Chapter 7: Thermochemistry

Before you can track energy in a chemical reaction, you need to define what you are tracking. In thermodynamics, we split the universe into two parts: the system (the specific reaction or process you are studying) and the surroundings (everything else). The boundary between them determines what can cross - energy, matter, or neither.

The Three System Types

Open system: Both energy and matter can cross the boundary. A boiling pot of water without a lid is an open system. Heat enters from the stove (energy in), and steam escapes into the air (matter out). Most biological systems are open - your body constantly exchanges heat and chemicals with the environment.

Closed system: Energy can cross the boundary, but matter cannot. A sealed pressure cooker on a stove is a closed system. Heat flows in through the walls, but the water and steam stay trapped inside. Most chemical reactions studied on the MCAT take place in closed systems.

Isolated system: Neither energy nor matter can cross the boundary. A perfect thermos is the closest everyday example. In reality, truly isolated systems do not exist (every thermos eventually leaks heat), but the concept is useful for theoretical calculations. The entire universe is sometimes treated as an isolated system because there is nothing outside it to exchange with.

Why System Classification Matters

The type of system you are working with determines which thermodynamic quantities you measure. In a constant-pressure system (like a coffee cup calorimeter open to the atmosphere), the heat flow equals the enthalpy change (q = ΔH). In a constant-volume system (like a sealed bomb calorimeter), the heat flow equals the internal energy change (q = ΔU). You will see these distinctions again in Section 7.4 on calorimetry.

The First Law of Thermodynamics

The first law states that energy cannot be created or destroyed, only transferred or converted between forms. Mathematically:

Energy enters a system as heat (q) or work (w). If you add heat to a gas and it does not expand, all that energy increases the internal energy (temperature goes up). If the gas expands against external pressure, some energy goes into doing work, and the internal energy increases by less.

Sign Conventions

Getting signs right is critical on the MCAT. The convention used in most chemistry contexts:

Imagine you are hiking from a trailhead at 1,000 feet elevation to a summit at 5,000 feet. Your elevation change is always 4,000 feet, whether you take the steep direct trail or the winding scenic route. But the distance you walk and the energy you burn depend entirely on which path you choose.

State Functions: Only Start and Finish Matter

A state function depends only on the current state of the system - its temperature, pressure, volume, and composition - not on how it got there. The change in a state function is calculated as:

ΔX = X(final) - X(initial)

The key state functions you need for the MCAT:

Path Functions: The Route Matters

A path function depends on how the process is carried out. Heat (q) and work (w) are path functions. You can transfer different amounts of heat and work to get between the same two states, depending on the process.

Consider heating water from 25 C to 100 C. You could:

• Heat it slowly at constant pressure (one amount of q and w)
• Compress it first, then heat it, then expand it (different q and w)

In both cases, ΔH, ΔU, ΔS, and ΔG are identical because the initial and final states are the same. But q and w differ because the path was different.

Why This Matters for the MCAT

The state function concept is the foundation for Hess’s law (Section 7.5). Because enthalpy is a state function, you can break a complex reaction into simpler steps, calculate ΔH for each step, and add them up. The total ΔH is the same regardless of which steps you use. This is enormously powerful for calculations.

Standard Conditions

When comparing thermodynamic values across different reactions, we need a common reference point. Standard conditions are defined as:

• Temperature: 25 C (298 K)
• Pressure: 1 atm (or 1 bar in newer conventions)
• Concentration: 1 M for solutions

Values measured under these conditions get the degree symbol: ΔH°, ΔS°, ΔG°. Do not confuse standard conditions (25 C, 1 atm) with STP (0 C, 1 atm), which is used for gas law calculations. They are different reference points.

Every chemical reaction involves energy. Some reactions release energy and feel hot to the touch. Others absorb energy and feel cold. Enthalpy is the thermodynamic quantity that tracks this heat flow at constant pressure - which is how most reactions in your body and in the lab actually occur.

What Is Enthalpy?

Enthalpy (H) is defined as:

In practice, you almost never calculate H directly. Instead, you work with the change in enthalpy:

Δ$H_{\text{rxn}}$ = H(products) - H(reactants)

Exothermic vs. Endothermic

Enthalpy Diagrams

An enthalpy diagram places reactants and products on a vertical energy axis. For an exothermic reaction, the products sit lower than the reactants - energy was released going “downhill.” For an endothermic reaction, the products sit higher - energy was absorbed going “uphill.”

