5.3 Bond Enthalpy

Bond Enthalpy

• Learn the Definition:

“The average bond enthalpy is the energy needed to ____________________ 1 mole of comparable bonds averaged cross a range of compounds, all in the ________________ state.”

1. We have to define whether we’re using bond breaking (endothermic) or bond forming (exothermic) as the sign depends on this. So we stick to one option.

2. Some bond energies are specific to certain molecules.

eg Cl-Cl bonds only exist in Cl₂, H-F bonds only in Hydrogen Fluoride.

So, the calculated values will be correct.

But C-H bonds exist in thousands of compounds and are not exactly the same strength in each one.

So the bond energies are calculated for a range of representative molecules and an average is taken.

This means the calculated value is always wrong (although close to correct), so all ∆H_Reaction calculated from bond energies will be wrong (although close enough to be a useful guide when there’s no other way to do the calculation)

3. If we didn’t use gaseous compounds the bond energies would not be comparable because some energy would go into breaking intermolecular forces as well as bonds.

But they also have more _________________ in those nuclei to attract the bond pair.

So, long bonds between big atoms might be stronger

Note: Hydrogen atoms are very small, so make very short bonds. But they have only one proton so they are sometimes an exception to this rule.

Bond Enthalpies measured this way are for covalent compounds.

Some bonds are perfectly covalent – the bond pair is exactly midway between the two atoms

Eg H-H, Cl-Cl, O=O etc. But this can only be true in elements

Most bonds are between different atoms, so the bond pair is shared unevenly.

• You should know that we call the of an element ability to attract the bond pair in a covalent bond its ___________________________

• A difference in electronegativity of around ______________ means that the bond is considered to be IONIC.

You could remember:

∆H_r = ∑ bonds broken - ∑ bonds formed

But it’s easy enough to use the same triangle method as before.

Note: Since the definition is for bond breaking the arrows point away from the compounds towards the single atoms (the bottom) where there are no bonds.

In this case,

H-H = + 436 kJ/mol O=O = + 498 kJ/mol O-H= = + 463 kJ/mol

So, ∆H_r = 436 + (0.5 x 498) – (2 x 463) = -241 kJ/mol

It’s usually easier to draw the displayed formulae of the reactants and products

So, ∆H_r = (4 x 413) + (2 x 498) – (2 x 805) - (4 x 464) = -818 kJ/mol

1 Calculate the enthalpy change for the following reaction

CH₃-CH_{3 (g)} + Cl_2(g) → CH₃-CH₂Cl_(g) + HCl_(g)

Bond enthalpies: C-C 348, C-H 412, Cl-Cl 242, C-Cl 338, H-Cl 431 kJ/mol

2 Hydrazine has the formula N₂H₄ and is used as a rocket fuel. It burns in the following reaction for which the enthalpy change is -583 kJ/mol.

N₂H_4(g) + O_2(g) → N_2(g) + 2 H₂O_(g)

Calculate the N-N bond enthalpy in Hydrazine given the bond enthalpies.

Bond enthalpies: N-H 388, O=O 498, NòN 944, O-H 463 kJ/mol

3 Use the following bond enthalpies to calculate ∆H for the following reactions. You may assume that all species are in the gaseous state.

Bond H–H O=O C–C C–H H–O Br–Br H–Br C=O

KJ mol^-1 436 496 348 412 463 193 366 743

a) H- H + Br -Br --> 2 H-Br

b) ∆H_c of Methane

c) ∆H_c of Propane

1. Calculate the average N-H bond energy in NH_3(g) using the data below.

∆H_f of NH_3(g) = -46 kJ mol^-1

Bond enthalpies: N≡N = 944; H-H = 436 kJ mol^-1

2. Calculate the C=C bond energy in Ethene using the data below.

CH₂=CH_2(g) + H_2(g) → CH₃CH_3(g) ∆H_r = -138 kJ mol^-1

Bond enthalpies: C-C = 348; H-H = 436; C-H = 412 kJ mol^-1

The concentration of Ozone in the atmosphere is held constant (in unpolluted air) by the following processes

It takes an energy input from UV light to destroy Ozone – making the process endothermic.

• This means that the bonds are stronger in ______________