🎯⭐ INTERACTIVE LESSON

Enthalpy and Calorimetry

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Enthalpy and Calorimetry - Complete Interactive Lesson

Part 1: Enthalpy & ΔH

🔥 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:

Term	Definition	Example
System	The part we are studying	The reacting chemicals in a beaker
Surroundings	Everything else	The beaker, the water, the air, the lab
Universe	System + Surroundings	Everything

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

Direction of Energy Flow	$q_{\text{system}}$	Temperature of Surroundings
Energy flows into system	Positive (+)	Decreases
Energy flows out of system	Negative (−)	Increases

📌 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

${CH}_{4} (g) + 2 O_{2} (g)$

📌 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

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

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 ✅

Part 2: Exothermic & Endothermic

🌡️ 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, is pressure, and is volume.

Part 3: Coffee Cup Calorimetry

☕ 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

$q = mc\Delta T$

Symbol	Meaning	Common Units
$q$	Heat absorbed or released	J or kJ

Part 4: Bomb 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:

The "bomb" — a rigid, sealed steel container where the reaction occurs
Water bath — surrounds the bomb, absorbs the released heat
Ignition wire — initiates combustion with an electric spark
Thermometer — measures the temperature change of the water
Stirrer — ensures uniform temperature in the water bath
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 )

Part 5: Hess\'s Law

🔄 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.

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

Part 6: Problem-Solving Workshop

🏗️ 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 reaction — without needing Hess's Law manipulations. This is the most powerful and commonly used method on the AP exam.

Part 7: Synthesis & AP Review

🎯 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

Concept	Key Equation	Notes
Heat transfer	$q = mc\Delta T$	Specific heat version
Coffee-cup calorimeter	$q_p = \Delta H$

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l) Δ H = - 890 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 $\text{NH}_4\text{NO}_3$

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

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

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):

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!

$\Delta H$	Meaning	Type
Negative (−)	Heat released	Exothermic
Positive (+)	Heat absorbed	Endothermic

🌡️ 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

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

🌡️ Standard Enthalpy

Standard conditions in thermochemistry use the symbol $°$ :

Parameter	Standard Value
Pressure	$1$ atm (or $1$ bar)
Concentration	$1$ M (for solutions)
Temperature	Usually $25°\text{C}$ ( $298$ K), but must be specified

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:

Substance	Standard State
Oxygen	$\text{O}_2(g)$
Carbon	$\text{C}(s, \text{graphite})$
Iron

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 ✅

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.

Substance	$c$ [J/(g·°C)]
Water (liquid)	4.184
Ice	2.09
Steam	2.01
Aluminum	0.897
Iron	0.449
Copper	0.385

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

Measure the initial temperature of the solution
Mix the reactants in the cup
Record the maximum (or minimum) temperature reached
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 ✅

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 ( $C_{\text{cal}}$ ):

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

Symbol	Meaning	Units
$q_{\text{cal}}$	Heat absorbed by calorimeter	kJ
$C_{\text{cal}}$

Important Distinction

Quantity	Symbol	Units	Usage
Specific heat	$c$	J/(g·°C)	Per gram
Heat capacity	$C$	J/°C or kJ/°C	For the whole calorimeter

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

$n = 1.50 \; \cancel{\text{g C}_7\text{H}_6\text{O}_2} \times \frac{1 \text{ mol C}_7\text{H}_6\text{O}_2}{122.12 \; \cancel{\text{g C}_7\text{H}_6\text{O}_2}} = 0.01228 \text{ mol C}_7\text{H}_6\text{O}_2$

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 ✅

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

Operation	Effect on $\Delta H$
Reverse the reaction	Change the sign
Multiply by a factor $n$	Multiply $\Delta H$ by $n$
Add reactions together	Add $\Delta H$ values

🛠️ Problem-Solving Strategy

Step-by-Step Approach

Write the target reaction — the one you need $\Delta H$ for
Examine the given reactions — look for each substance in your target
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
Add the manipulated reactions — substances on opposite sides cancel
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:

$\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5$ 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 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

Element	Standard State	$\Delta H°_f$
$\text{O}_2(g)$

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

Look up $\Delta H°_f$ for every compound in the reaction
Remember: $\Delta H°_f = 0$ for elements in their standard states
Multiply each by its coefficient

Worked Example

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

Substance	$\Delta H°_f$ (kJ/mol)	Coefficient
$\text{CH}_4(g)$

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

Formation Enthalpy Concept Quiz 🎯

Formation Enthalpy Calculations 🧮

Given:

Substance	$\Delta H°_f$ (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 ✅

Concept	Key Relationship	Notes
Exothermic	$\Delta H < 0$	System releases heat
Endothermic	$\Delta H > 0$	System absorbs heat
Reverse reaction	$\Delta H_{\text{rev}} = -\Delta H_{\text{fwd}}$	Sign change
Scaled reaction	$\Delta H_{\text{new}} = n \cdot \Delta H$	Linear scaling

