# 33 Standard Enthalpy of Formation (M6Q8)

## Introduction

Learning Objectives for Enthalpy of Formation Reactions

- Use standard enthalpies of formation to calculate a reaction enthalpy change.

| Standard Enthalpy of Formation | Hess’s Law with Standard Enthalpy of Formation |

| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |

## Standard Enthalpy of Formation

A **standard enthalpy of formation (ΔH° _{f})** is an enthalpy change for a reaction in which exactly one 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction using Hess’s law.

The standard enthalpy of formation of CO_{2}(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

C(s, graphite) + O_{2}(g) → CO_{2}(g) ΔH°_{f} = ΔH°_{298} = -393.509 kJ

starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO_{2}, also at 1 atm and 25 °C. For nitrogen dioxide, NO_{2}(g), ΔH°_{f} is 33.2 kJ/mol. This is the enthalpy change for the reaction:

½ N_{2}(g) + O_{2}(g) → NO_{2}(g) ΔH°_{f} = ΔH°_{298} = +33.18 kJ

A reaction equation with ½ mole of N_{2} and 1 mole of O_{2} is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO_{2}(g).

You will find a table of standard enthalpies of formation of many common substances in Appendix F. These values indicate that formation reactions range from highly exothermic (such as −2984.0 kJ/mol for the formation of P_{4}O_{10}) to strongly endothermic (such as +226.73 kJ/mol for the formation of acetylene, C_{2}H_{2}). By definition, the standard enthalpy of formation of an element in its most stable form (most stable allotrope) is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions. For example, carbon exists as several allotropes, two of them being graphite and diamond. Graphite is the more thermodynamically stable (ΔH°_{f } = 0 kJ/mole) whereas for diamond, ΔH°_{f } = 2.4 kJ/mole. While this table does not list the complete reaction that is represented by the enthalpy change, you can use the definition above to write these reactions when needed.

### Example 1

**Evaluating an Enthalpy of Formation**

Ozone, O_{3}(g), forms from oxygen, O_{2}(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, ΔH°_{f}, of ozone from the following information:

_{2}(g) → 2 O

_{3}(g) ΔH°

_{298}= +285.4 kJ

**Solution**

ΔH°_{f} is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, ΔH°_{f} for O_{3}(g) is the enthalpy change for the reaction:

_{2}(g) → O

_{3}(g)

For the formation of 2 mol of O_{3}(*g*), ΔH°_{298} = +285.4 kJ. This ratio, [latex](\frac{285.4\;\text{kJ}}{2 \;\text{mol O}_3})[/latex], can be used as a conversion factor to find the heat produced when 1 mole of O_{3}(g) is formed, which is the enthalpy of formation for O_{3}(g):

_{3}(g) = 1

_{3}

Therefore, ΔH°_{f }[O_{3}(g)] = +142.7 kJ/mol

**Check Your Learning**

Hydrogen gas, H_{2}, reacts explosively with gaseous chlorine, Cl_{2}, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H_{2}(g) with 1 mole of Cl_{2}(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.307 kJ/mol.

### Answer:

For the reaction:

H_{2}(g) + Cl_{2}(g) → 2 HCl(g) ΔH°_{298} = -184.61 kJ

### Example 2

**Writing Reaction Equations for ΔH° _{f}**

Write the enthalpy of formation reaction equations for:

(a) C_{2}H_{5}OH(l)

(b) Ca_{3}(PO_{4})_{2}(s)

**Solution**

Remembering that ΔH°_{f }reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

(a)

2 C(s, graphite) + 3 H_{2}(g) + ½ O_{2}(g) → C_{2}H_{5}OH(l)

(b)

3 Ca(s) + ½ P_{4}(s) + 4 O_{2}(g) → Ca_{3}(PO_{4})_{2}(s)

Note: The standard state of carbon is graphite, and phosphorus exists as P_{4}.

**Check Your Learning**

Write the enthalpy of formation reaction equations for:

(a) C_{2}H_{5}OC_{2}H_{5}(l)

(b) Na_{2}CO_{3}(s)

### Answer:

(a)

4 C(s, graphite) + 5 H_{2}(g) + [latex]\frac{1}{2}[/latex] O_{2}(g) → C_{2}H_{5}OC_{2}H_{5}(l)

(b)

2 Na(s) + C(s, graphite) + [latex]\frac{3}{2}[/latex] O_{2}(g) → Na_{2}CO_{3}(s)

## Hess’s Law with Standard Enthalpy of Formation

In the previous section, we learned about Hess’s law and how it can be useful for determining the ΔH of a given reaction since we can use other reactions as “steps” to achieve the overall reaction of interest and the ΔH values for the “steps” would sum to give the overall ΔH. We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. In this case, the stepwise reactions we consider are:

(i) Decompositions of the reactants into their component elements. If you were to write the reaction for the decomposition of a reactant (where you start with a compound and end with component elements), this would be related to the reverse reaction of enthalpy of formation (where you start with component elements and form a compound). So the enthalpy for these reactions would be proportional to the negative of the enthalpies of formation of the reactants.

This step would be followed by:

(ii) Recombinations of the elements to give the products. The enthalpy change of this step would then be proportional to the enthalpy of formation of the products since we are taking component elements and forming a compound.

The standard enthalpy change of the overall reaction is therefore equal to:

(ii) The sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and *n* standing for the stoichiometric coefficients:

_{reaction}= ∑

*n*× ΔH°

_{f }(products) – ∑

*n*× ΔH°

_{f }(reactants)

The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.

### Example 3

**Using Hess’s Law**

Find the standard enthalpy change for the following reaction:

_{2}(g) + H

_{2}O(l) → 2 HNO

_{3}(aq) + NO(g) ΔH° = ?

