Chemical Reactions And Energy (Thermochemistry) PDF

CHEMICAL REACTIONS AND ENERGY (THERMOCHEMISTRY) CLS3006-C Lectures 2.1-2.2 Dr. H M Sheldrake [email protected]  LEARNING OBJECTIVES Recall the Laws of Thermodynamics, and their consequences Define enthalpy changes associated with chemical reactions Iden...

CHEMICAL REACTIONS AND ENERGY (THERMOCHEMISTRY) CLS3006-C Lectures 2.1-2.2 Dr. H M Sheldrake [email protected]  LEARNING OBJECTIVES Recall the Laws of Thermodynamics, and their consequences Define enthalpy changes associated with chemical reactions Identify exothermic and endothermic reactions, using energy diagrams where appropriate Calculate enthalpy changes using calorimetry, bond energies and Hess’ Law. LEARNING OBJECTIVES CONT. Identify entropy changes in reactions Explain spontaneity of chemical reactions using entropy and Gibbs free energy, including application of ∆G = ∆H - T∆S Calculate entropy change values  ENERGY The ability to do work Force x Distance = Work Unit: joules (J) – 1 Joule = The amount of energy required to raise a 1- kg substance 10 cm against the force of gravity – 1 calorie = The amount of heat necessary to raise the temperature of exactly one gram of water by one degree Celsius 1 “food calorie” = 1 kcal = 1000 cal  TYPES OF ENERGY Potential energy E = mgh Kinetic energy E = ½ mv2 Electromagnetic Nuclear Chemical  LAWS OF THERMODYNAMICS I First Law of thermodynamics: The energy of the universe is constant Internal Energy (E): E = q + w E = change in system’s internal energy q = heat w = work  EXERCISE The gas in a piston is warmed and absorbs 670 J of heat. The expansion performs 350 J of work on the surroundings. What is the change in internal energy for the system?  ENTHALPY Internal energy A measure of the heat content of a substance at constant pressure Enthalpy CHANGE ∆Ho = heat released or absorbed during a chemical process ∆Ho = HProducts – Hreactants  STANDARD STATES Enthalpy values vary according to the conditions. For an element: The standard state of an element is the form in which the element exists under conditions of 1 atmosphere and 25 oC (standard conditions). For a compound: The standard state of a gaseous substance is a pressure of exactly 1 atmosphere. For a pure substance in a condensed state (liquid or solid), the standard state is the state it would have at 1 atmosphere. For a substance present in a solution, the standard state is a concentration of exactly 1 M. Use the correct subscript [(g), (l) or (s)] to indicate the physical state  ENTHALPY CHANGES Enthalpy of reactants > products Enthalpy of reactants < products ∆H negative ∆H positive EXOTHERMIC ENDOTHERMIC Heat given out Heat absorbed  EXERCISE II Identify the following as exothermic or endothermic: 1. Respiration 2. An electron being excited from the n = 1 to n = 3 shell 3. Production of quicklime CaO + H2O → Ca(OH)2 ΔHr = −63.7 kJ/mol 4. Thermal decomposition of limestone CaCO3 → CaO + CO2 (Requires temperatures > 900 oC)  DEFINING ENTHALPY CHANGES Eg. STANDARD ENTHALPY OF FORMATION Symbol : ∆Hfө The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. The standard enthalpy of formation of a pure element in its standard state is defined as zero. Reaction equation: 2C(s) + 2H2(g) → C2H4(g) ∆H=∆Hf(C2H4) N2(g) + 2O2(g) → 2NO2(g) ∆H =2∆Hf(NO2) MEASURING ENTHALPY CHANGES - CALORIMETRY q = mCp∆T  MEASURING ENTHALPY CHANGES – CALORIMETRY: EXAMPLE 22.0 g propane (C3H8) was burned in a bomb calorimeter containing 3.25 L water. The temperature change of the water was 88.5 °C. Assuming all the heat