Thermochemistry: Part of Thermodynamics

Thermochemistry: Part of Thermodynamics Dr. Vickie M. Williamson @vmwilliamson Student Version 1 Chemical Thermodynamics! Thermodynamics: study of the energy changes associated with physical and chemical processes! Thermochemistry: study of heat changes in chemical reactions and physical changes 2 A Chemical Reaction Copyright 1995 by Saunders College Publishing C 6 H 12 O 6(s) + 4KClO 3(l) --> 6CO 2(g) + 6H 2 O (g) + 4KCl (s) 3 vm williamson 1

Periodic Property: Electronegativity Na + Cl 2 à Copyright 1995 by Saunders College Publishing Energy Potential energy is stored energy Kinetic energy is motion energy 5 Heat and Temperature! Heat is energy, not matter! Heat is a type of Kinetic energyparticles (atoms/molecules) are moving.! Heat can be gained or lost! Temperature is a measure of heat or kinetic energy. heat flow 6 vm williamson 2

Heat Units: Joule (J) = kg-m 2 /s 2 =10 7 ergs calorie = 4.184 J British thermal unit (BTU) = 1055 J Conversion: 1 calorie = 4.184 J 1 calorie = amount of energy to raise the temperature of 1 g of water 1 o C 1 Calorie = 1 food calorie = 1 kcal = 1000 calories So for Snickers: 250.0 Calories This is 250,000 calories How many J? 7 Heat Transfer Copyright 1995 by Saunders College Publishing 8 Heat Transfer at the Molecular Level Copyright 1995 by Saunders College Publishing 9 vm williamson 3

Heat and Matter Copyright 1995 by Saunders College Publishing 10 Calculating Heat! To heat room temperature water for coffee, the amount of heat needed depends on??? 1. 2. 3.! q = 11 Characteristics of the Substance Heat capacity: amount of heat required to raise temperature Specific heat capacity: amount of heat required to raise temperature of 12 vm williamson 4

Comparative Heat Capacities 200 g of Cu 200 g H 2 O 1.4 x 10 6 g H 2 O 76 J/ o C 837 J/ o C 5.9 x 10 6 J/ o C 13 Some Specific Heat Capacities Substance aluminum 0.902 copper 0.385 gold 0.128 water(l) 4.184 water(s) 2.06 ethanol 2.46 O 2 (g) 0.917 N 2 (g) 1.04 Specific Heat (J/g. K) 14 Calculating Heat Units on q will depend on units of Specific Heat Specific heat of water = vm williamson 5

How many calories of heat is required to raise the temperature of g of liquid water from 25.0 to.0 C? (A) 1.0 e4 cal (B) 10.5 e3 cal (C) 43.9 e3 cal(d) 10,500. cal How many joules of heat? 16 System vs Surroundings 17 Thermodynamic Terms! System: Subject(s) involved in the change! Surroundings: Everything in the system s environment! Universe: 18 vm williamson 6

Thermodynamic Terms Surroundings Energy System Energy 19 20 21 vm williamson 7

Exothermic Reactions: Release of Stored Chemical Energy Copyright 1995 by Saunders College Publishing 22 Hot and Cold Packs Exothermic Endothermic 4Fe(s) + 3O 2 (g) --> 2Fe 2 O 3 (s) + Heat NH 4 NO 3 (s) + Heat --> NH 4 NO 3 (aq) 23 Endothermic and Exothermic Processes Energy Heat absorbed from surroundings (Endothermic) CO 2 gas CO 2 solid Heat released to surroundings (Exothermic) 24 vm williamson 8

A Chemical Reaction Copyright 1995 by Saunders College Publishing H 2 (g) + 1 / 2 O 2 (g) --> H 2 O(g) + 242 kj/mol 25 Endothermic and Exothermic Processes Energy H 2 (g) + 1 / 2 O 2 (g) Heat absorbed Heat from released surroundings to (Endothermic) surroundings H 2 O(g) (Exothermic) H 2 O(l) 26 Energy for the Space Shuttle Copyright 1995 by Saunders College Publishing 27 vm williamson 9

