heat is: Water Boils: H2O(l)  H2O(g) endothermic Steam Condenses: H2O(g)  H2O(l) exothermic Work occurs as the sample expands or contracts. Overall: ΔE = q + w

25. Conservation of Energy and Changes of State A liquid cools from 45°C to 30°C, transferring 911 J to the surroundings. No work is done on or by the liquid. What is ΔEliquid? ΔEliquid= qliquid + wliquid here wliquid= 0 Heat transfers from the liquid to the surroundings: qliquid = -911 J (qsurroundings = +911 J) ΔEliquid= -911J

26. Conservation of Energy and Changes of State A system does 50.2 J of work on its surroundings and there is a simultaneous 90.1 J heat transfer from the surroundings to the system. What is ΔEsystem? Work done on the surroundings by the system Heat transfers from the surroundings to the system wsystem = -50.2 J qsystem = +90.1 J ΔEsystem= qsystem + wsystem ΔEsystem= -50.2 J +90.1 J = +39.9 J

27. Enthalpy: Heat Transfer at Constant P Because ΔE = q + w: • At Constant V: ΔE = qV • subscript V shows fixed V • work requires motion against an opposing force. • constant V = no motion, so w = 0. • At Constant P: ΔE = qP + watm= ΔH + watm • Subscript P shows fixed P. • watm = work done to push back the atmosphere • H = enthalpy. ΔH = qp

28. Ice is melting. T remains at 0°C Water warms from 0 to 50°C Temperature (°C) -50 -25 0 25 50 Ice warms from -50 to 0°C 0 100 200 300 400 500 600 Quantity of energy transferred (J) Freezing and Melting (Fusion) During freezing (or melting) • Substance loses (or gains) E, but… • T remains constant. Example: Convert ice at -50°C to water at +50°C

29. Changes of State ΔHfusion = qP = heat to melt a solid. Change Name value for H2O (J/g) solid → liq enthalpy of fusion 333 liq → gas enthalpy of vaporization 2260 liq → solid enthalpy of freezing −333 gas → liq enthalpy of condensation −2260 Note: ΔHfusion = − ΔHfreezing etc. qfusion = −qfreezing

30. State Functions and Path Independence State functions Always have the same value whenever the system is in the same state. State functions HE P V T etc. • Two equal mass samples of water produced by: • Heating one from 20°C to 50°C. • Cooling the other from 100°C to 50°C. • have identical final H (and V, P, E…). State function changes are path independent. ΔH = Hfinal – Hinitial is constant.

31. Thermochemical Expressions ΔH = qP can be added to a balanced equation. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH° = −803.05 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)ΔH° = −890.36 kJ ΔH° is the standard enthalpy change • P = 1 bar. • T must be stated (if it isn’t, assume 25°C). • ΔH° is a molar value. Burn 1 mol of CH4 with 2 mol O2 to form 2 mol of liquid water and release 890 kJ of heat • Change a physical state, change ΔH° : H2O(l) vs. H2O(g)

32. Enthalpy Changes for Chemical Reactions Calculate the heat generated when 500. g of propane burns in excess O2. C3H8(l)+ 5 O2(g) 3 CO2(g)+ 4 H2O(l)ΔH° = – 2220. kJ Molar mass of C3H8 = 44.097 g/mol. nC3H8 = (500. g) / (44.097 g/mol) = 11.34 mol C3H8 Since ΔH° = qp= –2220. kJ/(1 mol C3H8) q = (11.34 mol C3H8) (–2220. kJ / mol C3H8 ) = –2.52 x 104 kJ

33. Where Does the Energy Come From? Bond Enthalpy (bond energy) • Equals the strength of 1 mole of bonds • Always positive • It takesE to break a bond • Separated parts are less stable than the molecule. • Less stable = higher E • E is always released when a bond forms • Product is more stable than the separated parts. • More stable = lower E

34. H2(g) + Cl2(g) 2 H(g) + 2 Cl(g) 2 HCl(g) endothermic ΔH= +678 kJ/mol exothermic ΔH= -862 kJ/mol Bond Enthalpies During a chemical reaction: Old bonds break: requires E (endothermic) New bonds form: releases E (exothermic) Both typically occur:

35. energy energy less stability less stability reactants products products reactants Bond Enthalpies Overall, heat may be absorbed or released: Exothermic reactions (ΔH < 0) • E is released. • New bonds are more stable than the old, or • More bonds are formed than broken. Endothermic reactions(ΔH > 0) • E is absorbed. • New bonds are less stable than the old, or • Fewer bonds are formed than broken

36. Measuring Enthalpy Changes Heat transfers are measured with a calorimeter. Common types: • Bomb calorimeter. • rigid steel container. • filled with O2(g) and a small sample to be burnt. • constant V, so qV = ΔE • Flame calorimeter. • samples burnt in an open flame. • constant P, so qp = ΔH • Coffee-cup calorimeter in lab (constant P).

