Mean Bond Enthalpy

Learning Objectives (OCR Spec 3.2.1)

By the end of this lesson you should be able to:

• Define bond enthalpy and mean bond enthalpy
• Explain why tabulated values are averages taken over a range of compounds
• Use mean bond enthalpies to estimate ΔrH° for gas-phase reactions
• Recognise the limitations of bond-enthalpy calculations
• Compare mean-bond-enthalpy values with exact calorimetric values

What Is Bond Enthalpy?

The bond dissociation enthalpy (often just "bond enthalpy") is the enthalpy change required to break one mole of a specified bond in the gas phase. It is always positive (endothermic) because breaking bonds requires input of energy.

Example: H2(g) → 2H(g), ΔH = +436 kJ mol⁻¹ - the bond enthalpy of H–H is +436 kJ mol⁻¹.

Why "Mean"?

The strength of, say, a C–H bond is not the same in every molecule. In CH4 the average C–H bond enthalpy is +413 kJ mol⁻¹, but:

• The first C–H bond to break in CH4 takes about +435 kJ mol⁻¹
• The last C–H in the sequence takes about +340 kJ mol⁻¹
• The C–H bonds in CHCl3, C6H6 or CH3CHO all differ slightly

The mean bond enthalpy is an average value taken over a representative set of molecules containing that bond. Tables quote one figure that you apply universally, accepting a modest loss of accuracy.

Bond	Mean bond enthalpy / kJ mol⁻¹
H–H	+436
C–H	+413
C–C	+347
C=C	+612
C≡C	+838
O=O	+498
C–O	+358
C=O	+805
O–H	+464
N≡N	+945
N–H	+391
Cl–Cl	+243
H–Cl	+432

These are typical OCR data-booklet values - always use the values provided in the exam paper.

Calculating ΔrH° from Mean Bond Enthalpies

For a gas-phase reaction:

ΔrH° = Σ(bonds broken in reactants) − Σ(bonds formed in products)

Bonds broken takes energy in (+), bonds formed releases energy out (−). The difference is the net enthalpy change. Note the sign convention: because bond enthalpies are tabulated as positive values (for breaking), we use reactant bonds minus product bonds, not the other way round.

Worked Example 1 – Combustion of Hydrogen

Q: Using mean bond enthalpies, estimate ΔrH° for 2H2(g) + O2(g) → 2H2O(g).

A:

Bonds broken (reactants):

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

Bonds formed (products):

• Each H2O contains 2 × O–H bonds, so 2H2O = 4 × O–H
• 4 × 464 = +1856 kJ mol⁻¹ (total bond energy released)

ΔrH° = 1370 − 1856 = −486 kJ mol⁻¹

The literature value for 2H2O(g) is approximately −484 kJ mol⁻¹, agreement to ~0.4%.

Important: The reaction is for gaseous water. If the question specified H2O(l) the calorimetric value would be ~−572 kJ mol⁻¹ (including condensation). Mean bond enthalpies always give gas-phase values.

Worked Example 2 – Hydrogenation of Ethene

Q: Use mean bond enthalpies to estimate ΔrH° for C2H4(g) + H2(g) → C2H6(g).

A:

Bonds broken:

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

Bonds formed:

• 1 × C–C = −347
• 6 × C–H = 6 × 413 = −2478
• Total = −2825 kJ mol⁻¹ (in magnitude)

ΔrH° = 2700 − 2825 = −125 kJ mol⁻¹

The Hess-cycle value (Lesson 4) was −137 kJ mol⁻¹ - a discrepancy of 12 kJ, about 9%. This is typical: mean bond enthalpies give approximate answers.

Shortcut: The four C–H bonds in ethene and four of the C–H bonds in ethane are essentially unchanged, so you can ignore them. The net change is "break C=C and H–H, form C–C and 2 × new C–H":

ΔH = [612 + 436] − [347 + 2 × 413] = 1048 − 1173 = −125 kJ mol⁻¹ ✓

Worked Example 3 – Hydrogen Chloride Formation

Q: Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g).

A: