Energy in Chemical Reactions — Print · Grade 11 Science

Chapter: Chemical Energy & Thermodynamics

Energy in Chemical Reactions

Learning goals

Readings for this unit

Before this unit

Energy conservation, bond changes, and system signs.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

How can a reaction release energy if breaking its reactant bonds requires energy?

Illustrative evidence case. The selected full account includes both bond breaking and product-bond formation.

Energy signs depend on which system is being describedEnergy signs depend on which system is being describedReaction systemHeat outExothermic system: q < 0 under the stated heat conventionSurroundings receive energy; total energy is accounted for.
Original selected outward heat-transfer diagram. It does not imply a temperature value or that heat is a material fluid stored inside a bond.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

How can a reaction release energy if breaking its reactant bonds requires energy?

Your first model

Draw energy inputs and outputs for the same reaction basis instead of labeling one bond as stored releasable heat.

Build the language from the idea

enthalpy change (ΔH)
The heat absorbed or released by a reaction at constant pressure; negative for reactions that release heat, positive for reactions that absorb it.
exothermic reaction
A reaction that releases energy to its surroundings; its ΔH is negative and the surroundings warm up.
endothermic reaction
A reaction that absorbs energy from its surroundings; its ΔH is positive and the surroundings cool down.
bond energy
The energy needed to break one mole of a particular chemical bond; the same amount is released when that bond forms.
activation energy (Ea)
The minimum energy colliding particles must have for a reaction to occur — the 'hump' reactants must climb before they can become products.
specific heat (c)
The energy needed to raise the temperature of 1 gram of a substance by 1°C; for liquid water it is 4.18 J/g°C.

Chemical reactions rearrange electron distributions and chemical bonds. Breaking a specified bond requires energy; forming that bond releases energy. A bond-energy calculation imagines separating reactants and then forming products as an accounting path, not a claim that every real reaction first breaks every bond completely. Net enthalpy depends on initial and final states, including phase and interactions.

An exothermic reaction has negative ΔH: the products have lower enthalpy than the reactants for the stated amounts and conditions. At constant pressure with only pressure-volume work, reaction heat equals ΔH and energy is transferred as heat to surroundings. An average bond-energy account can estimate the net difference, but phases and particle interactions also matter. Fuel combustion and many acid-base reactions are familiar examples; the temperature rise also depends on heat capacity and energy loss.

Energy signs depend on which system is being describedEnergy signs depend on which system is being describedReaction systemHeat outExothermic system: q < 0 under the stated heat conventionSurroundings receive energy; total energy is accounted for.
Original selected outward heat-transfer diagram. It does not imply a temperature value or that heat is a material fluid stored inside a bond.

An endothermic reaction has positive ΔH and, under the appropriate constant-pressure conditions, absorbs heat from surroundings. The observed temperature change depends on the surroundings and experimental arrangement. Some cold packs involve endothermic dissolution, a physical process rather than a new-substance reaction. Photosynthesis stores energy through a light-driven process; light uptake should not be presented as the same mechanism as heat absorption measured by a simple calorimeter. Identify both the process and the energy-transfer pathway.

This is why bond energy lets you predict a reaction's energy before ever running it. If you add up the energy needed to break every reactant bond and subtract the energy released when every product bond forms, the difference is ΔH: ΔH = (energy to break bonds) − (energy to form bonds). Take hydrogen burning in chlorine, H₂ + Cl₂ → 2HCl. Breaking one H–H bond (436 kJ) and one Cl–Cl bond (242 kJ) costs 678 kJ, while forming two H–Cl bonds releases 2 × 431 = 862 kJ. So ΔH = 678 − 862 = −184 kJ per mole — negative, and therefore exothermic.

Breaking bonds requires energy; formation can release more in the full accountBreaking bonds requires energy; formation can release more in the full accountSelected energy magnitude (kJ)Break reactant bonds500Form product bonds650ΔH estimate = energy in − energy out = 500 − 650= −150 kJ for the specified reaction amount
Original invented bond-energy magnitudes for one selected reaction basis, not measured values of particular named bonds.

Activation energy is the energy barrier along a reaction pathway. The transition state can contain partly broken and partly formed bonds; not every bond must be fully broken before reaction occurs. A potential-energy diagram distinguishes the barrier from the overall state difference ΔH. A catalyst changes the pathway and lowers a barrier while leaving the same reactant/product energy difference.

