Does breaking bonds release energy or absorb energy?

Breaking bonds always absorbs energy — never releases it. Bonded atoms are in a stable, low-energy state, so separating them requires an input. Reactions like combustion release energy overall because the new bonds formed in the products are stronger than the bonds broken, not because breaking bonds gave energy off.

How can I remember which sign goes with endothermic and which with exothermic?

Think of H as the system's bank balance. Endothermic means energy goes "endo," into the system, so the balance goes up and H is positive. Exothermic means energy exits, the balance goes down, and H is negative. The thermometer in the beaker reads the opposite of the sign, because it measures the surroundings.

Why is bond-energy calculation only an estimate of ΔH?

Tabulated bond energies are averages taken across many different molecules, and the strength of, say, a C–H bond varies slightly depending on what else is attached. Bond energies also apply to gas-phase species, so they ignore energy changes from phase changes and intermolecular forces. The result usually lands within a few percent of the measured value.

Can a reaction be exothermic and still need heating to start?

Yes, and this is extremely common. Every reaction has an activation energy barrier that must be crossed before products form. Lighting a match supplies that starting energy to a combustion reaction that then releases far more energy than it took to ignite. Activation energy determines whether the reaction gets going; H determines the net energy flow once it does.

CHEM-9.1

Endothermic & Exothermic Reactions

Learn to classify reactions as endothermic or exothermic using the sign of ΔH and bond-energy accounting — bonds broken absorb energy, bonds formed release it.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Endothermic & Exothermic Reactions, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Endothermic & Exothermic Reactions free

What this lesson covers

Strike a match and heat pours out. Snap a cold pack and it gets icy in your hand. Both are chemical changes, but energy flows in opposite directions. In this lesson you will learn to predict and explain that direction two different ways: from the sign of the enthalpy change

\Delta H

, and from a bond-by-bond accounting of what it costs to break bonds versus what you get back when new bonds form.

The big idea underneath everything is simple and worth memorizing right now: breaking bonds always absorbs energy, and forming bonds always releases energy. Whether a reaction warms its surroundings or cools them comes down to which of those two totals is bigger. Once you can run that comparison, energy diagrams,

\Delta H

signs, and thermochemical equations all start saying the same thing in different languages.

System, Surroundings, and the Sign of ΔH

Chemists split the universe into two parts. The system is the reacting chemicals themselves. The surroundings are everything else — the solution, the beaker, the air, your hand. Energy transferred as heat moves between these two.

An exothermic reaction releases energy from the system to the surroundings. The surroundings get warmer, so the thermometer in the beaker reads a higher temperature. Because the system lost energy, its enthalpy went down, and

\Delta H

is negative:

\Delta H < 0

.

An endothermic reaction absorbs energy from the surroundings into the system. The surroundings get colder, the thermometer reads lower, and the system gained enthalpy, so

\Delta H > 0

.

The single most common error in this unit is a sign flip caused by mixing up who is being described. If a beaker feels hot, that is a statement about the surroundings, and it means the system released energy, so

\Delta H

is negative. Students often reason "hot means energy went up, so positive" and get it backward. Anchor yourself with the rule: the sign of $\Delta H$ always describes the system, never your hand.

Type	Energy flow	Surroundings feel	Sign of $\Delta H$	Energy appears as
Exothermic	System → surroundings	Warmer	Negative	Product
Endothermic	Surroundings → system	Colder	Positive	Reactant

Combustion, neutralization of a strong acid with a strong base, and most oxidation reactions are exothermic. Photosynthesis, thermal decomposition of calcium carbonate, and the dissolving of ammonium nitrate in water are endothermic.

Bonds Broken Absorb, Bonds Formed Release

A chemical bond is a stable, low-energy arrangement. Pulling bonded atoms apart is like pulling apart two magnets stuck together — you have to put work in. So breaking a bond always requires an input of energy and is endothermic. Letting atoms snap together into a bond lets the system fall to lower energy and dumps that energy into the surroundings, so forming a bond always releases energy and is exothermic.

Every reaction does both. Reactant bonds must break, and product bonds must form. The overall sign of

\Delta H

is a competition:

\Delta H \approx \sum (\text{bond energies broken}) - \sum (\text{bond energies formed})

Bond energies are always tabulated as positive numbers — the energy needed to break one mole of that bond in the gas phase. The subtraction in the formula is what puts the negative sign on the energy released by bond formation.

If the products' bonds are stronger overall, more energy comes out than went in, the difference is negative, and the reaction is exothermic. If reactant bonds are stronger, you paid more than you got back,

\Delta H

is positive, and the reaction is endothermic.

