Endothermic and Exothermic Processes - AP Chemistry

SECTION 1

Historical Context & Motivation

The systematic study of heat exchange during chemical reactions emerged from centuries of inquiry into the nature of fire, heat, and energy. Before the nineteenth century, chemists and natural philosophers operated under the caloric theory, which treated heat as a weightless, self-repulsive fluid that flowed from hot bodies to cold ones. While this framework could account for certain thermal phenomena qualitatively, it failed to explain why some chemical reactions released heat while others absorbed it, and it could not connect thermal changes to a broader, conserved quantity we now call energy.

The transition from caloric theory to modern thermochemistry was driven by meticulous calorimetric experiments and the realization that heat is a form of energy transfer, not a substance. Pioneering scientists developed instruments to measure heat changes with increasing precision, laying the groundwork for the first law of thermodynamics and the classification of processes as exothermic or endothermic. This historical trajectory underscores a central theme in chemistry: macroscopic observations of temperature change reveal fundamental truths about the energetics of bond-breaking and bond-forming at the molecular level.

1780 • Lavoisier & Laplace's Ice Calorimeter

Antoine Lavoisier and Pierre-Simon Laplace constructed the first ice calorimeter to measure the heat released by combustion reactions, establishing quantitative thermochemistry and demonstrating that the heat evolved by a reaction equals the heat absorbed by its reverse.

1840 • Hess's Law of Constant Heat Summation

Germain Hess demonstrated that the total enthalpy change for a reaction is independent of the pathway taken, providing a powerful tool for calculating energy changes indirectly—a principle that remains central to AP Chemistry problem-solving.

1850 • Clausius & the First Law of Thermodynamics

Rudolf Clausius formalized the conservation of energy, distinguishing between heat (q) and work (w) as distinct modes of energy transfer, and established the internal energy (U) as a state function—providing the theoretical scaffold for classifying endothermic and exothermic processes.

1882 • Berthelot & Thomsen: Thermochemical Databases

Marcellin Berthelot and Julius Thomsen independently compiled extensive tables of heats of reaction, proposing (incorrectly) that all spontaneous reactions must be exothermic—a hypothesis later refined by Gibbs free energy considerations.

1923 • Lewis & Randall Standardize Thermodynamic Data

Gilbert N. Lewis and Merle Randall published their landmark text systematizing standard-state enthalpies, entropies, and free energies, enabling chemists to predict whether a process is endothermic or exothermic from tabulated formation data.

The central question that motivated all of this work remains the same one you face on the AP Chemistry exam: when a chemical or physical change occurs, does the system release energy to its surroundings, or does it absorb energy from them? Answering this question rigorously requires understanding enthalpy as a state function, interpreting the sign of ΔH, and connecting macroscopic calorimetric data to the microscopic energetics of bond dissociation and formation.

SECTION 2

Core Principles & Definitions

At the heart of thermochemistry lies the distinction between system and surroundings. The system is the specific reaction or process under study—be it a dissolving salt, a combustion reaction, or a phase transition—while the surroundings encompass everything else in the universe that can exchange energy with the system. Energy conservation dictates that any energy lost by the system is gained by the surroundings and vice versa, so the sign of the enthalpy change (ΔH) immediately tells us the direction of heat flow at constant pressure. A negative ΔH signifies an exothermic process in which the system releases heat, while a positive ΔH indicates an endothermic process in which the system absorbs heat from its surroundings.

Enthalpy (H) as a State Function

Enthalpy is defined as H = U + PV, where U is internal energy, P is pressure, and V is volume. Because H is a state function, ΔH depends only on the initial and final states, not on the reaction pathway—making Hess's Law possible.

Sign Convention for ΔH

ΔH < 0 means the system loses enthalpy (exothermic); ΔH > 0 means the system gains enthalpy (endothermic). The sign is always reported from the system's perspective, a convention that is critical for consistent thermochemical bookkeeping.

Bond Energy Perspective

Breaking bonds requires energy input (endothermic), while forming bonds releases energy (exothermic). The net ΔH of a reaction reflects the balance between the total energy invested in breaking reactant bonds and the total energy released in forming product bonds.

