Module 4: Thermochemistry

The Study of Energy Transfer in Chemical Systems

Module Goals: What You’ll Master

By the end of this module, you should be able to:

Distinguish between system and surroundings, and define energy flow direction (endothermic vs exothermic)
Interpret enthalpy (ΔH), entropy (ΔS), and Gibbs free energy (ΔG) in physical and biological contexts
Solve problems involving heat transfer, using the calorimetry equation q=mcΔTq = mc\Delta Tq=mcΔT
Analyze multistep reactions using Hess’s Law, and estimate energy changes from bond enthalpies
Predict whether a reaction is spontaneous, non-spontaneous, or temperature-dependent

How This Module Is Structured

This module is broken into five conceptual sections:

System vs. Surroundings
Understand what constitutes the system and surroundings, and how to define the direction of energy transfer.
State Functions: ΔH, ΔS, and ΔG
Explore the meaning of enthalpy, entropy, and free energy—and how they work together to determine spontaneity.
Calorimetry and Heat Capacity
Learn how temperature, specific heat, and mass determine heat flow, and how to use calorimeters to measure it.
Hess’s Law and Thermodynamic Cycles
Apply Hess’s Law to compute enthalpy changes from multistep reactions or formation reactions.
Bond Enthalpies and Reaction Energetics
Use bond dissociation energies to estimate the net energy released or absorbed in a reaction.

Why Is Thermochemistry Tested on the MCAT?

The MCAT doesn’t just test your ability to plug into formulas—it tests your conceptual grasp of how energy governs chemical and biological behavior. Thermochemistry is often integrated into:

Passages on metabolism (ΔG in biochemical reactions)
Physics-style calorimetry questions (q = mcΔT)
Reaction coupling in cellular processes (ATP hydrolysis)
Bond breaking/forming and enthalpy comparisons (ΔH of reaction)
Entropy and disorder in biological systems

Understanding these topics prepares you to reason through novel systems, not just recall definitions.

Introduction: What Is Thermochemistry and Why Does It Matter?

Thermochemistry is the branch of chemistry that deals with the energy changes—particularly heat transfer—that occur during chemical reactions and physical processes. On the MCAT and in real-world chemistry, thermochemistry allows us to quantitatively and qualitatively analyze how energy flows into or out of a system, which has direct implications for:

Reaction spontaneity
Metabolic energetics
Phase changes
Enzyme function and denaturation
Calorimetry and experimental design

From cellular respiration to combustion engines, thermochemical reasoning connects abstract concepts like enthalpy and entropy to physical realities like heat, temperature, and work.

Core Principle: Energy Is Conserved, But It Can Change Forms

At the heart of thermochemistry is the First Law of Thermodynamics, which states:

$$
\text{ΔE}_{\text{system}} = q + w
$$

Where:

q is the heat added to the system
w is the work done on the system
ΔE is the change in internal energy

The total energy of the universe is constant, but energy can be transferred between system and surroundings, or converted between forms—such as chemical potential energy becoming heat.

System vs. Surroundings

What Is a “System” in Thermochemistry?

In thermochemistry, the system refers to the specific part of the universe you’re focusing on—usually the chemicals involved in a reaction or a phase change. Everything else, the container, the water, the air in the lab, or even your calorimeter, is called the surroundings.

System = the thing you’re studying.
Surroundings = everything else that can exchange energy with the system.

Example:

Suppose you place a piece of sodium metal in water and a vigorous reaction occurs, releasing heat.

System: sodium + water (reactants and products)
Surroundings: the water bath, beaker, air, your hand (if nearby)

Types of Thermodynamic Systems

Systems can be categorized based on whether or not they exchange matter and/or energy with their surroundings:

Type of System	Can Exchange Matter?	Can Exchange Energy?	Example
Open	Yes	Yes	A cup of coffee (heat and steam escape)
Closed	No	Yes	A sealed flask in a water bath
Isolated	No	No	A thermos or perfectly insulated bomb calorimeter

MCAT Tip: Most problems assume closed systems, where no matter escapes, but heat or work can still transfer.

