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.

H-H (g) →H(g) + H(g) ∆H = 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.

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.

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 Enthalpy of reaction from bond enthalpies Step 1Energy is put in to break bonds to form separate, gaseous atoms Step 2The 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  r = Step 1 + Step 2

 H = ∑ (bond enthalpies of bonds broken) - ∑ (bond enthalpies of bonds made)

 H = Σ(bonds broken) − Σ(bonds formed) (worth 1 mark)

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Enthalpy of reaction from bond enthalpies  =  bond enthalpies –  bond enthalpies of reactants of products  =  bond enthalpies –  bond enthalpies of reactants of products Alternative view Step 1Energy is put in to break bonds to form separate, gaseous atoms. Step 2Gaseous atoms then combine to form bonds and energy is released; its value will be equal and opposite to that of breaking the bonds  r = 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. SUM OFTHE BOND ENTHALPIES OF THE REACTANTS REACTANTS PRODUCTS ATOMS  SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS

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

Calculate the enthalpy change for the hydrogenation of ethene Enthalpy of reaction from bond enthalpies  2 1 x C=C 611= 611 kJ 4 x C-H 413= 1652 kJ 1 x H-H 436= 436 kJ Total energy to break bonds of reactants= 2699 kJ

Calculate the enthalpy change for the hydrogenation of ethene Enthalpy of reaction from bond enthalpies  2 1 x C=C 611= 611 kJ 4 x C-H 413= 1652 kJ 1 x H-H 436= 436 kJ Total energy to break bonds of reactants= 2699 kJ  3 1 x C-C 346= 346 kJ 6 x C-H 413= 2478 kJ Total energy to break bonds of products= 2824 kJ Applying Hess’s Law  1 =  2 –  3 = (2699 – 2824) = – 125 kJ

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 Enthalpy of reaction from enthalpies of formation

Step 1Energy is released as reactants are formed from their elements. Step 2Energy is released as products are formed from their elements.  r = - 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. SUM OFTHE ENTHALPIES OF FORMATION OF THE REACTANTS REACTANTS PRODUCTS ELEMENTS  SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS Enthalpy of reaction from enthalpies of formation   =   f of products –   f of reactants