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Energetics

6. Energetics. 6.1 What is Energetics? 6.2 Enthalpy Changes Related to Breaking and Forming of Bonds 6.3 Standard Enthalpy Changes 6.4 Experimental Determination of Enthalpy Changes by Calorimetry 6.5 Hess’s Law 6.6 Calculations involving Standard Enthalpy Changes of Reactions.

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Energetics

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  1. 6 Energetics 6.1 What is Energetics? 6.2 Enthalpy Changes Related to Breaking and Forming of Bonds 6.3 Standard Enthalpy Changes 6.4 Experimental Determination of Enthalpy Changes by Calorimetry 6.5 Hess’s Law 6.6 Calculations involving Standard Enthalpy Changes of Reactions

  2. 6.1 What is energetics? (SB p.136) What is energetics? Energetics is the study of energy changes associated with chemical reactions. Thermochemistry is the study of heat changes associated with chemical reactions.

  3. Internal Energy (U) U = kinetic energy + potential energy

  4. Kinetic Energy Potential Energy heat  T (K) translational rotational vibrational Energy Relative position among particles Bond breaking  P.E.  Bond forming  P.E. 

  5. H – H(g)  H(g) + H(g) P.E.  H(g) + H(g)  H – H(g) P.E.  Bond breaking : - Bond forming : - Ionization : - Na(g)  Na+(g) + eP.E. 

  6. Internal energy bond breaking bond forming 2H2(g) + O2(g) U1 U = U2 – U1 = -(y-x) kJ 2H2O(l) U2 Reaction coordinate Q.1 4H(g) + 2O(g)

  7. enthalpy Internal energy 6.1 What is energetics? (SB p.137) Internal energy and enthalpy H = U + PV

  8. 6.1 What is energetics? (SB p.137) Internal energy and enthalpy e.g. Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) qv = U = -473 kJ mol1 qp = H = -470 kJ mol1 Mg Mg

  9. qv = U = -473 kJ mol1 qp = H = -470 kJ mol1 Work done against the surroundings qv = qp – 3 = qp – w = qp - PV PV  (Nm2)(m3) = Nm Force  displacement

  10. 6.1 What is energetics? (SB p.138) Internal energy and enthalpy H = U + PV = U + PV at constant P qp = qv + PV Heat change at fixed P Heat change at fixed V Work done

  11. 6.1 What is energetics? (SB p.138) Internal energy and enthalpy qp = qv + PV On expansion, PV > 0 Work done by the system against the surroundings System gives out less energy to the surroundings qp is less negative than qv (less exothermic)

  12. 6.1 What is energetics? (SB p.138) Internal energy and enthalpy qp = qv + PV On contraction, PV < 0 Work done by the surroundings against the system System gives out more energy to the surroundings qp is more negative than qv (more exothermic)

  13. H is more easily measured than H as • most reactions happen in open vessels. • i.e. at constant pressure. • The absolute values of H and H cannot • be measured.

  14. 6.1 What is energetics? (SB p.138) Exothermic and endothermic reactions An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)

  15. Check Point 6-1 6.1 What is energetics? (SB p.139) Exothermic and endothermic reactions An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)

  16. 6.1 What is energetics? (SB p.136) Law of conservation of energy The law of conservation of energy states that energy can neither be created nor destroyed, but can be exchanged between a system and its surroundings

  17. Exothermic : - P.E. of the system  K.E. of the surroundings Endothermic : - K.E. of the surroundings  P.E. of the system

  18. 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) Enthalpy changes related to breaking and forming of bonds CH4 + 2O2 CO2 + 2H2O

  19. 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) In an exothermic reaction, E absorbed to break bonds < E released as bonds are formed.

  20. 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) Enthalpy changes related to breaking and forming of bonds N2(g) + 2O2(g)  2NO2(g)

  21. Check Point 6-2 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) In an endothermic reaction, E absorbed to break bonds > E released as bonds are formed.

  22. For non-gaseous reactions, PV  0 H = U + PV  U For gaseous reactions, H = U + PV = U + (n)RT (PV = nRT)

  23. Q.2 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) Given : R = 8.314 J K1 mol1, T = 298 K U = H – (n)RT = 885 kJ mol1

  24. Enthalpy bond breaking bond forming CH4(g) + 2O2(g) H1 H = 890 kJ mol1 CO2(g) + 2H2O(l) H2 Reaction coordinate C(g) + 4H(g) + 4O(g)

  25. 373K 373K H / kJ mol1 At 298K CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) 890 At 373K CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 802 2H2O(l)  2H2O(g) +88 kJ

  26. Enthalpy CH4(g) + 2O2(g) H = 890 kJ mol1 CO2(g) + 2H2O(l) Reaction coordinate C(g) + 4H(g) + 4O(g) H1 298 K H2

  27. CH4(g) + 2O2(g) H = 890 kJ mol1 CO2(g) + 2H2O(l) C(g) + 4H(g) + 4O(g) Enthalpy 373 K H1’ H2’ Assume constant H Reaction coordinate

  28. Enthalpy CH4(g) + 2O2(g) H = 890 kJ mol1 CO2(g) + 2H2O(l) Reaction coordinate C(g) + 4H(g) + 4O(g) 373 K In fact, H depends on T

  29. Enthalpy CH4(g) + 2O2(g) H = 802 kJ mol1 CO2(g) + 2H2O(l) CO2(g) + 2H2O(g) H = +88 kJ mol1 Reaction coordinate C(g) + 4H(g) + 4O(g) 373 K