Energy Transfer: Endothermic vs. Exothermic

Once a system and its surroundings are defined, we can describe the direction of energy flow using standard conventions:

Process Type	Heat Flow (q)	System Behavior	Definition
Exothermic	q < 0	System releases heat	A reaction that transfers heat to the surroundings
Endothermic	q > 0	System absorbs heat	A reaction that draws heat from the surroundings into the system

Exothermic reactions feel warm or cause a temperature rise (e.g., combustion, neutralization)
Endothermic reactions feel cold or cause a temperature drop (e.g., photosynthesis, ice melting)

MCAT Mnemonic: HEAT OUT = EXO, HEAT IN = ENDO

EXO = EXit of heat
ENDO = ENters heat

Also:
“Exo feels hot, endo feels cold.”

Sign Conventions: Energy In vs. Energy Out

Thermochemistry follows the physics-based sign conventions for energy:

q > 0 → heat flows into the system
q < 0 → heat flows out of the system

Similarly:

w > 0 → work is done on the system (e.g., compression)
w < 0 → system does work on surroundings (e.g., expansion)

Quantity	Positive (+)	Negative (–)
q	Heat enters the system (endothermic)	Heat leaves the system (exothermic)
w	Work done on the system (compression)	Work done by the system (expansion)

Mnemonic for Signs:
“+q warms the system; –q gives it away.
+w work put in; –w system pushes out.”

Internal Energy and the First Law

The internal energy change of a system is:

$$
\Delta E = q + w
$$

$$
\text{(Change in energy = heat + work)}
$$

MCAT Strategy: In most chemistry-focused questions, w is negligible, and energy transfer is dominated by q (heat). But in physics-heavy passages involving gas expansion or compression, work will play a more prominent role.

MCAT Application: Why Does This Matter?

Defining the system correctly allows you to:

Determine whether a reaction releases or absorbs heat
Set up correct sign conventions in calorimetry problems
Understand the physical significance of ΔH (enthalpy), which we cover in the next section
Interpret energy diagrams (e.g., endothermic upward slopes vs. exothermic downward)

Quick Summary: Key Takeaways

System = chemical species undergoing a reaction or physical change
Surroundings = everything else in the universe
Most MCAT problems involve closed systems, where energy (but not matter) can flow
The direction of heat flow determines whether a reaction is endothermic (q > 0) or exothermic (q < 0)
The First Law of Thermodynamics is expressed as:

$$
\Delta E = q + w
$$

Enthalpy, Entropy, and Gibbs Free Energy

The Thermodynamic Triad for Predicting Reaction Behavior

Overview: What Do ΔH, ΔS, and ΔG Tell Us?

In thermochemistry, three state functions—enthalpy (ΔH), entropy (ΔS), and Gibbs free energy (ΔG)—work together to determine whether a chemical process is energetically favorable (i.e., spontaneous) or not.

ΔH (Enthalpy): Heat transferred at constant pressure
ΔS (Entropy): Change in disorder or dispersal of energy
ΔG (Gibbs Free Energy): Predicts spontaneity; incorporates both ΔH and ΔS

These quantities are all state functions, meaning they depend only on the initial and final states of the system—not on the path taken.

1. Enthalpy (ΔH): Heat at Constant Pressure

Definition:

Enthalpy is a measure of the heat content of a system at constant pressure. The change in enthalpy, ΔH, represents the net heat absorbed or released during a process.

ΔH Value	Interpretation	Example
ΔH < 0	Exothermic (releases heat)	Combustion, condensation
ΔH > 0	Endothermic (absorbs heat)	Melting, photosynthesis

Exothermic: Bonds formed in products are stronger (more stable) than those broken in reactants

Endothermic: Energy input is required to break bonds or overcome forces

MCAT Example:

Combustion of glucose:

$$
\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O} \quad \Delta H = -2800 \ \text{kJ/mol}
$$

Exothermic (releases energy to power biological work)

Entropy (ΔS): Disorder and Microstates

Definition:

Entropy is a measure of the dispersal of energy and disorder within a system. It reflects the number of possible microscopic arrangements (microstates).

