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IB Chemistry HL2 Thermochemistry (Energetics)

IB Chemistry HL2 Thermochemistry (Energetics). Topic 5 and 15. IB Standards. 15.1.1· Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion

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IB Chemistry HL2 Thermochemistry (Energetics)

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  1. IB Chemistry HL2 Thermochemistry (Energetics) Topic 5 and 15

  2. IB Standards • 15.1.1· Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion • 15.1.2· Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion Dr. Dura

  3. Temperature = Thermal Energy Standard Enthalpy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. greater thermal energy Dr. Dura

  4. Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. closed isolated open energy nothing Exchange: mass & energy Dr. Dura

  5. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. Dr. Dura

  6. DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature DE = Efinal - Einitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. Dr. Dura

  7. C3H8 + 5O2 3CO2 + 4H2O Chemical energy lost by combustion = Energy gained by the surroundings system surroundings First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. DEsystem + DEsurroundings = 0 or DEsystem = -DEsurroundings Exothermic chemical reaction! Dr. Dura

  8. Another form of the first law for DEsystem DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDVwhen a gas expands against a constant external pressure Dr. Dura

  9. At constant pressure: q = DH and w = -PDV Enthalpy and the First Law of Thermodynamics DE = q + w DE = DH - PDV DH = DE + PDV Dr. Dura

  10. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants Dr. Dura DH < 0 DH > 0

  11. H2O (s) H2O (l) DH = 6.01 kJ Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. Dr. Dura

  12. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. Dr. Dura

  13. 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 6.01 kJ DH = 2 x 6.01= 12.0 kJ Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance • If you reverse a reaction, the sign of DH changes • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. Dr. Dura

  14. P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ x H2O (l) H2O (g) H2O (s) H2O (l) 3013 kJ 1 mol P4 x DH = 44.0 kJ DH = 6.01 kJ 1 mol P4 123.9 g P4 Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? = 6470 kJ 266 g P4 Dr. Dura

  15. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial Dr. Dura

  16. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol Chemistry in Action: Fuel Values of Foods and Other Substances 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J Dr. Dura

  17. Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f DH0 (O2) = 0 DH0 (O3) = 142 kJ/mol DH0 (C, graphite) = 0 DH0 (C, diamond) = 1.90 kJ/mol f f f f Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? The standard enthalpy of formation of any element in its most stable form is zero. Dr. Dura

  18. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn mDH0 (reactants) dDH0 (D) nDH0 (products) cDH0 (C) aDH0 (A) bDH0 (B) f f f f f f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) Dr. Dura

  19. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn. Dr. Dura

  20. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 2973 kJ/mol C6H6 6DH0 (H2O) -5946 kJ f 2 mol mDH0 (reactants) nDH0 (products) f f Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. Dr. Dura

  21. Sample Problem Given the following thermochemical equations: B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol) H2O (l) → H2O (g) (ΔH = 44 kJ/mol) H2 (g) + (1/2) O2 (g) → H2O (l) (ΔH = -286 kJ/mol) 2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol) Determine the ΔHf of: 2 B (s) + (3/2) O2 (g) → B2O3 (s) Dr. Dura

  22. After the multiplication and reversing of the equations (and their enthalpy changes), the result is: B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) (ΔH = -2035 kJ/mol) 3 H2O (g) → 3 H2O (l) (ΔH = -132 kJ/mol) 3 H2O (l) → 3 H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol) 2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol) Adding these equations and canceling out the common terms on both sides, we get 2 B (s) + (3/2) O2 (g) → B2O3 (s) (ΔH = -1273 kJ/mol) Dr. Dura

  23. IB Standards • 15.2.1· Define and apply the terms lattice enthalpy and electron affinity • 15.2.2· Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds • 15.2.3· Construct a Born-Haber cycle for group 1 and group 2 oxides and chlorides, and use it to calculate an enthalpy change (9.1.12.A.1, 9.1.12.B.1) • 15.2.4· Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character Dr. Dura

  24. Experimental lattice energies cannot be determined directly. An energy cycle based on Hess’s Law, known as the Born-Haber cycle is used. The Key Concepts in Born-Haber Cycle are: • Heat of formation Δ Hf • Ionization Energy I.E. • Electron Affinity E.A. • Atomization (Sublimation) s  g • Bond enthalpy B.E. • Lattice Energy – is the enthalpy change that occurs when one mole of a solid ionic compound is separated into ions under standard conditions. Example: NaCl(s)  Na+ (g) + Cl -(g) (endothermic). • Note that in forming ionic compounds, the oppositely charged gaseous ions come together to form an ionic lattice – this is a very exothermic process as there is a strong attraction between the ions i.e. Na+ (g) + Cl -(g) NaCl(s) Dr. Dura