energy was transferred to the water, calculate the molar heat of combustion of propane. Cp of water = 4.18 kJ Kg-1 K-1. RMM C3H8 = 44 g q = mCp∆T = 3.25 kg × 4.18 kJ Kg-1 K-1 × 88.5 K = 1202 kJ for 22.0 g propane 22 g = 0.5 mol So ∆Hc = 1202kJ /0.5 mol= -2404 kJ/mol HESS’ LAW The overall enthalpy change accompanying a chemical reaction is independent of the route taken in going from reactants to products C6H12O6 + 6O2 → 6H2O + 6CO2  CALCULATING ENTHALPY CHANGES From bond energies Average bond energy/kcal/mol O=O 119 C-H 99 C, 4H, 4O O-H 111 Atoms C=O in CO2 192 Energy H = +634 H = +828 H O=O C H H O=O O H O H H CH4 + 2O2 C O O H H Reactants HR CO2 + 2H2O Products = -194 kcal/mol  BOND ENERGIES Single bonds I Br Cl S P Si F O N C H ∆HR = ΣHbroken +ΣHformed H 299 366 431 347 322 323 566 467 391 416 436 C 213 285 327 272 264 301 486 336 285 356 Remember: N - - 193 - 200 355 272 201 160 Breaking bonds is endothermic Positive value. O 201 - 205 - 340 368 190 146 Making bonds is exothermic F - - 255 326 490 582 158 Negative value Si 234 310 391 226 - 226 P 184 264 319 - 209 S - 213 255 226 Multiple bonds Cl 209 217 242 N=N 418 C=C 598 Br 180 193 N≡N 946 C≡C 813 I 151 C=N 616 C=O (CO2) 803 C≡N 866 C=O (carbonyl) 695 O=O 498 C≡O 1073  EXERCISE III Using the bond energies on the previous slide, estimate the enthalpy change for the following reaction H2 (g) + Cl2 (g) → 2 HCl (g) CALCULATING ENTHALPY CHANGES From enthalpies of combustion, formation etc. (Hess’ law cycle or Born-Haber cycle) Write a balanced reaction equation for the enthalpy you need to find A→B Write equations for the enthalpies you have been given A→X X→Y B→Y Arrange them as steps to what you need to find A→B X→Y  Example 1: What is the enthalpy change in the reaction C(s) + O2(g) CO2(g)? Given: C(s) + ½ O2(g) CO(g) H = – 110.5 kJ/mol CO(g) + ½ O2(g) CO2(g) H = – 283.0 kJ/mol Answer: -393.5 kJ/mol  Example 2 What is the enthalpy change in the reaction 4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ∆Hr = ? Given: ∆Hf NH3(g) -46.2 kJ/mol ∆Hf NO(g) 90.4 kJ/mol ∆Hf H2O(g) -241.8 kJ/mol ∆Hf NH3(g) -46.2 kJ/mol 1/2N2(g) +3/2H2(g) → NH3(g) ∆Hf NO(g) 90.4 kJ/mol 1/2N2(g) +1/2O2(g) → NO(g) ∆Hf H2O(g) -241.8 kJ/mol H2(g) +1/2O2(g) → H2O(g)  Example 2 4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ∆Hr = ? ∆Hf 4NH3(g) 4x-46.2 kJ/mol 2N2(g) +6H2(g) → 4NH3(g) ∆Hf 4NO(g) 4x90.4 kJ/mol 2N2(g) +2O2(g) → 4NO(g) ∆Hf 6H2O(g) 6x-241.8 kJ/mol 6H2(g) +3O2(g) → 6H2O(g) 4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ∆Hr = -4∆HfNH3 + 4∆Hf NO + ∆Hf 6H2O = -(4 x -46.2 kJ/mol) + 4 x 90.4 kJ/mol + 6x-241.8 kJ/mol = -904.4 kJ/mol 2N2(g) 6H2(g) +2O2(g) +3O2(g)  Example 2 4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ∆Hr = ? ∆Hf 4NH3(g) 4x-46.2 kJ/mol 2N2(g) +6H2(g) → 4NH3(g) ∆Hf 4NO(g) 4x90.4 kJ/mol 2N2(g) +2O2(g) → 4NO(g) ∆Hf 6H2O(g) 6x-241.8 kJ/mol 6H2(g) +3O2(g) → 6H2O(g) Flip first equation -∆Hf 4NH3(g) 4x+46.2 kJ/mol 4NH3(g) → 2N2(g) +6H2(g) Add the equations up 4NH3(g) + 2N2(g) +2O2(g) + 6H2(g) +3O2(g) → 2N2(g) +6H2(g) + 4NO(g)+ 6H2O(g)  ∆HR and ∆HF Hf (REACTANTS) ELEMENTS REACTANTS Hf (PRODUCTS) HR PRODUCTS In general, ∆HR = ΣHF(products) – ΣHF(reactants) Do not use this formula with bond energies, or if you do not have HF for all reactants and products!  