M6Q8: Standard Enthalpy of Formation

Introduction

Concluding Module 6, this section introduces the enthalpy of formation and how to perform calculations for enthalpy of formation reactions run under standard state conditions.  How to determine the enthalpy change of a reaction utilizing enthalpies of formation and Hess’s Law is also included. This section includes worked examples, sample problems, and a glossary.

Learning Objectives for Enthalpy of Formation Reactions

| 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 ΔH°f for an element in its standard state is 0 kJ/mol (you should memorize the standard states of the elements listed in Table 1).

Table 1. Select elements in their Standard States
Element  Standard State
most metals Li(s) for example
mercury Hg(ℓ)
carbon C(s, graphite) 
diatomic elements H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(ℓ), I2(s)
sulfur S8(s) (called S(s, rhombic) in Appendix F)
phosphorus P4(s, white)

The standard enthalpy of formation of CO2(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

C(s, graphite) + O2(g)  →  CO2(g)     ΔH°f = ΔH°298 = -393.509 kJ/mol

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 CO2, also at 1 atm and 25 °C. For nitrogen dioxide, NO2(g), ΔH°f is 33.2 kJ/mol. This is the enthalpy change for the reaction:

½ N2(g) + O2(g)  →  NO2(g)     ΔH°f = ΔH°298 = +33.18 kJ/mol

A reaction equation with ½ mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO2(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 P4O10) to strongly endothermic (such as +226.73 kJ/mol for the formation of acetylene, C2H2). 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/mol) whereas for diamond, ΔH°f = 2.4 kJ/mol. 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, O3(g), forms from oxygen, O2(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:

3 O2(g)  →  2 O3(g)     ΔH°298 = +285.4 kJ/mol

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 O3(g) is the enthalpy change for the reaction:

\frac{3}{2} O2(g)  →  O3(g)

For the formation of 2 mol of O3(g), ΔH°298  =  +285.4 kJ/mol. This ratio, \left(\dfrac{285.4\;\text{kJ}}{2 \;\text{mol O}_3}\right), can be used as a conversion factor to find the heat produced when 1 mole of O3(g) is formed, which is the enthalpy of formation for O3(g):

ΔH° for 1 mol of O3(g) = 1 mol O3 × \dfrac{285.4\;\text{kJ}}{2 \;\rule[0.5ex]{3.2em}{0.1ex}\hspace{-3.2em}\text{mol O}_3} = +142.7 kJ

Therefore, ΔH°f [O3(g)] = +142.7 kJ/mol

Check Your Learning
Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(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:

H2(g) + Cl2(g)  →  2 HCl(g)     ΔH°298 = -184.61 kJ/mol

Example 2

Writing Reaction Equations for ΔH°f
Write the enthalpy of formation reaction equations for:

(a) C2H5OH(ℓ)

(b) Ca3(PO4)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 H2(g) + ½ O2(g)  →  C2H5OH(ℓ)

(b)     3 Ca(s) + ½ P4(s) + 4 O2(g)  →  Ca3(PO4)2(s)

Note: The standard state of carbon is graphite, and phosphorus exists as P4.

Check Your Learning
Write the enthalpy of formation reaction equations for:

(a) C2H5OC2H5(ℓ)

(b) Na2CO3(s)

Answer:

(a)     4 C(s, graphite) + 5 H2(g) + ½ O2(g)  →  C2H5OC2H5(ℓ)

(b)     2 Na(s) + C(s, graphite) + \frac{3}{2} O2(g)  →  Na2CO3(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:

  1. 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:

  1. 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:

ΔH°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:

3 NO2(g) + H2O(ℓ)  →  2 HNO3(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 × ΔH°f (HNO3(aq)) + 1 × ΔH°f (NO(g))] – [3 × ΔH°f (NO2(g)) + 1 × ΔH°f (H2O(ℓ))]

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

=  -138.18 kJ/mol

Solution: Supporting Why the General Equation Is Valid
Alternatively, we can write this reaction as the sum of the decompositions of 3 NO2(g) and 1 H2O(ℓ) into their constituent elements, and the formation of 2 HNO3(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:

