Enthalpy and Calorimetry

Understand heat, enthalpy changes, calorimetry, Hess's law, and standard enthalpies of formation.

Written and reviewed by the Study Mondo Education TeamLast updated February 14, 2026

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Enthalpy and Calorimetry

Energy and Heat

System: Part of universe we're studying Surroundings: Everything else

Energy transfer:

Heat (q): Energy transferred due to temperature difference
Work (w): Energy transferred by force through distance

First Law of Thermodynamics:

$\Delta E = q + w$

Sign conventions:

+q: Heat absorbed by system (endothermic)
-q: Heat released by system (exothermic)
+w: Work done on system
-w: Work done by system

Enthalpy (H)

Enthalpy: Heat content at constant pressure

$\Delta H = q_p$

At constant pressure (most reactions): Heat change = enthalpy change

Exothermic reaction:

ΔH < 0 (negative)
Releases heat to surroundings
Products lower energy than reactants
Feels warm

Endothermic reaction:

ΔH > 0 (positive)
Absorbs heat from surroundings
Products higher energy than reactants
Feels cold

Calorimetry

Calorimetry: Measuring heat changes

Heat capacity equation:

$q = mc\Delta T$

Where:

q = heat (J)
m = mass (g)
c = specific heat capacity (J/g·°C)
ΔT = T_{final} - T_{initial}

Common specific heats:

Water: 4.18 J/(g·°C)
Metals: 0.1-1 J/(g·°C)

Coffee Cup Calorimeter

For solution reactions:

Constant pressure (open to atmosphere)
Measures q_p = ΔH
Simple styrofoam cup

Assumption: All heat goes to solution (usually aqueous)

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

Bomb Calorimeter

For combustion:

Constant volume
Measures ΔE (not ΔH)
More precise, sealed container

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

C_{cal} = calorimeter constant (J/°C)

Hess's Law

Hess's Law: ΔH is path independent - only depends on initial and final states

Use: Calculate ΔH for reaction from known ΔH values

Rules:

Reverse reaction → change sign of ΔH
Multiply reaction by n → multiply ΔH by n
Add reactions → add ΔH values

Example approach:

Manipulate given equations to match target
Add them to get target equation
Add corresponding ΔH values

Standard Enthalpy of Formation (ΔH°_f)

ΔH°_f: Enthalpy change to form 1 mole from elements in standard states

Standard conditions:

25°C (298 K)
1 atm pressure
1 M concentration

Key rules:

ΔH°_f of element in standard state = 0
ΔH°_f values tabulated for compounds

Examples:

C(graphite): ΔH°_f = 0
O₂(g): ΔH°_f = 0
H₂O(l): ΔH°_f = -285.8 kJ/mol
CO₂(g): ΔH°_f = -393.5 kJ/mol

Calculating ΔH°_{rxn}

From ΔH°_f values:

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

Steps:

Sum (coefficient × ΔH°_f) for products
Sum (coefficient × ΔH°_f) for reactants
Subtract: products - reactants

Shortcut: Products minus reactants (with coefficients)

Bond Enthalpies

Bond enthalpy: Energy to break 1 mole of bonds (always positive)

Estimating ΔH_{rxn}:

$\Delta H_{\text{rxn}} = \sum \text{bonds broken} - \sum \text{bonds formed}$

Energy required to break bonds (positive) Energy released forming bonds (negative)

Note: Bond enthalpies are averages, less accurate than ΔH°_f

📚 Practice Problems

1Problem 1easy

❓ Question:

When 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a coffee cup calorimeter, the temperature rises from 21.0°C to 27.5°C. Calculate ΔH for the reaction in kJ/mol. Assume solution density = 1.0 g/mL and c = 4.18 J/(g·°C).