Surroundings Temperature Change

In an exothermic process the surroundings warm up (q_surr > 0), while in an endothermic process the surroundings cool down (q_surr < 0). Calorimetry exploits this: the measured temperature change of the surroundings allows calculation of q_rxn.

Standard Enthalpy of Reaction (ΔH°rxn)

Standard conditions (25 °C, 1 atm, 1 M solutions) provide a reference point. ΔH°rxn can be calculated from standard enthalpies of formation using ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants), enabling predictions without performing the actual experiment.

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KEY TAKEAWAY

Think of a chemical reaction as a financial transaction between system and surroundings. In an exothermic process, the system is the spender—it releases energy (writes a check) and the surroundings' energy account grows, so the surroundings warm up. In an endothermic process, the system is the borrower—it withdraws energy from the surroundings, cooling them. The sign of ΔH is simply the system's balance sheet: negative means net payout, positive means net intake.

SECTION 3

Energy Diagrams: Visualizing Heat Flow

Energy diagrams—sometimes called enthalpy diagrams or reaction coordinate diagrams—are the most intuitive way to visualize whether a process is endothermic or exothermic. The vertical axis represents enthalpy (H), and horizontal bars represent the enthalpy levels of reactants and products. An arrow connecting the two bars indicates the direction and magnitude of ΔH. When the products sit lower on the enthalpy axis than the reactants, ΔH is negative and the process is exothermic. When the products sit higher, ΔH is positive and the process is endothermic.

Left: In an exothermic process, products reside at a lower enthalpy than reactants, so the system releases energy (ΔH < 0). Right: In an endothermic process, products reside at a higher enthalpy than reactants, so the system absorbs energy (ΔH > 0). The vertical separation between the reactant and product bars corresponds to the magnitude of ΔH.

Notice that the diagram does not depict the activation energy barrier (E_a) that the reaction must overcome; it focuses solely on the net enthalpy difference between initial and final states. This is deliberate: ΔH is a state function and is independent of the pathway, so the only thermochemically relevant information is the relative enthalpy of reactants versus products. When you see a question on the AP exam asking you to identify a process as exothermic or endothermic, mentally sketch this diagram—if the arrow points downward, the process releases heat; if it points upward, the process absorbs heat.

SECTION 4

Mathematical Framework

Quantifying energy changes requires a set of interconnected equations that relate heat transfer, enthalpy change, and calorimetric measurements. The mathematical framework for endothermic and exothermic processes rests on the first law of thermodynamics and extends through calorimetry equations and Hess's Law to standard enthalpy calculations.

FIRST LAW OF THERMODYNAMICS

ΔU = q + w

ΔU = change in internal energy, q = heat transferred to the system, w = work done on the system. At constant pressure with only PV work, q_p = ΔH.

ENTHALPY CHANGE AT CONSTANT PRESSURE

ΔH = q_p

At constant pressure, the heat exchanged (q_p) equals the enthalpy change. If q_p < 0, the process is exothermic; if q_p > 0, the process is endothermic.

CALORIMETRY EQUATION

q = m × c × ΔT

q = heat gained or lost (J), m = mass of substance (g), c = specific heat capacity (J·g⁻¹·°C⁻¹), ΔT = T_final − T_initial (°C). For the surroundings (solution in a coffee-cup calorimeter), a positive ΔT indicates the reaction is exothermic (q_rxn < 0).

STANDARD ENTHALPY OF REACTION (HESS'S LAW APPLICATION)

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)

ΔH°_f = standard enthalpy of formation (kJ/mol). Each ΔH°_f is multiplied by its stoichiometric coefficient. By convention, ΔH°_f of any element in its standard state is exactly 0.

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Sign Convention Pitfall

A common AP Chemistry error is confusing q_solution with q_rxn. In a coffee-cup calorimeter, q_rxn = −q_solution. If the solution temperature increases, the solution absorbed heat (q_solution > 0), which means the reaction released heat (q_rxn < 0, exothermic). Always include the negative sign when converting.

SECTION 5

Classifying Common Processes

In practice, the AP Chemistry exam expects you to classify a variety of chemical and physical processes as endothermic or exothermic, often without performing a calculation. The key organizing principle at the molecular level is straightforward: processes that result in the net formation of stronger or more numerous bonds tend to be exothermic, while those that require net bond-breaking or a transition to a higher-energy phase tend to be endothermic. The following table and diagram consolidate the most commonly tested examples, organized by process type.

