Energy calculation of Bond enthalpy

There is no denying the fact that Bond Enthalpy in the form of bond dissociation energy calculations for Enthalpy of Reaction is the energy required to break 1 mole of a specified bond for a gaseous species at 298K/25^oC.  i.e. A–B_(g) ==> A_(g) + B_(g)    ΔH_BE(A–B) = + ??? kJmol^–1

The energy output can be done  in terms of a dot and cross diagram which is the breaking of a C–Cl bond by homolytic bond fission. It may be noted that bond breaking is always endothermic (+) and bond formation is always exothermic (–). How are bond enthalpies determined can be reflected as follows:

1. Just as spectroscopy can be used to determine ionization energies  in the form of  spectroscopic techniques which can be used to determine bond energies i.e. it is possible to estimate the frequency of radiation required to cause bond fission.

2. Electron impact methods – in essence it is a method which measures the energy to fragment molecules.

3. Thermo chemical calculations – the method most appropriate to this pages level of study.

3. is illustrated by the diagram below to calculate the C=O bond enthalpy in carbon dioxide.

+498 is the bond enthalpy of the oxygen molecule (O=O) or twice the enthalpy of atomisation of oxygen.

+715 is the enthalpy of atomisation of carbon.

–393 is the enthalpy of combustion of carbon or the enthalpy of formation of carbon dioxide.

This becomes +393 in the Hess's Law cycle, arrow and sign reversed.

x is equal to twice the bond enthalpy of a C=O bond in a carbon dioxide molecule.

x is the only one of the four delta H values that cannot be determined by experiment.

However using Hess's Law

x = +393 +715 +498 = +1606 kJ mol–1

therefore the C=O bond energy in CO2 = 1606/2 = 803 kJ mol–1

A second example of using Hess's Law to calculate the average C–H bond enthalpy in the methane molecule

ΔHØf (methane) = –75 kJmol–1
C(s) + 2H2(g)  CH4(g)
ΔHθsub(C(s)) + 2 x ΔHθBE(H–H(g)) = (+715) + (2 x +436) both endothermic	C(g) +	4H(g)	–4 x avΔHθBE(C–H(g)) = –X exothermic formation of 4 C–H bonds
The cycle for the standard enthalpy of formation of methane, (ΔHθf), based on the enthalpy of sublimation of carbon (graphite) and the H–H and C–H bond enthalpies. From Hess's Law, add up the sequence of enthalpy changes via the lower 'staged route' to get the overall enthalpy change for the enthalpy of formation of methane from its elements in their normal stable states. ΔHθf (methane) = {ΔHθsub(C(s)) + 2 x ΔHθBE(H–H(g))} + {–4 x avΔHθBE(C–H(g))}  = –75 kJmol–1 –75 = +715 +872 –X X = (715 + 872 +75) = 1662 avΔHθBE(C–H(g)) = 1662/4 = +415.5 kJmol–1

  

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Bond enthalpy calculations using average bond enthalpies

• Bond enthalpy calculations using a Hess's Law cycle can be used to calculate unknown enthalpy changes for a reaction.
• BUT there are limitations to the results of these calculations since they are based on ..
• (i) average bond enthalpies (i.e. typical vales for a particular bond) and
• (ii) only valid gaseous reactants AND products (quite restrictive, this because bond enthalpies are defined, measured and based on gaseous species only).

Introductory example to illustrate the method of calculating the enthalpy of a reaction using bond enthalpies

A STARTER CALCULATION

methane + chlorine ==> chloromethane + hydrogen chloride

So, how can we theoretically calculate the energy change for this reaction using bond enthalpies?

CH4 + Cl2 ==> CH3Cl + HCl

Bond energies: The energy required to break or make 1 mole of a particular bond in kJ/mol

C–H = 412, Cl–Cl = 242 kJ/mol, C–Cl = 331 kJ/mol, H–Cl = 432

To appreciate all the bonds in the molecules its better to set out as follows ...

 + Cl–Cl ===>  + H–Cl

Then I'm using the diagram below to illustrate how you do the calculation and what its all about.

First, imagine which bonds must be broken to enable the reaction to proceed.

The energy absorbed equals that to break one C–H bond (in methane molecule) plus energy to break one Cl–Cl bond (in chlorine molecule), both endothermic changes – 'bond breaking'.

Theoretically imagine you've got these atomic or molecular fragments, put them together to form the products, in doing so, work out which bonds must be formed to give the products.

The energy released is that given out when C–Cl bond (in chloromethane molecule) is formed plus the energy released when one H–Cl bond (in hydrogen chloride molecule) is formed, both exothermic changes – 'bond making'.

Calculating the difference in the two sums gives the numerical energy change and since more heat energy is given out to the surroundings in forming the bonds than that absorbed in breaking bonds, the reaction must be exothermic.

Therefore ΔHreaction = –139 kJmol–1

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Further examples – how you might solve them in exams!

