3.2.1 (f) Bond Enthalpies

Syllabus

(i) explanation of the term average bond enthalpy (breaking of 1 mol of bonds in gaseous molecules)

(ii) explanation of exothermic and endothermic reactions in terms of enthalpy changes associated with the breaking and making of chemical bonds

(iii) use of average bond enthalpies to calculate enthalpy changes and related quantities (see also 2.2.2 f)

{Definition of average bond enthalpy not required.}

{Learners are expected to understand that an actual bond enthalpy may differ from the average value.}

What does this mean?

Defining a bond enthalpy

You are no longer expected to be able to define Average Bond Enthalpy.

This is a pity because you'll need to understand the definition anyway.

Firstly, the definition specified breaking bonds rather than making them.

If we consider the H-H bond in a Hydrogen molecule it is obvious (hopefully) that it would take some energy to break - so all bond enthalpies are endothermic and hence positive.

But how would you break one individual H-H bond?

So the definition was for breaking a whole mole of bonds. Which is why they are listed in kJmol^-1

And we'll see that bonds can break homolytically or heterolytically (see later) so the definition specified that the bond fission (breaking) should be homolytic.

And not everything is a gas. If we added energy to solid Iodine, for example, energy would be needed to overcome the van der Waals forces holding each I₂ molecule to its neighbours before you could break the I-I bonds.

This would make this a poor comparison to breaking bonds in F₂ - which would have started as a gas and wouldn't need this extra energy input.

So the definition also insisted that we had to break one mole of bonds homolytically in gaseous molecules.

H-H bonds only exist in Hydrogen molecules, F-F bonds only in Fluorine molecules and I-I bonds only in IOdine molecules.

This isn't true of most bonds: C-C bonds exist in nearly all Organic molecules, as do C-H bonds, O-H bonds, C-O bonds etc.

So the definition was "The average bond enthalpy term is the average amount of energy needed to break a specific type of bond homolytically in gaseous molecules , measured over a wide variety of different compounds."

Even if we broke all 4 C-H bonds in Methane measurements have shown that each bonds has a similar but different strength.

So the need for an average for most bonds is important but means that when we use bond-enthalpies to carry out a calculation they will nearly all be slightly wrong - hopefully not by more than 10%.

Using bond enthalpies.

Imagine burning Hydrogen in Oxygen to make water.

2 H₂ + O₂ → 2H₂O

We can use bond enthalpies to estimate the enthalpy change for the reaction - although it will not be precise.

Firstly we would need to know what bonds were in each molecule.

From the diagram we can see that the reaction above would require the breaking of 2 moles of H-H bonds and 1 mole of O=O bonds.

And that 4 moles of O-H bonds would be formed.

The enthalpy change for the reaction (ΔH_reaction) would be the difference between the energy input (bond breaking) and energy output (bond forming).

So we simply need to look up the H-H, O=O and O-H bond enthlapies.

And we could use this to estimate related quantities.

eg. ΔH_combustion (H₂) is the enthalpy change for burning one mole of Hydrogen which is half of the above ( -241 kJ/mol)

eg. ΔH_formation (H₂O) is the enthalpy change for makingone mole of water which is also half of the above ( -241 kJ/mol)

In reality ΔH_formation (H₂O) and ΔH_combustion (H₂) are both 286 kJ/mol which is a close but shows the limitation of these estimates.

Example.

If we take the example of Burning Methane in Oxygen.

We can simply total in the bonds to break and the bonds to form

Σ(bonds broken) = 4(C-H) + 2 (O=O) = (4 x 413) + ( 2 x 498) = + 2648 kJ/mol

Σ(bonds formed) = 4(O-H) + 2 (C=O) = (4 x 463) + ( 2 x 732) = -3316 kJ/mol

ΔH = +2648 - 3316 = - 668 kJ/mol

We could display the whole thing as an energy level diagram.

Although, for some reason, this genius has done his calculation in calories per mole rather than kJ/mol

Or you see the calculation set out as a triangle.

