1 / 46

Thermochemistry

Thermochemistry. Chapters 6 and 16. TWO Trends in Nature. Order  Disorder   High energy  Low energy . 2H 2 ( g ) + O 2 ( g ) 2H 2 O ( l ) + energy. H 2 O ( g ) H 2 O ( l ) + energy.

xanto
Download Presentation

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. Thermochemistry Chapters 6 and 16

  2. TWO Trends in Nature • Order  Disorder   • High energy  Low energy 

  3. 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. 6.2

  4. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. ΔH = H (products) – H (reactants) ΔH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants ΔH < 0 ΔH > 0 6.4

  5. H2O (s) H2O (l) ΔH = 6.01 kJ Thermochemical Equations Is ΔH negative or positive? System absorbs heat Endothermic ΔH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. 6.4

  6. ΔH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is ΔH negative or positive? System gives off heat Exothermic ΔH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. 6.4

  7. 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) ΔH = -6.01 kJ ΔH = 6.01 kJ/mol ΔH = 6.01 kJ ΔH = 2 mol x 6.01 kJ/mol= 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 ΔH changes • If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n. 6.4

  8. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? x H2O (l) H2O (g) H2O (s) H2O (l) 3013 kJ 1 mol P4 x ΔH = 44.0 kJ ΔH = 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. P4(s) + 5O2(g) P4O10(s) ΔHreaction = -3013 kJ = 6470 kJ 266 g P4 6.4

  9. ΔH0 (O2) = 0 ΔH0 (O3) = 142 kJ/mol ΔH0 (C, graphite) = 0 ΔH0 (C, diamond) = 1.90 kJ/mol f f f f Standard enthalpy of formation (ΔH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. 6.6

  10. 6.6

  11. The standard enthalpy of reaction (ΔH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = ΔH0 ΔH0 rxn rxn dΔH0 (D) cΔH0 (C) aΔH0 (A) bΔH0 (B) ΔH0 (products) f f f f f - ΔH0 (reactants) Σ Σ = 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.) 6.6

  12. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - Σ Σ = ΔH0 ΔH0 ΔH0 - [ ] [ + ] = rxn rxn rxn [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ = 12ΔH0 (CO2) 2ΔH0 (C6H6) f f = - 3267 kJ/mol C6H6 6ΔH0 (H2O) -6535 kJ f 2 mol ΔH0 (reactants) ΔH0 (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. 6.6

  13. C(graphite) + O2(g) CO2(g) ΔH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g) ΔH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g) ΔH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g) ΔH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g) ΔH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g) ΔH0 = +1072 kJ rxn ΔH0 = -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. 6.6

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

  15. The enthalpy of solution (ΔHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. ΔHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6.7

  16. The Solution Process for NaCl The lattice energy of an ionic solid is a measure of the strength of bonds in that ionic compound. It is given the symbol U and is equivalent to the amount of energy required to separate a solid ionic compound into gaseous ions. ΔHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7

  17. Energy Diagrams Exothermic Endothermic • Activation energy (Ea) for the forward reaction • Activation energy (Ea) for the reverse reaction • (c) Delta H

  18. S order disorder S H2O (s) H2O (l) Entropy (S) is a measure of the randomness or disorder of a system. If the change from initial to final results in an increase in randomness ΔS > 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 ΔS > 0 18.3

  19. 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: ΔSuniv = ΔSsys + ΔSsurr > 0 Equilibrium process: ΔSuniv = ΔSsys + ΔSsurr = 0 18.4

  20. The standard entropy of reaction (ΔS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD - Σ S0(reactants) Σ S0(products) = S0 ΔS0 ΔS0 ΔS0 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 (ΔSsys) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol 18.4

  21. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Entropy Changes in the System (ΔSsys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, ΔS0 > 0. • If the total number of gas molecules diminishes, ΔS0 < 0. • If there is no net change in the total number of gas molecules, then ΔS0 may be positive or negative BUT ΔS0 will be a small number. The total number of gas molecules goes down, ΔS is negative. 18.4

  22. 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 18.2

  23. Gibbs Free Energy Spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0 Equilibrium process: ΔSuniv = ΔSsys + ΔSsurr = 0 For a constant-temperature process: Gibbs free energy (G) ΔG = ΔHsys -TΔSsys ΔG < 0 The reaction is spontaneous in the forward direction. ΔG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. ΔG = 0 The reaction is at equilibrium. 18.5