A favorable energy difference can coexist with an activation barrierA favorable energy difference can coexist with an activation barrierEnergy (relative units)010203040506001234Reaction coordinate (schematic)Uncatalyzed pathCatalyzed path
Original qualitative path curves in arbitrary energy units, not a measured potential-energy surface or a rate-versus-time graph.

Calorimetry infers energy transfer from measured temperature changes and an energy account. For a material warming without a phase change, q ≈ mcΔT when its specific heat is approximately constant. Water’s commonly used value is 4.18 J g⁻¹ °C⁻¹ near ordinary conditions. In an insulated reaction setup, include the solution and calorimeter when relating their heat gain to reaction heat loss. Ignoring heat loss or cup heat capacity is an approximation; a temperature reading does not directly guarantee an exact reaction enthalpy.

Model check: Breaking a bond costs energy; the overall reaction can release more when product bonds form. ΔH=500−650=−150kJ for the represented amount; exothermic. Average bond values are approximate.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Choose the reaction system and energy sign

A system is the part of the world selected for an energy account; the rest is its surroundings. Exothermic means heat leaves the selected system, giving negative q for that system under the usual chemistry convention. Endothermic means it absorbs heat. At constant pressure with the relevant work assumptions, reaction heat connects with enthalpy change. A warmer thermometer in the surroundings can therefore accompany a negative reaction-system heat. State the boundary before assigning a sign or equating heat with temperature.

Energy signs depend on which system is being describedEnergy signs depend on which system is being describedReaction systemHeat outExothermic system: q < 0 under the stated heat conventionSurroundings receive energy; total energy is accounted for.
Original selected outward heat-transfer diagram. It does not imply a temperature value or that heat is a material fluid stored inside a bond.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

If the surroundings warm because the reaction releases heat, what sign is assigned to the reaction-system heat?

Visual reasoning — after your attempt

Negative under the stated convention. The surroundings receive positive heat from that transfer.

Reaction enthalpy ΔH is the products-minus-reactants enthalpy difference for a specified equation and conditions. At constant pressure under the appropriate work assumptions, qsystem corresponds to ΔH. Exothermic ΔH is negative for the reacting system; surroundings can gain positive heat.

OpenSciEd magnification template linking the observable heater to two particle-scale views before and after adding water; it is a model-building prompt, not a completed reaction diagram.
OpenSciEd 7.2’s flameless-heater template connects an observable device to two empty magnified views. It asks for a model and is not evidence of a particular completed reaction, a calorimeter record or a do-it-yourself construction plan. Pair it with this course’s complete system, bond-energy and heat-transfer accounts. Flameless-heater model at visible and particle scales · OpenSciEd · 7.2 Chemical Reactions & Energy, Lesson 1, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.

The thermometer’s warming describes the measured surrounding solution, not a positive ΔH for an exothermic reaction. Label both signs.

Pause and explain

Why can warming surroundings accompany negative reaction ΔH?

Check your reasoning after answering

The reaction system loses enthalpy as energy transfers out, while the surroundings gain energy. Signs refer to different chosen systems.

Part 2 · Include bond breaking and bond formation

Bond dissociation requires energy for the specified process, while formation of corresponding bonds releases energy relative to separated fragments. A reaction can release net energy when product formation more than offsets reactant bond-breaking costs. Introductory average bond-energy estimates use gas-phase approximations and are not exact enthalpies for every phase or molecule. Breaking a “high-energy bond” alone is not an energy-release explanation. Compare all reactants and products, their states and the chosen reaction amount.

Breaking bonds requires energy; formation can release more in the full accountBreaking bonds requires energy; formation can release more in the full accountSelected energy magnitude (kJ)Break reactant bonds500Form product bonds650ΔH estimate = energy in − energy out = 500 − 650= −150 kJ for the specified reaction amount
Original invented bond-energy magnitudes for one selected reaction basis, not measured values of particular named bonds.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why is the selected net estimate negative even though breaking reactant bonds needs positive energy?

Visual reasoning — after your attempt

The 650 kJ released in the modeled product formation exceeds the 500 kJ input, giving −150 kJ for the full account.

Breaking bonds requires input, while forming bonds releases energy. A bond-enthalpy estimate uses sums for bonds broken minus sums for bonds formed. Average bond enthalpies give approximate gas-phase estimates and are not interchangeable with exact solution reaction enthalpies.

If breaking costs 500 kJ and formation releases 700 kJ for the equation amount, the estimated ΔH is −200 kJ.

Pause and explain

Why is energy comes from breaking bonds incomplete?