  30. 6.3 Standard Enthalpy Changes

  31. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ mol-1at 373 K CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ mol-1 at 298 K 6.3 Standard enthalpy changes (SB p.141) Standard enthalpy changes

  32. Enthalpy change under standard conditions denoted by symbol: H ø 6.3 Standard enthalpy changes (SB p.141) Standard enthalpy changes As enthalpy changes depend on temperature and pressure, it is necessary to definestandard conditions: 1. elements or compounds in their normal physical states;2. a pressure of 1 atm (101325 Nm-2); and3. a temperature of 25oC (298 K)

  33. 2H2(g) + O2(g) 2H2O(l)  H = 572 kJ mol1 Standard enthalpy change of reaction The enthalpy change when the molar quantities of reactants as stated in the equation react under standard conditions. per mole of O2

  34. H2(g) + O2(g)  H2O(l)  H = 286 kJ mol1 4H2(g) + 2O2(g)  4H2O(l)  H  H = 1144 kJ depends on the equation Standard enthalpy change of reaction 2H2(g) + O2(g) 2H2O(l)  per mole of O2 H = 572 kJ mol1 per mole of H2 or H2O

  35. Hf H2(g) + O2(g)  H2O(l)    Hf [H2O] = 286 kJ mol1 Hf [O2] = 0 kJ mol1 Hf [element] = 0 kJ mol1 Standard enthalpy change of formation The enthalpy change when one mole of the substance is formed from its elements under standard conditions. Q.3 O2(g)  O2(g)

  36. Hf [diamond] = +1.9 kJ mol1 Most stable allotrope C(graphite)  C(diamond)

  37. (iii) Mg(s) + O2(g)  MgO(s) (iv) Na(s) + H2(g) + C(graphite) + O2(g)  NaHCO3(s) Q.4 (i) C(graphite) + O2(g)  CO2(g) (ii) C(graphite) + 2H2(g)  CH4(g) (v) 2C(graphite) + 2H2(g) + O2(g)  CH3COOH(l)

  38. H2(g) + O2(g)  H2O(l) Q.5 qv = U = 140.3 kJ per g of H2 n = 0 – 0.496 – 0.248 = -0.744 mol

  39. Q.5 H = U + nRT = -142.1 kJ Heat released for the formation of 0.496 mol of water Molar Hf[H2O] =

  40. Hc  Hc [C2H5OH(l)] = -1368 kJ mol1 Standard enthalpy change of combustion The enthalpy change when one mole of the substance undergoes complete combustion under standard conditions. C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)

  41. C(diamond) + O2(g) Enthalpy C(graphite) + O2(g) 1.9  395.4 Hf [diamond] 393.5 = +1.9 kJ mol1 CO2(g) Reaction coordinate 6.3 Standard enthalpy changes (SB p.147) ø Substance Hc(kJ mol-1) C (diamond) C (graphite) -395.4 -393.5

  42. (a) C(graphite) + O2(g)  CO(g)  = Hf [CO(g)]   Hc [graphite]  = 2  Hc [H2(g)]   H H  = 2  Hf [H2O(l)] Q.6 Incomplete combustion (b) 2H2(g) + O2(g)  2H2O(l)

  43.    = Hc [graphite] = Hf [CO2(g)] = Hc [CH4(g)]  Hf [CO2(g)]   H H Not formed from elements   2  Hf [H2O(l)] Check Point 6-3 Q.6 (c) C(graphite) + O2(g)  CO2(g) (d) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

  44. Standard enthalpy change of neutralization(Hneut) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions. ø e.g. H+(aq) + OH-(aq)  H2O(l) Hneut = -57.3 kJ mol-1 ø 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy changes of neutralization

  45. 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy changes of neutralization Enthalpy level diagram for the neutralization of a strong acid and a strong alkali

  46. Acid Alkali Hneu ø HCl HCl HCl HF NaOH KOH NH3 NaOH -57.1 -57.2 -52.2 -68.6 NH3(aq) + H2O(l) NH4+(aq) + OH(aq) H1 > 0 ø Hneu = H1 + H2 = 52.2 kJ mol1 6.3 Standard enthalpy changes (SB p.142) H+(aq) + OH-(aq) + Cl(aq) H2O(l)+ Cl(aq)H2 = -57.3 NH3(aq) + HCl(aq)  NH4Cl (aq)

  47. Acid Alkali Hneu ø HCl HCl HCl HF NaOH KOH NH3 NaOH -57.1 -57.2 -52.2 -68.6 HF(aq) H+(aq) + F(aq) H1 < 0 ø Hneu = H1 + H2 = 68.6 kJ mol1 6.3 Standard enthalpy changes (SB p.142) H+(aq) + OH-(aq) + Na+(aq)  H2O(l)+ Na+(aq)H2 = -57.3 HF(aq) + NaOH(aq)  NaF(aq) + H2O(l)

  48. Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved in a specified number of moles of solvent (e.g. water) under standard conditions. ø NaCl(s) + 10H2O(l)  Na+(aq) + Cl-(aq) Hsoln[NaCl(s)]= +2.008 kJ mol-1 ø dilution NaCl(aq) NaCl(aq) H > 0 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy change of solution

  49. Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions. ø concentration  0 NaCl(s) + water  Na+(aq) + Cl-(aq) Hsoln= +4.98 kJ mol-1 ø 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy change of solution

  50. e.g. NaCl(s) + water  Na+(aq) + Cl-(aq) Hsoln= +4.98 kJ mol-1 ø + 4.98 kJ mol1 Enthalpy level diagram for the dissolution of NaCl 6.3 Standard enthalpy changes (SB p.143) Standard enthalpy change of solution

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