high school Reaction Enthalpy

Reaction Enthalpy

A reaction is molecules climbing an activation-energy hill; products lower than reactants release heat, products higher absorb heat. Breaking bonds absorbs energy and forming bonds releases it, so their difference gives ΔH. Enthalpy is a state function, so the total change depends only on start and end—Hess’s law. Change ΔH on the energy diagram and connect bond energies with Hess’s law.

Crossing the Energy Hill

💡 Analogy: Hiking Over a Mountain

①A chemical reaction = molecules climbing an 'energy hill'

②Hill height = activation energy E_a — required to begin the reaction

③If the destination is lower than the start: heat is released (exothermic)

④If higher: heat is absorbed (endothermic)

⑤Hand warmer = exothermic (ΔH<0), instant cold pack = endothermic (ΔH>0)

Visualizing the Enthalpy Diagram

ΔH (kJ)-40

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①Negative ΔH → products lower than reactants = exothermic

②Positive ΔH → products higher = endothermic

③Either way, the transition state must be crossed

④E_a = minimum energy to start the reaction

Computing ΔH from Bond Energies

Bond separation (relative scale 0–10, not kJ)3

Bond-Energy Formula

ΔH = Σ(reactant bond energies) − Σ(product bond energies)

Energy absorbed in breaking − energy released in forming

💡 Why Does the Formula Hold?

①Every chemical reaction = 'break old bonds + form new bonds'

②Breaking bonds = absorbs energy (always +)

③Forming bonds = releases energy (always −)

④Stronger new bonds → release > absorb → exothermic (ΔH<0)

⑤This is the physical meaning of 'products are more stable'

Hess's Law

ΔH_total = ΔH₁ + ΔH₂ + ΔH₃ + ⋯

Independent of path: total enthalpy change depends only on initial and final states

Using Heats of Formation

ΔH = Σ(product ΔH_f) − Σ(reactant ΔH_f)

Compute ΔH of any reaction from heats of formation referenced to elements

💡 Why Path-Independent?

①Enthalpy is a state function — depends only on current state

②Seoul→Busan: highway or back roads, the altitude change is the same

③Even if a reaction's ΔH is unknown, combine others to find it

Worked Examples

Example 1

Find ΔH for H₂ + Cl₂ → 2HCl using bond energies. (H−H = 436, Cl−Cl = 243, H−Cl = 431 kJ/mol)

ΔH = Σ(bonds broken in reactants) − Σ(bonds formed in products).

ΔH = (436 + 243) − (2 × 431)

Compute.

ΔH = 679 − 862 = −183 kJ

▸ −183 kJ (exothermic)

Breaking bonds absorbs (+), forming releases (−). The new bonds (HCl) are stronger, so ΔH<0 → exothermic.

Example 2

Find ΔH for C + ½O₂ → CO from: (a) C + O₂ → CO₂, ΔH = −394 kJ; (b) CO + ½O₂ → CO₂, ΔH = −283 kJ.

Hess’s law: the target = (a) − (b).

C + ½O₂ → CO = (a) − (b)

Subtract the ΔH values.

ΔH = (−394) − (−283) = −111 kJ

▸ −111 kJ

Hess’s law: even hard-to-measure reactions get ΔH by combining (adding/subtracting) known reactions.

Summary

Key Formula

ΔH = Σ(bond energy_break) − Σ(bond energy_form)

Or ΔH = Σ(product ΔH_f) − Σ(reactant ΔH_f)

exam-style

In a reaction, the total bond energy of reactants is 1000 kJ and of products is 1200 kJ. What is ΔH and the type?

①+200 kJ, endothermic

②−200 kJ, exothermic

③+2200 kJ, endothermic

④−2200 kJ, exothermic

⑤0 kJ, no heat exchange

▸ ② −200 kJ, exothermic

Substitute into ΔH = Σ(reactant bonds) − Σ(product bonds).

ΔH = 1000 − 1200 = −200 kJ

Since ΔH < 0, the reaction is exothermic.

ΔH < 0 → exothermic

🎯 Exam Points

①ΔH < 0: exothermic — products more stable

②ΔH > 0: endothermic — products less stable

③Hess: path-independent, depends only on start/end

④ΔH_f: ΔH for forming 1 mol of compound from elements

⑤Activation energy E_a: minimum energy to start the reaction

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Properties of Solutions

Chemical Equilibrium

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