🔥 Energy, Systems, and Surroundings

Part 1 of 7 — Foundations of Thermochemistry

Every chemical reaction involves energy changes. Thermochemistry is the branch of chemistry that studies the heat absorbed or released during chemical reactions and physical changes. Before we can calculate enthalpy, we need to understand the language of energy flow.

System and Surroundings

In thermochemistry, we divide the universe into two parts:

Energy Transfer

Energy flows between the system and surroundings. The First Law of Thermodynamics states:

$\Delta E_{\text{universe}} = \Delta E_{\text{system}} + \Delta E_{\text{surroundings}} = 0$

Energy is conserved — it is neither created nor destroyed, only transferred.

Sign Conventions

Endothermic vs. Exothermic Processes

Exothermic ($q < 0$)

The system releases heat to the surroundings.

• The surroundings get warmer
• $\Delta H < 0$ (negative)
• Products have lower energy than reactants
• Energy is a product of the reaction

Examples: combustion, neutralization, condensation, freezing

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890 \text{ kJ}$

Endothermic ($q > 0$)

The system absorbs heat from the surroundings.

• The surroundings get cooler
• $\Delta H > 0$ (positive)
• Products have higher energy than reactants
• Energy is a reactant

Examples: photosynthesis, melting ice, evaporation, dissolving $\text{NH}_4\text{NO}_3$

$\text{6CO}_2(g) + 6\text{H}_2\text{O}(l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \quad \Delta H = +2803 \text{ kJ}$

Energy Diagrams

Energy diagrams visually show the energy change during a reaction.

Exothermic Diagram

• Reactants are at a higher energy level
• Products are at a lower energy level
• $\Delta H$ arrow points downward (negative)
• The difference = energy released to surroundings

Endothermic Diagram

• Reactants are at a lower energy level
• Products are at a higher energy level
• $\Delta H$ arrow points upward (positive)
• The difference = energy absorbed from surroundings

Key Relationship

$\Delta H_{\text{forward}} = -\Delta H_{\text{reverse}}$

If a reaction is exothermic in the forward direction, it is endothermic in reverse, and vice versa.

Energy Fundamentals Quiz 🎯

Classify the Process 🧮

Type "exothermic" or "endothermic" for each process:

1. Water freezing into ice

2. Dissolving ammonium nitrate in water (the solution feels cold)

3. Burning natural gas on a stove

System and Energy Flow 🔽

Exit Quiz — Energy Fundamentals ✅

🌡️ Enthalpy (ΔH) — The Heat of Reaction

Part 2 of 7 — State Functions and Standard Enthalpy

Enthalpy is the most commonly used thermodynamic quantity in chemistry. It tells us how much heat is absorbed or released during a reaction at constant pressure — which is how most reactions happen in the lab and in nature.

What Is Enthalpy?

Enthalpy ($H$) is defined as:

$H = E + PV$

where $E$ is internal energy, $P$ is pressure, and $V$ is volume.

We can never measure absolute enthalpy — only the change in enthalpy:

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

At Constant Pressure

At constant pressure (open beaker, atmospheric conditions):

$\Delta H = q_p$

The enthalpy change equals the heat transferred at constant pressure. This is why chemists love enthalpy — it directly corresponds to the heat you can measure!

Key Signs

Enthalpy Is a State Function

A state function depends only on the current state of the system, not on how it got there.

What This Means

• The enthalpy change $\Delta H$ depends only on the initial and final states
• It does not depend on the pathway or mechanism
• The same reaction will have the same $\Delta H$ regardless of how many steps it takes

Analogy

Think of altitude: if you climb a mountain, your change in altitude depends only on your starting and ending positions — not whether you took the steep trail or the winding road. Enthalpy works the same way.

Consequences

1. If a reaction can occur in one step or multiple steps, $\Delta H$ is the same
2. This is the foundation of Hess's Law (Part 5)
3. We can calculate $\Delta H$ for reactions we cannot directly measure

Standard Enthalpy

Standard conditions in thermochemistry use the symbol $°$:

Standard Enthalpy of Reaction ($\Delta H°_{\text{rxn}}$)

The enthalpy change when reactants in their standard states are converted to products in their standard states.

