Enthalpies of Formation

Caleb Thomas

8 min read

Study Guide Overview

This study guide covers standard enthalpy of formation (ΔH_f), including its definition, standard conditions, and the significance of zero ΔH_f for elements in their standard states. It explains how to calculate ΔH_rxn using the formula ΔH_rxn = ΣnΔH_f(products) - ΣmΔH_f(reactants), with examples and common mistakes to avoid. It includes an AP-style practice question on enthalpy of combustion and concludes with final exam focus, key takeaways, common question types, and last-minute tips.

#Thermochemistry: Enthalpy of Formation Study Guide

Hey there, future AP Chem master! Let's dive into enthalpy of formation – a key concept that pops up all over the exam. This guide is designed to make sure you're not just memorizing, but truly understanding this material. Let's get started!

#Standard Enthalpy of Formation (ΔH_f)

#What is it?

The standard enthalpy of formation (ΔH_f) is the change in enthalpy when one mole of a compound is formed from its elements in their most stable state at standard conditions (25°C and 1 atm). Think of it as the energy 'cost' of building a compound from scratch 🏗️.

Key Idea: It's always for one mole of product.
Standard Conditions: 25°C (298 K) and 1 atm.
Most Stable State: The form of an element that is most stable at standard conditions (e.g., O₂(g), C(s, graphite)).

Key Concept

For elements in their standard states, ΔH_f is always zero. This is a crucial point to remember!

#Visualizing ΔH_f

Let's break down CO₂ formation:

C(s, graphite) + O₂(g) → CO₂(g)

The ΔH for this reaction is the ΔH_f of CO₂. Notice how we're starting with the elements in their standard states? That's key!

Enthalpy of Formation

Image Courtesy of Google Sites

Memory Aid

Analogy: Think of ΔH_f as the energy needed to assemble a Lego set (the compound) from individual Lego bricks (the elements in their standard states). Elements are the basic building blocks, and their ΔH_f is zero because they are already in their simplest form.

#Calculating ΔH_rxn Using ΔH_f

#The Formula

Here's the magic formula to calculate the enthalpy change of a reaction (ΔH_rxn) using standard enthalpies of formation:

ΔH_rxn = ΣnΔH_f(products) - ΣmΔH_f(reactants)

Σ means 'sum of'.
n and m are the stoichiometric coefficients from the balanced equation.

Exam Tip

Products minus reactants – this is a common source of errors. Double-check! Also, remember that this formula is on the reference sheet. Use it to double-check your work and ensure you don't confuse it with bond enthalpy formula.

#Example 1

Let's calculate ΔH_rxn for:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l)

Given:

ΔH_f(CO₂(g)) = -393.5 kJ/mol
ΔH_f(H₂O(l)) = -241.8 kJ/mol
ΔH_f(C₂H₂(g)) = -147.3 kJ/mol
ΔH_f(O₂(g)) = 0 kJ/mol

ΔH_rxn = [4(-393.5) + 2(-241.8)] - [2(-147.3) + 5(0)] = -5271.4 kJ

Example Problem

#Example 2

Let's calculate ΔH_rxn for:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Substance	kJ/mol
CH₄(g)	-74.8
O₂(g)	0
CO₂(g)	-393.5
H₂O(g)	-241.8
H₂O(l)	-285.8

ΔH_rxn = [1(-393.5) + 2(-285.8)] - [1(-74.8) + 2(0)] = -890.3 kJ/mol

Common Mistake

State Symbols Matter! Note that we used the ΔH_f for liquid water (H₂O(l)) and not gaseous water (H₂O(g)). Always check the state symbols in the problem and the table. The AP exam loves to test this!

#Practice AP Question

Let's tackle a real AP-style question from the 2014 exam:

Question:

In a separate experiment, the student measures the enthalpies of combustion of propene and vinyl chloride. The student determines that the combustion of 2.00 mol of vinyl chloride releases 2300 kJ of energy, according to the equation below.