Process	Exothermic or Endothermic	Sign of ΔH	Molecular Rationale
Combustion (e.g., CH₄ + 2O₂ → CO₂ + 2H₂O)	Exothermic	ΔH < 0	Product bonds (C=O, O–H) are stronger than reactant bonds (C–H, O=O); net energy is released.
Neutralization (strong acid + strong base)	Exothermic	ΔH ≈ −57.1 kJ/mol	Formation of O–H bonds in water from H⁺ and OH⁻ releases energy.
Freezing / Condensation	Exothermic	ΔH < 0	Intermolecular attractions are established or strengthened; system moves to a lower energy state.
Melting / Vaporization	Endothermic	ΔH > 0	Energy is required to overcome intermolecular forces to move to a higher-energy phase.
Photosynthesis (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂)	Endothermic	ΔH > 0	Solar energy is absorbed to drive the formation of C–C and C–H bonds from thermodynamically stable CO₂ and H₂O.
Dissolving NH₄NO₃ in water	Endothermic	ΔH > 0	Lattice energy exceeds hydration energy; more energy is consumed separating ions than is released solvating them.

Common endothermic and exothermic processes tested on the AP Chemistry exam

Phase change energy flow diagram showing that transitions to higher-energy phases (solid → liquid → gas) are endothermic, while transitions to lower-energy phases (gas → liquid → solid) are exothermic. Sublimation and deposition are the direct solid–gas transitions that bypass the liquid phase.

A useful mnemonic for phase changes: moving "upward" on the energy ladder (solid → liquid → gas) always requires energy input and is endothermic, while moving "downward" (gas → liquid → solid) releases energy and is exothermic. For dissolution processes, the classification depends on the relative magnitudes of lattice energy (energy needed to separate ions in the solid) and hydration energy (energy released when ions are solvated by water molecules). When hydration energy exceeds lattice energy, dissolution is exothermic—as seen with NaOH in water. When lattice energy dominates, as with NH₄NO₃, the dissolution is endothermic, which is why cold packs use ammonium nitrate.

SECTION 6

Worked Example: Coffee-Cup Calorimetry

Consider the following problem, which mirrors the type of calorimetry calculation frequently tested on the AP Chemistry exam. When 50.0 mL of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M HCl in a coffee-cup calorimeter, the temperature of the combined solution rises from 22.0 °C to 28.9 °C. Assume the solution has the density of water (1.00 g/mL), the specific heat capacity of water (4.18 J·g⁻¹·°C⁻¹), and that no heat is lost to the calorimeter. Calculate the enthalpy of neutralization per mole of water formed and classify the process.

Coffee-Cup Calorimetry: HCl + NaOH Neutralization

1
Step 1 — Identify Given Values

Total volume of solution = 50.0 mL + 50.0 mL = 100.0 mL. Since density = 1.00 g/mL, mass = 100.0 g. Specific heat c = 4.18 J·g⁻¹·°C⁻¹. ΔT = 28.9 °C − 22.0 °C = 6.9 °C. Moles of HCl = moles of NaOH = (0.0500 L)(1.00 mol/L) = 0.0500 mol.

m = 100.0 g, c = 4.18 J·g⁻¹·°C⁻¹, ΔT = 6.9 °C, n = 0.0500 mol
2
Step 2 — Calculate Heat Absorbed by Solution (q_solution)

Using q = m × c × ΔT: q_solution = (100.0 g)(4.18 J·g⁻¹·°C⁻¹)(6.9 °C) = 2884.2 J ≈ 2880 J = 2.88 kJ. The positive value indicates the solution absorbed heat from the reaction.

q_solution = +2.88 kJ
3
Step 3 — Determine q_rxn (Sign Reversal)

By conservation of energy in an insulated calorimeter, q_rxn = −q_solution = −2.88 kJ. The negative sign confirms that the reaction released heat to the surroundings (the solution).

q_rxn = −2.88 kJ
4
Step 4 — Calculate ΔH per Mole

ΔH = q_rxn / n = (−2.88 kJ) / (0.0500 mol) = −57.6 kJ/mol. This value is consistent with the accepted enthalpy of neutralization for a strong acid–strong base reaction (≈ −57.1 kJ/mol), validating our experimental data.