A misconception worth killing early: some students say "bond breaking releases energy because gasoline burning releases energy." Burning gasoline releases energy because the new C=O and O–H bonds in carbon dioxide and water are much stronger than the C–H, C–C, and O=O bonds that broke. The release comes from bond formation, not bond breaking. Nothing in chemistry releases energy by breaking a bond.

Reading Energy Diagrams

An energy (enthalpy) diagram plots potential energy on the vertical axis against reaction progress on the horizontal axis. Reactants sit at one height on the left, products at another on the right, and a hump called the activation energy barrier sits between them.

For an exothermic reaction, products are drawn lower than reactants: the system ended up with less stored energy, and the difference was released.

\Delta H

is measured as the vertical drop from reactants to products, and it is negative.

For an endothermic reaction, products are drawn higher than reactants.

\Delta H

is the vertical climb, and it is positive.

Always measure

\Delta H

as products minus reactants:

\Delta H = H_{\text{products}} - H_{\text{reactants}}

The hump height is a separate quantity. Activation energy is measured from the reactants up to the peak and is always positive, whether the reaction is endothermic or exothermic. A very common mistake is reading the peak height as

\Delta H

, or concluding that a tall barrier means the reaction is endothermic. It does not. A reaction can be strongly exothermic and still have a huge barrier — a paper book sitting on a desk is not spontaneously bursting into flame even though combustion is very exothermic, because nothing has supplied the activation energy yet. The barrier controls how fast;

\Delta H

controls the net energy in or out.

Also note: reversing a reaction reverses the sign of

\Delta H

and swaps which side is higher. If forming a compound releases 92 kilojoules, decomposing it absorbs 92 kilojoules.

Thermochemical Equations and Energy as a Term

There are two equivalent ways to write energy into a balanced equation, and you should be fluent in both.

The first writes

\Delta H

after the equation:

\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)} \qquad \Delta H = -802\ \mathrm{kJ}

The second puts energy directly into the equation as if it were a chemical species. Because this reaction gives energy off, energy belongs on the product side:

\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) + 802\ kJ}

For an endothermic reaction, energy is consumed, so it is written as a reactant:

\mathrm{2NH_3(g) + 92\ kJ \rightarrow N_2(g) + 3H_2(g)}, \qquad \Delta H = +92\ \mathrm{kJ}

Notice the pattern: the side where energy appears in the written equation is the opposite of where the negative sign shows up in

\Delta H

notation. Exothermic means energy is a product and

\Delta H

is negative.

The

\Delta H

value is tied to the coefficients exactly as written. If you double all the coefficients, you double

\Delta H

. Burning two moles of methane releases 1604 kilojoules. This proportionality lets you scale to any amount: convert grams to moles, then multiply by the kilojoules per mole from the balanced equation. Watch units — kilojoules per mole of the substance the equation is written for, not per mole of just anything in the equation.

Key terms

Enthalpy change ( $\Delta H$ ).: The heat absorbed or released by a chemical system at constant pressure, calculated as enthalpy of products minus enthalpy of reactants.
Exothermic.: A process that releases energy from the system to the surroundings; the surroundings warm up and $\Delta H$ is negative.
Endothermic.: A process that absorbs energy from the surroundings into the system; the surroundings cool down and $\Delta H$ is positive.
System.: The specific chemicals undergoing the reaction — the part of the universe whose energy change $\Delta H$ describes.
Surroundings.: Everything outside the system, including the solvent, container, and air, whose temperature change we actually measure.
Bond energy.: The energy required to break one mole of a particular bond in the gas phase, always reported as a positive value.
Activation energy.: The minimum energy needed to reach the transition state, shown as the height of the hump on an energy diagram; always positive regardless of the sign of $\Delta H$ .
Thermochemical equation.: A balanced chemical equation that includes the energy change, either as a $\Delta H$ value or as an energy term on the reactant or product side.

Worked example

Use bond energies to estimate

\Delta H

for the combustion of hydrogen:

2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(g)}

. Bond energies in kilojoules per mole: H–H = 436, O=O = 498, O–H = 463. Then classify the reaction.

Step 1: Count the bonds broken (reactants). Two moles of

\mathrm{H_2}

contain 2 H–H bonds. One mole of

\mathrm{O_2}

contains 1 O=O bond.

Energy in

= 2(436) + 1(498) = 872 + 498 = 1370

kilojoules.

Step 2: Count the bonds formed (products). Each water molecule has two O–H bonds, and there are 2 moles of water, so 4 O–H bonds form.

Energy out

= 4(463) = 1852

kilojoules.

Step 3: Subtract, products from the correct side.