Check your reasoning after answering

Bond breaking takes energy. Net release requires the complete account, including the energy released when new bonds form.

Part 3 · Scale energy with the equation amount

Enthalpy is a state function, and a thermochemical reaction change is associated with a defined transformation and amount basis. Multiplying every coefficient scales the enthalpy change by the same factor; reversing the reaction reverses its sign. A number labeled kJ/mol must specify mol of what—product, reactant or reaction extent. Phase labels also matter because liquid and vapor have different enthalpy states. Keep these details when using a reference value in stoichiometry.

Reaction enthalpy belongs to the equation’s stated amount and directionReaction enthalpy belongs to the equation’s stated amount and directionA → B ΔH = −40 kJ per equation amount2A → 2B ΔH = −80 kJB → A ΔH = +40 kJScale coefficients and energy together; reversal changes the sign.
Original hypothetical transformation values, not real thermochemical data or an assertion about rate.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What is wrong with reversing A→B while keeping the same negative ΔH sign?

Visual reasoning — after your attempt

It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.

A stated ΔH belongs to the reaction as written. Reversing the equation changes its sign; multiplying coefficients multiplies ΔH. A per-mole label must specify per mole of which species or reaction extent. Do not transfer a numerical enthalpy to a differently scaled equation unchanged.

If A→B has ΔH=−50 kJ per written mole conversion, 2A→2B has −100 kJ and B→A has +50 kJ under matching conditions.

Pause and explain

What happens to ΔH when the equation is reversed?

Check your reasoning after answering

Its sign reverses because the products/reactants energy difference is reversed, provided conditions and species states match.

Part 4 · Keep energy and reaction rate separate

An activation barrier separates reactant and product states along a modeled pathway. A catalyst provides another pathway with a lower barrier but does not change the reactant/product endpoint difference for the same reaction conditions. A reaction can therefore be thermodynamically favored yet slow. Reaction coordinate is a schematic progress description, not elapsed time or distance traveled by a molecule across a real hill. Distinguish endpoint balance, barrier height and measured rate before explaining a temperature or catalyst effect.

A favorable energy difference can coexist with an activation barrierA favorable energy difference can coexist with an activation barrierEnergy (relative units)010203040506001234Reaction coordinate (schematic)Uncatalyzed pathCatalyzed path
Original qualitative path curves in arbitrary energy units, not a measured potential-energy surface or a rate-versus-time graph.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What remains unchanged between the two selected paths despite the lower catalytic peak?

Visual reasoning — after your attempt

The starting and final energy levels—and therefore the endpoint difference—remain unchanged in the selected account.

A favorable or energy-releasing reaction can be slow because of an activation barrier. A catalyst changes pathway and rate, not the thermodynamic difference between the same states. Energy release alone does not establish safe conditions, self-sustaining operation, or complete conversion.

A reaction may require initiation and lose energy to surroundings faster than it maintains a useful temperature. Thermodynamic and kinetic questions need different evidence.

Pause and explain

Does a negative ΔH establish a fast reaction?

Check your reasoning after answering

No. Rate depends on pathway, activation barriers, concentrations, temperature, and other conditions, not just net enthalpy.

Apply the whole unit · Interpret and scale a reaction energy record

Illustrative equation: A→B,ΔH=−80.0 kJ per 1 mol A converted under stated constant-pressure conditions. In the model 0.250 mol A converts completely.

Your task

Calculate reaction-system and ideal surroundings heat changes, reverse and double the equation’s enthalpy, and explain why neither the sign nor size establishes the reaction speed.

What your response should show

  • Choose the appropriate chemical model or relationship and state its conditions.
  • Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
  • Use supplied evidence to justify the conclusion; distinguish observation from inference.
  • Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
  1. Scale: 0.250×(−80.0)=−20.0 kJ for the reaction system.

  2. With no other energy destinations, surroundings gain +20.0 kJ.

  3. B→A has +80.0 kJ per written mole; 2A→2B has −160 kJ.

The energy changes are tied to the specified equation amount, states, and conditions. The surroundings result assumes a complete limited energy account. Rate and initiation require kinetic evidence; an exothermic label is not a proof of fast, complete, or self-sustaining conversion.

Scientific references and source-model foundations

Parent / teacher teaching notes

Have students label equation amount and system boundary before signs. Require a sentence distinguishing activation energy from net enthalpy. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.

Potential-Energy Diagram (Exothermic)Reaction progress →EaΔHReactantsProducts
Selected exothermic path with lower product enthalpy. Forward activation energy is the barrier height above the reactant level, not the absolute peak coordinate. Reaction coordinate is schematic rather than elapsed time; rate and net energy are distinct.