ΔH_neutralization ≈ −57.6 kJ/mol
5
Step 5 — Classify the Process

Since ΔH < 0, the neutralization of HCl by NaOH is an exothermic process. The temperature of the surroundings (the solution) increased, which is the observable hallmark of exothermicity in a calorimetry experiment.

Exothermic: ΔH < 0, T_surroundings increases

Temperature measurement during neutralization A temperature probe tracks the surroundings’ temperature during a reaction in a calorimeter.

SECTION 7

Exothermic vs. Endothermic: Side-by-Side Comparison

A direct comparison of exothermic and endothermic processes clarifies the symmetry and contrasts between them. The table below consolidates every distinguishing feature—from sign conventions and observable indicators to molecular-level explanations and real-world examples—into a single reference that mirrors the level of detail expected on the AP exam.

Feature	Exothermic Process	Endothermic Process
Sign of ΔH	Negative (ΔH < 0)	Positive (ΔH > 0)
Energy flow direction	System → Surroundings	Surroundings → System
Temperature of surroundings	Increases	Decreases
Enthalpy diagram	Products lower than reactants	Products higher than reactants
Bond energy interpretation	Energy released by bond formation > energy consumed by bond breaking	Energy consumed by bond breaking > energy released by bond formation
Common examples	Combustion, neutralization, freezing, condensation, deposition	Photosynthesis, melting, vaporization, sublimation, dissolving NH₄NO₃
Everyday application	Hand warmers (Fe oxidation), natural gas heating	Instant cold packs (NH₄NO₃ dissolution), cooking an egg

Comprehensive comparison of exothermic and endothermic processes

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KEY TAKEAWAY

The exothermic/endothermic classification describes the enthalpy ledger of a single process, but it does not by itself determine spontaneity. Many endothermic processes—such as the dissolution of NH₄NO₃ or the melting of ice above 0 °C—occur spontaneously because a favorable entropy increase (ΔS > 0) compensates for the unfavorable enthalpy change. The full criterion for spontaneity is ΔG = ΔH − TΔS < 0, which you will encounter in the Gibbs Free Energy unit. Understanding enthalpy is therefore a necessary but not sufficient condition for predicting reaction direction.

SECTION 8

Connection to Gibbs Free Energy & Entropy

Classifying a process as endothermic or exothermic provides crucial but incomplete thermodynamic information. On the AP Chemistry exam, you will be expected to integrate enthalpy data with entropy considerations to determine whether a reaction is spontaneous under a given set of conditions. The bridge between enthalpy and spontaneity is the Gibbs free energy equation: ΔG = ΔH − TΔS. When ΔG < 0, the process is thermodynamically favorable (spontaneous) under those conditions; when ΔG > 0, the reverse process is favored.

ΔH	ΔS	Spontaneity (ΔG < 0?)	Example
Negative (exothermic)	Positive	Spontaneous at all temperatures	Combustion of hydrocarbons
Negative (exothermic)	Negative	Spontaneous only at low T	Freezing of water below 0 °C
Positive (endothermic)	Positive	Spontaneous only at high T	Vaporization of water above 100 °C
Positive (endothermic)	Negative	Non-spontaneous at all temperatures	Decomposition of CaCO₃ at constant T (reverse is spontaneous)

Four sign combinations of ΔH and ΔS and their implications for spontaneity

This table reveals why the early hypothesis of Berthelot and Thomsen—that all spontaneous reactions must be exothermic—was incorrect. The entropy term (TΔS) can overcome an unfavorable enthalpy, particularly at high temperatures. When preparing for the AP exam, remember that enthalpy tells you about the energy preference of a process, entropy tells you about the disorder preference, and Gibbs free energy reconciles both into a single criterion for spontaneity. Mastering the sign of ΔH is therefore the essential first step toward the more complete thermodynamic analysis you will perform in later units.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student dissolves a salt in water in a coffee-cup calorimeter and observes that the temperature of the solution decreases by 8.2 °C. Which of the following correctly describes the dissolution process?