28 WHY?? WHY?? WHY?? 29 WHY?? The Laws of Thermodynamics Thermodynamics is based upon observations of common experience that have been formulated into Laws From these few Laws, all of the remaining Laws of Science are deducible by purely logical reasoning 30 vm williamson 10

The First Law of Thermodynamics 31 Change of Energy of the System System at State 1 with Energy E 1 ± Heat ± Work System at State 2 with Energy EE 2 2 Change in Energy = ΔE sys ΔE sys = E final E initial = E 2 E 1 If E final > E initial then If E final < E initial then 32 The First Law! ΔE universe = ΔE sys + ΔE surr = 0! ΔE system = q + w where: ΔE system internal energy change q heat w PV work = PΔV = (Δn)RT! At fixed n(gas) or V: w = 0 and ΔE system = q V 33 vm williamson 11

Sign Conventions for q and w Surroundings q q System w > 0 w < 0 34 P-V Work Copyright 1995 by Saunders College Publishing 35 Work in Chemical Systems! Recall: w = (Δn)RT = PΔV H 2 (g) + 1 / 2 O 2 (g) --> H 2 O(g) Δn = n prod n reac = = Therefore: work is (when work is done system)! Gummy Bear: C 6 H 12 O 6 (s) + 4KClO 3 (l) --> 6CO 2 (g) + 6H 2 O(g) + 4KCl(s) Δn = n prod n reac = = Therefore: work is (when work is done system) 36 vm williamson 12

! What is the internal energy of a system that does kj of work on the and releases kj of heat to the surrounding? (A) 250 kj (B) 150 kj (C) -250 kj (D) -150 kj 37 ΔE sys, q and w C 8 H 18 (l) + 25 / 2 O 2 (g) --> 8CO 2 (g) + 9H 2 O(g) initial state final state Energy Energy lost as heat C 8 H 18 (l) + 25/2O 2 (g) 8CO 2 (g) + 9H 2 O(g) Energy lost as heat and work 38 Sign Conventions for q and w System does work on surr Work done on system Heat absorbed by system (endothermic) Heat released by system (exothermic) Whenever energy (heat or work) is added to a system, the energy of a system 39 vm williamson 13

Enthalpy (H)! the total heat content of a system (H)! at constant pressure, the enthalpy change ΔH is: ΔH = q P = H final H initial! At constant P: w = PΔV and q P = ΔH ΔE system = q + w ΔE = q P PΔV ΔE = ΔH PΔV ΔH = ΔE + PΔV! For gases at fixed T: PΔV = (Δn)RT ΔH = ΔE + (Δn)RT 40! If no work ΔE system =! If no work and constant pressure then: ΔE system = =! ΔH is a state function and 1 factor in thermodynamics 41 State Function of the path taken to traverse from initial state to final state 42 vm williamson 14

State Functions P, V, T, H, E are state functions 43 Exothermic Reaction Energy Profile Heat Content H 2 (g) + 1 / 2 O 2 (g) Heat Evolved = 242 kj H 2 O(g) Progress of a Reaction 44 Energy Graphs System Exothermic = ΔH, IF not work, then also ΔE system = System Endothermic = ΔH, IF not work, then also ΔE system = 45 vm williamson 15

Internal Energy (E)! Defined as the sum of kinetic and potential energies of all particles in a system! It is a state function! In a chemical or physical process, where reactants are converted to products, the change in internal energy (ΔE) is given by: ΔE = E final E initial = E products E reactants 46 Predict the signs for the following: 1.H 2 O(l) à H 2 O(g) 2.Chocolate(l) à Chocolate (s) 3.CH 4 (g) + 2O 2 (g) à CO 2 (g) + 2H 2 O(g) 1 2 3 47 Heat Required for Phase Changes 48 vm williamson 16

