1 / 36

Chapter 6 - Thermochemistry

Chapter 6 - Thermochemistry. Prelab for Exp #2 due 10/7 or 10/9 Online HW Ch 6 Due before: 10/15 Exam #1 On 10/15; Quiz #1 on 10/14 (Monday) The exam will cover chapters 5 & 6; Fe & Unk Metal Labs. pgs 258-267 Suggested Problems. 6 11 17 18 19 25 26 38 43 47

carnig
Télécharger la présentation

Chapter 6 - Thermochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 - Thermochemistry Prelab for Exp #2 due 10/7 or 10/9 Online HW Ch 6 Due before: 10/15 Exam #1 On 10/15; Quiz #1 on 10/14 (Monday) The exam will cover chapters 5 & 6; Fe & Unk Metal Labs

  2. pgs 258-267 Suggested Problems 6 11 17 18 19 25 26 38 43 47 51 59 61 67 69 75 77 79 83 85 87 91 103 123 146 Plus: Calculate the ΔHr for: 1N2F2 (g) + 1F2 (g) ---) 1N2F4 (g) using the bond energy table 9.5 page 367. (answer = -1.30x102 kJ)

  3. I. Introduction A. Definitions • Thermodynamics deals with energy relationships. • Thermochemistrydeals with energy relationships in chemical reactions. • Thermochemistry allows us to: 1)predict the spontaneity of a reaction (Chapter 18) 2)calculate the heat lost or gained in a reaction (Ch. 6) 3) calculate the useful work involved in a reaction (Ch. 18) 4)calculate equilibrium constants (Chapters 18 & 19) 5)calculate voltages of redox reactions (Chapter 19) 6)calculate bond energies (Chapter 9 & 6)

  4. I. Introduction A. Definitions Energy = The ability to move matter - Are several forms of energy associated with chemical reactions: heat, electrical, light & mechanical energies. - These types of energy can be interconverted and are usually measured as if they were a) heat energy in calories or b) mechanical energy in joules. - Are two forms of energy: Kinetic & Potential. Potential Energy (Ep) – Stored energy due to position. - Ep can be energy stored in the chemical bonds, in the intermolecular forces, and in the subatomic particles.

  5. I. IntroductionA. Definitions & B. Examples Kinetic Energy – Energy associated with motion of an object Ek = ½ Mass x (Velocity)2 = ½ m x v2(know) If m is in kg & v in meters/sec then E in J; 1J = 1 kg m2 / sec2 B. Examples - Calculate Ek of 4 kg bowling ball going 2 meters per sec. Ek = ½ MV2 = ½ (4 kg)(2 m/s)2 = 8 kg m2 / s2 = 8 J (joule) - a joule is a small unit. - a watt (a measure of power or rate of using energy) = J/sec - a 60 watt light bulb uses 60 J of energy per second.

  6. I. Introduction B. Examples - A typical American home uses ~ 4 billion joules per month in electricity. Note: since types of energies can be interconverted, we can report electrical energy in terms of either mechanical energy (joules) or heat energy (calories). - Older unit of energy is calorie (cal) = amount of energy required to raise temperature of 1 g of water by 1 oC. - 1.000 cal = 4.184 J - How many J are in a large piece of apple pie which releases 500 Cal upon digestion? (Note: 1 dietary calorie = 1 Cal = 1 kcal; typical candy bar has 400 Cal = 400,000 cal !) 500. Cal 1 kcal 1000 cal 4.184 J = 2.09x106 J 1 Cal 1 kcal 1 cal

  7. I. Introduction C. Internal Energy • If we want to determine the total energy of a substance, then we must consider: 1) its position [Ep]; 2) its overall motion [Ek]; 3) the internal motion of its protons, neutrons and electrons [ek]; and 4) the potential energy of these subatomic particles [ep]. • ek + ep = U (Internal Energy) • ET = Ek + Ep + U • Internal Energy = sum of kinetic & potential energies of the subatomic particlesmaking up the substance.

  8. I. Introduction C. Internal Energy • ET = Ek+ Ep + U • EkEp & U can vary; however, ET remains constant = Law of Conservation of Energy • Law of Conservation of Energy = energy may be converted from one form to another, but the total quantity of energy remains constant. • We will not concern ourselves with U.