The vertical distance between reactants and products is |ΔH|. The direction tells you the sign.

Enthalpy and the PV Term

The full definition H = U + PV includes a pressure-volume term. For reactions involving only solids and liquids, volume changes are tiny, so ΔH is almost equal to ΔU. But for reactions involving gases, the PV term matters because gases expand or compress significantly.

For an ideal gas at constant temperature: PV = nRT

So ΔH = ΔU + Δ(nRT) = ΔU + ΔnRT

where Δn = moles of gaseous products - moles of gaseous reactants. If a reaction produces more moles of gas than it consumes (Δn > 0), ΔH > ΔU. If it produces fewer moles of gas (Δn < 0), ΔH < ΔU.

Reversibility of Enthalpy

Because enthalpy is a state function, the enthalpy change for the reverse reaction has the same magnitude but opposite sign:

• Forward: H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l), ΔH = -285.8 kJ
• Reverse: H₂O(l) → H₂(g) + $\frac{1}{2}$ O₂(g), ΔH = +285.8 kJ

If it costs 285.8 kJ to decompose water, then forming water releases exactly 285.8 kJ. This principle is essential for Hess’s law calculations.

Stoichiometry and Enthalpy

ΔH values are reported per mole of reaction as written. If you double all the coefficients, you double ΔH:

• H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l), ΔH = -285.8 kJ
• 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH = -571.6 kJ

Always check the stoichiometry when using ΔH values in calculations.

You know that mixing certain chemicals makes the solution hot or cold. But how do scientists actually measure the exact amount of heat released or absorbed? The answer is calorimetry - a technique that captures heat flow by measuring temperature changes in a known mass of water (or another substance).

Specific Heat and Heat Capacity

Before we can do calorimetry calculations, you need two definitions:

Specific heat (c): The amount of energy needed to raise the temperature of 1 gram of a substance by 1 C. Water’s specific heat is 4.18 J/(g·C) - one of the highest of any common substance. This is why water heats up and cools down slowly compared to metals.

Heat capacity (C): The total amount of energy needed to raise the temperature of a specific object by 1 C. Heat capacity = mass × specific heat (C = mc).

Coffee Cup Calorimeter (Constant Pressure)

The coffee cup calorimeter is simply an insulated cup (usually Styrofoam) with a thermometer. The reaction occurs in aqueous solution inside the cup, and since the cup is open to the atmosphere, the pressure is constant.

Key point: At constant pressure, q = ΔH. So the coffee cup calorimeter directly measures the enthalpy change of the reaction.

How it works:

1. Mix reactants in the calorimeter
2. Measure the temperature change of the solution
3. Calculate q using q = mcΔT (using the mass and specific heat of the solution)
4. The heat gained by the solution equals the heat released by the reaction (or vice versa): q(rxn) = -q(solution)

Bomb Calorimeter (Constant Volume)

The bomb calorimeter is a sealed, rigid steel container (“the bomb”) immersed in a known mass of water. The sample is placed inside the bomb with excess oxygen and ignited electrically.

Key point: Because the volume cannot change, no PV work is done (w = 0). All energy goes into heat. At constant volume, q = ΔU (not ΔH).

The temperature of the surrounding water rises, and we calculate:

q = C(calorimeter) × ΔT

where C(calorimeter) is the total heat capacity of the calorimeter (including the water and the bomb itself), often given in the problem in units of kJ/C.

Coffee Cup vs. Bomb: Summary

Assumptions in Calorimetry

For MCAT problems, you can usually assume:

• The calorimeter is perfectly insulated (no heat loss to the room)
• The solution has the density and specific heat of pure water [1 g/mL, 4.18 J/(g·C)]
• All heat from the reaction is absorbed by the solution (coffee cup) or by the calorimeter (bomb)

Worked Example

A student dissolves 5.0 g of NaOH (molar mass 40 g/mol) in 100 mL of water in a coffee cup calorimeter. The temperature rises from 22.0 C to 28.5 C. What is the molar enthalpy of dissolution?

1. q(solution) = mcΔT = (100 g)(4.18 J/g·C)(6.5 C) = 2,717 J = 2.72 kJ
2. q(rxn) = -q(solution) = -2.72 kJ (exothermic - temperature went up)
3. Moles NaOH = 5.0 g / 40 g/mol = 0.125 mol
4. ΔH = q(rxn) / moles = -2.72 kJ / 0.125 mol = -21.7 kJ/mol

Some reactions are impossible to measure directly. You cannot stick a thermometer into the formation of diamond from graphite because that reaction requires extreme pressure and takes geological time. But you CAN measure the combustion of graphite and the combustion of diamond separately. Hess’s law lets you combine those measurable reactions to calculate the enthalpy change for the unmeasurable one.