ΔS Value	Interpretation	Example
ΔS > 0	Entropy increases (more disorder)	Ice melting, gas formation
ΔS < 0	Entropy decreases (more order)	Water freezing, gas compression

MCAT Rule of Thumb:
Solids → liquids → gases = increasing entropy
Fewer particles → more particles = increasing entropy

MCAT-Relevant Processes with ΔS > 0:

Dissolution of salt in water
Evaporation
Decomposition reactions
Increasing number of gas molecules on product side

Gibbs Free Energy (ΔG): The Final Decider

Definition:

Gibbs free energy combines ΔH and ΔS to predict whether a process is spontaneous (can occur without input of external energy).

The defining equation:

$$
\Delta G = \Delta H – T\Delta S
$$

ΔG<0: Spontaneous
ΔG>0: Non-spontaneous
ΔG=0: Equilibrium

T = absolute temperature in Kelvin. Since T is always positive, the sign of ΔS plays a major role in determining spontaneity.

MCAT Mnemonic: “Goldfish Have Special Tanks”

G = H – S × T
This keeps the signs and variables straight during test-day calculations.

How to Analyze Reaction Spontaneity

Use this summary table to predict spontaneity based on ΔH and ΔS:

ΔH	ΔS	ΔG Prediction	Spontaneity
–	+	Always –	Spontaneous at all T
–	–	– at low T, + at high T	Spontaneous at low T
+	+	– at high T, + at low T	Spontaneous at high T
+	–	Always +	Non-spontaneous at all T

MCAT Strategy Tip: If both ΔH and ΔS are favorable, spontaneity is temperature-independent. If they conflict, temperature determines the outcome.

Biological and Real-World Relevance

ATP hydrolysis: Spontaneous (ΔG < 0), drives cellular work
Protein folding: ΔS < 0, but driven by favorable ΔH (bonding and hydrophobic effects)
Phase changes: Ice melting is endothermic but entropically favorable → becomes spontaneous at T > 0°C

Section Summary: Key Takeaways

ΔH = heat transfer at constant pressure
ΔS = entropy or disorder; increases with randomness, particle dispersion, and energy distribution
ΔG = master function predicting spontaneity:

$$
\Delta G = \Delta H – T\Delta S
$$

Negative ΔG → spontaneous process
Sign of ΔG depends on the competition between ΔH and TΔS
All three are state functions and extensively tested on the MCAT

Calorimetry, Heat Capacity, and q=mcΔT

Quantifying Heat Flow in Physical and Chemical Processes

What Is Calorimetry?

Calorimetry is the quantitative measurement of heat flow during chemical and physical processes. It allows us to calculate how much thermal energy (q) is absorbed or released by a substance when:

Its temperature changes, and/or
Its physical state changes (solid ↔ liquid ↔ gas)

On the MCAT, calorimetry problems often focus on:

q = mcΔT for heat during temperature changes
q = nΔH_phase for phase changes (like melting or boiling)
Distinguishing between specific heat and enthalpy of transition
Using energy conservation: what is lost by one part is gained by another

On the MCAT, calorimetry shows up in passages or discrete questions that involve q = mcΔT, ΔH = q/n, and interpreting heat flow in calorimeters (coffee cup or bomb).

The Core Equation: q=mcΔT

$$
q = mc\Delta T
$$

Where:

q = heat energy (in joules or calories)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g·°C)
ΔT = change in temperature = Tfinal−Tinitial

This equation applies to temperature changes, not phase changes (those require latent heat).

Worked Example (Temperature Change Only):

How much energy is needed to heat 100 g of water from 25°C to 75°C?

Use:

q = (100 \ \text{g})(4.18 \ \text{J/g·°C})(75 – 25) = 20,900 \ \text{J}

MCAT Tip: Know that specific heat capacity (c) varies between substances. Water has a high specific heat:

c_{\text{H}_2\text{O}} = 4.18 \ \text{J/g·°C}

This makes water excellent for buffering temperature changes.

Example Calculation:

“How much heat is required to raise the temperature of 200 g of water from 25°C to 45°C?”

Use:

q = mc\Delta T = (200)(4.18)(45 – 25) = 200 × 4.18 × 20 = 16,720 \ \text{J}

Key Insight: Why Specific Heat Matters

Water’s high specific heat (4.18 J/g·°C) explains why lakes resist rapid temperature changes—important in biological homeostasis.
Metals like iron or copper have low specific heats, meaning they heat up and cool down quickly.

