Mean Bond Enthalpy

Spec Mapping — OCR H432 Module 3.2.1 — Mean bond enthalpy, covering the definition of bond dissociation enthalpy in the gas phase, the distinction between exact bond dissociation enthalpies for specific bonds in specific molecules and mean bond enthalpies averaged across a representative sample of compounds, the algebraic shortcut $\Delta_r H^\ominus = \sum \text{(bonds broken in reactants)} - \sum \text{(bonds formed in products)}$Δr​H⊖=∑(bonds broken in reactants)−∑(bonds formed in products), and the limitations of bond-enthalpy calculations including their restriction to gas-phase species (refer to the official OCR H432 specification document for exact wording).

The Hess cycles of Lessons 3 and 4 require tabulated enthalpies of formation or combustion for every species in the reaction. When those data are not available — or when a quick mental estimate is needed — we can fall back on a third Hess-type calculation that uses gas-phase atoms as the common intermediate. Imagine taking every reactant molecule and breaking every bond, atomising the molecule into isolated gas-phase atoms; the enthalpy input is $\sum \text{(bond enthalpies broken)}$∑(bond enthalpies broken). Now reassemble those atoms into the product molecules: the enthalpy released is $\sum \text{(bond enthalpies formed)}$∑(bond enthalpies formed). The net is $\Delta_r H^\ominus = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$Δr​H⊖=∑(bonds broken)−∑(bonds formed). The catch is that "bond enthalpy" is not a single well-defined number for any given bond type — the strength of a C–H bond in CH$_4$4​ differs from the strength of a C–H bond in CHCl$_3$3​ or benzene by 10–30 kJ mol$^{-1}$−1 — so OCR data booklets quote mean bond enthalpies, averaged across a representative set of molecules. The result: bond-enthalpy calculations are intrinsically approximate, typically within 5–15% of the literature value, and they apply only to gas-phase reactions where no liquids or solids appear in either reactant or product side.

Key Equation: for a gas-phase reaction: $\Delta_r H^\ominus = \sum \text{bonds broken in reactants} - \sum \text{bonds formed in products}$Δr​H⊖=∑bonds broken in reactants−∑bonds formed in products Because mean bond enthalpies are tabulated as positive values (energy required to break), the algebra is reactant bonds minus product bonds — the reverse of the products-minus-reactants convention used with $\Delta_f H^\ominus$Δf​H⊖ in Lesson 3.

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What is bond enthalpy?

The bond dissociation enthalpy (often just "bond enthalpy" or BE) is the enthalpy change required to break one mole of a specified covalent bond in the gas phase, with the two product fragments separated to infinity at constant pressure. By convention bond enthalpies are positive — breaking bonds requires energy input.

Example: $\text{H}_2\text{(g)} \rightarrow 2\text{H(g)}$H2​(g)→2H(g), $\Delta H = +436$ΔH=+436 kJ mol$^{-1}$−1. The bond enthalpy of H–H is therefore $+436$+436 kJ mol$^{-1}$−1.

Conversely, forming a bond from gas-phase atoms is the reverse process and releases the same amount of energy: $2\text{H(g)} \rightarrow \text{H}_2\text{(g)}$2H(g)→H2​(g), $\Delta H = -436$ΔH=−436 kJ mol$^{-1}$−1.

Why "mean"?

The strength of, say, a C–H bond is not the same in every molecule. In methane the first C–H bond to break takes about $+439$+439 kJ mol$^{-1}$−1; the second $\sim +458$∼+458 kJ mol$^{-1}$−1; the third $\sim +396$∼+396 kJ mol$^{-1}$−1; the fourth $\sim +338$∼+338 kJ mol$^{-1}$−1. The geometric and electronic environment of each remaining C–H changes as we strip off the others. The C–H bonds in chloroform CHCl$_3$3​, ethanal CH$_3$3​CHO, or benzene C$_6$6​H$_6$6​ all differ slightly from each other and from methane.

The mean bond enthalpy for a bond type is an average value taken over a representative set of molecules containing that bond. The OCR data booklet quotes one figure that you apply universally; the price is a modest loss of accuracy compared to specific bond-dissociation enthalpies.

Bond	Mean bond enthalpy / kJ mol$^{-1}$−1
H–H	+436
C–H	+413
C–C	+347
C=C	+612
C≡C	+838
O=O	+498
C–O	+358
C=O (in CO$_2$2​)	+805
C=O (in aldehydes/ketones)	+745
O–H	+464
N≡N	+945
N–H	+391
N=N	+418
Cl–Cl	+243
H–Cl	+432
H–F	+568
H–Br	+366
H–I	+298
C–F	+484
C–Cl	+338
C–N	+286
C=N	+615
C≡N	+887

These are representative OCR data-booklet values — always use the values provided in the exam paper rather than memorised figures.