Standard State

The standard state of a substance is its most stable form at $1$ atm and the specified temperature:

Important Relationships

If you multiply a reaction by a factor $n$: $\Delta H_{\text{new}} = n \times \Delta H_{\text{original}}$

If you reverse a reaction: $\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}$

Enthalpy Concept Quiz 🎯

Enthalpy Scaling Practice 🧮

Given: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \quad \Delta H = -92 \text{ kJ}$

1. What is $\Delta H$ for $\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g)$? (in kJ)

2. What is $\Delta H$ for $2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g)$? (in kJ)

3. What is $\Delta H$ for $2\text{N}_2(g) + 6\text{H}_2(g) \rightarrow 4\text{NH}_3(g)$? (in kJ)

Enthalpy Properties 🔽

Exit Quiz — Enthalpy ✅

☕ Calorimetry — Measuring Heat

Part 3 of 7 — q = mcΔT and the Coffee-Cup Calorimeter

How do we actually measure enthalpy changes? We use calorimetry — the science of measuring heat flow. The basic idea is simple: if a reaction releases heat, the surrounding water gets warmer. By measuring that temperature change, we can calculate how much heat was transferred.

The Heat Equation

$q = mc\Delta T$

Specific Heat Capacity

The specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C.

Water has an unusually high specific heat, meaning it can absorb a lot of heat with only a small temperature change. This is why water is used as a coolant and why coastal climates are moderate.

The Coffee-Cup Calorimeter

A simple calorimeter made from a Styrofoam cup with a lid and thermometer.

How It Works

1. Measure the initial temperature of the solution
2. Mix the reactants in the cup
3. Record the maximum (or minimum) temperature reached
4. Calculate $q$ for the solution using $q = mc\Delta T$

Key Assumptions

• The calorimeter is perfectly insulated (no heat escapes)
• The solution has the same density and specific heat as pure water ($c = 4.184$ J/(g·°C), $d = 1.00$ g/mL)
• All heat from the reaction goes into the solution

Important Sign Convention

$q_{\text{rxn}} = -q_{\text{solution}}$

If the solution warms up ($q_{\text{solution}} > 0$), the reaction is exothermic ($q_{\text{rxn}} < 0$).

Constant Pressure

A coffee-cup calorimeter operates at constant pressure (open to the atmosphere), so:

$q_p = \Delta H$

Worked Example

Problem: When 50.0 mL of 1.00 M HCl is mixed with 50.0 mL of 1.00 M NaOH in a coffee-cup calorimeter, the temperature rises from 22.0°C to 28.9°C. Calculate $\Delta H$ per mole of water formed.

Step 1: Calculate total mass $m = 100.0 \text{ mL} \times 1.00 \text{ g/mL} = 100.0 \text{ g}$

Step 2: Calculate $\Delta T$ $\Delta T = 28.9 - 22.0 = 6.9°\text{C}$

Step 3: Calculate $q_{\text{solution}}$ $q_{\text{solution}} = mc\Delta T = (100.0)(4.184)(6.9) = 2887 \text{ J} = 2.89 \text{ kJ}$

Step 4: Find $q_{\text{rxn}}$ $q_{\text{rxn}} = -q_{\text{solution}} = -2.89 \text{ kJ}$

Step 5: Calculate moles of water formed $n = 0.0500 \text{ L} \times 1.00 \text{ M} = 0.0500 \text{ mol}$

Step 6: Calculate $\Delta H$ per mole $\Delta H = \frac{-2.89 \text{ kJ}}{0.0500 \text{ mol}} = -57.8 \text{ kJ/mol}$

The accepted value is $-55.8$ kJ/mol — our measurement is close!