37. A constant for a calorimeter Measuring Enthalpy Changes Bomb Calorimeter Measure ΔT of the water. Constant V: qV =ΔE Conservation of E: qreaction + qbomb + qwater = 0 or −qreaction = qbomb + qwater with qbomb = mcalccalΔT = CcalΔT

38. Measuring Enthalpy Changes Octane (0.600 g) was burned in a bomb calorimeter containing 751 g of water. T increased from 22.15°C to 29.12°C. Calculate the heat evolved per mole of octane burned. Ccal = 895 J°C-1. 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l) −qreaction = qbomb + qwater qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C = +6238 J qwater = m c ΔT = 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C = +2.190 x 104 J So −qreaction = +6238 + 2.190 x 104 J = 2.81 x 104 J = 28.1 kJ qreaction = −28.1 kJ

39. Heat evolved /mol octane= −28.1 kJ 0.00525 mol Measuring Enthalpy Changes Octane (0.600 g) was burned in a bomb calorimeter… Calculate the heat evolved per mole of octane burned Molar mass of C8H18 = 114.23 g/mol. nC8H18 = (0.600 g) / (114.23 g/mol) = 0.00525 mol C8H18 = −5.35 x 103 kJ/mol = −5.35 MJ/mol

40. Measuring Enthalpy Changes Coffee-cup calorimeter Nested styrofoam cups prevent heat transfer with the surroundings. Constant P. ΔTmeasured. q = qp = ΔH Assume the cups do not absorb heat. (Ccal = 0 and qcal = 0)

41. 1 mol 24.31 g 0.4 mol 1 L nMg = 0.800 g = 0.03291 mol nHCl = 0.250 L = 0.100 mol Measuring Enthalpy Changes 0.800g of Mg was added to 250. mL of 0.40 M HCl in a coffee-cup calorimeter at 1 bar. Tsolution increased from 23.4 to 37.9°C. Assume csolution = 4.184 J g-1°C-1 and complete the equation: Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq) ΔH = ? 1Mg ≡ 2HClMg is limiting

42. ΔH 1 mol Mg So = or ΔH= – 462 kJ –15.22 kJ 0.03291mol Measuring Enthalpy Changes … 0.800g of Mg was added to 250. mL of 0.40 M HCl. T increased from 23.4 to 37.9°C. Assume the solution has c = 4.184 J g-1°C-1.ΔH = ? qsolution = msolutioncΔT(msolution = macid + mMg ) = 250.8 g (4.184 J g-1 °C-1)(37.9 − 23.4)°C = 15,220 J Heat from the reaction went into the solution. So: qsolution = – qreaction qreaction = –15.22 kJ = ΔH exothermic

43. Hess’s Law “If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔH° for the 1st reaction must be the sum of the ΔH° values of the other reactions.” Another version: “ΔH° for a reaction is the same whether it takes place in a single step or several steps.” H is a state function

44. Hess’s Law Multiply a reaction, multiply ΔH. Reverse a reaction, change the sign of ΔH. 2 CO(g) + O2(g) → 2 CO2 (g) ΔH = −566.0 kJ Then 2 CO2(g) → 2 CO(g) + O2(g) ΔH = –1(–566.0 kJ) = + 566.0 kJ 4 CO2(g) → 4 CO(g) + 2 O2(g) ΔH = –2(–566.0 kJ) = +1132.0 kJ

45. Example It is difficult to measure ΔH for: 2 C(graphite) + O2(g) 2 CO(g) Some CO2 always forms. Calculate ΔH given: C(graphite) + O2(g) CO2(g) ΔH = −393.5 kJ 2 CO(g) + O2(g) 2 CO2(g) ΔH = −566.0 kJ Hess’s Law Use Hess’s Law to find ΔH for unmeasured reactions.

46. 2 C + 2 O2 2 CO2ΔH° = −787.0 kJ 2 CO2 2 CO + O2 ΔH° = +566.0 kJ 2 C + O2 2 CO ΔH° = −221.0 kJ Hess’s Law Calculate ΔH for the reaction: 2C(graphite) + O2(g) → 2CO(g) Rearrange: +2[C + O2 CO2] +2(−393.5) = −787.0 −1[2 CO + O2 2 CO2] −1(−566.0) = +566.0 or: Add, then cancel: 2 C + 2 O2 + 2 CO2 2 CO2 + 2 CO + O2−221.0

47. 2 C(s) + 2 H2O(g) CH4(g) + CO2(g) C(s) + H2O(g) CO(g) + H2(g) ΔH° = 131.3 kJ CO(g) + H2O(g) CO2(g) + H2(g) ΔH° = −41.2 kJ CH4(g) + H2O(g) CO(g) + 3 H2(g) ΔH° = 206.1 kJ Hess’s Law Determine ΔH° for the production of coal gas: Using: A B C

48. 2 C on left, use 2 x A 2 C(s) + 2 H2O(g) 2 CO(g) + 2 H2(g) +262.6 1 CH4 on right, use −1 x C CO(g) + 3 H2(g) CH4(g) + H2O(g) −206.1 1 CO2 on right, so use 1 x B CO(g) + H2O(g) CO2(g) + H2(g) −41.2 Add and cancel: 2C + 3H2O + 2CO + 3H2 2CO + 3H2 + CH4 + H2O + CO2 15.3 kJ change to 2 H2O Hess’s Law Want:2 C(s) + 2 H2O(g) CH4(g) + CO2(g) 2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ

49. H2 combustion: 2 H2(g) + O2(g) 2 H2O(l)ΔH° = −571.66 kJ but the formation reaction is: H2(g) + ½ O2(g) H2O(l)ΔHf° = −285.83 kJ 1 Standard Molar Enthalpy of Formation Hess’s law problems often use a combustion or … Formation reaction Make 1 mol of compound from its elements in their standard states. f = formation

50. Standard Molar Enthalpy of Formation Standard state = most stable form of the pure element at P = 1 bar. • e.g. C standard state = graphite (not diamond) ΔHf° for any element in its standard state is zero. (take 1 mol of the element and make… 1 mol of element) ΔHf° (Br2(l) ) = 0 at 298 K ΔHf° (Br2(g) ) ≠ 0 at 298 K