Calculating Heat Transferred in a Process - Relationships For temperature changes: Heat(J) = [Mass(g)] [Sp Heat(J/g C)] [Temp change( C)] q = For phase changes: Heat (J) = [Mass(g)] [ H phasechange (J/g)] q = For a specific amount of material or equipment: Heat (J) = [Heat Capacity(J/ C)] [Temp change( C)] q = 49 Calculate the heat required to transform 25.00 g of ice at -10.00 C to water at 30.00 C. S solid = 2.04 J/g C S gas = 1.40 J/g C S liquid = 4.184 J/g C H vap = 2200 J/g H fus = 335 J/g! Solid -10.00 to 0.00 q =! Solid 0.00 to Liquid 0.00 phase change q =! Liquid 0.00 to 30.00 q = TOTAL = Or J in sig figs 50 Calorimetry! Specific amount of substance in calorimeter undergoes chemical and/or physical change; resulting energy transfer is monitored by temperature changes! For an exothermic reaction in solution: q rxn = q cal + q sol = K T + ms T! If q rxn is losing heat (neg q), then q cal + q sol are gaining heat (pos q).! Work in absolute, then assign negative OR apply a neg to one side. 51 vm williamson 17

Coffee-Cup Calorimeter! Pressure is constant! Measures q p ( H) 52 Bomb Calorimeter! Volume is constant, so no work! Measures q V ( E) 53! When 2.00 g of NaOH are dissolved in 500. ml of water in a cup calorimeter, the water changes from 25.00 C to 37.00 C. The heat capacity of the calorimeter is 470. J/ C. What is the heat involved per mole? (molar enthalpy)! q p =! =! = g (4.184J/g C) (12.00 C) + 470. J/ C (12.00 C)! = +! = J given off for 2.00g NaOH! Mole NaOH = 2.00/40.0g = 0.0500mol! Heat/mole NaOH =! H/mol NaOH = = -6.17 e5 J/mol or -617 kj/mol 54 vm williamson 18

Thermochemical Equations: Constructing Unit Factors! For the reaction C 2 H 5 OH(l) + 3O 2 (g) --> 2CO 2 (g) + 3H 2 O(l) + 1367 kj the following unit factors may be written: 1367 kj mol O 2 mol O 2 1367 kj 1367 kj mol C 2 H 5 OH 1367 kj mol CO 2 1367 kj 1 mol rxn 1 mol C 2 H 5 OH 1 mol rxn Heat is proportional!! 55 Thermochemical Equations: Reversibility! Reversing the equation changes the sign but not the magnitude of H H 2 (g) + 1 / 2 O 2 (g) --> H 2 O(g) ΔH = 242 kj H 2 O(g) --> H 2 + 1 / 2 O 2 (g) ΔH = kj 56 Thermochemical Equations: Proportionality! If equation is multiplied by a factor n, H must be multiplied by same factor H 2 (g) + 1 / 2 O 2 (g) --> H 2 O(g) ΔH = 242 kj 2H 2 (g)+ O 2 (g) --> 2H 2 O(g) ΔH = kj 57 vm williamson 19

Thermochemical Equations: Combinations of Known H Values! H for individual steps of a reaction sequence may be added to give H for overall reaction C(s) + O 2 (g) --> CO 2 (g) CO 2 (g) --> CO(g) + 1 / 2 O 2 (g) ΔH = 394 kj/mol ΔH = +283 kj/mol 58 Hess Law! Enthalpy change of a process depends only on the initial and final states; it is independent of the path or the number of steps involved in the process! If several reactions are added (or subtracted) to give an overall (net) reaction, H terms are also added (or subtracted) accordingly! If reaction is reversed, the sign of H must be changed accordingly! If reaction is multiplied,, the value of H must be multiplied accordingly 59 Hess Law: Law of Heat Summation ΔH 2 ΔH 1 ΔH 3 Reactants ΔH Products ΔH 4 ΔH 5 ΔH = ΔH 1 + ΔH 2 + ΔH 3 = ΔH 4 + ΔH 5 60 vm williamson 20