Calorimetry Concept Quiz 🎯

Calorimetry Calculations 🧮

1. How much heat is needed to raise the temperature of 200.0 g of water from 20.0°C to 45.0°C? (answer in kJ, to 3 significant figures; $c_{\text{water}} = 4.184$ J/(g·°C))

2. A 50.0 g piece of metal at 95.0°C is placed in 150.0 g of water at 20.0°C. The final temperature is 23.0°C. What is the specific heat of the metal? (in J/(g·°C), to 3 significant figures)

3. When 100.0 mL of 0.500 M HCl and 100.0 mL of 0.500 M NaOH are mixed, the temperature rises by 3.4°C. What is $q_{\text{rxn}}$ in kJ? (to 3 significant figures, include sign)

Calorimetry Concepts 🔽

Exit Quiz — Calorimetry ✅

💣 Bomb Calorimetry

Part 4 of 7 — Constant-Volume Calorimetry

While coffee-cup calorimeters work at constant pressure, some reactions — especially combustion — release enormous amounts of gas and energy. For these, we use a bomb calorimeter, which operates at constant volume. This distinction has important thermodynamic consequences.

Bomb Calorimeter Structure

A bomb calorimeter consists of:

1. The "bomb" — a rigid, sealed steel container where the reaction occurs
2. Water bath — surrounds the bomb, absorbs the released heat
3. Ignition wire — initiates combustion with an electric spark
4. Thermometer — measures the temperature change of the water
5. Stirrer — ensures uniform temperature in the water bath
6. Insulated jacket — minimizes heat loss to the environment

Key Feature: Constant Volume

The bomb is sealed and rigid — the volume cannot change. This means:

• No $PV$ work is done ($w = 0$ since $\Delta V = 0$)
• At constant volume: $q_v = \Delta E$ (internal energy change)
• This is different from coffee-cup calorimetry where $q_p = \Delta H$

Relationship Between $\Delta H$ and $\Delta E$

$\Delta H = \Delta E + \Delta(PV)$

For reactions involving only solids and liquids, $\Delta H \approx \Delta E$.

For reactions involving gases:

$\Delta H = \Delta E + \Delta n_{\text{gas}} RT$

where $\Delta n_{\text{gas}}$ = moles of gaseous products − moles of gaseous reactants.

Heat Capacity of the Calorimeter

For a bomb calorimeter, we use the heat capacity of the entire calorimeter ($C_{\text{cal}}$):

$q_{\text{cal}} = C_{\text{cal}} \Delta T$

Important Distinction

The heat capacity $C_{\text{cal}}$ is determined by calibration — burning a substance with a known heat of combustion.

Finding $q_{\text{rxn}}$

$q_{\text{rxn}} = -q_{\text{cal}} = -C_{\text{cal}} \Delta T$

The negative sign reflects that heat released by the reaction is absorbed by the calorimeter.

Worked Example

Problem: A 1.50 g sample of benzoic acid ($\text{C}_7\text{H}_6\text{O}_2$, molar mass = 122.12 g/mol) is burned in a bomb calorimeter with $C_{\text{cal}} = 10.34$ kJ/°C. The temperature rises from 22.45°C to 25.71°C. Calculate the molar heat of combustion.

Step 1: Calculate $\Delta T$ $\Delta T = 25.71 - 22.45 = 3.26°\text{C}$

Step 2: Calculate $q_{\text{cal}}$ $q_{\text{cal}} = C_{\text{cal}} \Delta T = (10.34)(3.26) = 33.71 \text{ kJ}$

Step 3: Find $q_{\text{rxn}}$ $q_{\text{rxn}} = -q_{\text{cal}} = -33.71 \text{ kJ}$

Step 4: Calculate moles of benzoic acid $n = \frac{1.50}{122.12} = 0.01228 \text{ mol}$

Step 5: Calculate molar heat of combustion $\Delta E = \frac{-33.71}{0.01228} = -2745 \text{ kJ/mol}$

Note: This gives $\Delta E$ (internal energy), not $\Delta H$, because the bomb calorimeter operates at constant volume. For this reaction, $\Delta H \approx \Delta E$ because $\Delta n_{\text{gas}}$ is small.