  9. II. Heat/Enthalpy A. Introduction • System = the part of the universe you are studying (rxn). • Surroundings = the rest of the adjacent universe which interacts with the system (calorimeter). • Heat = energy that flows because of a difference in temperature. • Heat spontaneously flows from hot to cold substances because of higher molecular motion in hotter substance. • The amount of heat in chemical reactions is related to the amount, type & temperature change of the substances; temperature is related to molecular motion of the particles of the substance.

  10. II. Heat/Enthalpy A. Introduction • Generally we use Δq or ΔH for heat; Δ = change. H is also called enthalpy & is the heat involved under constant P. • Δq or ΔH = + when heat absorbed by a chemical reaction. • Δq or ΔH = - when heat given up by a chemical reaction. Changes that absorb heat (surroundings cooler) = endothermic(Δq = +) Changes that give upheat (surroundings hotter) = exothermic(Δq = - ) • When one burns wood, what is the sign of Δq for this rxn? • When one dissolves NH4NO3 in water, the water drops in temperature. What is sign of Δq for this dissolving? • KNOW these sign conventions well; many points in both chemistry 122 & 123 depend on knowing these signs.

  11. II. Heat/Enthalpy B. Enthalpy • ΔHr = heat of reaction = change in enthalpy for a reaction. Be careful, since ΔHrmay be expressed as either 1) kJ per mol of a given chemical or 2) total kJ for the balanced reaction (rxn). Example: 2H2(g) + 1O2(g) 2H2O(l) ∆Hr = 1) -286 kJ for 1 mol of H2O or ∆Hr = 2) -572 kJ for the balanced rxn • ∆Hr = Hfinal - Hinitial = ∑Hproducts - ∑Hreactants Note: This equation works because ΔH is a state function. State Function – property whose value depends on the initial & final states; not how that change was reached.

  12. II. Heat/Enthalpy B. Enthalpy (state function) Example of a State Function = Altitude change from B to A is 3600 ft regardless of path taken.

  13. III. Heats of Reaction, ΔHrA. Introduction – Signs, Rxn & ΔH • We frequently show ΔHr (heat of reaction) after the reaction: 2 H2 + 1 O2 ---) 2 H2O ΔHr = - 572 kJ for rxn(exothermic) - 572 kJ for rxn; - 572 kJ/1 mol O2; - 286 kJ/1 mol H2; - 286 kJ/1 mole H2O • Note: For reverse reaction ΔH has same magnitude, but changes sign = + 572 kJ (+ = heat absorbed). 2 H2O ---) 2 H2 + 1 O2 ΔHr = + 572 kJ(endothermic) • Calculate the ΔH in the top rxn when 5.1g of H2 reacts. 5.1 g H21.0 mol H2 – 572 kJ = - 729 kJ 2.0 g H22.0 mol H2

  14. III. Heats of Reaction, ΔHrA. Intro – Method Summary 1) Reactions in Calorimeters (exp): ΔH = C x ΔT 2) Reactions in Calorimeters (exp): ΔH = g x s x ΔT 3) Hess’s Law: Add & Subtract Reactions 4) Bond Energies: (table 9.5, p 367. a) Draw Lewis Structures; b) Consider mols of bonds broken & formed); c) use sign (+) from Table. ∆Hr = ∑ ∆Hbonds broken -∑ ∆Hbonds formed 5)Heats of Formation: ΔHof (table 6.2, p 250; a) use sign (±) from table; b) consider mols of compound) ΔHr = ∑ ΔHof products- ∑ ΔHofreactants

  15. III. Heats of Reaction, ΔHrB. Determining H experimentally1) Reactions in Calorimeters: ΔH = C x ΔT • Heat Capacity = C = Heat needed to raise T of system 1 oC. ΔH = C x ΔT where ΔT = Tf – TiC in J/oC or kJ/oC(know) Be careful: C may be for a) empty or b) full calorimeter Example: Where C=11 J/oCfor full calorimeter. Calculate ΔHr in J for a rxn in this calorimeter in which the T goes from 25.1 to 30.3 oC. ΔH = C x ΔT = 11 J/oC x 5.2 oC = 57 J Example: Calculate C for a full calorimeter if the calorimeter is calibrated and found to absorb 69 J when the T goes from 30.0 to 37.2 oC. C = ΔH / ΔT = 69J / 7.2 oC = 9.6 J/oC