💡 Show Solution

Given:

V_{HCl} = 50.0 mL, [HCl] = 1.0 M
V_{NaOH} = 50.0 mL, [NaOH] = 1.0 M
T_i = 21.0°C, T_f = 27.5°C
Density = 1.0 g/mL, c = 4.18 J/(g·°C)

Reaction: HCl + NaOH → NaCl + H₂O

Step 1: Calculate heat absorbed by solution

Total volume = 50.0 + 50.0 = 100.0 mL

Mass = 100.0 mL × 1.0 g/mL = 100.0 g

ΔT = 27.5 - 21.0 = 6.5°C

$q_{\text{solution}} = mc\Delta T$ $q_{\text{solution}} = (100.0)(4.18)(6.5)$ $q_{\text{solution}} = 2717 \text{ J} = 2.72 \text{ kJ}$

Step 2: Calculate heat of reaction

$q_{\text{reaction}} = -q_{\text{solution}} = -2.72 \text{ kJ}$

(Negative because reaction releases heat - exothermic)

Step 3: Calculate moles reacted

Moles HCl = (1.0 M)(0.050 L) = 0.050 mol Moles NaOH = (1.0 M)(0.050 L) = 0.050 mol

Limiting reactant: Both 0.050 mol (1:1 ratio) → 0.050 mol reacts

Step 4: Calculate ΔH per mole

$\Delta H = \frac{q_{\text{reaction}}}{\text{moles}} = \frac{-2.72 \text{ kJ}}{0.050 \text{ mol}}$

$\Delta H = -54.4 \text{ kJ/mol}$

Answer: ΔH = -54 kJ/mol (exothermic)

Note: Literature value is -57.1 kJ/mol - our answer is close!

2Problem 2medium

❓ Question:

A 50.0 g sample of aluminum is heated from 20.0°C to 95.0°C. (a) Calculate the heat absorbed by the aluminum (specific heat of Al = 0.900 J/g°C). (b) If this heat came from burning methane (CH₄), which releases 890 kJ/mol, how many grams of methane were burned?

💡 Show Solution

Solution:

(a) Heat absorbed by aluminum: q = mcΔT where m = mass, c = specific heat, ΔT = temperature change

q = (50.0 g)(0.900 J/g°C)(95.0 - 20.0)°C q = (50.0)(0.900)(75.0) q = 3,375 J = 3.38 kJ

(b) Mass of methane burned: Energy from CH₄: 890 kJ/mol

Moles of CH₄ = 3.38 kJ / 890 kJ/mol = 0.00380 mol

Molar mass of CH₄ = 12.01 + 4(1.008) = 16.04 g/mol Mass = 0.00380 mol × 16.04 g/mol = 0.0609 g

3Problem 3medium

❓ Question:

Given: (1) C(s) + O₂(g) → CO₂(g), ΔH° = -393.5 kJ, (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH° = -285.8 kJ, (3) C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH° = -1367 kJ. Use Hess's law to find ΔH° for: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

💡 Show Solution

Target equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

Given equations:

C(s) + O₂(g) → CO₂(g), ΔH° = -393.5 kJ
H₂(g) + ½O₂(g) → H₂O(l), ΔH° = -285.8 kJ
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH° = -1367 kJ

Strategy: Manipulate equations to match target

Target needs:

2 C(s) on left → multiply equation 1 by 2
3 H₂(g) on left → multiply equation 2 by 3
C₂H₅OH(l) on right → reverse equation 3

Equation 1 × 2: 2C(s) + 2O₂(g) → 2CO₂(g) ΔH° = 2(-393.5) = -787.0 kJ

Equation 2 × 3: 3H₂(g) + 3/2 O₂(g) → 3H₂O(l) ΔH° = 3(-285.8) = -857.4 kJ

Equation 3 reversed: 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g) ΔH° = -(-1367) = +1367 kJ

Add all three:

2C(s) + 2O₂(g) → 2CO₂(g) 3H₂(g) + 3/2 O₂(g) → 3H₂O(l) 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g)

Cancel species on both sides:

2CO₂(g): appears as product (eq 1) and reactant (eq 3) → cancel
3H₂O(l): appears as product (eq 2) and reactant (eq 3) → cancel
O₂(g): 2 + 3/2 on left, 3 on right → net ½ on left

Net equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) ✓

Add ΔH° values:

$\Delta H° = -787.0 + (-857.4) + 1367$ $\Delta H° = -787.0 - 857.4 + 1367$ $\Delta H° = -277.4 \text{ kJ}$

Answer: ΔH° = -277 kJ

This is ΔH°_f of ethanol!

4Problem 4hard

❓ Question:

Given the following thermochemical equations:

C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJ
H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -285.8 kJ
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH° = -3119.6 kJ

Use Hess's Law to calculate ΔH°_f for C₂H₆(g).

