5.4 Bond Enthalpies. CH 4(g) + O 2(g) CO 2(g) + 2H 2 O (g) 1 P a g e

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1 5.4 Bond Enthalpies Bond breaking is endothermic and bond making is exothermic. Bond making produces greater stability because the resulting products have a lower energy state. bond making bond breaking CH 4(g) + O 2(g) CO 2(g) + 2H 2 O (g) 1 P a g e

2 During every chemical reaction sufficient energy needs to first be absorbed (endothermic) to break the bonds. This energy is called the activation energy and is defined as the minimum amount of energy required by the reactants before products will form. Some reactions have small activation energies, because there is enough energy in the surroundings at room temperature to get them started. Other reactions have higher activation energies and need to absorb much more heat energy in order to break the bonds. Burning / combusting a fuel is an example. Not all the bonds of the reactants need to break at the same time before products form. Once one or two bonds have broken new bonds can start to form and this usually releases enough energy to keep the reaction going. Recall that a covalent bond is an electrostatic attraction between positive protons in the nucleus (called the positive nuclear charge) of the bonded atoms and the negative shared electrons in the bond. To break this bond sufficient energy needs to be absorbed by the bond to overcome the attraction between the protons and electrons. The stronger the bond the greater the amount of energy required to break it. 2 P a g e

3 The factors that affect the strength of a bond are: 1. When the number of shared electrons in the bond increases the attraction between them and the protons in the nuclei of the bonded atoms increases. Triple bonds are stronger than double which are stronger than single bonds. (C-C, H = 348 kjmol -1 ; C=C, H = 612 kjmol -1 and C C, H = 837 kjmol -1 ) 2. When the radius of the bonded atoms decreases, the protons and electrons of the bond are closer to each other so the attraction between them increases, making the bond weaker. 3. When the number of protons in the nuclei of the bonded atoms increases the attraction between them and the bonding electrons increases, increasing the strength of the bond. So far we have calculated enthalpy changes the following ways: 1. H = (m x c x T) n 2. Hess s Law 3 P a g e

4 Another way to calculate enthalpy changes is using average bond enthalpies. Average bond enthalpy is defined as the average energy required to break one mole of gaseous covalent bonds. They are calculated by taking the average of the enthalpies for that specific bond obtained from a number of similar compounds. The average bond enthalpy provides an indication of the strength of a chemical bond. Average bond enthalpies values are not particularly accurate because they are average values. They are most accurate and most useful when only a few bonds are made and broken, when specific bond energies are used instead of averages and when the reactants and products are gases. Calculating Enthalpy Changes from Average Bond Enthalpies H = H (bonds broken in reactants ) - H (bonds made in products ) (kj mol -1 ) Tables of average bond enthalpies are found in the Chemistry data booklet. 4 P a g e

5 Questions Use the average bond enthalpy data above to calculate the enthalpy change, H for questions 1 to 8. For each calculation show all the steps and final units. Express your answer using the correct number of significant figures. Since addition and subtraction is involved follow the decimal place rule. Each question is worth 3 points. 1. CH 4 (g) + 4 Cl 2 (g) CCl 4 (g) + 4 HCl(g) 2. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) 3. CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) 4. H 2 (g) + Br 2 (g) 2 HBr(g) 5. C 2 H 6 (g) + 3 ½ O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) 6. C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O(g) 7. C 2 H 2 (g) + 2 ½ O 2 (g) 2 CO 2 (g) + H 2 O(g) 8. 3 C(g) + 4 H 2 (g) C 3 H 8 (g) 9. Identify the enthalpy of combustion reactions in questions 1 to 8. Which is the best fuel and why? 10. Account for the differences in the strength of single, double and triple bonds by comparing the average bond enthalpies of the C-O, C=O and C N bonds. 11. Account for the differences in the enthalpies of combustion for ethane, C 2 H 6 (g), ethane, C 2 H 4 (g) and ethyne, C 2 H 2 (g) 12. (M99) Define the term average bond enthalpy. [3] 13. Determine the bond enthalpy for the HF bond in kjmol -1 in the reaction. H 2 (g) + F 2 (g) 2 HF (g) H = -521 kj 14. Given that the bond enthalpy of the carbon-oxygen bond in carbon monoxide and carbon dioxide are 1073 kjmol -1 and 743 kjmol -1 respectively and that of the bond in the oxygen molecule is 496 kjmol -1. Calculate the enthalpy change for the combustion of one mole of carbon monoxide. [3] 5 P a g e