Example 3: What is the enthalpy of formation of hexane? Given: Hfө CO2 = -394.4 kJ/mol Hfө H2O = -285.8 kJ/mol Hcө C6H14 = -4163.0 kJ/mol Want Hfө = Hrө for 6C(s) + 7H2(g) C6H14(l) Know: Hfө CO2 = -394.4 kJ/mol C(s) + O2(g) CO2(g) Hfө H2O = -285.8 kJ/mol H2(g) + ½O2(g) H2O(l) Hcө C6H14 = -4163.0 kJ/mol C6H14(l) + 9½O2(g) 6CO2(g) + 7H2O(l)  Example 3 Hf 6C(s) + 7H2(g) + 9.5O2(g) C6H14(l) + 9.5O2(g) Hc(hexane) 6Hf (CO2) + 7Hf (H2O) 6CO2(g) + 7H2O(l) ΔHfө (hexane) = 6ΔHfө (CO2) + 7ΔHfө (H2O) – ΔHcө (hexane) = 6 x -394.4 + 7 x -285.8 - (-4163.0 ) = -204.0 kJ/mol  Born-Haber cycles Hformation = HatomNa + HatomCl + H1st IE + H1st EA + Hlattice  LAWS OF THERMODYNAMICS II Second Law of thermodynamics: Entropy will always increase in a spontaneous process. Third Law of thermodynamics: Entropy is temperature dependent. Absolute Zero = 0 Kelvin = -273.15° Celsius  ENTROPY Measures the amount of energetic disorder in a system – Symbol S – Unit J/mol/K DS = Sfinal - Sinitial DS = Dq/T DStotal = DSsystem + DSsurroundings – In a spontaneous process DStotal is positive  INCREASES IN ENTROPY All spontaneously occurring chemical and physical changes involve an overall increase in entropy. Examples of increases in entropy: solids melting  INCREASES IN ENTROPY II Liquids boiling Number of molecules increases Ionic solids dissolving Temperature increasing Increases in molecular movement ENTROPY SUMMARY Solid Liquid Gas Fewer moles More moles of reactant of product  ENTROPY CALCULATIONS What is the entropy change of the universe in the reaction H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(l) at 25 ˚C Data: Species ΔHf (kJ/mol) S (J/K/mol) H2CO3(aq) -698.7 191 HCO3-(aq) -691.11 95 OH-(aq) -229.94 -10.54 H2O(l) -285.83 69.91  CALCULATION: SYSTEM H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(l) at 25 ˚C Species ΔHf (kJ/mol) S (J/K/mol) H2CO3(aq) -698.7 191 HCO3-(aq) -691.11 95 OH-(aq) -229.94 -10.54 H2O(l) -285.83 69.91 DStotal = DSsystem + DSsurroundings DSSYST = Sfinal - Sinitial = (95+69.91) – (191+ -10.54) = -15.6J/K/mol CALCULATION: SURROUNDINGS H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(l) at 25 ˚C Species ΔHf (kJ/mol) S (J/K/mol) H2CO3(aq) -698.7 191 HCO3-(aq) -691.11 95 OH-(aq) -229.94 -10.54 H2O(l) -285.83 69.91 DSSURR = Dq/T ΔHR = (-691.11 + -285.83)-(-698.7+ -229.94) = -48.3 kJ/mol DSSURR = +48.3 x 1000 J/mol/(25+273)K = +162.1 J/K/mol  CALCULATION: UNIVERSE H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(l) at 25 ˚C Species ΔHf (kJ/mol) S (J/K/mol) H2CO3(aq) -698.7 191 HCO3-(aq) -691.11 95 OH-(aq) -229.94 -10.54 H2O(l) -285.83 69.91 DStotal = DSsystem + DSsurroundings = -15.6+162.1 = +146.5 J/mol/K  GIBBS FREE ENERGY Gibbs free energy = Energy from a reaction free to do work ΔG = ΔH – TΔS NB: T in K (+273 to convert from ˚C) ΔG < 0 – exergonic; reaction will be spontaneous ΔG > 0 – endergonic; reaction needs energy input to happen ΔG = 0 – the system is in equilibrium  GIBBS FREE ENERGY II ΔG = ΔH – TΔS ΔH ΔS When is ΔG negative? - + - - + + + -  GIBBS FREE ENERGY III Catabolic reactions High energy compounds → simple molecules Glucose + 6O2 → 6CO2 + 6H2O ΔG = -2870 kJ/mol ATP + H2O → ADP + Pi ΔG = -30.5 kJ/mol  Coupling of reactions in metabolism OH O OH HO HO OH HO O HO O OH HO + HO HO O OH O HO OH OH OH glucose + fructose → sucrose ΔGo +29.3 kJ/mol HO OH OH OH O O HO + ATP HO + ADP HO HO OH OH OH glucose + ATP → glucose-p + ADP OP HO ΔGo = -16.7 kJ/mol OH HO OP O O OH + ATP OH + ADP fructose + ATP → fructose-p + ADP HO OH HO OH ΔGo = -14.2 kJ/mol OH OH O HO HO HO O glucose-p + fructose-p → sucrose + OH OP HO HO 2 Pi HO OP HO O O + 2P ΔGo = -0.8 kJ/mol O OH OH OH HO OH HO  SUMMARY Energy Enthalpy Calculate ∆HR From calorimetry From bond energies From enthalpies of formation From other enthalpies (draw a cycle) Entropy Gibbs free energy and spontaneous reactions