MCAT Angle: Use q = mcΔT when heating or cooling within the same phase.
Don’t use it during phase transitions (next section).

Calorimetry Setups: Constant Pressure vs. Constant Volume

Type	Used For	Assumptions	MCAT Name
Coffee cup	Solution reactions, phase changes	Constant pressure (open to atmosphere)	q = ΔH directly
Bomb calorimeter	Combustion reactions	Constant volume (no expansion work)	Use ΔU ≈ q, not ΔH

MCAT Rule: For constant pressure systems (like coffee cup), heat q is equal to the enthalpy change ΔH. For constant volume systems (like a bomb calorimeter), heat is equal to internal energy change ΔU.

Related Concepts: Molar Heat and Heat Per Mole

Sometimes MCAT questions give heat in joules but ask about ΔH in kJ/mol.

Use this relationship:

$$
\Delta H = \frac{q}{n}
$$

Where:

q: total heat absorbed or released
n: number of moles of substance involved.

Example:
If 33.6 kJ of heat is released when 2.00 mol of methane combusts:

$$
\Delta H = \frac{-33.6}{2.00} = -16.8 \ \text{kJ/mol}
$$

Heat Transfer from Phase Changes

When a substance undergoes a phase change, its temperature does not change, but energy is still absorbed or released to break or form intermolecular forces.

Use:

$$
q = n\Delta H_{\text{phase}}
$$

Where:

n: number of moles
ΔH_phase: molar enthalpy of the phase change (kJ/mol)

Melting (Fusion) and Freezing

Melting (fusion): solid → liquid
- Requires energy input: endothermic

$$
q = n\Delta H_{\text{fus}}, \quad \Delta H_{\text{fus}} > 0
$$

Freezing: liquid → solid
- Releases energy: exothermic

$$
q = -n\Delta H_{\text{fus}}, \quad \Delta H_{\text{fus}} < 0
$$

MCAT Tip: Ice at 0°C melts at constant temperature. The energy you add doesn’t change T—it breaks hydrogen bonds.

Vaporization and Condensation

Vaporization (boiling): liquid → gas
- Requires significant energy: endothermic

$$
q = n\Delta H_{\text{vap}}
$$

Condensation: gas → liquid
- Releases heat: exothermic

$$
q = -n\Delta H_{\text{vap}}
$$

Water’s ΔH_vap ≈ 40.7 kJ/mol, much higher than ΔH_fus ≈ 6.0 kJ/mol — it takes far more energy to vaporize than melt water.

Sublimation and Deposition

Sublimation: solid → gas
- Direct transition without passing through liquid (e.g., dry ice, iodine)

$$
q = n\Delta H_{\text{sub}}, \quad \text{endothermic}
$$

Deposition: gas → solid
- Releases heat, reverse of sublimation

$$
q = -n\Delta H_{\text{sub}}
$$

Summary Table: Enthalpies of Phase Change

Phase Change	Direction	ΔH sign	Process Type	MCAT Classification
Melting (Fusion)	Solid → Liquid	+	Endothermic	Absorbs heat
Freezing	Liquid → Solid	–	Exothermic	Releases heat
Vaporization	Liquid → Gas	+	Endothermic	Absorbs heat
Condensation	Gas → Liquid	–	Exothermic	Releases heat
Sublimation	Solid → Gas	+	Endothermic	Absorbs heat
Deposition	Gas → Solid	–	Exothermic	Releases heat

Understanding Phase Diagrams

A phase diagram is a graphical representation of the physical states of a substance under various combinations of pressure and temperature. It enables prediction of the stable phase (solid, liquid, or gas) under specific conditions — and how phase changes occur in response to changes in those variables.