  16. III. Heats of Reaction, ΔHrB. Determining H experimentally2) Reactions in Calorimeters: ΔH = g x s x ΔT • Specific Heat Capacity = s = Heat needed to raise T of 1 g of substance by 1 oC; s is characteristic of substance. ΔH = g x s x ΔT s in units of: J / goC or kJ / goC (know) Example: How much heat (kJ) is must be added to raise 225 g of H2O from 25.0 to 100.0 oC? Given: s = 4.18x10-3 kJ/(goC) for water ΔH = g x s xΔT ΔH = 225 gx4.18x10-3 kJ/(goC) x 75.0 oC = 70.5 kJ Example: 5.00 g of Ag absorbs 25.0 J of heat when heated by 21.3 oC. Calculate s for Ag. ΔH = g x s x ΔT s = ΔH / [g x ΔT] s = 25.0 J / [5.00 g x 21.3 oC] = 0.235 J/goC

  17. III. Heats of Reaction, ΔHrB. Determining H experimentally; notes for methods 1 & 2: 1) C may be for empty calorimeter or fullcalorimeter – caution 2) Small s = good heat conductor; Large s = good insulator 3) s = 1.00 cal/goC or 4.18 J/goCor 4.18x10-3 kJ/goCfor water 4) May use one or both equations (1&2)in a problem 5) These two are Experimental methods (rest of methods are theoretical) 6) When using ΔH = g x s x ΔTg= mass of solution; get g from density (1.00 g/mL unless told otherwise) & from volume.

  18. III. Heats of Reaction, ΔHrB. Measuring - Using s & C Example: 15.5 g of metal (Me) is heated to 98.9 oC and dropped into 25.0 g of water at 22.5 oC in a calorimeter. The water temperature goes to 25.7 oC. Calculate the specific heat capacity (s) of the unknown Me. Note: ΔT water = (25.7- 22.5)oC = 3.2 oC for water ΔTmetal = (25.7- 98.9)oC = -73.2 oC for metal a) qwater = g x s x ∆T = 25.0g x 4.18 J/(goC) x3.2oC =3.3x102 J b) qwater = - qMe & -qMe = - 3.3x102J = gMexsMexΔTMe sMe = - 3.3x102J/ [gxΔT] = - 3.3x102J / [15.5g x -73.2oC] = 0.29 J / (goC) Note: a) When perform a reaction in a calorimeter, the q lost by Rxn = q gained by calorimeter -- will have opposite signs. b) s must always be a positive value.

  19. III. Heats of Reaction, ΔHrB. Measuring – Calorimeter Example given C for entire system • 0.56 g of carbon is placed in a bomb calorimeter at 25.00 oCalong with excess O2. After ignition the T is 25.89oC. The heat capacity for the filled calorimeter, C = 20.7 kJ/oC.Calculate theΔHreaction(ΔHr) in kJ/mol of C. Note: ΔT = (Tf – Ti) = 0.89 oC. • Note: Use only heat capacity equation:ΔH = C x ΔT ΔHcal= CxΔT = 20.7 kJ/oC x 0.89oC = +18 kJ (for cal) ΔHr = - ΔHcal = -18 kJ for burning 0.56 g of C -18 kJ 12.0 g C = - 3.9x102kJ/mol of C 0.56 g C 1.00 mol C 1C + 1O2 -----) 1CO2ΔHr = - 3.9x102 kJ/mol of C