The key idea

A reaction energy account needs a system, defined states and amount. Bond breaking requires energy, product formation can offset it, and reaction enthalpy is distinct from activation barrier and rate. Calorimetry uses a complete heat-capacity/boundary account.

Heat Transfer (Calorimetry)

q = mcΔT

q is heat energy in joules (J); m is mass in grams; c is the specific heat (4.18 J/g°C for liquid water); ΔT = T final − T initial in °C.

Enthalpy from Bond Energies

ΔH = (energy to break bonds) − (energy to form bonds)

a negative ΔH means the reaction releases energy overall (exothermic); a positive ΔH means it absorbs energy overall (endothermic).

Worked examples

Example 1

How much heat (in joules) is absorbed when 200. g of water is heated from 20.0°C to 50.0°C? The specific heat of water is 4.18 J/g°C.
  1. Identify the values: m = 200 g, c = 4.18 J/g°C, ΔT = T final − T initial = 50.0 − 20.0 = 30.0°C.
  2. Apply q = mcΔT = 200 × 4.18 × 30.0.
  3. Multiply: 4.18 × 30.0 = 125.4, and 125.4 × 200 = 25,080 J.
Answer: 25,080 J (about 25.1 kJ) absorbed

Example 2 · Worked application

A reaction has ΔH=−80 kJ per stated reaction amount. Interpret direction and reverse-reaction value.
  1. Negative ΔH means the system's enthalpy decreases and heat is released at constant pressure under the model.
  2. Reversing the reaction reverses the sign.
  3. The magnitude refers to the stated stoichiometric amount.
Answer: Forward releases 80 kJ; reverse ΔH=+80 kJ.
Check the reasoning

Do not omit the reaction-amount basis when scaling.

Example 3

For the reaction H₂ + Cl₂ → 2HCl, the bond energies are H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, and H–Cl = 431 kJ/mol. Calculate ΔH and state whether the reaction is exothermic or endothermic.
  1. Energy to break the reactant bonds: one H–H (436) + one Cl–Cl (242) = 678 kJ absorbed.
  2. Energy released forming the product bonds: two H–Cl = 2 × 431 = 862 kJ released.
  3. ΔH = (bonds broken) − (bonds formed) = 678 − 862 = −184 kJ/mol. The negative sign means energy is released.
Answer: ΔH = −184 kJ/mol; the reaction is exothermic

Example 4 · Find and fix the mistake

Context: A reaction has ΔH=−80 kJ per stated reaction amount. Interpret direction and reverse-reaction value. A student claims: “A negative enthalpy means no activation energy is required.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Do not omit the reaction-amount basis when scaling.
  2. Heat release and activation barrier are different properties; an exothermic reaction may still need initiation.
Answer: Heat release and activation barrier are different properties; an exothermic reaction may still need initiation.
Check the reasoning

Do not omit the reaction-amount basis when scaling.

Example 5 · Reason from the visual model

Reaction enthalpy belongs to the equation’s stated amount and directionReaction enthalpy belongs to the equation’s stated amount and directionA → B ΔH = −40 kJ per equation amount2A → 2B ΔH = −80 kJB → A ΔH = +40 kJScale coefficients and energy together; reversal changes the sign.
Original hypothetical transformation values, not real thermochemical data or an assertion about rate.
What is wrong with reversing A→B while keeping the same negative ΔH sign?
  1. Enthalpy is a state function, and a thermochemical reaction change is associated with a defined transformation and amount basis. Multiplying every coefficient scales the enthalpy change by the same factor; reversing the reaction reverses its sign. A number labeled kJ/mol must specify mol of what—product, reactant or reaction extent. Phase labels also matter because liquid and vapor have different enthalpy states. Keep these details when using a reference value in stoichiometry.
  2. It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.
Answer: It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.
Check the reasoning

Original hypothetical transformation values, not real thermochemical data or an assertion about rate.

Avoid these mistakes

Check your understanding

Breaking costs 500kJ and forming releases 650kJ. Find modeled ΔH.

Discuss the answer

ΔH=500−650=−150kJ for the represented amount; exothermic. Average bond values are approximate.