Axes and Key Features

Component	Meaning
X-axis (Temperature, T)	Usually in °C or K — increasing rightward
Y-axis (Pressure, P)	Often in atm or mmHg — increasing upward
Curved lines	Indicate phase boundaries — i.e., where phase transitions occur
Triple point	The only set of conditions where all three phases coexist in equilibrium
Critical point	The temperature and pressure beyond which a liquid and gas cannot be distinguished (supercritical fluid region)

Phase Changes and Boundaries

Phase Change	Boundary	Thermodynamics	Directionality
Melting / Freezing	Solid–Liquid line	Endo ↔ Exo	Increase ↔ Decrease T
Vaporization / Condensation	Liquid–Gas curve	Endo ↔ Exo	Heat ↔ Cool
Sublimation / Deposition	Solid–Gas curve	Endo ↔ Exo	Often at low P
Supercritical Transition	Above critical point	No distinct phase	Fluid behaves as both liquid & gas

MCAT Triggers and Conceptual Pitfalls

Which phase is favored?
- ↑ Temperature = favors gas
- ↑ Pressure = favors solid (except water, where ice has lower density)
What happens at the triple point?
- All three phases are in equilibrium.
- Sublimation/melting/boiling occur simultaneously.
What does the slope of the solid–liquid line tell you?
- Most substances: positive slope (solid more dense than liquid)
- Water: negative slope (ice less dense — floats!)
If you’re above the critical point:
- Liquid and gas are indistinguishable.
- No more boiling — it’s a supercritical fluid.

MCAT Example Question Types

“Which of the following changes would result in deposition?”
→ Move downward and leftward on the phase diagram (gas → solid)
“A sample of CO₂ is suddenly depressurized and cooled…”
→ Locate new P–T point to predict if sublimation will occur.
“At which temperature and pressure would solid, liquid, and gas exist together?”
→ The triple point.

Mnemonics

SLG → Solid, Liquid, Gas = Increasing kinetic energy
Sublimation & Vaporization = Endothermic (heat in)
Deposition & Condensation = Exothermic (heat out)
Triple = Trio of phases
Critical = Can’t tell gas from liquid anymore

Comprehensive MCAT Example:

“How much energy is required to convert 18.0 g of ice at –10 °C to steam at 110 °C?”

Steps:

Warm ice to 0°C → use q = mcΔT
Melt ice → use q = nΔH_fus
Heat water to 100°C → use q = mcΔT
Boil water → use q = nΔH_vap
Heat steam to 110°C → use q = mcΔT

Add all five q values to find total energy required.

Strategic Mnemonics and Notes

Flat means phase, slope means temp
- Use q = mcΔT on sloped sections, and q = nΔH on flat regions of heating curves.
Melting and boiling absorb, freezing and condensing release
To break bonds, pay energy (endothermic); to form bonds, you get paid (exothermic).
MCAT Sign Traps
- Energy in → q is positive
- Energy out → q is negative
- Confusing this in condensation/freezing problems is a classic test trap!

Common Pitfalls to Avoid

Mistake	Fix
Forgetting to convert units (e.g., g ↔ kg, °C ↔ K)	Use °C for ΔT in q = mcΔT; only convert to K in PV = nRT
Mixing up signs for q	Heat absorbed = positive q; heat released = negative q
Using q = mcΔT for phase changes	Use q = nΔH_fus or q = nΔH_vap for phase transitions

MCAT Mnemonics & Strategy Tips

“Mass × c × ΔT gives the heat you’ll see” → for temperature change problems
“Flat lines need phase time” → on heating curves, q = mcΔT only applies to sloped sections; flat parts use q = nΔH

Section Summary: Key Takeaways

q=mcΔT is the go-to equation for heat flow during temperature changes
Use ΔH = q/n to relate total heat to per-mole enthalpy
Coffee cup calorimeters assume constant pressure: q=ΔH
Bomb calorimeters = constant volume: q=ΔU
For phase changes, use q =nΔH_fus or q = nΔH_vap (not q = mcΔT)

Hess’s Law and Bond Enthalpies

Solving for ΔH by Rearranging Reactions or Counting Bonds

What Is Hess’s Law?

Hess’s Law states that the total enthalpy change for a reaction is the same regardless of the number of steps or the path taken — because enthalpy is a state function.

This means: if you can break a complex reaction into simpler ones, you can sum their ΔH values to get the overall ΔH.

MCAT Framing of Hess’s Law

You’ll often be given a target equation and a set of other reactions with known ΔH values. Your job: manipulate and sum the equations to match the target reaction.