The Principle

The Three Manipulation Rules

When combining reactions to reach a target, you have three tools:

Rule 1 - Reverse: If you flip a reaction (swap products and reactants), change the sign of ΔH.

• Forward: C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ
• Reverse: CO₂(g) → C(s) + O₂(g), ΔH = +393.5 kJ

Rule 2 - Multiply: If you multiply all coefficients by a factor n, multiply ΔH by the same factor.

• Original: H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l), ΔH = -285.8 kJ
• Doubled: 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH = -571.6 kJ

Rule 3 - Add: Stack the manipulated reactions and add them. Species that appear on both sides cancel out (like algebra). Add the ΔH values.

Worked Example

Target reaction: C(s) + O₂(g) → CO₂(g), ΔH = ?

Given:

1. C(s) + $\frac{1}{2}$ O₂(g) → CO(g), ΔH₁ = -110.5 kJ
2. CO(g) + $\frac{1}{2}$ O₂(g) → CO₂(g), ΔH₂ = -283.0 kJ

Solution: Both reactions are already in the correct direction. Add them:

C(s) + $\frac{1}{2}$ O₂(g) → CO(g)
CO(g) + $\frac{1}{2}$ O₂(g) → CO₂(g)

CO(g) appears on both sides and cancels. The $\frac{1}{2}$ O₂ terms combine: $\frac{1}{2}$ + $\frac{1}{2}$ = 1 O₂.

Result: C(s) + O₂(g) → CO₂(g), ΔH = -110.5 + (-283.0) = -393.5 kJ

Strategy for Hess’s Law Problems

1. Write the target reaction clearly
2. Scan the given reactions - identify which ones contain your target reactants and products
3. Reverse or multiply reactions as needed so intermediates cancel
4. Add everything up, including the ΔH values
5. Verify: check that the final equation matches the target

Hess’s law is powerful, but manipulating multiple reactions can be tedious. There is a shortcut: if you know the standard enthalpy of formation (ΔHf°) for every compound in a reaction, you can calculate ΔH° in a single step using a simple “products minus reactants” formula.

What Is a Formation Reaction?

A formation reaction produces exactly one mole of a compound from its elements in their standard states (the most stable form at 25 C and 1 atm).

Examples:

• C(graphite) + O₂(g) → CO₂(g), ΔHf° = -393.5 kJ/mol
• H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l), ΔHf° = -285.8 kJ/mol
• $\frac{1}{2}$ N₂(g) + $\frac{1}{2}$ O₂(g) → NO(g), ΔHf° = +90.3 kJ/mol

The Critical Rule: Elements Have ΔHf° = 0

Elements in their standard states are the reference point. Their ΔHf° is defined as zero:

• ΔHf°[O₂(g)] = 0
• ΔHf°[N₂(g)] = 0
• ΔHf°[C(graphite)] = 0
• ΔHf°[Fe(s)] = 0

This makes sense: you do not need to “form” an element from itself.

Calculating ΔH°rxn from ΔHf° Values

Worked Example

Calculate ΔH° for the combustion of methane:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

Given ΔHf° values (kJ/mol): CH₄(g) = -74.8, O₂(g) = 0, CO₂(g) = -393.5, H₂O(l) = -285.8

Products: (1)(-393.5) + (2)(-285.8) = -393.5 + (-571.6) = -965.1 kJ

Reactants: (1)(-74.8) + (2)(0) = -74.8 kJ

ΔH°rxn = -965.1 - (-74.8) = -965.1 + 74.8 = -890.3 kJ

The reaction is strongly exothermic, which makes sense - methane combustion powers your stove.

Standard Enthalpy of Combustion

The standard enthalpy of combustion (ΔH°comb) is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. Combustion reactions are always exothermic (ΔH°comb < 0).

Common values:

• Methane (CH₄): -890 kJ/mol
• Glucose (C₆H₁₂O₆): -2,803 kJ/mol
• Ethanol (C₂H₅OH): -1,367 kJ/mol

What if you do not have ΔHf° values for the compounds in your reaction? There is another way to estimate ΔH: use the energies of the individual bonds being broken and formed. This approach is less precise than using ΔHf° (because bond energies are averages), but it gives you a fast estimate - and the MCAT tests it regularly.