A. The process is exothermic because the surroundings lost heat.
B. The process is endothermic because the system absorbed heat from the surroundings, causing the solution temperature to decrease.
C. The process is exothermic because ΔH is positive when temperature decreases.
D. The process is endothermic because the lattice energy is less than the hydration energy.

Show answer (B)

The correct answer is B. When the solution temperature decreases, it means the surroundings (the solution) lost energy to the system (the dissolving salt). The system absorbed heat (q_rxn > 0), making ΔH > 0, which defines an endothermic process. Choice A incorrectly labels the process as exothermic despite the temperature decrease. Choice C confuses the sign convention—a positive ΔH corresponds to endothermic, not exothermic. Choice D has the energetic comparison backwards: in an endothermic dissolution, the lattice energy exceeds the hydration energy, meaning more energy is needed to separate the ions than is released upon solvation.

Question 2: BASIC CALCULATION

When 3.50 g of NH₄NO₃ (molar mass = 80.04 g/mol) dissolves in 50.0 g of water in a calorimeter, the temperature drops from 25.0 °C to 21.2 °C. The specific heat of the solution is 4.18 J·g⁻¹·°C⁻¹. What is the molar enthalpy of dissolution (ΔH_diss) of NH₄NO₃? Which of the following is closest to the correct value?

A. −19.4 kJ/mol
B. +12.8 kJ/mol
C. +19.4 kJ/mol
D. −12.8 kJ/mol

Show answer (C)

The correct answer is C. When NH₄NO₃ dissolves in water, the temperature decreases, indicating an endothermic process (ΔH > 0). First, calculate the heat absorbed by the reaction. The total mass of the solution is 3.50 g + 50.0 g = 53.5 g. The temperature change is ΔT = 21.2 − 25.0 = −3.8 °C. The heat gained by the solution is q_solution = mcΔT = (53.5 g)(4.18 J·g⁻¹·°C⁻¹)(−3.8 °C) = −849.6 J. Since q_rxn = −q_solution, q_rxn = +849.6 J = +0.8496 kJ. The moles of NH₄NO₃ = 3.50 g / 80.04 g/mol = 0.04372 mol. Therefore, ΔH_diss = +0.8496 kJ / 0.04372 mol = +19.4 kJ/mol ≈ +19.4 kJ/mol. Among the choices, +19.4 kJ/mol is the correct value. Choice A (−19.4 kJ/mol) is incorrect because the negative sign would indicate an exothermic process, which is inconsistent with the observed temperature drop. Choice B (+12.8 kJ/mol) is too low. Choice D (−12.8 kJ/mol) is both the wrong sign and the wrong magnitude.

Question 3: INTERMEDIATE

Given the following standard enthalpies of formation: ΔH°f[CO₂(g)] = −393.5 kJ/mol, ΔH°f[H₂O(l)] = −285.8 kJ/mol, and ΔH°f[C₂H₆(g)] = −84.7 kJ/mol, calculate the standard enthalpy of combustion of ethane: C₂H₆(g) + 7⁄2 O₂(g) → 2CO₂(g) + 3H₂O(l). Classify the reaction and explain the molecular basis for the sign of ΔH.

A. −1559.7 kJ/mol; exothermic because stronger C=O and O−H bonds form in the products
B. +1559.7 kJ/mol; endothermic because all combustion reactions absorb heat
C. −1644.4 kJ/mol; exothermic because CO₂ and H₂O are very stable
D. −702.3 kJ/mol; exothermic because bond-breaking in ethane releases energy

Show answer (A)

The correct answer is A. Using ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants): ΔH°rxn = [2(−393.5) + 3(−285.8)] − [(−84.7) + 7⁄2(0)] = [−787.0 + (−857.4)] − [−84.7] = −1644.4 − (−84.7) = −1644.4 + 84.7 = −1559.7 kJ/mol. The large negative value confirms this is strongly exothermic: the C=O bonds in CO₂ and O−H bonds in H₂O that form in the products are collectively much stronger than the C−H, C−C, and O=O bonds broken in the reactants, resulting in a net release of energy. Choice B is wrong because combustion is always exothermic, not endothermic. Choice C has an arithmetic error (fails to subtract the enthalpy of formation of ethane). Choice D has an incorrect magnitude and a conceptual error: bond-breaking requires energy input, not release.

