Energetics

1. Energetics Bond Enthalpies 5.4

2. Bond Enthalpies • Chemical reactions involve the breaking and making of bonds. • To understand the energy changes in a chemical reaction, we need to look at the energy needed to break the bonds that hold the atoms together in the reactants and the energy released when new bonds are formed in the products.

3. Breaking bonds is an endothermic process • Energy is needed to separate the atoms in a bond. • The bond enthalpy is the energy needed to break one mole of bonds in gaseous molecules under standard conditions.

4. Bond enthalpies • For example, the energy change during the formation of two moles of chlorine atoms from one mole of chlorine molecules can be represented as: Cl2(g)  2Cl(g) Δ Ho = +242 kJ/mol

5. Bond enthalpy • The situation is complicated in molecules which contain more than two atoms. • For example, the C-H bond energies differ slightly from compound to compound, as in CH4 (435 kJ) and CH3F (452 kJ). • However, their bond energies are sufficiently constant to be useful. • So, for polyatomic molecules, we speak of average bond enthalpies. • Multiple bonds generally have higher bond enthalpies and are shorter than single bonds.

6. Average Bond Enthalpies (H) • Average bond enthalpies are positive, because bond breaking is an endothermic process. NOTE: These are average bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.

7. Forming bonds is an exothermic process • The same amount energy that is absorbed when a bond is broken, is released when a bond is made. • For example: Δ Ho = -242 kJ/mol

8. Energy changes • We can now understand how energy changes occur in chemical reactions. • Consider the complete combustion of methane: CH4(g) + 2O2(g)  CO2(g) and 2H2O(g) 1. Draw the Lewis structures for the molecules involved. + 2O=O  O=C=O + 2H-O-H

9. Energy changes 2. Determine what type of bonds need to be broken. 4 C-H and 2 O=O bonds need to be broken 3. Add up all the bond enthalpies to obtain an estimate of how much energy is required to break the bonds in the molecules. (obtain bond enthalpies from the average bond enthalpies table). For C-H = 4 (412kJ/mol) =1648 kJ/mol For O=O = 2 (496 kJ/mol)= 992 kJ/mol Total energy required to break bonds = 2640 kJ/mol

10. Energy changes 4. Determine the bonds that will form during the reaction with their corresponding energy changes. 2 C=O and 4 O-H bonds will form For C=O, 2 (743 kJ/mol) = 1486 kJ/mol For O-H = 4 (463 kJ/mol) = 1852 kJ/mol Total energy released by the formation of the C=O and O-H bonds = 3338 kJ/mol

11. Enthalphies of Reaction 5. To estimate H for the reaction, simply compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. • Hrxn = (bond enthalpies of bonds broken)  (bond enthalpies of bonds formed) • So, Hrxn = 2640 kJ/mol- 3338 kJ/mol= -698 kJ/mol • This value should be compared with the experimental value of -890 kJ/mol measured under standard conditions given in the enthalpies of combustion table. The values are different because the standard state of water is liquid and the bond enthalpy calculation assumes that the reaction occurs in the gaseous state.

12. Enthalpies of Reaction (H ) • Comparing Fuels • The enthalpy change of combustion of methane (CH4 ) is -890 kJ/mol while that of methanol (CH3OH) is -715 kJ/mol. • Methanol has a lower enthalpy of combustion, because when it reacts only three O-H bonds are formed, compared with the combustion of methane, in which 4 O-H bonds are formed.

13. Bond enthalpy • Exercises: • 1. Calculate the H for the reaction: C2H4(g) + H2(g) C2H6(g) • 2. Calculate the H for the reaction: 2H2(g)+ O2(g)  2H2O(g)