  24. ΔG = ΔH - TΔS 18.5

  25. The standard free-energy of reaction (ΔG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD - [ + ] [ + ] = - ΔG0 (reactants) Σ Σ = f Standard free energy of formation (ΔG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. ΔG0 ΔG0 rxn rxn f ΔG0 of any element in its stable form is zero. f dΔG0 (D) cΔG0 (C) aΔG0 (A) bΔG0 (B) ΔG0 (products) f f f f f 18.5

  26. - ΔG0 (reactants) Σ Σ = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) ΔG0 ΔG0 ΔG0 - [ ] [ + ] = rxn rxn rxn [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ = Is the reaction spontaneous at 25 0C? 12ΔG0 (CO2) 2ΔG0 (C6H6) f f 6ΔG0 (H2O) f ΔG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? ΔG0 = -6405 kJ < 0 spontaneous 18.5

  27. Recap: Signs of Thermodynamic Values

  28. Gibbs Free Energy and Chemical Equilibrium ΔG = ΔG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium Q = K ΔG = 0 0 = ΔG0 + RT lnK ΔG0 = -RT lnK 18.6

  29. Gibbs Free Energy and Chemical Equilibrium Some concepts to remember Equilibrium constant (K) and Reaction Quotient (Q) http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_Equilibrium_Constants.htm http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Reaction_Quotient.htm K is the concentrations of products raised to their coefficient over the concentrations of the reactants raised to their coefficient. At equilibrium

  30. ΔG = ΔGΕ + RT ln (Q) Define terms: ΔG = free energy not at standard conditions ΔGΕ = free energy at standard conditions R = universal gas constant 8.3145 J/molK T = temp. in Kelvin ln = natural log Q = reaction quotient: (for gases this is the partial pressures of the products divided by the partial pressures of the reactants—all raised to the power of their coefficients) Q = [products] [reactants]

  31. “RatLink”: ΔG° = -RTlnK Terms: basically the same as above --- however, here the system is at equilibrium, so ΔG = 0 and K represents the equilibrium constant under standard conditions. K = [products] [reactants] still raised to power of coefficient

  32. “nFe”: ΔG° = - nFE° remember this!! Terms: ΔG° = just like above—standard free energy n = number of moles of electrons transferred (look at ½ reactions) F = Faraday’s constant 96,485 Coulombs/mole electrons E° = standard voltage ** one volt = joule/coulomb** BIG MAMMA, verse 3: ΔG°rxn = ΣGΕ (products) – Σ GΕ (reactants)

  33. ΔG0 = -RT lnK 18.6

  34. The specific heat (s) [most books use lower case c] 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 = ms Heat (q) absorbed or released: q = msΔT q = CΔT ΔT = Tfinal - Tinitial 6.5

  35. How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g •0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x 0.444 J/g •0C x –890C = -34,000 J 6.5

  36. Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msΔt qcal = Ccal/Δt Cp= ΔH/ΔT Reaction at Constant P ΔH = qrxn No heat enters or leaves! 6.5

  37. 6.5

  38. Phase Changes The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm. 11.8

  39. The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature. 11.8

  40. Can you find… The Triple Point? Critical pressure? Critical temperature? Where fusion occurs? Where vaporization occurs? Melting point (at 1 atm)? Boiling point(at 6 atm)? Where’s Waldo? Carbon Dioxide

  41. H2O (s) H2O (l) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium Freezing Melting 11.8

  42. H2O (s) H2O (g) Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid. Sublimation Deposition DHsub = DHfus + DHvap ( Hess’s Law) 11.8

  43. Molar heat of fusion (ΔHfus) is the energy required to melt 1 mole of a solid substance. 11.8

  44. 11.8

  45. Sample Problem • How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C? Step 1: Heat the ice Q=mcΔT Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ Step 2: Convert the solid to liquid ΔH fusion Q = 2.0 mol x 6.01 kJ/mol = 12 kJ Step 3: Heat the liquid Q=mcΔT Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ

  46. Sample Problem • How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C? Step 4: Convert the liquid to gas ΔH vaporization Q = 2.0 mol x 44.01 kJ/mol = 88 kJ Step 5: Heat the gas Q=mcΔT Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ Now, add all the steps together 0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ

More Related