Question 4: APPLIED

10 pts

A chemical engineer is designing an industrial cold pack that must lower the temperature of 200.0 g of water from 25.0 °C to 5.0 °C. The engineer plans to use NH₄NO₃, which has a molar enthalpy of dissolution of +25.7 kJ/mol and a molar mass of 80.04 g/mol. (a) Calculate the mass of NH₄NO₃ required. (b) Explain why this process is endothermic at the molecular level, referencing lattice energy and hydration energy. (c) Would increasing the initial water temperature affect the mass of NH₄NO₃ needed? Justify your answer. (d) The engineer considers using CaCl₂ (ΔH_diss = −81.3 kJ/mol) instead. Explain whether CaCl₂ would be suitable for a cold pack and why.

Show answer

Rubric (10 points): (a) Calculation of mass of NH₄NO₃ (4 points): • 1 pt: Calculate q_water = mcΔT = (200.0 g)(4.18 J·g⁻¹·°C⁻¹)(5.0 − 25.0 °C) = −16,720 J = −16.72 kJ • 1 pt: Recognize q_rxn = −q_water = +16.72 kJ (positive because the dissolution absorbs heat) • 1 pt: Calculate moles: n = q_rxn / ΔH_diss = 16.72 kJ / 25.7 kJ/mol = 0.651 mol • 1 pt: Calculate mass: m = (0.651 mol)(80.04 g/mol) = 52.1 g (b) Molecular-level explanation (2 points): • 1 pt: State that lattice energy (energy required to separate NH₄⁺ and NO₃⁻ ions from the crystal lattice) exceeds the hydration energy (energy released when ions are solvated by water dipoles). • 1 pt: Explain that because more energy is consumed in breaking ion-ion attractions than is released in forming ion-dipole interactions, the net process absorbs energy from the surroundings. (c) Effect of initial water temperature (2 points): • 1 pt: Yes, if the initial temperature is higher, ΔT increases (larger temperature drop needed to reach 5.0 °C), so more heat must be absorbed. • 1 pt: Since q = mcΔT increases, more moles (and thus more mass) of NH₄NO₃ are required. Note: this assumes c and ΔH_diss are approximately constant over the temperature range. (d) Suitability of CaCl₂ (2 points): • 1 pt: CaCl₂ would NOT be suitable for a cold pack because its dissolution is exothermic (ΔH_diss = −81.3 kJ/mol), meaning it releases heat and would warm the water. • 1 pt: CaCl₂ is instead used in hot packs. A cold pack requires an endothermic dissolution process (ΔH > 0) to absorb heat from the surroundings and lower the temperature.

Question 5: CRITICAL THINKING

10 pts

A student performs two calorimetry experiments to determine whether dissolving NaOH and NH₄Cl in water are endothermic or exothermic. The data are summarized below. Experiment 1 (NaOH): 4.00 g NaOH (M = 40.00 g/mol) dissolved in 100.0 g water. T_initial = 23.0 °C, T_final = 33.4 °C. Experiment 2 (NH₄Cl): 5.35 g NH₄Cl (M = 53.49 g/mol) dissolved in 100.0 g water. T_initial = 23.0 °C, T_final = 20.1 °C. Assume the specific heat capacity of each solution is 4.18 J·g⁻¹·°C⁻¹ and that the calorimeter absorbs no heat. (a) For each experiment, calculate q_solution, q_rxn, and ΔH_diss per mole of solute. Show all work clearly. (b) Classify each dissolution as endothermic or exothermic. Justify each classification using both the sign of ΔH_diss and the observed temperature change. (c) The student hypothesizes that all ionic compounds dissolve exothermically. Using the experimental data and your understanding of lattice energy versus hydration energy, evaluate this hypothesis. (d) Identify one source of systematic error in a coffee-cup calorimetry experiment and explain whether it would cause the calculated |ΔH| to be too high or too low.

