Topic 6 · page 14 of 15

Limits of bond enthalpy calculations, and finding a mean bond enthalpy

Bond enthalpies give a close estimate, not an exact answer, because the values are averages and the method only works for gases. Run the calculation backwards and a known ΔH gives a mean bond enthalpy.

14 min read6.96.10

Read first: Bond enthalpies and enthalpy changes

Why the answer is only an estimate 6.9

Compare the two ways of finding ΔH for the burning of methane to gaseous water, $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$ .

Worked exampleΔH from bond enthalpies

Use C–H 413, O=O 498, C=O (in CO₂) 805 and O–H 464 kJ mol⁻¹.

	Bonds	Energy / kJ
Broken	4 × C–H = 4 × 413	1652
	2 × O=O = 2 × 498	996
	Total	+2648
Made	2 × C=O = 2 × 805	1610
	4 × O–H = 4 × 464	1856
	Total	−3466

ΔH = 2648 − 3466 = −818 kJ mol⁻¹.

From formation data, with $\Delta_\text{f}H$ (kJ mol⁻¹) CH₄(g) −74.8, CO₂(g) −393.5 and H₂O(g) −241.8:

\Delta_\text{r}H = [-393.5 + 2(-241.8)] - (-74.8) = -802.3\ \text{kJ mol}^{-1}

The bond enthalpy answer, −818, is close but not equal: it is about 16 kJ mol⁻¹, or 2%, too exothermic. The formation data come from measurements on the actual compounds, so they are more reliable.

The second limitation explains why the liquid-water value for methane is −890 kJ mol⁻¹, not −818. Condensing 2 mol of water vapour to liquid gives out about 2 × 44 = 88 kJ more:

-802 - 88 \approx -890\ \text{kJ mol}^{-1}

The 44 kJ mol⁻¹ is the energy released when 1 mol of water vapour condenses. You do not need to quote it; it shows what the bond enthalpy method leaves out.

Calculating a mean bond enthalpy 6.10

The same relationship runs both ways. If ΔH is known and one bond enthalpy is unknown, rearrange:

\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds made})

MethodFinding a mean bond enthalpy

Write the equation and the displayed formulae.
Add up the energy of the known bonds broken and made.
Put the unknown as x multiplied by the number of those bonds, and substitute ΔH.
Solve for x. Divide by the number of bonds if you found the total for several.

Not to scale: the net change (−92) is much smaller than the bonds broken and made. The unknown is the downward arrow.

Worked exampleMean N–H bond enthalpy in ammonia

$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ has ΔH = −92 kJ mol⁻¹. N≡N = 945 and H–H = 436 kJ mol⁻¹. Find the mean N–H bond enthalpy.

Broken: 945 + 3 × 436 = 945 + 1308 = +2253 kJ.
ΔH = broken − made, so −92 = 2253 − made, which gives made = 2253 + 92 = 2345 kJ.
Two moles of NH₃ have six N–H bonds. Mean N–H = 2345 ÷ 6 = 391 kJ mol⁻¹.

Worked exampleA mean bond enthalpy from enthalpy changes of formation and atomisation

Calculate the mean C–H bond enthalpy in methane from these data (kJ mol⁻¹): $\Delta_\text{f}H$ (CH₄(g)) = −74.8, $\Delta_\text{at}H$ (C) = +717 and $\Delta_\text{at}H$ (H) = +218.

The enthalpy change for $\ce{CH4(g) -> C(g) + 4H(g)}$ is the energy to break four C–H bonds. Go round a cycle through the elements, C(s) and 2H₂(g).

Elements to gaseous atoms: $\Delta_\text{at}H$ (C) + 4 × $\Delta_\text{at}H$ (H) = 717 + 4(218) = 1589.
Elements to CH₄: −74.8. The route from CH₄ to atoms goes against this arrow, then with the atomisation.
ΔH(CH₄ → C + 4H) = −(−74.8) + 1589 = +1663.8 kJ.
Four C–H bonds break, so the mean C–H bond enthalpy = 1663.8 ÷ 4 = 416 kJ mol⁻¹.

The result, 416, is close to the table value of 413. That is as expected: a mean bond enthalpy calculated from one molecule is close to the average over many.

Check yourself

State two limitations of calculating ΔH from mean bond enthalpies.
Calculate ΔH for $\ce{C2H4(g) + H2(g) -> C2H6(g)}$ using C=C 612, C–C 347, C–H 413 and H–H 436 kJ mol⁻¹. The value from combustion data is −136.8 kJ mol⁻¹. Suggest why the two differ.
For $\ce{2H2(g) + O2(g) -> 2H2O(g)}$ , ΔH = −484 kJ mol⁻¹. H–H = 436 and O=O = 498 kJ mol⁻¹. Calculate the mean O–H bond enthalpy.
For $\ce{N2(g) + 2H2(g) -> N2H4(g)}$ , ΔH = +95 kJ mol⁻¹. The molecule is H₂N–NH₂. Using N≡N 945, H–H 436 and N–H 391 kJ mol⁻¹, calculate the N–N bond enthalpy.
Explain why a calculation using bond enthalpies gives a poor value of ΔH for the combustion of ethanol to liquid water.
Calculate the mean O–H bond enthalpy in water from these data (kJ mol⁻¹): $\Delta_\text{f}H$ (H₂O(g)) = −241.8, $\Delta_\text{at}H$ (H) = +218 and $\Delta_\text{at}H$ (O) = +249.

Answers cover these until you have tried the questions

Any two: bond enthalpies are mean values, not exact for the bonds in this molecule; the method assumes gaseous species, so changes of state are ignored; intermolecular forces are ignored.
Broken: C=C + H–H = 612 + 436 = 1048. Made: C–C + 2 × C–H = 347 + 826 = 1173. ΔH = 1048 − 1173 = −125 kJ mol⁻¹. It differs from −136.8 by about 12 kJ because mean bond enthalpies are averages, so they are not exact for the bonds in these molecules.
Broken: 2 × 436 + 498 = 872 + 498 = 1370. −484 = 1370 − made, so made = 1854 kJ. There are four O–H bonds (2 H₂O), so mean O–H = 1854 ÷ 4 = 463.5 kJ mol⁻¹ (about 464).
Broken: 945 + 2 × 436 = 945 + 872 = 1817. Made: N–N + 4 × N–H = x + 4 × 391 = x + 1564. +95 = 1817 − (x + 1564), so x = 1817 − 1564 − 95 = 158 kJ mol⁻¹.
Ethanol burns to liquid water, but bond enthalpies are for gases. The energy given out when water vapour condenses (about 44 kJ per mole of water) is not included, so the answer is not the enthalpy change for liquid water, and mean bond enthalpies are further approximations.
$\ce{H2O(g) -> 2H(g) + O(g)}$ is the reverse of formation followed by atomisation. ΔH = +241.8 + 2(218) + 249 = 241.8 + 436 + 249 = +926.8 kJ. Two O–H bonds break, so mean O–H = 926.8 ÷ 2 = 463 kJ mol⁻¹ (463.4).