Standard **enthalpy of formation **(or heat of formation), **Δ***H*^{o}_{f }, is the enthalpy change when 1 mol of the substance is formed from its constituent elements in their standard states.

For example, the formation of 1 mol ammonia from H_{2} and N_{2} gases releases 46.0 kJ heat:

1.5H_{2}(*g*) + 0.5N_{2}(*g*) ⇆ NH_{3}(*g*)** Δ H^{o}_{f }= 46.0 kJ**

A reminder about the s**tandard states; **depending on the topic, standard states are defined as follows:

It is the pure gas at a pressure of exactly 1 atm.*Gases:*It is the pure substance in its most stable form at a pressure of 1 atm and at the temperature of interest (usually at 25*Liquids and Solids:*^{o}C).It is when the concentration is exactly 1*A Substance in Solution:**M*.

Let’s answer one question before moving on.

**Why do we not talk about the enthalpy of a compound? **Why enthalpy of formation of CO_{2} and not enthalpy of CO_{2}?

Well, the problem is it is **impossible to measure the absolute value of the enthalpy** of a substance. So, what we do, is take a reference point like the sea level for the height of mountains, and measure the enthalpy change when a given compound or a substance is formed from its elements.

This **reference point** is the **standard enthalpy of formation of any element** in its most stable form which is set to **zero**.

* *

For example, the standard enthalpy of formation for O_{2} is zero, and that of Na, C (graphite), Br_{2}, etc. is zero.

Why did we mention graphite for carbon? Remember, carbon has a few allotropes, and the most common ones are graphite and diamond. Graphite is the most stable (standard state) of carbon and therefore, its enthalpy of formation is zero (), while that of diamond is not. Diamond is less stable and its conversion to graphite is an exothermic (downhill – thermodynamically favorable) process:

C(diamond) → C(graphite) Δ*H*°* = – *1.9 kJ/mol

Fortunately, this process does not happen overnight, and we get to keep our diamonds.

Going back to the enthalpy of formation. So, imagine the elements in their standard form are lined up on the sea level (Δ*H *°_{f} = 0). Now, when a compound is formed by combining any elements, there is an enthalpy change which is the standard enthalpy of formation for that compound:

**For example**, the standard enthalpy of formation for methane from carbon, and hydrogen is -74.6 kJ/mol:

C(graphite) + 2H_{2}(*g*) → CH_{4}(*g*) Δ*H *°_{f} = -74.6 kJ/mol

The comprehensive list of the heat of formation can be found in your textbook or by a quick google search.

**How do We use the Enthalpies of Formation?**

There are countless reactions that can be carried out and it is not always that we can or need to measure their enthalpy change. However, by having a library of values for the enthalpies of formations, we can determine the enthalpy of a given reaction. There are two approaches to doing this.

**1) The Direct Approach**

If the enthalpies of formations for all the components of ac chemical reaction are known, then the following formula is used to determine the enthalpy of this reaction:

We will discuss this approach in the next post called “Enthalpy of Reaction from Enthalpies of Formation”.

**2) The Indirect Approach**

If the enthalpies of formation of all the components in chemical reactions are not known, then Hess’s law is used. You can go to the link for detailed coverage of the Hess’s law, but let’s use it here to address one more question you may be wondering about:

**How is Enthalpy of Formation Measured?**

We talked about the enthalpy of formation and, as an example, showed that the enthalpy of formation of methane is -74 kJ/mol, but **how do we measure this?**

C(graphite) + 2H_{2}(*g*) → CH_{4}(*g*) Δ*H *°_{f} = -74.6 kJ/mol

When we say the “reaction” forming a substance from its constituent elements, it is, very often, a hypothetical reaction. We can’t combine carbon and hydrogen in the laboratory to make methane. So, to determine its enthalpy of formation, we measure the enthalpy of the combustion reaction of methane using bomb calorimetry and then, apply the Hess’s law to obtain the **Δ***H ***°**** _{f}** of methane.

The combustion reaction of methane is the sum of three reactions of formations of methane, CO_{2}, and H_{2}O:

C(*s*) + O_{2}(*g*) → CO_{2}(*g*) Δ*H *°_{f }(CO_{2})* *= -393 kJ

+

2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*l*) Δ*H *°_{f }(H_{2}O)* *= -572 kJ

+

C(graphite) + 2H_{2}(*g*) → CH_{4}(*g*) **–Δ H **

**°**

_{f }(CH_{4})**= ?**

**____________________________________________**

CH_{4}(*g*) + 2O_{2}(*g*) → CO_{2}(*g*) + 2H_{2}O(*l*) Δ*H *° = -891 kJ

Notice that there is a negatice sign in from of Δ*H *°_{f }(CH_{4}) becasue the *reaction is reversed*.