The Core Principle

Every chemical bond is a store of energy. Breaking a bond requires energy input (endothermic). Forming a bond releases energy (exothermic). The overall enthalpy change of a reaction depends on the balance between these two processes.

The Bond Energy Formula

Important: This formula uses the convention that bond dissociation energies (BDEs) are always positive numbers (the energy required to break). The subtraction handles the sign:

• If you form stronger bonds than you break → ΔH < 0 (exothermic)
• If you break stronger bonds than you form → ΔH > 0 (endothermic)

Common Bond Energies

You do not need to memorize these - the MCAT will provide them in a table. But knowing the trends helps:

Key trend: Triple bonds > double bonds > single bonds in energy. Stronger bonds = more stable molecules = more energy released when formed.

Worked Example

Estimate ΔH for the combustion of methane: CH₄ + 2 O₂ → CO₂ + 2 H₂O

Bonds broken (reactants):

• 4 × C-H = 4 × 413 = 1,652 kJ
• 2 × O=O = 2 × 498 = 996 kJ
• Total broken = 2,648 kJ

Bonds formed (products):

• 2 × C=O (in CO₂) = 2 × 799 = 1,598 kJ
• 4 × O-H (in 2 H₂O) = 4 × 463 = 1,852 kJ
• Total formed = 3,450 kJ

ΔH ≈ 2,648 - 3,450 = -802 kJ

The actual value is -890 kJ. The estimate is off by about 10% because bond energies are averages across many molecules. For the MCAT, this level of accuracy is expected and acceptable.

Why Combustion Is Always Exothermic

In combustion, you break relatively weak C-H, C-C, and O=O bonds and form very strong C=O and O-H bonds. The bonds formed are stronger than the bonds broken, so energy is released. This is why all combustion reactions are exothermic - the products (CO₂ and H₂O) contain some of the strongest bonds in chemistry.

Enthalpy tells you about heat flow, but it does not tell the whole story of whether a reaction will happen on its own. Some endothermic reactions DO happen spontaneously - ice melts at room temperature even though it absorbs heat. The missing piece of the puzzle is entropy.

What Is Entropy?

Entropy (S) is a measure of the disorder or randomness in a system. More precisely, it counts the number of microstates - the different ways particles can be arranged while still looking the same from the outside.

A neatly stacked deck of cards (one specific arrangement) has low entropy. A shuffled deck (one of trillions of possible arrangements) has high entropy. Nature overwhelmingly favors shuffled decks because there are so many more disordered arrangements than ordered ones.

The Second Law of Thermodynamics

The Third Law of Thermodynamics

The third law states that the entropy of a perfect crystal at absolute zero (0 K) is exactly zero. This provides an absolute reference point for entropy - unlike enthalpy, where we can only measure changes. This is why elements in their standard states do NOT have S° = 0 (unlike ΔHf° = 0). Every substance at temperatures above 0 K has some molecular motion and therefore some entropy.

Predicting Entropy Changes

You can often predict the sign of ΔS without any calculation by asking: “Did things become more or less disordered?”

ΔS > 0 (entropy increases) when:

• Solids melt into liquids
• Liquids vaporize into gases
• A reaction produces more moles of gas
• A solid dissolves in a solvent
• Temperature increases

ΔS < 0 (entropy decreases) when:

• Gases condense into liquids
• Liquids freeze into solids
• A reaction produces fewer moles of gas
• Molecules combine into a larger molecule
• A gas dissolves in a liquid

Entropy Ranking by Phase

S(gas) >> S(liquid) > S(solid)

A gas at any temperature has enormously more entropy than the same substance as a liquid, which has more entropy than the solid. Phase changes are the biggest drivers of entropy change.

Calculating ΔS° for a Reaction

Just like enthalpy, you can calculate the standard entropy change using tabulated values:

Entropy and Temperature

The relationship between entropy change and heat transfer at a given temperature is:

ΔS = q(rev) / T

where q(rev) is the heat transferred in a reversible process and T is the absolute temperature in Kelvin. This equation tells you two things:

1. Adding heat to a system increases its entropy (q > 0 → ΔS > 0)
2. The same amount of heat causes a LARGER entropy change at low temperatures than at high temperatures

You now know two pieces of the puzzle: enthalpy (does the reaction release heat?) and entropy (does the reaction increase disorder?). The entropy concepts from physics thermodynamics apply directly here. But neither alone tells you whether a reaction will happen spontaneously. An exothermic reaction with a large entropy decrease might not be spontaneous. An endothermic reaction with a large entropy increase might be. You need one equation that combines both factors to give you the final answer.