Show answer

Rubric (10 points): (a) Calculations (5 points): Experiment 1 (NaOH): • 1 pt: Correct calculation of q_solution: mass of solution = 100.0 g + 4.00 g = 104.0 g; ΔT = 33.4 − 23.0 = 10.4 °C; q_solution = (104.0 g)(4.18 J·g⁻¹·°C⁻¹)(10.4 °C) = 4,521 J = +4.52 kJ. • 1 pt: Correct calculation of q_rxn and ΔH_diss for NaOH: q_rxn = −q_solution = −4.52 kJ; moles of NaOH = 4.00 g / 40.00 g·mol⁻¹ = 0.100 mol; ΔH_diss = −4.52 kJ / 0.100 mol = −45.2 kJ/mol. Experiment 2 (NH₄Cl): • 1 pt: Correct calculation of q_solution: mass of solution = 100.0 g + 5.35 g = 105.35 g; ΔT = 20.1 − 23.0 = −2.9 °C; q_solution = (105.35 g)(4.18 J·g⁻¹·°C⁻¹)(−2.9 °C) = −1,277 J = −1.28 kJ. • 1 pt: Correct calculation of q_rxn and ΔH_diss for NH₄Cl: q_rxn = −q_solution = +1.28 kJ; moles of NH₄Cl = 5.35 g / 53.49 g·mol⁻¹ = 0.1000 mol; ΔH_diss = +1.28 kJ / 0.100 mol = +12.8 kJ/mol. • 1 pt: Correct signs and units reported consistently for q_solution, q_rxn, and ΔH_diss in both experiments. (b) Classification (2 points): • 1 pt: NaOH dissolution is exothermic because ΔH_diss is negative (−45.2 kJ/mol) and the solution temperature increased from 23.0 °C to 33.4 °C, indicating the dissolving process released heat to the surroundings (the solution). • 1 pt: NH₄Cl dissolution is endothermic because ΔH_diss is positive (+12.8 kJ/mol) and the solution temperature decreased from 23.0 °C to 20.1 °C, indicating the dissolving process absorbed heat from the surroundings (the solution). (c) Evaluation of hypothesis (2 points): • 1 pt: The hypothesis is incorrect. The experimental data show that NaOH dissolves exothermically (ΔH = −45.2 kJ/mol) while NH₄Cl dissolves endothermically (ΔH = +12.8 kJ/mol), directly contradicting the claim that all ionic compounds dissolve exothermically. • 1 pt: Whether dissolution is exothermic or endothermic depends on the relative magnitudes of the lattice energy (energy required to separate ions in the solid) and the hydration energy (energy released when gaseous ions are solvated by water). For NaOH, the combined hydration energies of Na⁺ and OH⁻ exceed the lattice energy, so the overall process releases energy and is exothermic. For NH₄Cl, the lattice energy is large relative to the combined hydration energies of NH₄⁺ and Cl⁻, so the overall process absorbs energy and is endothermic. (d) Source of systematic error (1 point): • 1 pt: Heat exchange between the calorimeter and the surrounding environment is a significant source of systematic error. In an exothermic reaction, the solution is warmer than the surroundings, so heat is lost through the walls and lid of the coffee-cup calorimeter, making the measured temperature change smaller than the true value and causing the calculated |ΔH| to be underestimated (too low). In an endothermic reaction, the solution is cooler than the surroundings, so heat flows into the calorimeter from the environment, reducing the observed temperature drop and again causing the calculated |ΔH| to be underestimated (too low). In both cases, imperfect insulation leads to a systematic underestimate of the magnitude of the enthalpy of dissolution. Total: 5 + 2 + 2 + 1 = 10 points.

SUMMARY

Summary: Endothermic and Exothermic Processes

Summary

Every chemical and physical change involves energy transfer between the system and its surroundings. An exothermic process releases heat to the surroundings (ΔH < 0, surroundings warm up), while an endothermic process absorbs heat from the surroundings (ΔH > 0, surroundings cool down). At the molecular level, the net balance of bond-breaking (energy input) and bond-forming (energy output) determines the sign of ΔH.

Quantitatively, calorimetry (q = mcΔT) measures heat flow via temperature changes, and the critical sign reversal q_rxn = −q_solution connects the observable surroundings measurement to the system's enthalpy change. Standard enthalpies of formation allow prediction of ΔH°_rxn without performing experiments, via Hess's Law. Remember: enthalpy alone does not determine spontaneity—integration with entropy (ΔS) through the Gibbs free energy equation (ΔG = ΔH − TΔS) provides the complete thermodynamic picture.