Grade 11-12 (age 16-18)

high school Reaction Enthalpy

A hand warmer heats up from the reaction of iron rusting, and a cold pack chills as ammonium nitrate dissolves in water and absorbs heat. Whether a reaction gives off or takes in heat is expressed by the enthalpy difference between products and reactants, ΔH: negative means exothermic, positive means endothermic. Breaking the reactants' bonds costs energy and forming the products' bonds releases it, so bond energies give an estimate of ΔH; and because enthalpy depends only on the starting and ending states, ΔH can also be found by adding and subtracting other reactions (Hess's law). The diagrams show the energy hill as ΔH changes, the energy taken in and given out as H₂ and Cl₂ become HCl, and two routes from carbon to carbon dioxide.

Climbing the Energy Hill

💡 Think of a Route over a Mountain

①To become products, reactants must climb over a high-energy hill called the transition state

②The height of the hill is the activation energy Eₐ, the least energy needed for the reaction to start

③If the destination (products) lies lower than the start (reactants), the difference is released as heat: an exothermic reaction, ΔH < 0

④If the destination lies higher, the difference is absorbed as heat: an endothermic reaction, ΔH > 0

⑤ΔH is only the difference in height between start and finish, so it is set independently of the height of the hill (Eₐ)

Reading the Enthalpy Diagram

ΔH (kJ)−40 kJ

🔍 What to Check with the Slider

①Set ΔH negative and the products sit below the reactants: exothermic

②Set ΔH positive and the products rise above: endothermic

③Either way, the reactants must first climb to the top of the transition state

④In an endothermic reaction Eₐ is larger than ΔH, because the products are already higher than the reactants and the top of the hill has to be higher still

Finding ΔH from Bond Energies

Reaction progress (breaking bonds → forming bonds)30%

ΔH from Bond Energies

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

Subtract the energy released by the new bonds from the energy absorbed by the bonds broken

💡 Why This Formula Works

①Every reaction can be split into two steps: breaking the reactants' bonds and forming the products' bonds

②Breaking bonds always costs energy; in the first half of the diagram, H–H (436 kJ) and Cl–Cl (243 kJ) take in 679 kJ in all

③Forming bonds always releases energy; in the second half, two H–Cl bonds form and give out 2 × 431 = 862 kJ

④The energy released exceeds the energy absorbed by 183 kJ, so ΔH = −183 kJ and the reaction is exothermic

⑤The stronger the new bonds are compared with the broken ones, the more heat comes out

Hess's Law

ΔH_total = ΔH₁ + ΔH₂ + ΔH₃ + ⋯

Whatever the route, the overall enthalpy change depends only on the starting and ending states

Using Enthalpies of Formation

ΔH = Σ(enthalpies of formation of products) − Σ(enthalpies of formation of reactants)

With a table of enthalpies of formation, taking elements as zero, the ΔH of any reaction can be calculated

💡 Why the Route Does Not Matter

①Enthalpy is a state function, set only by the current state

②The drop in height from a mountaintop to the shore is the same whether you take the main road or a side path

③In the diagram, carbon reacting with O₂ in one step gives −394 kJ, and going through CO in two steps gives −111 + (−283) = −394 kJ

④So even the ΔH of a reaction that is hard to measure directly, such as burning carbon only as far as CO, can be found by adding and subtracting other reactions

Worked Examples

Example 1

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

Subtract the energy of the bonds formed (two H–Cl) from the energy of the bonds broken (one H–H and one Cl–Cl).

ΔH = (436 + 243) − (2 × 431)

Work out the brackets.

ΔH = 679 − 862 = −183 kJ

▸ −183 kJ (exothermic)

The two new H–Cl bonds are stronger than the two bonds broken, so heat comes out. This is the value at the end of the bond-energy diagram.

Example 2

Use these two reactions to find ΔH for C + ½O₂ → CO. (a) C + O₂ → CO₂, ΔH = −394 kJ (b) CO + ½O₂ → CO₂, ΔH = −283 kJ

The target reaction is (a) minus (b); subtracting cancels the CO₂.

C + ½O₂ → CO = (a) − (b)

Subtract the ΔH values the same way.

ΔH = (−394) − (−283) = −111 kJ

▸ −111 kJ

The middle level (CO + ½O₂) in the Hess's law diagram is exactly this −111 kJ. A reaction that is hard to measure was found from two that can be measured.

Summary

Key Formulas

ΔH = Σ(bond energies broken) − Σ(bond energies formed)

or ΔH = Σ(enthalpies of formation of products) − Σ(enthalpies of formation of reactants)

CSAT-style problem, adapted

In a reaction, the bond energies of the reactants add up to 1000 kJ and those of the products to 1200 kJ. What are ΔH and the type of reaction?

①+200 kJ, endothermic

②−200 kJ, exothermic

③+2200 kJ, endothermic

④−2200 kJ, exothermic

⑤0 kJ, no heat exchanged

▸ ② −200 kJ, exothermic

Substitute into ΔH = Σ(bond energies broken) − Σ(bond energies formed).

ΔH = 1000 − 1200 = −200 kJ

ΔH < 0, so the reaction is exothermic; the products' bonds are stronger.

ΔH < 0 → exothermic

🎯 Exam Points

①ΔH < 0 means an exothermic reaction, with products lower in energy and more stable than reactants

②ΔH > 0 means an endothermic reaction

③Hess's law: ΔH depends only on the starting and ending states; adding or subtracting equations adds or subtracts their ΔH

④Enthalpy of formation: the ΔH when 1 mol of a compound forms from its elements

⑤Activation energy is the least energy needed for a reaction to occur, set independently of ΔH

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Properties of Solutions

Chemical Equilibrium

Last updated 2026-10-17

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