\Delta H \approx \sum \text{broken} - \sum \text{formed} = 1370 - 1852 = -482\ \mathrm{kJ}

Step 4: Interpret. The result is negative, so the reaction is exothermic. More energy was released forming the four strong O–H bonds than was spent breaking the H–H and O=O bonds. The extra 482 kilojoules flows out to the surroundings, which is why a hydrogen flame is so hot.

Step 5: Check with a diagram. On an energy diagram, the products (water) sit 482 kilojoules below the reactants. Written the other way, the thermochemical equation is

2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(g)} + 482\ \mathrm{kJ}

.

A quick sanity check on your bond counting: forgetting that each water molecule has two O–H bonds gives

1370 - 926 = +444

and the wrong classification entirely. Always draw the structures if you are unsure.

Practice questions

A student dissolves ammonium nitrate in water in a foam cup and the temperature of the water drops from 22.0 °C to 14.5 °C. Which statement correctly describes the process?

The process is exothermic and $\Delta H$ is negative, because the water lost energy.
The process is endothermic and $\Delta H$ is positive, because the system absorbed energy from the water.
The process is endothermic and $\Delta H$ is negative, because the temperature decreased.
The process is exothermic and $\Delta H$ is positive, because energy left the dissolving salt.

Answer: The process is endothermic and $\Delta H$ is positive, because the system absorbed energy from the water.

The water is the surroundings, and it got colder, which means energy left the water and went into the dissolving salt (the system). Energy flowing into the system is the definition of endothermic, and a system that gains enthalpy has

\Delta H > 0

. The tempting wrong answer pairs "temperature went down" with a negative sign, but the sign of

\Delta H

describes the system's energy, not the thermometer reading in the surroundings.

For the reaction

\mathrm{H_2(g)} + \mathrm{Cl_2(g)} \rightarrow 2\mathrm{HCl(g)}

, the bond energies are H–H = 436, Cl–Cl = 242, and H–Cl = 431 kilojoules per mole. Calculate

\Delta H

, classify the reaction, and explain the result in terms of relative bond strengths.

Answer: $\Delta H \approx -184$ kJ; the reaction is exothermic because the two H–Cl bonds formed are collectively stronger than the H–H and Cl–Cl bonds broken.

Bonds broken: one H–H and one Cl–Cl, so

436 + 242 = 678

kJ absorbed. Bonds formed: two H–Cl bonds, so

2(431) = 862

kJ released. Then

\Delta H = 678 - 862 = -184

kJ. The negative sign means the system ended at lower enthalpy and gave 184 kJ to the surroundings. Physically, the electrons end up in a more stable arrangement in HCl than they were in the separate elements, so the payback from bond formation exceeds the investment in bond breaking. A complete answer states the number, the sign, the classification, and the bond-strength reasoning.

On an energy diagram, a reaction shows reactants at 150 kJ, a peak at 310 kJ, and products at 60 kJ. State the activation energy and

\Delta H

, and explain why a large activation energy does not make a reaction endothermic.

Answer: Activation energy = 160 kJ; $\Delta H = -90$ kJ, so the reaction is exothermic.

Activation energy is measured from reactants up to the peak:

310 - 150 = 160

kJ. The enthalpy change is products minus reactants:

60 - 150 = -90

kJ, negative and therefore exothermic. These are independent quantities. Activation energy is the temporary hill the system must climb to reach the transition state, and any energy spent climbing is recovered on the way down; it controls the rate.

\Delta H

compares only the starting and ending heights, so it controls the net energy transferred.

FAQ

Does breaking bonds release energy or absorb energy?: Breaking bonds always absorbs energy — never releases it. Bonded atoms are in a stable, low-energy state, so separating them requires an input. Reactions like combustion release energy overall because the new bonds formed in the products are stronger than the bonds broken, not because breaking bonds gave energy off.
How can I remember which sign goes with endothermic and which with exothermic?: Think of $\Delta H$ as the system's bank balance. Endothermic means energy goes "endo," into the system, so the balance goes up and $\Delta H$ is positive. Exothermic means energy exits, the balance goes down, and $\Delta H$ is negative. The thermometer in the beaker reads the opposite of the sign, because it measures the surroundings.
Why is bond-energy calculation only an estimate of ΔH?: Tabulated bond energies are averages taken across many different molecules, and the strength of, say, a C–H bond varies slightly depending on what else is attached. Bond energies also apply to gas-phase species, so they ignore energy changes from phase changes and intermolecular forces. The result usually lands within a few percent of the measured value.
Can a reaction be exothermic and still need heating to start?: Yes, and this is extremely common. Every reaction has an activation energy barrier that must be crossed before products form. Lighting a match supplies that starting energy to a combustion reaction that then releases far more energy than it took to ignite. Activation energy determines whether the reaction gets going; $\Delta H$ determines the net energy flow once it does.

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