  20. III. Heats of Reaction, ΔHrB. Measuring – Calorimeter Example given s for soln • Mix 50.0 mL 0.250 M HCl & 50.0 mL 0.250 M NaOH together. The T goes from 19.50 oC to 21.21 oC. Calculate ΔHr in kJ/mol. Given the dsoln = 1.00 g/mL & ssoln = 4.18x10-2 kJ/(goC). • notes: a) 100.0 g soln; b) use ΔH = s x g x ΔT c) ΔT = 1.71 oC (exo; q = - ) 0.250 mol/L x 0.0500L = 0.0125mol of either HCl & NaOH 1HCl + 1NaOH -----) 1NaCl + 1H2O ΔH = s x g x ∆T ΔHsoln = 4.18x10-3 kJ/(goC) x100.0g x 1.71oC = 0.715 kJ ΔHr = - 0.715 kJ & is for 0.0125 mol (convert to kJ/mol) ΔHr = - 0.715 kJ / 0.0125 mol= - 57.2 kJ / mol

  21. III. Heats of Reaction, ΔHrB. Measuring – Example using both s & C Some calorimeter problems use both ΔH = s x g x ΔT (given s for contents) and ΔH = C x ΔT (given C for empty calorimeter) • AnEmptycalorimeterhas a heat capacity of 0.0111 kJ/oC. A chemical reaction takes place in 32 mL of water inside the calorimeter, and the temperature goes from 25oC to 46oC. Calculate the ΔHr in kJ. ssoln = 4.18x10-3 kJ/goC ; d = 1.00 g/mL • ΔHgainedbysystem = ΔHcalorimeter + ΔHsolution ΔHE cal = C x ΔT = 0.0111 kJ/oC x 21 oC = 0.23 kJ ΔHsoln = s x g x ΔT = 4.18x10-3 kJ/goC x 32 g x 21 oC = 2.8 kJ ΔH = ΔHcalorimeter + ΔHwater = 0.23 kJ + 2.8 kJ = 3.03 kJ Since the reaction is exothermic: ΔHr = - 3.0 kJ

  22. III. Heats of Reaction, ΔHr B. Measuring – using both s & C (doing in lab) Calculate ΔHr in kJ\mol Mg for: 1Mg + 2HCl -----) 1MgCl2 + 1H2 102 g of soln (s= 3.93 J/goC) contains 1 mol of HCl. You add 0.01036 mol of Mg to the soln in a calorimeter (C= 15 J/ oC for the empty calorimeter )& T raises by 12.47 oC. Note: Mg is limiting reagent = 0.01036 m ΔH = Hsoln + Hcalorimeter= (s x g x ΔT) + (C x ΔT) ΔH = (3.93 J/goCx 102g x 12.47oC) + (15J/oC x 12.47oC) = 5000+190 = 5190 J Calorimeter system absorbs 5190 J or 5.19 kJ (-5.19 kJ for reaction) ΔHrxn = - 5.19 kJ / 0.01036 mol Mg = - 501 kJ/mol Mg

  23. III. Heats of Reaction, ΔHrC. Hess’s Law • Hess’s Law:The heat of reaction is the same, whether the reaction is carried out directly in one step or indirectly through a number of steps – add & subtract heats. (Use given rxn as a template.) • Example 1: Calculate ΔHr for: 2S + 3O2 -----) 2SO3 Given: a) S + O2 -----) SO2ΔH = - 297 kJ b) 2SO3 -----) 2SO2 + O2ΔH = + 198 kJ Solve: Multiply # a) by 2; reverse # b); ADD results together 2 x [S + O2 -----) SO2ΔH = -297] = - 594 kJ 2SO2 + O2 -----) 2SO3ΔH = - 198 kJ 2SO2 + 2S + 3O2 ----) 2SO2 + 2SO3ΔHr = - 594 - 198 = -792 kJ 2S + 3O2 -----) 2SO3ΔHr = - 792 kJ

  24. III. Heats of Reaction, ΔHrC. Hess’s LawExample 2 – from lab #4 Calculate ΔHr for: 2 Mg + 1 O2 -----) 2MgO From following 3 reactions. 1) Mg + 2HCl -----) MgCl2 + H2 H1 = - 501 kJ x 2 2) MgO + 2 HCl -----) MgCl2 + H2O H2 = - 151 kJ x - 2 3) 2 H2 + O2 -----) 2 H2O H3 = - 572 kJ OK 2 Mg + 4 HCl -----) 2 MgCl2 + 2 H2 H = - 1002 kJ 2 MgCl2 + 2 H2O -----) 2 MgO + 4 HCl H = + 302 kJ 2 H2 + O2 -----) 2 H2O H3 = - 572 kJ 2Mg + 4HCl + 2MgCl2 + 2H2O + 2H2 + O2 --) 2MgCl2 + 2H2 + 2MgO + 4HCl + 2H2O 2Mg + O2 --) 2MgO ΔHr = -1002 +302 -572 = -1272 kJ