Vinyl Chloride Combustion

Image Courtesy of College Board

Using the table of standard enthalpies of formation below, determine whether the combustion of 2.00 mol of propene releases more, less, or the same amount of energy that 2.00 mol of vinyl chloride releases. Justify your answer with a calculation.

The balanced equation for the combustion of 2.00 mol of propene is 2 C₃H₆(g) + 9 O₂(g) → 6 CO₂(g) + 6 H₂O(g).

Enthalpies of Formation Table

Image Courtesy of College Board

Solution:

Calculate ΔH_rxn for propene combustion:

ΔH_rxn = [6(-242) + 6(-394)] - [2(21) + 9(0)] = -3858 kJ/mol
Compare:
- Combustion of 2 mol of vinyl chloride releases 2300 kJ.
- Combustion of 2 mol of propene releases 3858 kJ.
Conclusion:

The combustion of 2.00 mol of propene releases more energy than the combustion of 2.00 mol of vinyl chloride.

Exam Tip

Show Your Work! Always show your calculations and substitutions, especially when the question asks you to "justify your answer with a calculation." This is how you secure those points!

#Final Exam Focus

#Key Takeaways

ΔH_f Definition: Know what it represents and that elements in their standard state have a ΔH_f of zero.
ΔH_rxn Calculation: Master the formula: ΔH_rxn = ΣnΔH_f(products) - ΣmΔH_f(reactants)
State Symbols: Always pay attention to the state symbols (s, l, g, aq).
Units: Make sure you understand the units (kJ/mol) and use them correctly.

#Common Question Types

Multiple Choice: Expect questions that test your understanding of the definition of ΔH_f and the formula for calculating ΔH_rxn. Also, expect questions where you need to identify the correct state of the reactants and products.
Free Response: FRQs often combine ΔH_f calculations with other concepts like Hess's Law or calorimetry. Be prepared to justify your answers with calculations.

#Last-Minute Tips

Time Management: Practice doing these calculations quickly. It's all about speed and accuracy.
Common Pitfalls: Double-check your math, pay attention to state symbols, and make sure you're using the correct formula (products - reactants).
Stay Calm: You've got this! Take a deep breath and approach each question systematically.

#Practice Questions

Practice Question

#Multiple Choice Questions

The standard enthalpy of formation of which of the following substances is NOT zero at 298 K and 1 atm?

(A) Br₂(l) (B) Fe(s) (C) H₂(g) (D) NaCl(s)
Given the following standard enthalpies of formation:

ΔH_f°[Al₂O₃(s)] = -1676 kJ/mol ΔH_f°[Fe₂O₃(s)] = -824 kJ/mol

What is the standard enthalpy change for the following reaction?

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

(A) -2500 kJ/mol (B) -852 kJ/mol (C) +852 kJ/mol (D) +2500 kJ/mol

#Free Response Question

Consider the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Given the following standard enthalpies of formation:

ΔH_f°[NH₃(g)] = -46 kJ/mol

a) Calculate the standard enthalpy change (ΔH°) for the reaction. b) If the reaction is carried out at a higher temperature, how would this affect the value of ΔH°? Explain your reasoning. c) If the reaction is carried out in the presence of a catalyst, how would this affect the value of ΔH°? Explain your reasoning.

Scoring Breakdown:

a) (2 points)

 -   1 point for correctly applying the formula: ΔH° = ΣnΔH<sub>f</sub>(products) - ΣmΔH<sub>f</sub>(reactants)
 -   1 point for the correct calculation: ΔH° = [2(-46)] - [1(0) + 3(0)] = -92 kJ/mol

b) (2 points)

 -   1 point for stating that the value of ΔH° would not be affected.
 -   1 point for explaining that enthalpy change is independent of temperature

c) (2 points)

 -   1 point for stating that the value of ΔH° would not be affected.
 -   1 point for explaining that a catalyst does not change the enthalpy of the reaction, it only changes the rate.

You've got this! Remember to stay confident, trust your preparation, and tackle the exam with a clear mind. You're ready to ace it! 🚀