Bomb Calorimetry Concept Quiz 🎯

Bomb Calorimetry Calculations 🧮

1. A bomb calorimeter has $C_{\text{cal}} = 8.50$ kJ/°C. If the temperature rises by 4.20°C, what is $q_{\text{rxn}}$? (in kJ, include sign)

2. When 0.500 g of sugar ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$, molar mass = 342.3 g/mol) is burned in a bomb calorimeter ($C_{\text{cal}} = 9.20$ kJ/°C), the temperature rises by 1.23°C. What is the energy released per mole? (in kJ/mol, round to nearest whole number, report as positive)

3. A calibration experiment burns 1.000 g of benzoic acid (heat of combustion = 26.38 kJ/g) and the temperature rises by 2.55°C. What is $C_{\text{cal}}$? (in kJ/°C, to 3 significant figures)

Bomb vs. Coffee-Cup Calorimetry 🔽

Exit Quiz — Bomb Calorimetry ✅

🔄 Hess's Law — Adding Enthalpy Changes

Part 5 of 7 — The Power of State Functions

Some reactions are impossible to carry out directly in a calorimeter. How do we find $\Delta H$ for them? Hess's Law gives us the answer: since enthalpy is a state function, we can add up the enthalpy changes of individual steps to get the total.

Hess's Law

Hess's Law: If a reaction can be expressed as the sum of two or more other reactions, the enthalpy change of the overall reaction is the sum of the enthalpy changes of the individual reactions.

$\Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots$

Why It Works

Because enthalpy is a state function, the total enthalpy change depends only on the initial and final states, not on the path. Whether a reaction occurs in one step or ten steps, $\Delta H$ is the same.

Rules for Manipulating Equations

Problem-Solving Strategy

Step-by-Step Approach

1. Write the target reaction — the one you need $\Delta H$ for
2. Examine the given reactions — look for each substance in your target
3. Manipulate given reactions so that when added, they equal the target:
   • Reverse reactions if a reactant needs to be a product (or vice versa)
   • Multiply reactions to match the coefficients in the target
4. Add the manipulated reactions — substances on opposite sides cancel
5. Add the adjusted $\Delta H$ values to get $\Delta H_{\text{overall}}$

Worked Example

Find $\Delta H$ for: $\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$

Given:

1. $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5$ kJ
2. $\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2 = -283.0$ kJ

Solution:

• Keep reaction 1 as written (has C as reactant ✓)
• Reverse reaction 2 (need CO as product): $\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H = +283.0$ kJ

Add:

$\text{C}(s) + \text{O}_2(g) + \text{CO}_2(g) \rightarrow \text{CO}_2(g) + \text{CO}(g) + \frac{1}{2}\text{O}_2(g)$

Cancel $\text{CO}_2$ and simplify $\text{O}_2$:

$\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$

$\Delta H = -393.5 + 283.0 = -110.5 \text{ kJ}$

Hess's Law Concept Quiz 🎯

Hess's Law Calculations 🧮

Given:

• (1) $\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \quad \Delta H_1 = -296.8$ kJ
• (2) $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g) \quad \Delta H_2 = -197.8$ kJ

Find $\Delta H$ for: $2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)$

1. What must you multiply reaction (1) by? (enter the number)

2. What must you multiply reaction (2) by? (enter the number)

3. What is $\Delta H$ for the target reaction? (in kJ, to 3 significant figures)

Hess's Law Strategy 🔽

Exit Quiz — Hess's Law ✅

🏗️ Standard Enthalpies of Formation

Part 6 of 7 — The Master Equation

Standard enthalpies of formation ($\Delta H°_f$) provide a systematic way to calculate $\Delta H°_{\text{rxn}}$ for any reaction — without needing Hess's Law manipulations. This is the most powerful and commonly used method on the AP exam.

Standard Enthalpy of Formation ($\Delta H°_f$)

The enthalpy change when one mole of a compound is formed from its elements in their standard states.

