thermo part 3.pdf - THERMO PART 3 BOND ENERGIES IN COVALENT MOLECULES NOTES(Refer to section 9.11 in text Covalent Bonds \u2013 bonds resulting from the

thermo part 3.pdf - THERMO PART 3 BOND ENERGIES IN COVALENT...

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25 THERMO - PART 3 BOND ENERGIES IN COVALENT MOLECULES (Refer to section 9.11 in text) Covalent Bonds – bonds resulting from the sharing of electrons between atoms. e.g ., :Cl– Cl: < < < < < < < < , H–C C–H, H–(C=O)–H The distance between the atoms is the bond length (depends on the atoms). Not all covalent bonds are equal – they involve different nuclei and different electron distributions, ( e.g. , C–C, C=C, C C, etc.). Hence the energy to break bonds can differ. Bond Enthalpy (or Energy) (BE) or Bond Dissociation Enthalpy (or Energy) (BDE), H BD - energy required to BREAK a mole of bonds in the gas phase (does not hold for liquid or solid state). Note: Products are neutral species NOT ions! - is ALWAYS POSITIVE , since enthalpy change is associated with breaking a covalent bond (energy must be supplied). - is a measure of the strength of a covalent bond. BE can be defined as the enthalpy change ( H BD ) for the dissociation of the bond into its constituent atoms or free radicals in the gas phase. e.g ., H–H ( g ) 2H• ( g ) ; H BD = +436 kJ i.e. , to disrupt, break, pull apart, 1 mole of H–H bonds, we must provide 436 kJ. We can then say that the bond enthalpy of H 2 is 436 kJ. Note: H BD = BE(H–H) = +436 kJ 2 ×∆ H f [H, g ] ∴∆ H f [H , g ] = +436 kJ/mol ÷ 2 = 218 kJ/mol H BD also 2 × H Atomisation for H Generally, BE are not as exact as H f values, since BE values come from average values. What about F 2 ( g ) , Cl 2 ( g ) , Br 2 ( g ) , N 2 ( g ) , O 2 ( g ) ? No Problem! Cl 2 ( g ) 2 Cl ( g ) , H BD = 242 kJ/mol O 2 ( g ) 2 O ( g ) , H BD = 498 kJ/mol What about an O–H bond? Use: H 2 O ( g ) 2H ( g ) + O ( g ) ; H Rxn = 926 kJ ONE mole of O–H bonds would require 926kJ/2 or ~_________ kJ. NOTES:
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26 What about C–H? Use: CH 4 ( g ) C ( g ) + 4H ( g ) , H Rxn = +1656 kJ i.e. , to break FOUR moles of C–H bonds costs 1656 kJ, therefore bond enthalpy of C–H 1656 kJ / 4 414 kJ Using BE’s ( H BD ) and their Relationship to H f EXAMPLE: Estimate the H ° f for NH 3 given H BD (N N) is 945 kJ, H BD (H–H) is 436 kJ, and H BD (N–H) is 391 kJ. (Aside: Table 9–5 provides several values.) Strategy: What equation do we want? What we need to do is break the N N & H–H bonds and reform 3(N–H) bonds. H ° f (NH 3 ) = H 1 + H 2 H ° f (NH 3 ) = [ ½ BE(N N) + 3/2BE(H–H)] + [– 3BE(N–H)] H ° f (NH 3 ) = H ° f (NH 3 ) = – 46.5 kJ/mol (compare H ° f (NH 3 ) = – 45.90 kJ from Appendix C) GENERALISATION: H Rxn = (energy needed ( + ) to break bonds) + (energy released ( ) when bonds formed) OR H Rxn = BE Reactant Bonds BE Product Bonds NOTES: Atoms (g) N (g) , 3H (g) Form NH 3 : – 3(N-H) bonds Break H 2 : 3 × [ ½ H 2 (g) H (g) ] Break N 2 : ½ N 2 (g) N (g) ½ N 2 (g) + 3/2 H 2 (g) Elements (most stable form) NH 3 (g) Product (1 mole) H 1 H 2 H ° f = ?
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27 ________________________________________________________________________________ ASIDE: Same question but with different data: Estimate the H ° f for NH 3 given: N( g ) H ° f = 472 kJ/mol H( g ) H ° f = 218 kJ/mol BE(N–H) = H BD = 391 kJ/mol Rxns for values given: H ° f ( N, g ) = 472 kJ/mol; Rxn: H ° f ( H, g ) = 218 kJ/mol; Rxn: NH 3 ( g ) N( g ) + 3H( g ); H Rxn = H ° f (NH 3 ) = – 47 kJ/mol ___________________________________________________________________________________________________ What about C–C? (there are no C–C molecules!) Given: H ° f (C
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