6 15. Given that the enthalpy change for the reaction below is +688 kjmol -1. Calculate the bond enthalpy of the N-Cl given that the bond enthalpy in the nitrogen molecule and the chlorine molecule are 944 kjmol -1 and 242 kjmol -1 respectively. [3] N 2(g) + 3 Cl 2(g) 2 NCl 3(g) 16. Use bond energy data to calculate the enthalpy when cyclopropane, C 3 H 6 reacts with hydrogen to form propane, C 3 H 8. The actual value found is -159 kjmol -1. State one reason why you think this differs from the value you have calculated. [3] 17. (M04) Consider the following reaction. HL Only N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Use the values in the Chemistry data booklet calculate the enthalpy change, Hº for the reaction. [3] 18. (M04) Enthalpies of reactions, for example combustion, can be calculated using average bond enthalpies or enthalpies of formation. The two methods give closer results for cyclohexane, C 6 H 12 than they do for benzene, C 6 H 6. Explain this difference. [3] 19. The bond enthalpy of the N-O bond in nitrogen dioxide is 305 kjmol -1. Those of the bonds in the oxygen molecule and the nitrogen molecule are 496 kjmol -1 and 944 kjmol -1 respectively. a) What will be the enthalpy change for the reaction? [3] N 2 (g) + 2 O 2 (g) 2 NO 2 (g) b) Explain why experimental results show the N-O bond lengths are the same in NO 2. 6 P a g e

7 Bibliography Clark, Jim. Chem Guide < Clugston, Michael and Rosalind Flemming. Advanced Chemistry. Oxford: Oxford University Press, Derry, Lanna, Maria Connor and Carol Jordan. Chemistry for use for the IB Diploma Standard level. Melbourne: Pearson Education, Green, John and Sadru Damji. Chemistry for use with the International Baccalaureate Programme. Melbourne: IBID Press, Neuss, Geoffrey. IB Diploma Programme Chemistry Course Companion. Oxford: Oxford University Press, IB Study Guides, Chemistry for the IB Diploma. Oxford: Oxford University Press, Organisation, International Baccalaureate. Online Curriculum Centre. < "Chemistry Data Booklet." International Baccalaureate Organisation, March "Chemistry Guide." International Baccalaureate Organisation, March "IB Chemistry Examination Papers." Cardiff: International Baccalaureate Organisation, P a g e

8 6.4 Bond Enthalpy ANSWERS kj mol kj mol kj mol kj mol ethane kj mol ethene kj mol ethyne kj mol C(g) + 4 H 2 (g) C 3 H 8 (g) H = 4(H-H) - 2(C-C) + 8(C-H) = kj mol Combustion reactions are those where a substance burns/combusts in oxygen. The combustion reactions are: 2, 5, 6, 7. A fuel is a substance that when combusted releases heat energy. Ethane is he best fuel as it releases more kilojoules of energy per mole. 10. When the number of shared electrons in the bond increases the attraction between them and the protons in the nucleus of the bonded atoms increases, increasing the amount bond dissociation energy. Triple bonds (6 e) are stronger than double (4 e) which are stronger than single bonds (2 e). 11. The enthalpy change increases from C 2 H 6 (C-C) to C 2 H 4 (C=C) to C 2 H 2 (C C) because the bond enthalpy to break the carbon carbon bond increases. 12. The energy needed to break ; one mole of covalent bonds; in their gaseous state; (bolded words must be used to get the mark) 8 P a g e

9 13. H = (H-H) + (F-F)) 2(HF) -521 = ( ) 2(HF); 2HF = 1115 kj mol -1 ; HF = = 558 kj mol -1 ; (Divide by two because the bond enthalpy of HF calculated is for two moles of bonds) 14. CO (g) + ½ O 2(g) CO 2(g) H(CO) = (C O) + ½ (O=O) 2(C=O); = (1073) + ½ (496) 2(743); = -165 kj mol -1 ; 15. H = 1(N N) + 4(Cl-Cl) 6(N-Cl); +688 = 1(944) + 3(242) H 6(N-Cl) ; H 6(N-Cl) = = 982 kj mol -1 ; H (N-Cl) = = 164 kj mol -1 ; 16. H = 3(C-C) + 6(C-C) + 1(H-H) 2(C-C) + 8(C-H)); = 3(348) + 6(412) + (436) - 2(348) + 8(412) = = - 40 kj mol -1 ; 17. The H has been calculated using average bond enthalpies. H = (N N) + 3(H-H) 6(N-H); = (436) 6(388); = - 76 (kj mol -1 ); 9 P a g e

10 HL 18. cyclohexane has single bonds of the same length / strength ; benzene has delocalized / resonance structure ; with intermediate length / strength / 1.5 bond order ; So average bond enthalpies for cyclohexane are more accurate / closer to the expected value ; a) H = (N N) + 2(O=O) 4(N-O); = (496) 4(305); = 716 (kj mol -1 ); b) NO 2 exists as a resonance hybrid due to it having delocalized π electrons. 10 P a g e