Examples

$\text{C}(s, \text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H°_f = -393.5 \text{ kJ/mol}$

$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H°_f = -285.8 \text{ kJ/mol}$

$\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta H°_f = -45.9 \text{ kJ/mol}$

Critical Rule

$\Delta H°_f$ of any element in its standard state = 0

This makes sense: an element doesn't change to form itself!

The Master Equation

$\Delta H°_{\text{rxn}} = \sum n \cdot \Delta H°_f(\text{products}) - \sum m \cdot \Delta H°_f(\text{reactants})$

where $n$ and $m$ are the stoichiometric coefficients.

How to Use It

1. Look up $\Delta H°_f$ for every compound in the reaction
2. Remember: $\Delta H°_f = 0$ for elements in their standard states
3. Multiply each $\Delta H°_f$ by its coefficient
4. Subtract the sum of reactants from the sum of products

Worked Example

Calculate $\Delta H°_{\text{rxn}}$ for: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

$\Delta H°_{\text{rxn}} = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)]$ $= [-393.5 - 571.6] - [-74.8]$ $= -965.1 + 74.8 = -890.3 \text{ kJ}$

Formation Enthalpy Concept Quiz 🎯

Formation Enthalpy Calculations 🧮

Use these $\Delta H°_f$ values (kJ/mol):

• $\text{CO}_2(g) = -393.5$, $\text{H}_2\text{O}(l) = -285.8$, $\text{C}_2\text{H}_6(g) = -84.7$
• $\text{NH}_3(g) = -45.9$, $\text{NO}(g) = +90.3$, $\text{O}_2, \text{N}_2, \text{H}_2 = 0$

1. Calculate $\Delta H°_{\text{rxn}}$ for: $\text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$ (in kJ, to 3 significant figures)

2. Calculate $\Delta H°_{\text{rxn}}$ for: $4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(l)$ (in kJ, to 3 significant figures)

Formation Enthalpy Concepts 🔽

Exit Quiz — Formation Enthalpies ✅

🎯 Synthesis & AP Review — Enthalpy and Calorimetry

Part 7 of 7 — Bringing It All Together

This final part integrates everything: energy flow, calorimetry, Hess's Law, and formation enthalpies. Master these connections and you'll be ready for any AP-level thermochemistry question.

Complete Concept Map

Energy and Heat

Enthalpy

Hess's Law & Formation

AP Exam Strategies

Common AP Question Types

1. Calorimetry calculation — given mass, specific heat, ΔT → find q → find ΔH per mole
2. Hess's Law — manipulate 2-3 reactions to find ΔH for a target reaction
3. Formation enthalpy — use the master equation with a table of $\Delta H°_f$ values
4. Conceptual — identify exo/endothermic, explain sign conventions, predict temperature changes

Common Mistakes to Avoid

• Forgetting to flip the sign of ΔH when reversing a reaction
• Using specific heat ($c$) when heat capacity ($C$) is given (or vice versa)
• Forgetting that $\Delta H°_f = 0$ for elements in their standard states
• Mixing up $q_{\text{rxn}}$ and $q_{\text{solution}}$ (they have opposite signs)
• Not converting between J and kJ

Comprehensive AP Review Quiz 🎯

Integration Problems 🧮

1. 150.0 mL of 2.00 M HCl reacts with excess NaOH in a coffee-cup calorimeter. The temperature rises by 13.4°C. Assume the solution's mass is 150.0 g and $c = 4.184$ J/(g·°C). What is $\Delta H$ per mole of HCl? (in kJ/mol, to 3 significant figures, include sign)

2. Using $\Delta H°_f$ (kJ/mol): CO₂(g) = −393.5, H₂O(l) = −285.8, C₃H₈(g) = −103.8. Calculate $\Delta H°_{\text{rxn}}$ for $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$. (in kJ)

Comprehensive Concept Review 🔽

Final Exit Quiz — Thermochemistry Mastery ✅