  25. III. Heats of Reaction, ΔHrD. Bond Energies • Bond Energy = Energy needed to break a bond in gas phase. ∆Hbreak bond= + value (adding energy) (Table 9.5, pg 367 + values) ∆Hform bond= - value (releasing energy) - Know signs • Typical bond energies ~ 150 to1000 kJ/mol • Can use bond energies to calculate heats of reaction (∆Hr). • ∆Hr = ∑ ∆Hbonds broken ∑ ∆Hbonds formed(Know) 1) Need to know # bonds & type of bonds involved; use Lewis Structures and stoichiometry: examples: a) 2BF3 = six B-F bonds b) CO is CΞO (not C=O & not C-O) 2) The above equation works USING SIGNS GIVEN IN TABLE since ∆Hbonds broken = add energy (+) ∆Hbonds formed = releases energy (-).

  26. III. Heats of Reaction, ΔHrD. Bond Energies • ∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed] • Example 1: calculate the ∆Hrxn for following: 1 H2 + 1 Cl2 ------) 2 HCl - Given these bond energies from Table 9.5; pg 367 [ H-H = 432 ; Cl-Cl = 240; H-Cl = 428 kJ/mole ] 1 H2 + 1 Cl2 ------) 2 HCl 1x432 1x240 2x428 ∆Hr = [432 + 240] - [856] = - 184 kJ 1 H2 + 1 Cl2 ------) 2 HCl∆Hr = - 184 kJ

  27. III. Heats of Reaction, ΔHrD. Bond Energies • Example 2: Bond energy problem (may not be single bonds). Calculate the ΔHr for: 1C2H2 + 2Br2 -----) 1CHBr2CHBr2 1) Draw the Lewis Structures; 2) determine bond energies; 3) calculate ΔHr. Br Br 1) H-C≡C-H + 2 Br-Br -----) H-C - C-H Br Br 2) C-H= 411 C≡C = 835 Br-Br= 190 C-Br= 285 C-C= 346 3) ∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed] Bonds BrokenBonds Formed 2C-H 1C≡C 2Br-Br 4C-Br 2C-H 1C-C ∆Hr = [2x411 + 1x835 + 2x190] - [4x285 + 2x411 + 1x346] = -271 kcal

  28. III. Heats of Reaction, ΔHrE. Summary & Standard Ethalpies of Formation: ΔHof • We have five ways of calculating heats: 1) Reactions in Calorimeters 1: q = C x ΔT 2) Reactions in Calorimeters 2: q = s x g x ΔT 3) Hess’s Law: Adding & Subtracting Rxns 4) Bond Energies: ∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed] 5)Heats of Formation: ΔHof ΔHr = ∑ ΔHof products– ∑ ΔHofreactants • In order to use the last way we need some values of ΔHof - Chemists devised a relative scale of enthalpies since path to answer not important (ΔH is state function).

  29. III. Heats of Reaction, ΔHrE. Heats of Formation, ΔHof • ΔHofis a relative number defined using a standard state. - The Standard State = 25 oC and 1.0 atm; represented by o • ΔHof = enthalpy change for formation of 1 mol of substance in its standard state from its elements in their standard states. KNOW This Definition (several pts on Ex 1 & Final involved) - Examples of Heat of formation reactions (use definition): 1) for H2O(l): 1H2(g) + 1/2O2(g) 1H2O(l)ΔHr = ΔHofH2O(l) 2) for N2O4(g) : 1N2(g) + 2O2(g) 1N2O4(g)ΔHr = ΔHofN2O4(g) 3) for AgCl(s) : 1Ag(s) + 1/2Cl2(g) 1AgCl(s)ΔHr = ΔHofAgCl(s)