Ex 1. Calculating the enthalpy of a reaction

Given the following bond enthalpies in kJ mol–1: Cl–Cl = 242, H–H = 436, H–Cl = 431

Calculate the enthalpy of the reaction of forming 2 moles of hydrogen chloride from its elements in their standard states ... the Hess's Law cycle for this is ...

H2(g) + Cl2(g)  2HCl(g)

 2H(g) + 2Cl(g) 

For simple Q's like this, unless asked for, you can solve easily without drawing a cycle, either way ...

ΔHθreaction = {endothermic ΔH for H2 and Cl2 bonds broken} + {exothermic ΔH for HCl bonds formed}

ΔHθreaction = {+436 +242} + {2 x –431}

ΔHθreaction = +(436 + 242 –862) = –184 kJmol–1

Note that ΔHθreaction /2 = –92 kJmol–1 = ΔHθf(HCl(g))

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Ex 2. Given the following bond dissociation enthalpies at 298K in kJmol–1 ...

Bond	C–C (single)	C–H	C=O in CO2	O–H	O=O
Bond enthalpy	+348	+412	+805	+463	+496

... calculate the enthalpy of combustion of butane. You construct a Hess's Law Cycle from the 'normal' combustion equation and 'atomise' the reactant molecules to give ALL of the 'theoretical' intermediate atoms. From this you can  theoretically calculate the enthalpy of the reaction:

ΔHcomb, 298 (butane) = ??? kJmol–1
(g) + 61/2O=O(g)  4O=C=O(g) + 5H–O–H(g)
(3 x ΔHBE(C–C)) + (10 x ΔHBE(C–H)) + (6.5 x ΔHBE(O=O)) = (3 x +348) + (10 x +412) + (6.5 x +496)	endothermic	4C(g) + 10H(g) + 13O(g)	exothermic	–(8 x ΔHBE(C=O)) – (10 x ΔHBE(O–H)) = –(8 x 743) – (10 x 463)
ΔHcomb, 298 (butane) = 1044 + 4120 + 3224 –6440 –4630 = –2682 kJmol–1 The true thermodynamic value from very accurate calorimetry experiments is –2877 kJmol–1 So why the significant difference/error? The difference is the theoretical value from the bond energy calculation is less exothermic by 195 kJ. All enthalpy calculations done by this method will always be in error to some extent because the bond enthalpies quoted in data information are based on average values for that particular bond. Each 'A–B' bond will differ slightly depending on the 'ambient' electronic situation e.g. The C–H bond in methane, , will be slightly different than in ethane,  etc. There is a 2nd reason which for some reason many textbooks don't bother to mention. Since only gaseous species can be considered, if any reactant or product is a liquid or solid, then the enthalpy value for any state change is NOT taken into account. For example, in the above example, five molecules of water are formed which at 298K would condense to water and this is an exothermic process. ΔHvap for water is 41 kJmol–1, so if water condensation takes place then 5 x 41 = 205 kJ extra would be released and this actually accounts for most of the error!

example of bond enthalpy calculation method

Liquid Hydrazine N2H4 and hydrogen peroxide H2O2 have both been used as rocket fuels.

Hydrazine has been used as monopropellant in rocket engines because it can catalysed to decompose into nitrogen and hydrogen gas very rapidly and exothermically.

(1)  N2H4 ==> N2 + 2H2

Hydrazine can also be used to power rockets in combination with hydrogen peroxide (a bipropellant fuel).

(2)  N2H4 + 2H2O2 ==> N2 + 4H2O

From the list of bond enthalpies in kJmol–1, given below, calculate the enthalpy changes for reactions (1) and (2)

Bond	N–H	NN	N–N	H–H	O–H	O–O
Bond enthalpy	+388	+944	+163	+436	463	+146

For (1)  N2H4 ==> N2 + 2H2

H2N–NH2  NN + 2H–H
ΔHBE(N–N) + (4 x ΔHBE(N–H)) = (+163) + (4 x +388)	endothermic	2N + 4H	exothermic	–(ΔHBE(NN)) – (2 x ΔHBE(H–H)) = –(+944) –(2 x 436)

ΔHreaction(decomp N2H4) = +163 + (4 x 388) –944 – (2 x 436)

= +163 +1552 –944 –872 = –101 kJmol–1

For (2)  N2H4 + 2H2O2 ==> N2 + 4H2O

H2N–NH2 + 2H2O2  NN + 4H–O–H
ΔHBE(N–N) + (4 x ΔHBE(N–H)) + (2 x ΔHBE(O–O)) + (4 x ΔHBE(O–H))   = (+163) + (4 x +388) + (2 x +146) + (4 x +463)	endothermic	2N + 8H + 4O	exothermic	–(ΔHBE(NN)) – (8 x ΔHBE(O–H)) = –(+944) –(8 x +463) watch the – signs!

ΔHreaction(combustion N2H4) = +163 +1552 +292 +1852 –944 –3704

ΔHreaction(combustion N2H4) = –789 kJmol–1

which is considerably more exothermic than the catalyzed thermal decomposition of hydrazine into nitrogen and hydrogen,

but the mixture is more costly and more dangerous to handle for the which we should be more careful about this.