AP Chemistry 6.5: Reaction Enthalpy and Bond Enthalpies

The enthalpy of reaction, written ΔHrxn, is the enthalpy change associated with a chemical reaction.

At constant pressure:

• ΔHrxn = qreaction,p
• Khan Academy's current lesson emphasizes that ΔHrxn depends on how the balanced chemical equation is written and is often expressed per mole of reaction.

Consider:

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
• Suppose:
• ΔHrxn = −890 kJ/molrxn
• This means the reaction releases 890 kJ when the reaction occurs exactly as written:
• 1 mol CH₄ reacts with 2 mol O₂ to produce 1 mol CO₂ and 2 mol H₂O.
• Because ΔH is negative, the reaction is exothermic.

Enthalpy is an extensive property for a reaction: the energy change depends on how much reaction occurs.

• Suppose:
• A → B ΔH = +50 kJ

If you double the equation:

• 2A → 2B
• then:
• ΔH = +100 kJ

If you reverse it:

• B → A
• then:
• ΔH = −50 kJ
• This idea becomes central in Hess's law.

A balanced chemical equation that includes its enthalpy change is called a thermochemical equation.

• For example:
• 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = −572 kJ
• This equation tells us that formation of 2 mol H₂O(l) releases 572 kJ.
• Therefore formation of 1 mol H₂O(l) would correspond to:
• −286 kJ/mol H₂O
• assuming the same conditions.

Enthalpy depends on physical state.

• For example:
• H₂O(l)
• and:
• H₂O(g)
• do not have the same enthalpy.
• Why?
• Converting liquid water into water vapor requires energy.
• Therefore, a reaction producing liquid water can have a different ΔH than the same reaction producing water vapor.

Always pay attention to:

• (s), (l), (g), and (aq)
• when comparing thermochemical equations.

Coffee-cup calorimetry can be used to measure ΔHrxn for reactions in solution.

• Suppose a reaction causes the surrounding solution to warm.
• The solution gained heat:
• qsolution > 0
• Therefore:
• qreaction < 0

At constant pressure:

• ΔHrxn = qreaction

If the reaction amount is known, divide by the appropriate moles of reaction to express the result in:

• kJ/molrxn

For an exothermic reaction:

• reactants higher → products lower
• ΔH < 0

For an endothermic reaction:

• reactants lower → products higher
• ΔH > 0
• Again, this overall vertical difference is different from the activation-energy barrier from Unit 5.

Do not ignore state symbols in thermochemical equations.

Do not treat ΔH as unrelated to stoichiometric coefficients.

If you multiply a reaction, multiply ΔH by the same factor.

If you reverse a reaction, reverse the sign of ΔH.

ΔHrxn belongs to the reaction exactly as written.

• Change the equation → change ΔH accordingly.

A bond enthalpy, sometimes called bond dissociation energy, is the energy required to break one mole of a particular type of bond in the gas phase.

• Because breaking a bond requires energy:
• bond breaking → positive energy
• Khan Academy's current Unit 6 describes bond enthalpy as positive for bond breaking and uses the relationship between bonds broken and bonds formed to estimate ΔHrxn.

Consider:

• Cl₂(g) → 2Cl(g)
• The Cl—Cl bond must be broken.
• Suppose its bond enthalpy is:
• 242 kJ/mol
• Then:
• +242 kJ
• must be supplied per mole of Cl—Cl bonds broken.
• This is endothermic.

The reverse process is:

• 2Cl(g) → Cl₂(g)
• Forming the Cl—Cl bond releases the same magnitude:
• −242 kJ/mol
• Therefore:
• breaking bonds = energy absorbed
• forming bonds = energy released
• This is one of the most important thermodynamics ideas.

The approximate reaction enthalpy can be calculated using:

• ΔHrxn ≈ Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)
• A common shorthand is:
• ΔH ≈ broken − formed

Why subtract formed bonds?

• Bond enthalpies are tabulated as positive bond-breaking energies. Forming those same bonds releases energy, so their contribution has the opposite sign.

Consider:

• H₂(g) + Cl₂(g) → 2HCl(g)
• Suppose:
• H—H = 436 kJ/mol
• Cl—Cl = 242 kJ/mol
• H—Cl = 431 kJ/mol

Break:

• 1 H—H
• 1 Cl—Cl
• Energy required:
• 436 + 242 = 678 kJ

Form:

• 2 H—Cl
• Energy associated with breaking those bonds would be:
• 2(431) = 862 kJ
• But because the bonds are being formed, that amount is released.

ΔH = broken − formed

• ΔH = 678 − 862
• ΔH = −184 kJ
• The negative result means the reaction is exothermic.

Suppose relatively weak bonds are broken and very strong bonds are formed.

• Breaking the weak bonds requires some energy, but forming the stronger bonds releases a larger amount.
• Therefore:
• energy released > energy absorbed
• and:
• ΔH < 0
• The reaction is exothermic.

If much stronger bonds must be broken than are formed:

• energy absorbed > energy released
• and the reaction may be endothermic.

This is often the hardest part of bond-enthalpy problems.

You must count the bonds that actually disappear and the bonds that actually appear.

• Consider a molecule with:
• 4 C—H bonds
• 1 C—C bond
• 1 O—H bond

If there are two molecules, multiply every relevant bond count by two.

Lewis structures are often useful because they make the bonds visible.

A C—C single bond and C=C double bond do not have the same bond enthalpy.

• Likewise:
• C—C
• C=C
• C≡C
• must be treated as different bond types.

Generally:

• higher bond order → stronger bond → larger bond enthalpy
• although exact values depend on molecular environment.

Most bond enthalpy tables contain average bond enthalpies.

• A C—H bond does not have exactly the same strength in every molecule because its chemical environment changes.
• Therefore calculations using average bond enthalpies provide an:
• estimate of ΔHrxn
• rather than an exact value.

Standard enthalpies of formation can often give more precise reaction enthalpies when accurate tabulated data are available.

Do not confuse bond enthalpy with activation energy.

Bond enthalpy describes the energy needed to break a particular chemical bond.

Activation energy describes the energy barrier associated with an entire reaction pathway or elementary step.

• They are related to molecular energy, but they are not the same quantity.

Do not use:

formed − broken

The formula is:

broken − formed

Do not forget to count every bond.

Do not treat single, double, and triple bonds as identical.

Remember that bond enthalpy calculations using average values usually provide approximate reaction enthalpies.

Use:

• ΔH ≈ bonds broken − bonds formed
• Breaking costs energy.
• Forming pays energy back.