The Master Equation

Unit Warning

Watch units carefully. ΔH is usually given in kJ/mol, but ΔS is usually given in J/(mol·K). Before plugging into ΔG = ΔH - TΔS, convert them to the same units. Either divide ΔS by 1,000 to get kJ/(mol·K), or multiply ΔH by 1,000 to get J/mol.

The Four Scenarios

This is one of the most tested concepts in MCAT thermochemistry. Depending on the signs of ΔH and ΔS, there are four possible combinations:

The Crossover Temperature

For the two “depends on T” cases, you can find the exact temperature where spontaneity switches by setting $\Delta G = 0$:

$0 = \Delta H - T\Delta S \quad \rightarrow \quad T = \dfrac{\Delta H}{\Delta S}$

Below this temperature, one factor dominates; above it, the other does. This crossover temperature is the phase transition temperature for phase changes (0 C for water freezing/melting at 1 atm, 100 C for boiling/condensing).

Exergonic vs. Endergonic

Do not confuse these with exothermic/endothermic. Exothermic/endothermic refer to ΔH (heat). Exergonic/endergonic refer to ΔG (free energy). A reaction can be exothermic but endergonic (if entropy decreases enough), or endothermic but exergonic (if entropy increases enough).

Spontaneity Does Not Mean Fast

A critical concept: ΔG tells you WHETHER a reaction is thermodynamically favorable, not HOW FAST it occurs. Diamond converting to graphite has ΔG < 0 (spontaneous), but it takes billions of years. The rate of a reaction depends on kinetics (activation energy, catalysts), not thermodynamics. ΔG says “will it?”; kinetics says “how fast?”

In the previous section, you learned that ΔG determines spontaneity. But which ΔG? There is a critical distinction between ΔG° (standard free energy change) and ΔG (actual free energy change under real conditions). Confusing the two is one of the most common mistakes on the MCAT.

ΔG° and the Equilibrium Constant

Interpreting ΔG° and K

The sign of ΔG° tells you which side of the reaction is favored at equilibrium:

ΔG Under Non-Standard Conditions

Real reactions rarely occur under standard conditions. The actual free energy change depends on the current concentrations via the reaction quotient Q:

How ΔG Changes as a Reaction Proceeds

When a reaction starts:

• If Q < K: ΔG < 0 (reaction proceeds forward to make more products)
• If Q > K: ΔG > 0 (reaction proceeds backward to make more reactants)
• If Q = K: ΔG = 0 (equilibrium - no net change)

As the reaction approaches equilibrium, ΔG approaches zero. At equilibrium, Q = K and:

ΔG = ΔG° + RTlnK = 0

This is exactly where the equation ΔG° = -RTlnK comes from - it is the ΔG = 0 condition rearranged.

Connecting Everything

Here is how the three big thermodynamic equations fit together:

1. ΔG = ΔH - TΔS (relates free energy to enthalpy and entropy)
2. ΔG° = -RTlnK (relates standard free energy to equilibrium)
3. ΔG = ΔG° + RTlnQ (relates actual free energy to current conditions)

These three equations are the thermodynamic backbone of the MCAT. Know them cold.

Temperature and K

Because ΔG° depends on temperature (through the TΔS° term in ΔG° = ΔH° - TΔS°), and K depends on ΔG° (through ΔG° = -RTlnK), the equilibrium constant changes with temperature.

For an exothermic reaction (ΔH° < 0): increasing temperature makes ΔG° less negative (or more positive), which decreases K. Heat shifts equilibrium toward reactants.

For an endothermic reaction (ΔH° > 0): increasing temperature makes ΔG° more negative, which increases K. Heat shifts equilibrium toward products.

This connects directly to Le Chatelier’s principle from Chapter 6: heat acts as a reactant (endothermic) or product (exothermic).

If a reaction has a positive ΔG (nonspontaneous), does that mean it can never happen? Not at all. Nature has an elegant solution: couple the nonspontaneous reaction with a highly spontaneous (exergonic) one. If the combined ΔG is negative, the overall process is spontaneous. This is how your body drives thousands of thermodynamically unfavorable reactions every second.