💡 Show Solution

Solution:

We want: 2C(s) + 3H₂(g) → C₂H₆(g) ΔH°_f = ?

Strategy: Manipulate given equations to get target equation.

From equation 3: 4CO₂(g) + 6H₂O(l) → 2C₂H₆(g) + 7O₂(g) ΔH = +3119.6 kJ (reversed)

From equation 1 (×4): 4C(s) + 4O₂(g) → 4CO₂(g) ΔH = 4(-393.5) = -1574.0 kJ

From equation 2 (×6): 6H₂(g) + 3O₂(g) → 6H₂O(l) ΔH = 6(-285.8) = -1714.8 kJ

Adding all three: 4CO₂ + 6H₂O → 2C₂H₆ + 7O₂ +3119.6 kJ 4C + 4O₂ → 4CO₂ -1574.0 kJ 6H₂ + 3O₂ → 6H₂O -1714.8 kJ

4C + 6H₂ → 2C₂H₆ -169.2 kJ

For 1 mole of C₂H₆: 2C(s) + 3H₂(g) → C₂H₆(g) ΔH°_f = -169.2 kJ / 2 = -84.6 kJ/mol

5Problem 5hard

❓ Question:

Calculate ΔH°_{rxn} for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) using ΔH°_f values: CH₄(g) = -74.8 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(l) = -285.8 kJ/mol, O₂(g) = 0.

💡 Show Solution

Reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Given ΔH°_f values:

CH₄(g): -74.8 kJ/mol
O₂(g): 0 (element in standard state)
CO₂(g): -393.5 kJ/mol
H₂O(l): -285.8 kJ/mol

Formula:

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

Products:

1 mol CO₂: 1(-393.5) = -393.5 kJ 2 mol H₂O: 2(-285.8) = -571.6 kJ

Sum products: -393.5 + (-571.6) = -965.1 kJ

Reactants:

1 mol CH₄: 1(-74.8) = -74.8 kJ 2 mol O₂: 2(0) = 0 kJ

Sum reactants: -74.8 + 0 = -74.8 kJ

Calculate ΔH°_{rxn}:

$\Delta H°_{\text{rxn}} = -965.1 - (-74.8)$ $\Delta H°_{\text{rxn}} = -965.1 + 74.8$ $\Delta H°_{\text{rxn}} = -890.3 \text{ kJ}$

Answer: ΔH°_{rxn} = -890 kJ

Interpretation:

Highly exothermic (negative ΔH)
This is combustion of methane (natural gas)
Releases 890 kJ per mole CH₄ burned
Why natural gas is good fuel

Check:

Products more negative than reactants → exothermic ✓
Magnitude makes sense for combustion ✓

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections

Section	Format	Questions	Time	Weight	Calculator
Multiple Choice	MCQ	60	90 min	50%	✅
Free Response (Long)	FRQ	3	69 min	30%	✅
Free Response (Short)	FRQ	4	36 min	20%	✅

📊 Scoring: 1-5

Extremely Qualified

~12%

Well Qualified

~16%

Qualified

~24%

Possibly Qualified

~24%

No Recommendation

~24%

💡 Key Test-Day Tips

✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws

⚠️ Common Mistakes: Enthalpy and Calorimetry

Avoid these 3 frequent errors

🌍 Real-World Applications: Enthalpy and Calorimetry

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

2Determine the limiting reagent

3Calculate moles of product

4Convert moles to grams

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Entropy and the Second Law

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❓ Frequently Asked Questions

What is Enthalpy and Calorimetry?▾

Understand heat, enthalpy changes, calorimetry, Hess's law, and standard enthalpies of formation.

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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Enthalpy and Calorimetry

Try the Interactive Version!

Enthalpy and Calorimetry

Energy and Heat

Enthalpy (H)

Calorimetry

Coffee Cup Calorimeter

Bomb Calorimeter

Hess's Law

Standard Enthalpy of Formation (ΔH°_f)

Calculating ΔH°_{rxn}

Bond Enthalpies

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

📋 AP Chemistry — Exam Format Guide

📊 Scoring: 1-5

💡 Key Test-Day Tips

⚠️ Common Mistakes: Enthalpy and Calorimetry

🌍 Real-World Applications: Enthalpy and Calorimetry

📝 Worked Example: Stoichiometry — Limiting Reagent

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Thermodynamics

❓ Frequently Asked Questions