Return to the phenomenon

Return to the initial case. How can a reaction release energy if breaking its reactant bonds requires energy? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Bond breaking costs energy, while formation can release more. Net change depends on all species and conditions, with a separate activation-barrier question for rate.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. How can a reaction release energy if breaking its reactant bonds requires energy? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the chemical mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep species identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Bond breaking costs energy, while formation can release more. Net change depends on all species and conditions, with a separate activation-barrier question for rate.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A selected reaction absorbs 720 kJ in the bond-breaking account and releases 880 kJ on product formation. Estimate net enthalpy, then predict the corresponding value when the entire equation is doubled.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.

Words to know

enthalpy change (ΔH)
The heat absorbed or released by a reaction at constant pressure; negative for reactions that release heat, positive for reactions that absorb it.
exothermic reaction
A reaction that releases energy to its surroundings; its ΔH is negative and the surroundings warm up.
endothermic reaction
A reaction that absorbs energy from its surroundings; its ΔH is positive and the surroundings cool down.
bond energy
The energy needed to break one mole of a particular chemical bond; the same amount is released when that bond forms.
activation energy (Ea)
The minimum energy colliding particles must have for a reaction to occur — the 'hump' reactants must climb before they can become products.
specific heat (c)
The energy needed to raise the temperature of 1 gram of a substance by 1°C; for liquid water it is 4.18 J/g°C.

Lavoisier's ice calorimeter: measuring heat with a melting snowball

Long before electronic sensors, Antoine Lavoisier and mathematician Pierre-Simon Laplace built a clever device in the 1780 s to measure the heat a reaction releases. They nested a reacting sample inside a chamber packed with ice; the heat given off melted a measurable amount of that ice, and the mass of meltwater told them exactly how much energy had flowed out. In one famous run they even placed a live guinea pig inside and showed that respiration warms an animal by the same kind of slow combustion that burns charcoal — an early hint that the chemistry of a candle and the chemistry of a living body are cousins. Every modern calorimeter, and the equation q = mcΔT you use with it, descends directly from that icy experiment.

The misconception that 'exothermic' means 'starts on its own'

It is tempting to assume that because an exothermic reaction releases energy, it must happen spontaneously the moment the reactants touch. But gasoline can sit safely in a can beside oxygen for years, and a log will not burst into flame just because burning is exothermic. The reason is the activation energy — the hump every reaction must climb before it can release its energy — so a spark, a flame, or a hot surface is needed to push those first molecules over the barrier. Once over, the energy released can supply the activation energy for the next molecules, which is why a fire, once lit, sustains itself. Exothermic tells you where a reaction ends up, not whether it will start without a push.

Where the Calories on a food label come from

The 'Calories' printed on a cereal box are literally combustion measurements. Food scientists burn a dried sample inside a sealed bomb calorimeter submerged in water, then use q = mcΔT to convert the water's temperature rise into energy. One food Calorie (capital C) is actually a kilocalorie — 1000 of the small calories used in chemistry, and about 4184 joules. So the same equation that sizes a power-plant cooling system also decides whether a snack is labeled 100 Calories or 300, connecting the abstract idea of enthalpy to the very concrete question of how much energy your dinner will give you.

Self-heating meals and instant cold packs

Engineers put endothermic and exothermic chemistry to work in your hands. A flameless ration heater tucked into a soldier's or camper's meal uses the exothermic reaction of magnesium metal with water to boil a pouch of food without a stove. An instant cold pack does the reverse: snapping its inner barrier lets ammonium nitrate dissolve endothermically, pulling heat out of a sprained ankle within seconds. Both are designed around ΔH — chemists deliberately chose reactions with the right sign and size of energy change so the product warms or cools by just the amount a person needs.

Why ΔH is only half the story

A common surprise for chemistry students is that some endothermic reactions happen readily anyway — an instant cold pack works without any push, even though it absorbs energy. Energy release is not the only thing driving a reaction; nature also favors an increase in disorder (called entropy). When a solid dissolves into freely moving ions, the jump in disorder can be enough to make an energy-absorbing process go forward on its own. Chemists combine both factors — the energy change ΔH and the disorder change — into a single quantity called free energy, which you will meet in a later course. For now, remember that ΔH tells you about the heat, but it is only one of the two forces steering whether a reaction actually proceeds.

In the real world

Chemical engineers use calorimetry and ΔH to design everything from safe rocket propellants to the flameless self-heating pouches that warm a soldier's or a camper's meal.

Exit check

Breaking costs 500kJ and forming releases 650kJ. Find modeled ΔH.