How Coupling Works

Two reactions can be coupled if they share a common intermediate and are catalyzed by the same enzyme (or occur in the same cellular pathway). The overall ΔG is simply the sum of the individual ΔG values:

ΔG(overall) = ΔG₁ + ΔG₂

If ΔG₁ is positive (nonspontaneous) and ΔG₂ is sufficiently negative (spontaneous), the sum can be negative, making the overall process spontaneous.

ATP: The Universal Energy Currency

The most common coupling partner in biology is ATP hydrolysis:

ATP + H₂O → ADP + Pi, ΔG° = -30.5 kJ/mol

This reaction is strongly exergonic because:

• The products (ADP + Pi) have less electrostatic repulsion than ATP (which has four negative charges clustered on its phosphate groups)
• The products are stabilized by resonance and hydration

Example: Glutamine Synthesis

The synthesis of glutamine from glutamate and ammonia is nonspontaneous:

Glutamate + NH₃ → Glutamine + H₂O, ΔG° = +14.2 kJ/mol

But when coupled with ATP hydrolysis:

Glutamate + NH₃ + ATP → Glutamine + ADP + Pi, ΔG° = +14.2 + (-30.5) = -16.3 kJ/mol

The coupled reaction is spontaneous. The enzyme glutamine synthetase catalyzes this coupled reaction in a single active site.

Beyond ATP

Other high-energy molecules can serve as coupling partners:

Key Principles for the MCAT

1. You can add ΔG values for coupled reactions because ΔG is a state function
2. The coupled ΔG must be negative for the overall process to be spontaneous
3. Enzymes facilitate coupling by bringing both reactions together in one active site, but they do NOT change the thermodynamics
4. ATP hydrolysis provides about -30.5 kJ/mol under standard conditions (and even more under cellular conditions, roughly -50 to -54 kJ/mol)

When you heat a substance, the temperature rises steadily - until you hit a phase change. Then something strange happens: you keep adding heat, but the temperature stops rising. All that energy goes into breaking intermolecular forces rather than speeding up molecules. Understanding this behavior is essential for interpreting heating curves on the MCAT.

Phase Change Terminology

Key rule: Breaking intermolecular forces requires energy (endothermic). Forming intermolecular forces releases energy (exothermic). Going from more ordered → less ordered absorbs heat.

The Heating Curve

A heating curve plots temperature (y-axis) vs. heat added (x-axis) for a substance being heated from solid to gas at constant pressure. It has five distinct regions:

1. Solid warming (sloped line): Temperature rises as heat increases the kinetic energy of molecules in the solid. q = mcΔT using c(solid).

2. Melting plateau (flat line at melting point): Temperature stays constant while the solid melts. All added heat goes into breaking intermolecular forces. q = n × ΔH(fus).

3. Liquid warming (sloped line): Temperature rises again. q = mcΔT using c(liquid).

4. Boiling plateau (flat line at boiling point): Temperature stays constant while the liquid vaporizes. q = n × ΔH(vap).

5. Gas warming (sloped line): Temperature rises. q = mcΔT using c(gas).

Key Formulas for Heating Curves

ΔH(vap) >> ΔH(fus)

The enthalpy of vaporization is always much larger than the enthalpy of fusion for the same substance. This is because vaporization completely separates molecules from each other (overcoming all remaining intermolecular forces), while melting only loosens the rigid crystal structure.

For water:

• ΔH(fus) = 6.01 kJ/mol (melting ice)
• ΔH(vap) = 40.7 kJ/mol (boiling water)

The boiling plateau on a heating curve is much wider than the melting plateau because more energy is needed.

Why the Boiling Plateau Matters

Calculating Total Heat for a Complete Heating Process

To find the total heat needed to convert ice at -20 C to steam at 120 C, you must add five separate terms:

1. Heat the ice from -20 C to 0 C: q₁ = mc(ice)ΔT
2. Melt the ice at 0 C: q₂ = nΔH(fus)
3. Heat the water from 0 C to 100 C: q₃ = mc(water)ΔT
4. Boil the water at 100 C: q₄ = nΔH(vap)
5. Heat the steam from 100 C to 120 C: q₅ = mc(steam)ΔT

q(total) = q₁ + q₂ + q₃ + q₄ + q₅

Entropy and Phase Changes

Phase changes also involve entropy changes:

• Melting: ΔS = ΔH(fus) / T(melting) > 0
• Boiling: ΔS = ΔH(vap) / T(boiling) > 0

At the phase transition temperature, ΔG = 0 (the two phases are in equilibrium). This is consistent with ΔG = ΔH - TΔS = 0, so T = ΔH/ΔS at the phase boundary.