Calculate the enthalpy change of a reaction using bond energies

Chemical Energetics – Enthalpy Change Using Bond Energies (IGCSE 0620)

Learning objective

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

• write the enthalpy change of a reaction as ΔH = H_products − H_reactants,
• use bond‑energy values to calculate ΔH for a balanced chemical equation,
• state whether the reaction is exothermic or endothermic, and
• recognise when phase‑change corrections (ΔH_fus, ΔH_vap, ΔH_sub) are required.

Key definitions

• Enthalpy (H) – the heat content of a system at constant pressure (kJ).
• Enthalpy change (ΔH) – heat absorbed or released during a reaction.
  • ΔH = H_products − H_reactants
  • ΔH < 0 → exothermic (heat released)
  • ΔH > 0 → endothermic (heat absorbed)
• Bond energy – the energy required to break one mole of a specific bond in the gas phase (kJ mol⁻¹). The same amount of energy is released when that bond is formed.
• Activation energy (Eₐ) – the minimum energy needed to reach the transition state; it appears as the “hill” on a reaction‑profile diagram and does **not** affect the sign of ΔH.

Why bond‑energy calculations give ΔH

During a reaction two opposite processes occur:

1. All bonds in the reactants must be broken – an endothermic step that **absorbs** energy.
2. New bonds are formed in the products – an exothermic step that **releases** energy.

The net enthalpy change is therefore the total energy required to break bonds minus the total energy released on forming bonds:

\[ \Delta H = \sum\text{(bond energies of bonds broken)} \;-\; \sum\text{(bond energies of bonds formed)} \]

Step‑by‑step procedure

1. Write and **balance** the chemical equation.
2. List every bond that must be broken in the reactants; write the corresponding bond‑energy value for each.
3. List every bond that is formed in the products; write the corresponding bond‑energy value for each.
4. Multiply each bond‑energy by the number of identical bonds and add them to obtain:
   • Σ_broken – total energy required to break bonds.
   • Σ_formed – total energy released on forming bonds.
5. Calculate ΔH using the formula \[ \Delta H = \Sigma_{\text{broken}} - \Sigma_{\text{formed}} \] Keep the sign produced by the calculation.
6. If any reactant or product is a liquid or solid, add the appropriate phase‑change enthalpy (ΔH_fus, ΔH_vap, ΔH_sub) to the result.
7. Interpret the sign:
   • ΔH < 0 → exothermic
   • ΔH > 0 → endothermic

Bond‑energy table (kJ mol⁻¹) – gas‑phase values

Bond	Energy	Bond	Energy	Bond	Energy
C–C	347	O–O	498	C–H	413
C=C	614	O=O	498	C–O	358
C≡C	839	O–H	463	C=O	799
C–Cl	327	H–H	436	H–Cl	432
Cl–Cl	242	F–F	158	N≡N	945
N–N	607	N=O	607	Cl–O	200

All values are for gaseous molecules. Use phase‑change corrections when a substance is a liquid or solid.

Reaction‑profile diagram (hand‑drawn style)

          Eₐ
          ^                ΔH
          |                |
          |                |
          |                |
Reactants |________________| Transition state
          |                |
          |                |
          |                |
          |________________| Products

Key:

• Vertical distance from reactants to products = ΔH.
• Peak above reactants = activation energy (Eₐ).
• ΔH < 0 → products lower than reactants (exothermic).
• ΔH > 0 → products higher than reactants (endothermic).

Worked examples

Example 1 – Exothermic: combustion of methane

\[ \mathrm{CH_4(g) + 2\,O_2(g) \rightarrow CO_2(g) + 2\,H_2O(g)} \]

Bonds broken
4 C–H	4 × 413 = 1652 kJ
2 O=O	2 × 498 = 996 kJ
Σ broken	2648 kJ

Bonds formed
2 C=O (in CO₂)	2 × 799 = 1598 kJ
4 O–H (in 2 H₂O)	4 × 463 = 1852 kJ
Σ formed	3450 kJ

ΔH = 2648 − 3450 = −802 kJ per mole of CH₄.

Negative sign ⇒ the reaction releases 802 kJ (exothermic).

Example 2 – Endothermic: thermal decomposition of calcium carbonate

\[ \mathrm{CaCO_3(s) \;\xrightarrow{\Delta}\; CaO(s) + CO_2(g)} \]

Only the gaseous CO₂ requires bond‑energy data; the solid lattice contribution is supplied as a correction (ΔH_lattice = +178 kJ mol⁻¹).

Bonds broken (in CO₃²⁻)
1 C=O	1 × 799 = 799 kJ
2 C–O	2 × 358 = 716 kJ
Σ broken	1515 kJ

Bonds formed (in CO₂)
2 C=O	2 × 799 = 1598 kJ
Σ formed	1598 kJ

Bond‑energy part: ΔH_bond = 1515 − 1598 = −83 kJ.

Overall ΔH:  ΔH = ΔH_bond + ΔH_lattice = (−83 kJ) + (+178 kJ) = +95 kJ.

Positive sign ⇒ the decomposition absorbs heat (endothermic).

Practice questions

1. Calculate ΔH for:

   \[ \mathrm{H_2(g) + Cl_2(g) \rightarrow 2\,HCl(g)} \]
   Given: H–H = 436 kJ mol⁻¹, Cl–Cl = 242 kJ mol⁻¹, H–Cl = 432 kJ mol⁻¹.

2. Determine the sign of ΔH and its magnitude for:

   \[ \mathrm{N_2(g) + O_2(g) \rightarrow 2\,NO(g)} \]
   Bond energies: N≡N = 945 kJ mol⁻¹, O=O = 498 kJ mol⁻¹, N=O = 607 kJ mol⁻¹.

3. What is the enthalpy change per gram of CH₄ for the combustion reaction above?

   
   Use the ΔH = −802 kJ obtained in Example 1. (Molar mass of CH₄ = 16.04 g mol⁻¹.)

4. Explain why bond‑energy calculations give only an approximate value for ΔH.

Common mistakes to avoid

• Forgetting to multiply a bond‑energy value by the correct number of identical bonds.
• Using bond energies for the wrong physical state – the table values apply only to gases.
• Mixing up the sign convention – the energy released on forming bonds must be **subtracted** from the energy required to break bonds.
• Neglecting phase‑change corrections when a reactant or product is a liquid or solid.
• Omitting the lattice‑energy or sublimation‑energy term for ionic solids (e.g., CaCO₃, NaCl).

Summary

• ΔH = H_products − H_reactants. A negative ΔH means the reaction is exothermic; a positive ΔH means it is endothermic.
• Bond‑energy method:
  1. List every bond broken → Σ_broken (energy absorbed).
  2. List every bond formed → Σ_formed (energy released).
  3. ΔH = Σ_broken − Σ_formed.
• All bond‑energy values are for gaseous molecules; add ΔH_fus, ΔH_vap or lattice‑energy corrections when solids or liquids are involved.
• Use the calculated ΔH together with the activation‑energy concept to sketch and interpret reaction‑profile diagrams.