Find Hrxn from Combinations of Known H Values! Problem: Calculate ΔH o rxn for: CO(g) + NO(g) --> CO 2 (g) + 1 / 2 N 2 (g) given the following data: CO(g) + 1 / 2 O 2 (g) --> CO 2 (g) ΔH o A = 283 kj O 2 (g) + N 2 (g) --> 2NO(g) ΔH o B = 180.6 kj! Solution: CO(g) + 1 / 2 O 2 (g) --> CO 2 (g) ΔH o A = NO(g) --> 1 / 2 O 2 (g) + 1 / 2 N 2 (g) ΔH o C = ΔH o rxn = 61 Thermochemical Equations: Guidelines! H values change significantly with moderate changes in temperature 62 Standard Molar Enthalpy of Formation (ΔH f )! ΔH f is the amount of heat released or absorbed when of a compound in a specified state is formed from its elements also in their standard states; also referred to as heat of formation 2C (graphite) + 3H 2(g) + 1 / 2 O 2(g) --> C 2 H 5 OH (l) H o f = 277.7 kj/mol 63 vm williamson 21

Standard State! Defined as the most stable state of a substance at atm and at some specified temperature, almost always 25 C (298 K)! Liquids =! Gases =, 8 th family! Solids = rest H 2 O(l) Br 2 (l) C(s) NO 2 (g) Hg(l) 64 Standard States Hg(l) NO 2 (g) S(s) HCl(1 M)! Pure substance (liquid or solid): standard state is the pure liquid or solid! Gases: standard state is the gas at atm pressure! Solutions: standard state is M concentration 65 ΔH f for Some Elements! The standard molar enthalpy of formation for any element in its standard state is Substance ΔH o f (kj/mol) Br 2 (l) 0 Br 2 (g) 30.91 C(diamond) 1.897 C(graphite) 0 O 2 (g) 0 Na(s) 0 66 vm williamson 22

ΔH f for Some Compounds Substance ΔH o f (kj/mol) CO 2 (g) 393.5 CaCO 3 (s) 635.5 HBr(g) 36.4 H 2 O(l) 285.8 H 2 O(g) 241.8 SO 2 (g) 296.8 SiO 2 (s) 910.9 67 Enthalpies of Formation 68 11/1/17 Thermodynamics Thermochemical Equations: Guidelines! Physical states of all reactants must be specified; these determine the magnitude of the energy changes H 2 (g) + 1 / 2 O 2 (g) --> H 2 O(g) ΔH = 242 kj H 2 (g) + 1 / 2 O 2 (g) --> H 2 O(l) ΔH = 286 kj 69 vm williamson 23

Stand. Enthalpies of Rxn ( H rxn ) and ΔH o f 2C (graphite) + 3H 2(g) + 1 / 2 O 2(g) --> C 2 H 5 OH (l) ΔH o rxn = 277.0 kj/mol ΔH o rxn ΔHo f 2C (graphite) + 3H 2(g) + 1 / 2 O 2(g) --> C 2 H 5 OH (g) = 235.3 kj/mol ΔH o rxn ΔH o rxn ΔH o f C 2 H 4(g) + H 2 O (l) -->C 2 H 5 OH (l) ΔH o rxn = 510.36 kj/mol ΔH o rxn ΔH o f 70 For which is the =? A = B = C(s)+ 2H 2 (g)--> CH 4 (g) 2H 2 (g)+ O 2 (g)--> 2H 2 O(g) 1/2 N 2 (l)+ 3/2 H 2 (g)--> NH 3 (g) 2Ca(s)+ 2C(s) + 3O 2 (g)--> 2CaCO 3 (g) H 2 (g)+ ½ O 2 (g)--> H 2 O(g) 71 Hess Law: Schematic Representation Elements n H o f (reactants) Reactants H o rxn n H o f (products) Products 72 vm williamson 24