The advantage of these methods is that they provide a visual prompt that is missing from the purely mathematical method.

If the examiner may give you some but not all of the bond enthalpies and also ΔH _reaction

You could work out the missing bond enthalpy.

It wouldn't matter which method you used because the examiner doesn't have a preferred method.

They will simply give you full marks for a correct answer.

And partial marks if they can see some correct work but an incorrect answer.

So set out which ever method you use carefully.

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Exam-style Questions

1. (a) The Propane gas in the tank is used as a fuel in the factory. The equation for its combustion is:

C₃H_8(g) + 5O₂(g) ® 3CO2(g) + 4H2O(l) DH = –2220 kJ mol–1

Calculate the amount of heat energy, in kJ, produced during the combustion of 30.0 kg of Propane gas.

(2)

(b) (i) Explain what is meant by the term average bond enthalpy.

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.........................................................................................................................................................................(3)

(ii) Use the average bond enthalpy data below to calculate the molar enthalpy change for the following reaction.

C₃H_8(g) + 5O₂(g) ® 3CO2(g) + 4H2O(g)

(3)

(iii) The value obtained in (b)(ii) is different from the standard molar enthalpy change of combustion of Propane gas given in (a). State two reasons for this.

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.........................................................................................................................................................................(2)

(Total 10 marks)

2. The table below contains some mean bond enthalpy data.

(a) Explain the term mean bond enthalpy.

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.........................................................................................................................................................................(2)

(b) (i) Write an equation for the formation of one mole of ammonia, NH₃, from its elements.

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(ii) Use data from the table above to calculate a value for the enthalpy of formation of ammonia.

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.........................................................................................................................................................................(4)

(c) Use the following equation and data from the table above to calculate a value for the C–H bond enthalpy in ethane.

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.........................................................................................................................................................................(3)

(Total 9 marks)

Answers

1. (a) 2220 × 30,000/44 (1)

=1 510 000 kJ (unit must be correct for value) (1)

ignore –ve sign

do not allow kJ mol–1 for unit

penalise significant figures (other than 3 or 4) once in (a) and (b) together 2

(b) (i) the energy / enthalpy needed to break 1 mol of bonds (1)

averaged over a number of molecules containing that bond

or ‘different environments for the bond’ (1)

reference to gas phase or to produce gaseous atoms (1)

ignore reference to s.t.p. or standard conditions 3

(ii) bonds broken: (2 × 348) + (8 × 412) + (5 × 496) = 6472 (1)

(condone wrong sign)

bonds formed: (6 × 743) + (8 × 463) = 8162 (1)

(condone wrong sign)

give one mark for correct 2:8:5 or 6:8 ratio if both totals above are wrong

enthalpy change = –1690 (ignore units) (1)

(iii) value in (c)(ii) used average / non-specific values (1)

H₂O is

in different states in the two equations / is a gas in (c) / is a liquid in (b) (1)

2. (a) Enthalpy (Energy) to break a (covalent) bond (1) OR dissociation energy

Varies between compounds so average value used (1) QL mark

OR average of dissociation energies in a single molecule / e.g. CH4

Do not allow mention of energy to form bonds

but with this case can allow second mark otherwise 2nd mark

consequential on first

2

(b) (i) 1/2 N₂ + 3/2 H₂ ® NH₃ (1)

Ignore s s

(ii) DH = (S)bonds broken – (S)bonds formed (1)

= 1/2 × 944 + 3/2 × 436 – 3 × 388 (1)

= –38 kJ mol–1 (1)

Ignore no units, penalise wrong units

Score 2/3 for -76

1/3 for +38

Allow 1/3 for +76

4

(c) 4 (C–H) + (C=C) + (H–H) – (6 (C–H) + (C–C)) = –136 (1)

(C=C) + (H–H) – ((C–C) + 2 (C–H)) = –136

2 (C–H) = 836 (1)

(C–H) = 418 (kJ mol–1) (1)

Note: allow (1) for –836

another (1) for –418

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