3 NO2(g)  →  \frac{3}{2} N2(g) + 3 O2(g)
⇒ ΔH°1 = -3 × ΔH°f (NO2(g)) = -99.54 kJ/mol

and

H2O(ℓ)  →  H2(g) + ½ O2(g)
⇒ ΔH°2 = -ΔH°f (H2O(l)) = +285.83 kJ/mol

and

H2(g) + N2(g) + 3 O2(g)  →  2 HNO3(aq)
⇒ ΔH°3 = 2 × ΔH°f (HNO3(aq)) = -414.72 kJ/mol

and

½ N2(g) + ½ O2(g)  →  NO(g)
⇒ ΔH°4 = ΔH°f (NO(g)) = +90.25 kJ/mol

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

3 NO2(g) + H2O(ℓ)  →  2 HNO3(aq) + NO(g)

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

ΔH°rxn = ΔH°1 + ΔH°2 + ΔH°3 + ΔH°4 = (-99.54 kJ/mol) + (+285.83 kJ/mol) + (-414.72 kJ/mol) + (+90.25 kJ/mol) = -138.18 kJ/mol

So the standard enthalpy change for this reaction is ΔH° = -138.18 kJ/mol.

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, C2H5OH(ℓ), when H2O(ℓ) and CO2(g) are formed. Use the following enthalpies of formation: C2H5OH(ℓ), −277.69 kJ/mol; H2O(ℓ), −285.83 kJ/mol; and CO2(g), −393.509 kJ/mol.

Answer:

The combustion reaction is: C2H5OH(ℓ) + 3 O2(g)  →  3 H2O(ℓ) + 2 CO2(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 (ΔHf°)
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

  1. Calculate the standard molar enthalpy of formation of NO(g) from the following data:
    N2(g) + 2O2(g) ⟶ 2NO2(g) ΔH°298 = 66.4 kJ/mol
    2NO(g) + O2(g) ⟶ 2NO2(g) ΔH°298 = -114.1 kJ/mol
  2. Calculate the enthalpy change for the following reaction:
    2C4H10(g) + 13O2(g)  →  10H2O(g) + 8CO2(g)

    Determine the enthalpy change for the complete combustion of 22.3 g butane, C4H10(g).

  3. Propane, C3H8, is a hydrocarbon that is commonly used as a fuel.
    1. Write a balanced equation for the complete combustion of propane gas.
    2. 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 O2 by volume. (Hint: Use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O2 per liter.)
    3. The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, ΔH°f, of propane given that ΔH°f of H2O(ℓ) = −285.8 kJ/mol and ΔH°f of CO2(g) = −393.5 kJ/mol.
    4. 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.
  4. Use the enthalpy of formation values at 1 bar (~1 atm) and 25 °C to answer the following questions. Hint: be specific about what is happening at the particle level and refer to energy released or required in any chemical reactions or physical processes.
    ΔfH° (kJ/mol)
    Molecular hydrogen gas, H2(g) 0
    Atomic hydrogen gas, H(g) +218
    1. State why the enthalpy of formation of molecular hydrogen gas is 0 kJ/mol.
    2. Rationalize why the enthalpy of formation of atomic hydrogen gas is not 0 kJ/mol and why it is a positive value.

Answers to Chemistry End of Section Exercises

  1. 90.3 kJ/mol of NO
  2. -5313.8 kJ/mol; -1020 kJ
  3. (a) C3H8(g) + 5 O2(g)  ⟶  3 CO2(g) + 4 H2O(ℓ)
    (b) 330 L
    (c) −104.5 kJ/mol
    (d) 75.4 °C
  4. (a) Molecular hydrogen gas (H2­) is an element in the standard state, or the most stable state, so we arbitrarily set it as 0 kJ/mol.
    (b) H(g) atoms are not in the standard state for elemental hydrogen at 1 bar and 25°C because H-H bonds need to be broken to form H atom from H2 and breaking bonds is endothermic.
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