  30. III. Heats of Reaction, ΔHrE. Heats of Formation, ΔHof(Most Common Method for Determining ΔHr) • ΔHof of a pure element in it’s standard state is 0.00 kJ/mol - Which are not elements in their standard states?: H2(g) Br2(l) Ag(s) KF(s) Hg(l)Fe+3(aq) O3(g) ΔHof = 0.00 kJ/mol for each except: KF(s) Fe+3(aq) O3(g) • Table 6.2, pg250 & appendix C have tables of ΔHof values. • ΔHr = ∑ ΔHof products– ∑ ΔHofreactants (use signs given in Table & Table values are kJ/1Mole)

  31. III. Heats of Reaction, ΔHrE. Heats of Formation, ΔHof • Calculate ΔHr for: 2 NO2(g) ----) 1 N2O4(g) Table 6.2:ΔHof = 33.2 for NO2& 9.16 for N2O4(kJ/1 mol) 2 NO2(g) ----) 1 N2O4(g) 2 x 33.21 x 9.16 ΔHr = ∑ ΔHof products– ∑ ΔHofreactants ΔHr = [1 x 9.16] - [2 x 33.2] = - 57.2 kJ

  32. III. Heats of Reaction, ΔHrE. Heats of Formation, ΔHof • Calculate ΔHr for: 2Cdiamond + 1O2(g) -----) 2CO(g) Pg 250: ΔHof = 2 kJ/m For Cdiamond & - 111 kJ/m for CO(g) 2Cdiamond + 1O2(g) -----) 2CO(g) 2 x 2 1 x 0 2 x -111 4 0 -222 ΔHr = ∑ ΔHof products – ∑ ΔHofreactants ΔHr = [ - 222] - [ 4 + 0] = - 226 kJ Note: ΔHof = 0.00 for O2 (g) & will not be given on exams

  33. III. Heats of Reaction, ΔHrE. Heats of Formation, ΔHof(Note: I like the following set-up format for these problems) • Calculate ΔHr for following at 25oC & 1 atm pressure. GetHof from Table 6.2, pg 250 (element in Std. St. = 0); Values in kJ/mol. 3 CH4(g) + 2 H2O(l) + 1 CO2(g) -----) 4 CO(g) + 8 H2(g) Hof : 3(-74.9) 2(-286) 1(-394) 4(-111) 8(0) -225 -572 -394 -444 0 ΔHr = ∑ ΔHof products– ∑ ΔHofreactants ΔHr = [-444 + 0] - [-225 + -572 + -394] = + 747 kJ

  34. IV. Summary for Calculation of ∆Hreaction 1) Reactions in Calorimeters: q = C x ∆T 2) Reactions in Calorimeters: q = s x g x ∆T 3) Hess’s Law: Add & Subtract Rxns 4) Bond Energies: [take # mols & # bonds (Lewis Structure) into account; use Table 9.5, pg 367] ∆Hr = ∑ ∆Hbonds broken- ∑ ∆Hbonds formed 5)Heats of Formation, ΔHof: (from tables: pg 250 or pg A-8; take # mols ito account & ΔHof = 0 for element in it’s standard state) ∆Hr = ∑ ∆Hof products– ∑ ∆Hofreactants

  35. Brief Summary of Lewis Structures Note: Rules 1 and 4 are double checks on your structure. 1. Calculate total # of valence electrons - take into account the charge if it is a polyatomic ion. 2. Place atom that forms most bonds at center (Closest to Group IVA/14 & Lowest if in same group). If is a charge, then take into account on central atom. 3. Arrange other atoms around central atom & allow sharing so that each atom has stable electron configuration. Show bonding pairs as lines & nonbonding valence e- as dots. 4. Double check rule: a) see that each atom has a stable electron configuration and b) check that you still have the same number of valence electrons that you started with in step #1. Useful generalities: For neutral, Stable Compounds: H, F, Cl, Br have 1 bond; C has 4 bonds; N has 3 bonds; O has 2 bonds. B stable with 6 electrons/3 bonds; H stable with 2 electrons. Try with: H2Se AlF4- H2CO N2F2 C2H2 C2H4 C2H6

More Related