You can:

Flip a reaction → change sign of ΔH
Multiply a reaction → multiply ΔH by the same factor
Then add reactions → sum all ΔH values

Example: Hess’s Law Calculation

Given:

$$
\ce{C(s) + O2(g) -> CO2(g)} \quad \Delta H = -393.5 \ \text{kJ}
$$

$$
\ce{H2(g) + \tfrac{1}{2}O2(g) -> H2O(l)} \quad \Delta H = -285.8 \ \text{kJ}
$$

$$
\text{Target:} \quad \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}
$$

Also given:

$$
\ce{C(s) + 2H2(g) -> CH4(g)} \quad \Delta H = -74.8 \ \text{kJ}
$$

Strategy:

Rearrange equation 4 to isolate CH₄ on reactant side → flip and change sign
Add equations 1, 2×2, and flipped 4

Final ΔH:

$$
\Delta H_{\text{rxn}} = [-393.5 + 2(-285.8)] + (+74.8) = -890.3 \ \text{kJ}
$$

Standard Enthalpy of Formation (ΔHf°)

Definition:

The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (1 atm, 298 K).

Notation:
ΔHf∘ = formation under standard conditions

MCAT Shortcut Equation:

To find the ΔH of a reaction:

$$
\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f(\text{products}) – \sum \Delta H^\circ_f(\text{reactants})
$$

Multiply each ΔHf° by its stoichiometric coefficient
Subtract total ΔHf° of reactants from that of products

Example: Combustion of Methane

Reaction:

$$
\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}
$$

ΔHf° values:

$$
\Delta H_f^\circ \left( \ce{CH4(g)} \right) = -74.8 \ \text{kJ/mol}
$$

$$
\Delta H_f^\circ \left( \ce{CO2(g)} \right) = -393.5 \ \text{kJ/mol}
$$

$$
\Delta H_f^\circ \left( \ce{H2O(l)} \right) = -285.8 \ \text{kJ/mol}
$$

$$
\Delta H_f^\circ \left( \ce{O2(g)} \right) = 0 \ \text{kJ/mol} \quad \text{(standard element)}
$$

Apply formula:

$$
\Delta H^\circ_{\text{rxn}} = \left[(-393.5) + 2(-285.8)\right] – \left[(-74.8) + 0\right] = -890.3 \ \text{kJ}
$$

Bond Enthalpies: Using Bonds Broken – Bonds Formed

Definition:

A bond enthalpy (or bond dissociation energy) is the energy required to break one mole of a particular bond in the gas phase.

All bond enthalpies are positive because bond breaking is endothermic

MCAT Bond Strategy Equation:

$$
\Delta H_{\text{rxn}} = \sum \text{Bond Energies of Bonds Broken} – \sum \text{Bond Energies of Bonds Formed}
$$

Breaking bonds = input energy (+)
Forming bonds = release energy (–)
So net ΔH depends on whether more energy is spent breaking or gained forming

Example: H₂ + Cl₂ → 2 HCl

Bonds involved:

Break 1 H–H (436 kJ/mol)
Break 1 Cl–Cl (243 kJ/mol)
Form 2 H–Cl (431 kJ/mol × 2)

Apply equation:

$$
\Delta H = [436 + 243] – [2 \times 431] = 679 – 862 = -183 \ \text{kJ}
$$

Reaction is exothermic: more energy is released forming bonds than required to break them

When to Use Which Method (MCAT Decision Tree)

Given…	Use…
ΔH for component reactions	Hess’s Law (add and manipulate equations)
ΔHf° values for all species	ΔH = ΣΔH_f(products) – ΣΔH_i(reactants)
Only bond energies for gas-phase species	Bond Enthalpy Method

Section Summary: Key Takeaways

Hess’s Law allows you to combine ΔH values algebraically when adding or flipping reactions
ΔHf° values let you calculate ΔH directly from tabulated formation enthalpies
Bond enthalpies give an approximation when formation data is not available – apply Bonds Broken – Bonds Formed
Know that standard elemental forms have ΔHf° = 0 (e.g., O₂, H₂, N₂, graphite)