Bond Enthalpy

1. Bond Enthalpy L.O.: Explain exothermic and endothermic reactions in terms of enthalpy changes associated with the breaking and making of chemical bonds. Define and use the term average bond enthalpy. Calculate an enthalpy change of reaction from average bond enthalpies.

2. H-H (g) →H(g) + H(g) ∆H = + 436 KJ mol-1 The bond enthalpy is the change that takes place when breaking one mole of a given bond in the molecules of a gaseous species.

3. Average bond enthalpy is the average enthalpy change that takes place when breaking by homolytic fission 1 mole of a given type of bond in the molecules of a gaseous species.

4. Enthalpy of reaction from bond enthalpies TheoryImagine that, during a reaction, all the bonds of reacting species are broken and the individual atoms join up again but in the form of products. The overall energy change will depend on the difference between the energy required to break the bonds and that released as bonds are made. energy released making bonds > energy used to break bonds ...EXOTHERMIC energy used to break bonds > energy released making bonds ...ENDOTHERMIC Step 1 Energy is put in to break bonds to form separate, gaseous atoms Step 2 The gaseous atoms then combine to form bonds and energy is released its value will be equal and opposite to that of breaking the bonds Applying Hess’s Law DHr=Step 1 + Step 2

5. DH = ∑ (bond enthalpies of bonds broken) - ∑ (bond enthalpies of bonds made)

10. DH = Σ(bonds broken) − Σ(bonds formed) (worth 1 mark)

13. Page 134. Q1-3

14. ATOMS SUM OFTHE BOND ENTHALPIES OF THE REACTANTS SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS REACTANTS DH PRODUCTS Enthalpy of reaction from bond enthalpies Alternative view Step 1 Energy is put in to break bonds to form separate, gaseous atoms. Step 2 Gaseous atoms then combine to form bonds and energy is released; its value will be equal and opposite to that of breaking the bonds DHr=Step 1 - Step 2 Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of bond enthalpy, it’s value is subtracted. DH =  bond enthalpies –  bond enthalpies of reactants of products

15. Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene

16. Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DH2 1 x C=C bond @ 611 = 611 kJ 4 x C-H bonds @ 413 = 1652 kJ 1 x H-H bond @ 436 = 436 kJ Total energy to break bonds of reactants = 2699 kJ

17. Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DH2 1 x C=C bond @ 611 = 611 kJ 4 x C-H bonds @ 413 = 1652 kJ 1 x H-H bond @ 436 = 436 kJ Total energy to break bonds of reactants = 2699 kJ DH3 1 x C-C bond @ 346 = 346 kJ 6 x C-H bonds @ 413 = 2478 kJ Total energy to break bonds of products = 2824 kJ Applying Hess’s LawDH1 = DH2 – DH3= (2699 – 2824) = – 125 kJ

18. Enthalpy of reaction from enthalpies of formation If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements. Enthalpy of formation tends to be an exothermic process

19. ELEMENTS SUM OFTHE ENTHALPIES OF FORMATION OF THE REACTANTS SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS REACTANTS DH PRODUCTS Enthalpy of reaction from enthalpies of formation Step 1 Energy is released as reactants are formed from their elements. Step 2 Energy is released as products are formed from their elements. DHr =-Step 1 +Step 2 orStep 2-Step 1 In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, it’s value is subtracted. DH =  DHfof products – DHfof reactants