Check your explanation after answering

ΔH=500−650=−150kJ for the represented amount; exothermic. Average bond values are approximate.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

In an endothermic reaction, the sign of ΔH is…
  1. negative
  2. positive
  3. always zero
  4. impossible to determine
Answer and explanation

positive

An endothermic reaction absorbs energy into the system, so ΔH is positive.

Problem 2

How much heat (in J) is absorbed when 50.0 g of water is warmed from 15.0°C to 35.0°C? (c = 4.18 J/g°C)
Answer and explanation

4180 J

ΔT = 35.0 − 15.0 = 20.0°C, so q = mcΔT = 50 × 4.18 × 20 = 4180 J.

Problem 3

On a potential-energy diagram, the activation energy is the difference in energy between…
  1. the two dashed helper lines
  2. the reactants and the products
  3. the reactants and the peak (the activated complex)
  4. the products and the x-axis
Answer and explanation

the reactants and the peak (the activated complex)

Activation energy is the height of the peak above the reactant energy level — the energy barrier the reaction must climb.

Problem 4

In an insulated, constant-pressure reaction calorimeter, the measured solution temperature falls. Assume negligible heat loss and no other process. Is the reaction endothermic or exothermic, what is the sign of ΔH, and why?
Answer and explanation

Under the stated assumptions, the reaction absorbs heat from the solution, so it is endothermic and ΔH is positive. A temperature fall without these conditions would not uniquely prove reaction enthalpy; other energy transfers or processes could contribute.

Problem 5

A reaction releases 6270 J of heat into 100. g of water (c = 4.18 J/g°C). By how many °C does the water's temperature rise?
Answer and explanation

15 °C

Rearranging q = mcΔT gives ΔT = q ÷ (m × c) = 6270 ÷ (100 × 4.18) = 6270 ÷ 418 = 15°C.

Problem 6

If ΔH=−50 kJ for one reaction amount, find ΔH for twice that amount. Show the reasoning, not just the conclusion.
Hints

Do not omit the reaction-amount basis when scaling.

Answer and explanation

−100 kJ under identical conditions.

Problem 7

Why does bond-energy estimation subtract energy released in forming product bonds from energy used breaking reactant bonds?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Net reaction energy balances input to break bonds against release from forming new bonds; breaking alone is not the energy source.

Problem 8

Reaction enthalpy belongs to the equation’s stated amount and directionReaction enthalpy belongs to the equation’s stated amount and directionA → B ΔH = −40 kJ per equation amount2A → 2B ΔH = −80 kJB → A ΔH = +40 kJScale coefficients and energy together; reversal changes the sign.
Original hypothetical transformation values, not real thermochemical data or an assertion about rate.
Interpret the supplied visual. What is wrong with reversing A→B while keeping the same negative ΔH sign?
Answer and explanation

It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.

Enthalpy is a state function, and a thermochemical reaction change is associated with a defined transformation and amount basis. Multiplying every coefficient scales the enthalpy change by the same factor; reversing the reaction reverses its sign. A number labeled kJ/mol must specify mol of what—product, reactant or reaction extent. Phase labels also matter because liquid and vapor have different enthalpy states. Keep these details when using a reference value in stoichiometry.

It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 9

A selected reaction absorbs 720 kJ in the bond-breaking account and releases 880 kJ on product formation. Estimate net enthalpy, then predict the corresponding value when the entire equation is doubled.
Answer and explanation

ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.

ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 10

OpenSciEd magnification template linking the observable heater to two particle-scale views before and after adding water; it is a model-building prompt, not a completed reaction diagram.
OpenSciEd’s empty observable/particle-scale heater template is a model-building prompt, not a completed reaction or a calorimeter record. Flameless-heater model at visible and particle scales · OpenSciEd · 7.2 Chemical Reactions & Energy, Lesson 1, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
What would a complete explanation need to add to the empty source heater template? Distinguish system boundary, reaction species, bond-energy account and thermometer response, and name one conclusion the template alone cannot establish.
Answer and explanation

Define the selected reaction system and surroundings, identify species from appropriate evidence, conserve material and account for bond breaking plus formation. Connect transferred energy with heat capacity and measured temperature under stated conditions. The empty template alone cannot identify the reaction, quantify enthalpy or prove a construction method.

Identify the source’s visible labels, scales and unfinished or omitted information.

Use the local complete explanations to connect the observation with an explicit mechanism.

State assumptions and a limit before extending the model.

Response criteria

  • Interpret the source symbols and boundary correctly.
  • Explain the relevant mechanism using quantities/species consistently.
  • State a meaningful omission, assumption or evidence limitation.

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