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# Applications of Hess’s Law Mean Bond Enthalpies

Applications of Hess’s Law Mean Bond Enthalpies. The mean bond enthalpy is defined as the energy required to break a covalent bond or bonds in the gaseous state. H 2 (g)  2H(g) H = +435 kJmol -1. Or it can be the energy released when 1 mole of a particular type of covalent bond is formed.

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## Applications of Hess’s Law Mean Bond Enthalpies

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1. Applications of Hess’s Law Mean Bond Enthalpies

2. The mean bond enthalpy is defined as the energy required to break a covalent bond or bonds in the gaseous state H2(g)  2H(g) H = +435 kJmol-1 Or it can be the energy released when 1 mole of a particular type of covalent bond is formed 2H(g)  H2(g) H= – 435 kJmol-1

3. Bond HH O=O CC CH CO C=O mean B.E. (kJ mol-1) 432 497 346 414 365 798 Since the precise energy of a bond is subject to variation due to the effects of neighbouring atoms or groups, data books always quote an average or mean. These are found on page 9 of Higher / Advanced Higher data book

4. The idea of measuring and calculating the bond enthalpies for various bonds allows another application of Hess’s law It allows the enthalpy changes for a variety of chemical reactions or other bond enthalpies to be calculated that might otherwise be difficult. The values for these are in the data book and can be used as follows to calculate enthalpy changes:-

5. 1. The bond enthalpies for all of the bond breaking and bond making steps are calculated as a “profit and loss” account. 2. The total enthalpies for each side are added together i.e. H =  bond breakingsteps + bond making steps Note : it is helpful when doing these to draw any relevant molecules so as to count the number of bonds being broken and formed accurately.

6. e.g. 1 Calculate the enthalpy change for the reaction H2(g) + Cl2(g) 2HCl(g) H  H 1 x 436 kJ 2HCl 2 x  431 kJ Cl  Cl 1 x 243 kJ Total  862kJ Total + 679 kJ

7. Hence the enthalpy change, H = 679 + ( 862) = 183kJ Since two moles of product are formed H = 183  2 =  91.5kJmol-1

8. e.g. 2 Calculate the enthalpy change for the formation of ethene, C2H4 2C(s) + 2H2(g) C2H4(g) C=C 1 x  607 kJ 2C(s) 2C(g) 2 x 715 kJ 2HH 2 x 436 kJ 4CH 4 x  414 kJ Total  2263 kJ Total + 2302 kJ

9. Hence the enthalpy of formation of ethene = 2302 + ( 2263) = + 39 kJmol-1

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