**Solution: Using the Equation**

Use the special form of Hess’s law given previously:

ΔH°_{reaction} = ∑*n* × ΔH°_{f }(products) – ∑*n* × ΔH°_{f }(reactants)

= [2 mol HNO_{3} × [latex]\frac{-207.36\;\text{kJ}}{\text{mol HNO}_{3}}[/latex] + 1 mol NO × [latex]\frac{+90.25\;\text{kJ}}{\text{mol NO}}[/latex]] - [3 mol NO_{2} × [latex]\frac{+33.18\;\text{kJ}}{\text{mol NO}_{2}}[/latex] + 1 mol H_{2}O × [latex]\frac{-285.83\;\text{kJ}}{\text{mol H}_{2}\text{O}}[/latex]]

= [2(-207.36 kJ) + 1(+90.25 kJ)] - [3(+33.18 kJ) + 1(-285.83 kJ)]

= -138.18 kJ

**Solution: Supporting Why the General Equation Is Valid**

Alternatively, we can write this reaction as the sum of the decompositions of 3 NO_{2}(g) and 1 H_{2}O(l) into their constituent elements, and the formation of 2 HNO_{3}(aq) and 1 NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the ΔH°_{f} values for these compounds (from Appendix F), we have:

_{2}(g) → [latex]\frac{3}{2}[/latex] N

_{2}(g) + 3 O

_{2}(g)

⇒ ΔH°

_{1}= -3 × ΔH°

_{f }(NO

_{2}(g)) = -99.54 kJ

_{2}O(l) → H

_{2}(g) + [latex]\frac{1}{2}[/latex] O

_{2}(g)

⇒ ΔH°

_{2}= -ΔH°

_{f}(H

_{2}O(l)) = +285.83 kJ

_{2}(g) + N

_{2}(g) + 3 O

_{2}(g) → 2 HNO

_{3}(aq)

⇒ ΔH°

_{3}= 2 × ΔH°

_{f }(HNO

_{3}(aq)) = -414.72 kJ

_{2}(g) + [latex]\frac{1}{2}[/latex] O

_{2}(g) → NO(g)

⇒ ΔH°

_{4}= ΔH°

_{f }(NO(g)) = +90.25 kJ

Summing these reaction equations gives the reaction we are interested in:

_{2}(g) + H

_{2}O(l) → 2 HNO

_{3}(aq) + NO(g)

Summing their enthalpy changes gives the value we want to determine:

_{rxn}= ΔH°

_{1 }+ ΔH°

_{2 }+ ΔH°

_{3 }+ ΔH°

_{4}= (-99.54 kJ) + (+285.83 kJ) + (-414.72 kJ) + (+90.25 kJ) = -138.18 kJ

So the standard enthalpy change for this reaction is ΔH° = −138.18 kJ.

Note that this result was obtained by (1) multiplying the ΔH°_{f} of each product by its stoichiometric coefficient and summing those values, (2) multiplying the ΔH°_{f} of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.

**Check Your Learning**

Calculate the heat of combustion of 1 mole of ethanol, C_{2}H_{5}OH(l), when H_{2}O(l) and CO_{2}(g) are formed. Use the following enthalpies of formation: C_{2}H_{5}OH(l), −277.69 kJ/mol; H_{2}O(l), −285.83 kJ/mol; and CO_{2}(g), −393.509 kJ/mol.

### Answer:

The combustion reaction is: C_{2}H_{5}OH(l) + 3 O_{2}(g) → 3 H_{2}O(l) + 2 CO_{2}(g)

−1366.8 kJ/mol

## Key Concepts and Summary

The standard enthalpy of formation, ΔH°_{f}, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar (standard state). Many of the calculations are carried out at 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.

## Key Equations

- ΔH°
_{reaction}= ∑*n*× ΔH°_{f }(products) - ∑*n*× ΔH°_{f }(reactants)

## Glossary

- standard enthalpy of formation (ΔH
_{f}°) - enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions

### Chemistry End of Section Exercises

- Calculate the standard molar enthalpy of formation of NO(
*g*) from the following data:

N _{2}(*g*) + 2O_{2}(*g*)⟶ 2NO _{2}(*g*)Δ *H*°_{298}= 66.4 kJ2NO( *g*) + O_{2}(*g*)⟶ 2NO _{2}(*g*)Δ *H*°_{298}= -114.1 kJ - Calculate the enthalpy change for the following reaction: 2C
_{4}H_{10}(*g*) + 13O_{2}(g) → 10H_{2}O(*g*) + 8CO_{2}(*g*). Determine the enthalpy change for the complete combustion of 22.3 g butane, C_{4}H_{10}(g). - Propane, C
_{3}H_{8}, is a hydrocarbon that is commonly used as a fuel.- Write a balanced equation for the complete combustion of propane gas.
- Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O
_{2}by volume. (Hint: Use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O_{2}per liter.) - The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, [latex]\Delta H^{\circ}_\text{f}[/latex], of propane given that [latex]\Delta H^{\circ}_\text{f}[/latex] of H
_{2}O(*l*) = −285.8 kJ/mol and [latex]\Delta H^{\circ}_\text{f}[/latex] of CO_{2}(*g*) = −393.5 kJ/mol. - Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.

**Answers to Chemistry End of Section Exercises**

- 90.3 kJ/mol of NO
- -5313.8 kJ/mol; -1020 kJ
- (a) C
_{3}H_{8}(*g*) + 5O_{2}(*g*) ⟶ 3CO_{2}(*g*) + 4H_{2}O(*l*); (b) 330 L; (c) −104.5 kJ/mol; (d) 75.4 °C