  25. + Na+(g) Cl-(g) Na(g) Cl(g) IE EA DHsub ½ BDE Lattice E. = 786.5 kJ/mol (p.180 of text) DHf Na(s) + ½ Cl2(g) NaCl(s) Born-Haber Cycle for NaCl

  26. Do NOW Construct the Born-Haber cycle for KF and using appropriate data, calculate the electron affinity of fluorine. Thermodynamic data K(s) → K(g) ΔH°a = 90 kJ mol-1 K(g) → K+(g) + e- IE = 419 kJ mol-1 ½ F2(g) → F(g) ΔH°a = 79.5 kJ mol-1 K(s) + F(g) → KF(s) ΔH°f = -569 kJ mol-1 K+(g) + F-(g) → KF(s) ΔH°lattice = -821 kJ mol-1 Choose the correct answer: a) EA = 336.5 kJ mol-1 b) EA = -336.5 kJ mol-1 c) EA = -840.5 kJ mol-1 Dr. Dura

  27. Practice problem Draw the Born-Haber cycle for potassium sulfide K2S. Thermodynamic data Potassium K Sulfur S Potassium Sulfide K2S IEI = 419 kJ mol-1 ΔH°a= 279 kJ mol-1 ΔH°f = -257 kJ mol-1 ΔH°a = 78 kJ mol-1 EAI= -199.5 kJ mol-1 ΔH°lattice = -1979 kJ mol EAII - 648.5 kJ mol-1 Dr. Dura

  28. nonspontaneous spontaneous Spontaneous Physical and Chemical Processes • A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm, water freezes below 0 0C and ice melts above 0 0C • Heat flows from a hotter object to a colder • object • A gas expands in an evacuated bulb • Iron exposed to oxygen and water • forms rust

  29. CH4(g) + 2O2(g) CO2(g) + 2H2O (l)DH0 = -890.4 kJ H+(aq) + OH-(aq) H2O (l)DH0 = -56.2 kJ H2O (s) H2O (l)DH0 = 6.01 kJ H2O NH4NO3(s) NH4+(aq) + NO3-(aq)DH0 = 25 kJ Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions

  30. IB Standards • 15.3.1 State and explain the factors that increase entropy in a system15.3.2· Predict whether the entropy change for a given reaction or process is positive or negative· 15.3.3 Calculate the standard entropy change for a reaction using standard entropy values (5.2.12.D.2, 9.1.12.A.1) Dr. Dura

  31. S order disorder S H2O (s) H2O (l) Entropy (S) is a measure of the randomness or disorder of a system. DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas DS > 0

  32. Processes that lead to an increase in entropy (DS > 0)

  33. How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 600C to 800C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0)

  34. Entropy State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, enthalpy, pressure, volume, temperature , entropy Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

  35. First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0

  36. aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD - S mS0(reactants) S nS0(products) = DS0 DS0 DS0 DS0 rxn rxn rxn rxn What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol Entropy Changes in the System (DSsys) The standard entropy of reaction (DS0 rxn ) is the entropy change for a reaction carried out at 1 atm and 250C. S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol

  37. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Entropy Changes in the System (DSsys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, DS0 > 0. • If the total number of gas molecules diminishes, DS0 < 0. • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. The total number of gas molecules goes down, DS is negative.

  38. IB Standards 15.4.1 Predict whether a reaction or process will be spontaneous using the sign of ΔG15.4.2· Calculate ΔG for a reaction using the equation with enthalpy, temperature, and entropy and by using values of free energy change of formation, ΔGf (9.1.12.A.1) 15.4.3 Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and enthalpy changes and the equation Dr. Dura

  39. Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

  40. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD - [ + ] [ + ] = - mDG0 (reactants) S S = f Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DG0 DG0 rxn rxn f . DG0 of any element in its stable form is zero f dDG0 (D) nDG0 (products) cDG0 (C) aDG0 (A) bDG0 (B) f f f f f

  41. - mDG0 (reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ = 12DG0 (CO2) 2DG0 (C6H6) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? Is the reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 spontaneous

  42. DG = DH - TDS

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