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Calculating $\Delta_r H^\ominus$Δr​H⊖ from mean bond enthalpies

For a gas-phase reaction:

$\boxed{\Delta_r H^\ominus = \sum \text{(bonds broken in reactants)} - \sum \text{(bonds formed in products)}}$Δr​H⊖=∑(bonds broken in reactants)−∑(bonds formed in products)​

Bonds broken take energy in ($+$+); bonds formed release energy out ($-$−). The difference is the net enthalpy change. The sign convention follows from the convention that bond enthalpies are tabulated positive: we use reactant bonds minus product bonds — the reverse of the products-minus-reactants rule used with $\Delta_f H^\ominus$Δf​H⊖ in Lesson 3.

ΔrH° = Σ bonds broken − Σ bonds formed

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Worked example 1 — combustion of hydrogen

Q: Using mean bond enthalpies, estimate $\Delta_r H^\ominus$Δr​H⊖ for $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$2H2​(g)+O2​(g)→2H2​O(g).

A:

Bonds broken (reactants — endothermic input):

• 2 × (H–H) = 2 × 436 = $+872$+872 kJ
• 1 × (O=O) = $+498$+498 kJ
• Total = $+1370$+1370 kJ

Bonds formed (products — exothermic output):

• Each H$_2$2​O contains 2 × O–H bonds, so 2H$_2$2​O = 4 × O–H
• 4 × 464 = $+1856$+1856 kJ (energy released on bond formation)

$\Delta_r H^\ominus = 1370 - 1856 = \mathbf{-486 \text{ kJ mol}^{-1}}$Δr​H⊖=1370−1856=−486 kJ mol−1

The literature value for $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O(g)}$2H2​+O2​→2H2​O(g) is $-484$−484 kJ — agreement to $0.4\%$0.4%. Note the explicit specification of gaseous water: mean bond enthalpies give gas-phase values only. For $2\text{H}_2\text{O(l)}$2H2​O(l) the literature value is $-572$−572 kJ mol$^{-1}$−1, the extra $-88 = 2 \times -44$−88=2×−44 kJ being the enthalpy of condensation.

Worked example 2 — hydrogenation of ethene

Q: Estimate $\Delta_r H^\ominus$Δr​H⊖ for $\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}$C2​H4​(g)+H2​(g)→C2​H6​(g) using mean bond enthalpies.

A:

Ethene contains: 1 × C=C, 4 × C–H. Hydrogen contains: 1 × H–H. Ethane contains: 1 × C–C, 6 × C–H.

Bonds broken:

• 1 × C=C = $+612$+612
• 4 × C–H = 4 × 413 = $+1652$+1652
• 1 × H–H = $+436$+436
• Total = $+2700$+2700 kJ

Bonds formed:

• 1 × C–C = $+347$+347
• 6 × C–H = 6 × 413 = $+2478$+2478
• Total = $+2825$+2825 kJ

$\Delta_r H^\ominus = 2700 - 2825 = \mathbf{-125 \text{ kJ mol}^{-1}}$Δr​H⊖=2700−2825=−125 kJ mol−1

The combustion-cycle value from Lesson 4 was $-137$−137 kJ mol$^{-1}$−1 — a discrepancy of $12$12 kJ ($\sim 9\%$∼9%), typical of mean bond enthalpy calculations.

Shortcut: four of the C–H bonds in ethene and four of the C–H bonds in ethane are identical (both occupy sp$^3$3-like environments after the addition) and cancel from the calculation. The net change is "break C=C and H–H, form C–C and two new C–H":

$\Delta H = [612 + 436] - [347 + 2 \times 413] = 1048 - 1173 = -125 \text{ kJ mol}^{-1} \checkmark$ΔH=[612+436]−[347+2×413]=1048−1173=−125 kJ mol−1✓

This cancellation trick is invaluable in exam time-pressure: the unchanged bonds simply do not contribute.

Worked example 3 — formation of hydrogen chloride

Q: Estimate $\Delta H$ΔH for $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$H2​(g)+Cl2​(g)→2HCl(g).

