1. Topic 5.3 and 5.4 Hess’s Law and Bond Enthalpies

2. Hess’s Law Topic 5.3 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l)  KJ

3. shows three different pathways: A  B A  C  B A  D  E  B enthalpy change from reactants to products for all of these is the same

6. if a series of reactions are added together, the enthalpy change for the net reaction (  H final ) will be the sum of the enthalpy change for the individual reactions (  H ind +  H ind +  H ind ….) – the change in enthalpy is the same whether the reaction takes place in one step, or in a series of steps –  H is independent of the reaction pathway – depends only on the difference between the enthalpy of the products and the reactants  H = H products − H reactants provides a way to calculate enthalpy changes even when the reaction cannot be performed directly 6

7. energy in reactants energy in products

8. Problem-Solving Strategy – work backwards from the final reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal – if a reaction is reversed the sign on ΔH is reversed – N 2 (g) + 2O 2 (g) → 2NO 2 (g) Δ H = 68kJ – 2NO 2 (g) → N 2 (g) + 2O 2 (g) Δ H = - 68kJ – multiply reactions to give the correct numbers of reactants and products in order to get the final reaction. – the value of Δ H is also multiplied by the same integer – identical substances found on both sides of the summed equation cancel each other out

9. Example 1 Given: N 2 (g) + O 2 (g)  2 NO (g)  H 1 = +181 kJ 2 NO (g) + O 2 (g)  2 NO 2 (g)  H 2 = -113 kJ Find the enthalpy change for: N 2 (g) + 2 O 2 (g)  2 NO 2 (g )  H =  H 1 +  H 2 = +181 kJ +(-113) = + 68 kJ 9

10. Example 2 ΔH = (- 184 kJ) + (+ 1452 kJ) + 1268 kJ

11. Example 3 Given: C (s) + O 2 (g)  CO 2 (g)  H 1 = - 393 kJ mol -1 H 2 (g) + ½O 2 (g)  H 2 O (g)  H 2 = - 286 kJ mol -1 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g)  H 2 = - 890 kJ mol -1 Find the enthalpy change for: C (s) + 2H 2 (g)  CH 4 (g ) 11 This equation needs to be “flipped”. The CH 4 is on the wrong side of the equation

12. Example 3 Given: C (s) + O 2 (g)  CO 2 (g)  H 1 = - 393 kJ mol -1 H 2 (g) + ½O 2 (g)  H 2 O (g)  H 2 = - 286 kJ mol -1 CO 2 (g) + 2H 2 O (g)  CH 4 (g) + 2O 2 (g)  H 2 = + 890 kJ mol -1 Find the enthalpy change for: C (s) + 2H 2 (g)  CH 4 (g ) (- 393 kJ mol -1 ) + (- 572 kJ mol -1 ) + (+ 890 kJ mol -1 ) = - 75 kJ mol -1 1 22- 572

13. Calculate  H° for the reaction 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s) Use the following data: 2 Al (s) + 6 HCl (aq)  2 AlCl 3 (aq) + 3 H 2 (g)  H° = -1049. kJ mol -1 HCl (g)  HCl (aq)  H° = -74.8 kJ mol -1 H 2 (g) + Cl 2 (g)  2 HCl (g)  H° = -1845. kJ AlCl 3 (s)  AlCl 3 (aq)  H° = -323. kJ mol -1 = - 6,981.8 kJ 33 6 - 5,535 kJ mol -1 Example 4

14. can be used to calculate the enthalpy change for a chemical reaction if we know the energy necessary to break or form bonds in the gaseous state – breaking bonds energy is required so enthalpy is positive (endothermic) the molecule was stable so energy was necessary to break apart the molecule – forming bonds energy is released so enthalpy is negative (exothermic) the new molecule is more stable than the individual atoms so energy is released Bond Enthalpies. Topic 5.4

15. a molecule with strong chemical bonds generally has less tendency to undergo chemical change than does one with weak bonds – SiO bonds are among the strongest ones that silicon forms it is not surprising that SiO 2 and other substances containing SiO bonds (silicates) are so common it is estimated that over 90 percent of Earth's crust is composed of SiO 2 and silicates

16. we use average bond enthalpies – again, in the gaseous state – different amount of energy can be required to break the same bond example- methane, CH 4 – if you took methane to pieces, one hydrogen at a time, it needs a different amount of energy to break each of the four C-H bonds – every time you break a hydrogen off the carbon, the environment of those left behind changes, and the strength of the remaining bonds is affected – therefore, the 412 kJ mol -1 needed to break C-H is just an average value, may vary, and is not very accurate

17. The average bond enthalpies for several types of chemical bonds are shown in the table below: 17

18. Page 326 in text Also see Chemistry Data Booklet p. 11

19. Bonds broken (requires energy) 1 N N = 945 kJ 3 H-H 3(435) = 1305 kJ Total = 2250 kJ Bonds formed (releases energy) 2x3 = 6 N-H 6 (390) = - 2340 kJ Net enthalpy change = (+ 2250) + (- 2340) = - 90 kJ (exothermic) Calculate the enthalpy change for the reaction. Is it endo or exothermic? N 2 + 3 H 2  2 NH 3 Bond Enthalpy Calculations Example 1: H-H

20. Example 2 energy course of reaction 2H 2 + O 2 2H 2 O

21. Working out ∆H Show all the bonds in the reactants energy course of reaction 2H 2 O H―H O=O+

22. Working out ∆H Show all the bonds in the products energy course of reaction H―H O=O + HH O HH O

23. Working out ∆H Show the bond energies for all the bonds energy course of reaction 436 O=O+ HH O HH O

24. Working out ∆H Show the bond energies for all the bonds energy course of reaction 436 498+ HH O HH O

25. Working out ∆H Show the bond energies for all the bonds energy course of reaction 436 498+ HH O 464 +

26. Working out ∆H Show the bond energies for all the bonds energy course of reaction 436 498+ 464 + +

27. Working out ∆H Add the reactants’ bond energies together energy course of reaction 464 + + 1370

28. Working out ∆H Add the products’ bond energies together energy course of reaction 1370 1856

29. Working out ∆H ∆H = energy in ― energy out energy course of reaction 1370 1856 1370 1856 - +

30. Working out ∆H ∆H = energy in ― energy out energy course of reaction 1370 1856 1370 1856 - 486 - +

31. Working out ∆H ∆H = energy in ― energy out energy course of reaction 1370 1856 ∆H = -486 exothermic

32. The average bond enthalpies for several types of chemical bonds are shown in the table below: 32

33. Page 326 in text Also see Chemistry Data Booklet p. 11

34. Example 3 Calculate the enthalpy change for the following reaction : CH 4 + Cl 2  CH 3 Cl + HCl 4(414) + 243 -3(414) & (-397) + (-431) 1,899 - 2,070 - 171 kJ mol -1 (exothermic) Cl-Cl  H-Cl  