Bond Enthalpy Calculator

Results

A rough estimate for gas-phase reactions: average bond energies ignore the surroundings of each bond, so expect differences of tens of kJ/mol from a calculation with enthalpies of formation.

How to Use the Bond Enthalpy Calculator

1. 1. Draw the structures. Write the Lewis structure of every reactant and product so you can see each bond. The tool does not guess structures from formulas, because the same formula can hold different bonds.
2. 2. List the bonds broken and formed. Type them in the two boxes — 4 C–H, 2 O=O — counting one for every molecule in the balanced equation. Use - for single, = for double and ≡ for triple bonds. You can also press “+ broken” or “+ formed” in the table, or choose one of the example reactions.
3. 3. Check the list. A bond that is not in the table is flagged by name. Bonds that appear unchanged on both sides can be left out: they cancel.
4. 4. Read ΔH. The answer is Σ broken − Σ formed, with both sums, each bond's subtotal and the working. A negative ΔH means the reaction is exothermic.

The Method

Breaking a bond absorbs energy and forming one releases it. If the bonds you break in the reactants cost more than the bonds you make in the products give back, the reaction takes in heat overall. Written as an equation, with D for the average bond energy:

ΔH ≈ Σ D(bonds broken) − Σ D(bonds formed)

ΔH < 0: exothermic.  ΔH > 0: endothermic.

kJ per mole of reaction as written; 1 kcal = 4.184 kJ

The sign convention is the one that surprises people: bond energies are positive numbers, and the products' sum is subtracted. Do not write ΔH = Σ products − Σ reactants here, which is the rule for enthalpies of formation. Use whichever matches the data you are holding, and don't mix them in one calculation.

Worked Example: Burning Methane

For CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g), methane has four C–H bonds, each O₂ has one O=O bond, carbon dioxide has two C=O bonds and each water has two O–H bonds. With the table's average values:

• Broken: 4 × 415 (C–H) + 2 × 498 (O=O) = 1660 + 996 = 2656 kJ
• Formed: 2 × 741 (C=O) + 4 × 464 (O–H) = 1482 + 1856 = 3338 kJ
• ΔH ≈ 2656 − 3338 = −682 kJ per mole of CH₄ burned

Why It Differs from Formation-Enthalpy Results

• A bond energy is an average. The C–H bond in methane is not the same as the C–H bond in ethanol or benzene. The table gives one value taken over many compounds, so every molecule is a little off. The table's C=O of 741 kJ/mol suits aldehydes and ketones, while the C=O bonds of CO₂, at about 799 kJ/mol, are stronger. That one difference accounts for most of the gap in the example above.
• It describes gases only. Bond energies refer to separate gaseous molecules, so the answer is the gas-phase ΔH. If a product is liquid water, you must add the enthalpy of condensation, about −44.0 kJ for each mole of H₂O (285.8 − 241.8 from the CRC formation values).
• It cannot see delocalisation or strain. Resonance in benzene or carbonate, ring strain and hydrogen bonding make real molecules more stable or less stable than the sum of their bonds suggests.
• Enthalpies of formation are measured for the actual compounds. ΔH°_rxn = ΣΔH°_f(products) − ΣΔH°_f(reactants) uses real, tabulated values at 298 K, so it is the better answer whenever data exist. Use bond energies when they do not, or to see which bonds drive the energy change. The Enthalpy of Reaction Calculator does the formation-enthalpy route.

Frequently Asked Questions

How do you calculate ΔH using bond enthalpies?

Add up the energy needed to break every bond in the reactants, add up the energy released by forming every bond in the products, and subtract: ΔH ≈ Σ D(broken) − Σ D(formed). Count each bond once per molecule in the balanced equation. For H₂ + Cl₂ → 2 HCl, that is (436 + 243) − 2(432) = −185 kJ.

Why is breaking a bond endothermic and forming one exothermic?

Atoms in a bond sit at a lower energy than they would apart, so pulling them apart takes energy and letting them bond gives that energy back. The same amount is involved in each direction, which is why bond energies are quoted as positive numbers for breaking.

Why doesn't my bond-energy answer match the enthalpy of formation answer?

Bond energies are averages for gas-phase bonds, so they ignore each molecule's particular surroundings, and they cannot account for liquids, resonance or strain. Differences of tens of kJ/mol are normal. Treat the enthalpy-of-formation result as the more accurate one when you have the data.

Do I need to list every bond in the molecules?

Only those that change. Bonds that are identical in a reactant and a product appear in both sums and cancel. For C₂H₄ + H₂ → C₂H₆, you can list just C=C and H–H broken and C–C and two C–H formed, and the answer, −128 kJ, is the same as counting every bond.

How do I count bonds in a double bond like O=O or C=C?

A double bond is one bond with its own, larger energy, so enter it as a single item: O=O is 498 kJ/mol, not two O–O bonds. Triple bonds are the same: N≡N is 946 kJ/mol.

Can I use bond enthalpies for reactions involving liquids or solids?

Only after converting everything to gases. Add the enthalpy of vaporisation or sublimation for each species that is not a gas, and treat the result with extra caution. For reactions in solution, enthalpies of formation or calorimetry are the sound route.