A:

$\Delta H = [\text{BE}(\text{H–H}) + \text{BE}(\text{Cl–Cl})] - [2 \times \text{BE}(\text{H–Cl})]$ΔH=[BE(H–H)+BE(Cl–Cl)]−[2×BE(H–Cl)] $= [436 + 243] - [2 \times 432] = 679 - 864 = \mathbf{-185 \text{ kJ mol}^{-1}}$=[436+243]−[2×432]=679−864=−185 kJ mol−1

Compare with $\Delta_f H^\ominus(\text{HCl(g)}) = -92.3$Δf​H⊖(HCl(g))=−92.3 kJ mol$^{-1}$−1, so for $2$2 mol HCl the literature value is $-184.6$−184.6 kJ — agreement within $0.3\%$0.3%. The closeness here reflects the fact that H–H, Cl–Cl and H–Cl are well-defined diatomic bond strengths, not averaged over a range of environments.

Worked example 4 — combustion of methane (cautionary)

Q: Estimate $\Delta_c H^\ominus(\text{CH}_4\text{(g)})$Δc​H⊖(CH4​(g)) using mean bond enthalpies. Compare with the literature value.

A:

Equation: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(g).

Bonds broken:

• 4 × C–H = 4 × 413 = $+1652$+1652
• 2 × O=O = 2 × 498 = $+996$+996
• Total = $+2648$+2648 kJ

Bonds formed:

• 2 × C=O (in CO$_2$2​) = 2 × 805 = $+1610$+1610
• 4 × O–H = 4 × 464 = $+1856$+1856
• Total = $+3466$+3466 kJ

$\Delta H = 2648 - 3466 = \mathbf{-818 \text{ kJ mol}^{-1}}$ΔH=2648−3466=−818 kJ mol−1

Compare with the literature $\Delta_c H^\ominus$Δc​H⊖ of methane: $-890$−890 kJ mol$^{-1}$−1 for H$_2$2​O(l), or $-802$−802 kJ mol$^{-1}$−1 for H$_2$2​O(g). The mean-bond-enthalpy estimate of $-818$−818 kJ mol$^{-1}$−1 deviates by $\sim 2\%$∼2% from the H$_2$2​O(g) value — better than the $9\%$9% ethene–ethane discrepancy. The remaining gap arises because the C=O bond in CO$_2$2​ is unusually strong (the molecule is cumulenic, with O=C=O extended $\pi$π system) — using a "specific" C=O for CO$_2$2​ rather than the generic value reduces the error.

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Limitations of mean bond enthalpy calculations

1. Mean values are averages. The tabulated value for, say, C–H is averaged across CH$_4$4​, CHCl$_3$3​, C$_6$6​H$_6$6​, CH$_3$3​OH, and many others. The specific C–H in your molecule may differ by 10–30 kJ mol$^{-1}$−1.
2. Gas phase only. Mean bond enthalpies assume gaseous reactants and products. State changes (vaporisation, condensation, dissolution) carry their own enthalpy that bond-enthalpy methods do not include.
3. Not applicable to ionic compounds. Bond enthalpies describe covalent bonds; NaCl(s) and Fe(s) cannot be treated with this method (use Born–Haber, OCR Year 13).
4. Not applicable to aqueous reactions. For dissolved species, the solvation enthalpy (Hess: hydration enthalpy) dominates and bond enthalpies miss it entirely.
5. No information about reaction rate. Bond enthalpies give the net $\Delta H$ΔH, not the activation energy $E_a$Ea​. Bond breaking and bond making happen in concert at the transition state, and $E_a$Ea​ is generally much less than the sum of broken bonds.

For high-precision work, always prefer Hess-cycle calculations (Lessons 3–4) using tabulated $\Delta_f H^\ominus$Δf​H⊖ or $\Delta_c H^\ominus$Δc​H⊖ data. Mean bond enthalpies are an estimation tool, not a precision tool.

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Comparison — mean bond enthalpy vs Hess-cycle values

Reaction	Mean-BE estimate / kJ mol$^{-1}$−1	Hess-cycle literature / kJ mol$^{-1}$−1	Discrepancy
$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O(g)}$2H2​+O2​→2H2​O(g)	$-486$−486	$-484$−484	$0.4\%$0.4%
$\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$C2​H4​+H2​→C2​H6​	$-125$−125	$-137$−137	$9\%$9%
$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$H2​+Cl2​→2HCl	$-185$−185	$-184$−184	$0.5\%$0.5%
$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O(g)}$CH4​+2O2​→CO2​+2H2​O(g)	$-818$−818	$-802$−802	$2\%$2%
$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$N2​+3H2​→2NH3​	$-103$−103	$-92$−92	$12\%$12%

The methane and ammonia discrepancies are dominated by the C=O (in CO$_2$2​ specifically) and N–H bonds, both of which deviate substantially from their tabulated averages.

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Bond-enthalpy trends — qualitative

Understanding the structural trends in the data table helps make sense of bond strength and reactivity: