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5.3 Bond Enthalpy. Understanding Bond forming releases energy and bond breaking requires energy. Bonds form because it is energetically favorable to do.

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Presentation on theme: "5.3 Bond Enthalpy. Understanding Bond forming releases energy and bond breaking requires energy. Bonds form because it is energetically favorable to do."— Presentation transcript:

1 5.3 Bond Enthalpy

2 Understanding Bond forming releases energy and bond breaking requires energy. Bonds form because it is energetically favorable to do so (releasing energy) Bonds are stable and it takes energy to force the separation

3 Understanding/Application & Skills Calculation of the enthalpy changes from known bond enthalpy values and comparison of these with experimentally measured values. Values found in section 11 of Data Booklet  H Θ =∑  (BE bonds broken)-∑  (BE bonds formed) These values will vary from  H Θ =∑  H f Θ (products)-∑  H f Θ (reactants) because the bond enthalpies are average values Average Bond enthalpy is the energy needed to break 1 mol of a bond in a gaseous molecule averaged over similar compounds. Values are positive because they correlate to breaking bonds Positive enthalpy values Do not take into account intermolecular forces in liquids

4 Using Bond enthalpies Using the data from section 11 of the data booklet, find the enthalpy change of the reaction C 2 H 4 (g) + HBr (g)  C 2 H 5 Br (g)  H Θ =∑  (BE bonds broken)-∑  (BE bonds formed) =[4BE C-H + BE C=C + BE H-Br ] –[5BE C-H + BE C-C + BE C-Br ] =[(4 x 415) + (614) + (366)] – [(5 x 414) + (346) + (285)] =2636 – 2701 = -65 kJ mol -1  H Θ =∑  H f Θ (products)-∑  H f Θ (reactants) Comparison = (-90.0)-(52.0+-36.3) = -105.7 kJ

5 Applications and skills Sketching and evaluation of potential energy profiles in determining whether reactants or products are more stable and if the reaction is endothermic or exothermic. Materials are more stable when they have lower energy.

6 Applications and skills Discussion of the bond strength in ozone relative to oxygen in its importance to the atmosphere. Double bonds are stronger than single bonds The resonance structures of ozone show about 1.5 bonds between each oxygen Oxygen oxygen bond in O 2 takes 489 kJ mol -1 but oxygen oxygen bond in O 3 takes 364 kJ mol -1 This means ozone molecules is decomposed more readily than oxygen

7 End of Chapter #14 Methanol is made in large quantities as it is used in the production of polymers and in fuels. CH 3 OH (l) + 1 ½ O 2(g)  CO 2 (g) + 2 H 2 O (g) a.Using the information in the data booklet, determine the theoretical enthalpy of combustion of methanol. (3) b.A burner containing methanol was weighed (80.557g) and used to heat the water (20.000g) from 21.5 to 26.4 C. 80.034 g of methanol remained. a.Calc # mol of methanol burned (2) b.Calc heat absorbed, in kJ by the water (3) c.Determine the enthalpy change in kJ mol -1 for the combustion of 1 mol methanol. (2) c.Data booklet value for enthalpy is -726 kJ mol-1. Suggest why this value differs from the values calculated in parts a and b a.Part a (1) b.Part b (1)


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