Calculating ΔH rxn from ΔH f! H o f may be used to calculate Ho rxn : H o rxn = n Ho f (products) n H o f (reactants) n = stoichiometric coefficient! For the general reaction: aa + bb +... --> xx + yy +... H o rxn = [x Ho f (X) + y Ho f (Y) +...] [a H o f (A) + b Ho f (B) +...] 73 Calculating ΔH rxn from ΔH f CO(g) + NO(g) --> CO 2 (g) + 1 / 2 N 2 (g) H o (reactants) = 1mol H o f (CO) + 1mol Ho f (NO) Reactants = H o (products) = 1 H o f (CO 2) + 1/2 H o f (N 2) Products = H o rxn = products reactants = NOTE: same as before (or very close) 74 Hess Law Using ΔH f Values! Calculate ΔH o rxn at 298 K: SiH 4 (g)+ 2O 2 (g)--> SiO 2 (s)+ 2H 2 O(l) ΔH o react = [2ΔHo f,o 2 + ΔH o f,sih 4 ] ΔH o prod = [ + ] ΔH o rxn = ΔHo f,prod ΔHo f,react ΔH o = [ ] [ ] = kj! ΔH o = kj 75 vm williamson 25

Bond Energy! Energy necessary to break one mole of bonds in a gaseous covalent substance at constant temperature and pressure! Also referred to as bond enthalpy! Always positive in the table.! Bond-breaking is an endothermic process; bond-forming is exothermic A B(g) --> A(g) + B(g) ΔH o rxn = Be break - BE make 76 Bond Energy Copyright 1995 by Saunders College Publishing 77 Bond Energy 78 vm williamson 26

Bond Energy and ΔH rxn Reactants (g) H o rxn Atoms (g) Energy to break bonds Products(g) Energy released in bond formation In the gas-phase: H o rxn = BE (break) BE (make) 79 Bond Energy and ΔH rxn H 2 + Br 2 à 2HBr 2H + 2Br 2H + Br 2 2 ΔH Br Br = 192 kj 3 ΔH H H = 435 kj H 2 + Br 2 ΔH rxn = 109 kj 1-2ΔH H Br = 736 kj 2HBr 80 Carbon-Carbon Bond Energy Copyright 1995 by Saunders College Publishing 81 vm williamson 27

Selected Single Bond Energies* H C N O F P Cl 436 414 389 464 569 318 431 H 347 293 351 439 264 330 C 159 201 272 209 201 N 138 184 351 205 O 159 490 255 F 213 331 P *These are average values in kj/mol of bonds 82 Selected Multiple Bond Energies* N=N 418 N N 946 C=N 615 C N 891 O=O 498 C=C 611 C C 837 C=O 745 C=O 799 in CO 2 C O 1070 *These are average values in kj/mol of bonds Note trend with single, double, triple bonds: C-C single bond is 347 kj/mol 83 Bond Energy and ΔH rxn H 2 (g) + Cl 2 (g) à 2HCl(g)! ΔH o rxn = BE b BE m ΔH o rxn = [(BE H H + BE Cl Cl ) 2BE H Cl ] ΔH o rxn = ΔH o rxn =! ΔH o rxn = kj Copyright 1995 by Saunders College Publishing 84 vm williamson 28

Using a bond energy table, the change in enthalpy is closer to: CO + H 2 O à CO 2 + H 2 (A)-40kJ (B) -2000kJ (C) -400kJ (D) ΔH is positive C 2 H 4 + 3O 2 à 2CO 2 + 2H 2 O Break: Make: ΔH o rxn = = kj 85 Summary Calculating H (enthalpy) 1.from heating curve: " q = ms T and q = m H fus or vap! m = mass! s = specific heat of substance! T = T final T original 2. from calorimeter: " q rxn = q water + q cal " q rxn = ms T + K T! K = calorimeter constant 3. from Proportionality: " Given equation + H rxn, then find H for a certain amount of one of the reactants or products 86 4. from Combining Equations (Hess Law): " Given equations + their H values " Use heat is reversible, additive, proportional to H find for a Target equation 5. from standard H f values: " Find H for an equation, given H f values " H rxn = Sum of product H f sum of reactant H f 6. from Bond Energies: " Find H for an equation, given bond energy values " H rxn = Sum of bonds broken sum of bonds made 87 vm williamson 29