Hess's Law & Formation

Method	Equation
Hess's Law	$\Delta H_{\text{total}} = \sum \Delta H_{\text{steps}}$
Formation enthalpies	$\Delta H°_{\text{rxn}} = \sum n \cdot \Delta H°_f(\text{products}) - \sum m \cdot \Delta H°_f(\text{reactants})$

🎯 AP Exam Strategies

Common AP Question Types

Calorimetry calculation — given mass, specific heat, ΔT → find q → find ΔH per mole
Hess's Law — manipulate 2-3 reactions to find ΔH for a target reaction
Formation enthalpy — use the master equation with a table of $\Delta H°_f$ values
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 the $\Delta H°_f$ values below, 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)

Substance	$\Delta H°_f$ (kJ/mol)
$\text{CO}_2(g)$

Comprehensive Concept Review 🔽

Final Exit Quiz — Thermochemistry Mastery ✅

(100.0)(4.184)(6.9)=

122.12g C7H6O21 mol C7H6O2=

\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2 = -283.0

Subtract the sum of reactants from the sum of products

= [-393.5 - 571.6] - [-74.8]

= -965.1 + 74.8 = -890.3 \text{ kJ}

Enthalpy and Calorimetry

Endothermic ( $q > 0$ )

At Constant Pressure

Key Signs

🌡️ Enthalpy Is a State Function

What This Means

Analogy

Consequences

🌡️ Standard Enthalpy

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

Standard State

Important Relationships

Specific Heat Capacity

📌 The Coffee-Cup Calorimeter

How It Works

Key Assumptions

Important Sign Convention

Constant Pressure

🧪 Worked Example

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

📌 Heat Capacity of the Calorimeter

Important Distinction

Finding $q_{\text{rxn}}$

🧪 Worked Example

Why It Works

Rules for Manipulating Equations

🛠️ Problem-Solving Strategy

Step-by-Step Approach

Worked Example

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

Examples

Critical Rule

📌 The Master Equation

How to Use It

Worked Example

Enthalpy

Hess's Law & Formation

🎯 AP Exam Strategies

Common AP Question Types

Common Mistakes to Avoid

Enthalpy and Calorimetry

Enthalpy and Calorimetry - Complete Interactive Lesson

Part 1: Enthalpy & ΔH

🔥 Energy, Systems, and Surroundings

📌 System and Surroundings

Energy Transfer

Sign Conventions

📌 Endothermic vs. Exothermic Processes

Exothermic (q<0q < 0q<0)

📌 Energy Diagrams

Exothermic Diagram

Endothermic Diagram

Key Relationship

Part 2: Exothermic & Endothermic

🌡️ Enthalpy (ΔH) — The Heat of Reaction

📖 What Is Enthalpy?

Part 3: Coffee Cup Calorimetry

☕ Calorimetry — Measuring Heat

📌 The Heat Equation

Part 4: Bomb Calorimetry

💣 Bomb Calorimetry

🏗️ Bomb Calorimeter Structure

Key Feature: Constant Volume

Part 5: Hess\'s Law

🔄 Hess's Law — Adding Enthalpy Changes

📏 Hess's Law

Part 6: Problem-Solving Workshop

🏗️ Standard Enthalpies of Formation

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review — Enthalpy and Calorimetry

📌 Complete Concept Map

Energy and Heat

Endothermic (q>0q > 0q>0)

At Constant Pressure

Key Signs

🌡️ Enthalpy Is a State Function

What This Means

Analogy

Consequences

🌡️ Standard Enthalpy

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

Standard State

Important Relationships

Specific Heat Capacity

📌 The Coffee-Cup Calorimeter

How It Works

Key Assumptions

Important Sign Convention

Constant Pressure

🧪 Worked Example

Relationship Between ΔH\Delta HΔH and ΔE\Delta EΔE

📌 Heat Capacity of the Calorimeter

Important Distinction

Finding qrxnq_{\text{rxn}}qrxn​

🧪 Worked Example

Why It Works

Rules for Manipulating Equations

🛠️ Problem-Solving Strategy

Step-by-Step Approach

Worked Example

🌡️ Standard Enthalpy of Formation (ΔH°f\Delta H°_fΔH°f​)

Examples

Critical Rule

📌 The Master Equation

How to Use It

Worked Example

Enthalpy

Hess's Law & Formation

🎯 AP Exam Strategies

Common AP Question Types

Common Mistakes to Avoid

Exothermic ( $q < 0$ )

Endothermic ( $q > 0$ )

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

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

Finding $q_{\text{rxn}}$

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