Therefore,

Δ*H *° = Δ*H *°_{f }(CO_{2}) + Δ*H *°_{f }(H_{2}O) + **(–Δ H °_{f }(CH_{4})) **= -891 kJ

Solving for Δ*H *°_{f }(CH_{4}), we get:

Δ*H *°_{f }(CH_{4}) = Δ*H *°_{f }(CO_{2}) + Δ*H *°_{f }(H_{2}O) + 891

**Δ H °_{f }(CH_{4}) **= -393 – 572 + 891 =

**-74 kJ/mol**

**Check Also**

- Energy Related to Heat and Work
- Endothermic and Exothermic Processes
- Heat Capacity and Specific Heat
- Heat Capacity Practice Problems
- What is Enthalpy
- Constant-Pressure Calorimetry
- Bomb calorimeter – Constant Volume Calorimetry
- Stoichiometry and Enthalpy of Chemical Reactions
- Hess’s Law and Enthalpy of Reaction
- Hess’s Law Practice Problems
- Enthalpy of Reaction from Enthalpies of Formation
**Thermochemistry Practice Problems**

#### Practice

Using the standard heats of formation given below, calculate the heat of reaction for the combustion of ammonia:

4NH_{3}(*g*) + 5O_{2}(*g*) → 4NO(*g*) + 6H_{2}O(*g*), Δ*H*°_{rxn} = ?

Δ*H*_{f}° for NH_{3}(*g*) = –46.2 kJ/mol

Δ*H*_{f}° for NO(*g*) = 90.4 kJ/mol

Δ*H*_{f}° for H_{2}O(*g*) = –241.8 kJ/mol

Δ*H*°_{rxn} = -904 kJ

To calculate 𝚫*H*°_{rxn}, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

Where n_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Δ*H*°_{rxn} = [4 x Δ*H*^{o}_{f} (NO) + 6 x Δ*H*^{o}_{f} (H_{2}O)] – [4 x Δ*H*^{o}_{f} (NH_{3}) + 5 x Δ*H*^{o}_{f} (O_{2})]

Δ*H*°_{rxn} = [4 x 90.4 kJ + 6 x (–241.8)] – [4 x (–46.2 kJ) + 5 x 0]

Δ*H*°_{rxn} = -904.4 kJ

Δ*H*°_{rxn }is negative which means it is an exothermic reaction releasing 904.40kJ of heat.

Combustion of butane (C_{4}H_{10}) releases 5755 kJ of energy according to the following chemical equation.

2C_{4}H_{10}(*g*) + 13O_{2}(*g*) → 8CO_{2}(*g*) + 10H_{2}O(*l*), Δ*H*°_{rxn} = -5755 kJ

Calculate the molar enthalpy of formation of butane using the information given below:

Δ*H*_{f}° for CO_{2}(*g*) = –393.5 kJ/mol

Δ*H*_{f}° for H_{2}O(*l*) = –285.8 kJ/mol

-126 kJ

To calculate Δ*H*^{o}_{f} (C_{4}H_{10}), we are going to use the equation for the heat of reaction based on the standard enthalpies of formation:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

Where n_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Let’s add the components for our reaction:

Δ*H*°_{rxn} = [8 x Δ*H*^{o}_{f} (CO_{2}) + 10 x Δ*H*^{o}_{f} (H_{2}O)] – [2 x Δ*H*^{o}_{f} (C_{4}H_{10}) + 13 x Δ*H*^{o}_{f} (O_{2})] = -5755 kJ

Δ*H*^{o}_{f} (C_{4}H_{10}) is the unknown that we need to solve for in this equation:

Δ*H*°_{rxn} = [8 x (–393.5 kJ) + 10 x Δ*H*^{o}_{f} (–285.8 kJ)] – [2 x Δ*H*^{o}_{f} (C_{4}H_{10}) + 13 x 0] = -5755 kJ

-3148 kJ – 2858 kJ – 2 Δ*H*^{o}_{f} (C_{4}H_{10}) = -5755 kJ

-3148 kJ – 2858 kJ + 5755 kJ = 2 Δ*H*^{o}_{f} (C_{4}H_{10})

Δ*H*^{o}_{f} (C_{4}H_{10}) = -125.5 kJ/mol

Rounding off to three significant figures, we get -126 kJ/mol.

Zinc is recovered from ZnS by first oxidizing it to ZnO. Calculate the enthalpy of this oxidation reaction using the data given below:

2ZnS(*s*) + 3O_{2}(*g*) → 2ZnO(*s*) + 2SO_{2}(*g*)

Δ*H*_{f}° for SO_{2}(*g*) = -296.8 kJ/mol

Δ*H*_{f}° for ZnS(*s*) = -206.0 kJ/mol

Δ*H*_{f}° for ZnO(*s*) = -350.5 kJ/mol

-883 kJ

To calculate 𝚫*H*°_{rxn}, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

Where n_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Δ*H*°_{rxn} = [2 x Δ*H*^{o}_{f} (ZnO) + 2 x Δ*H*^{o}_{f} (SO_{2})] – [2 x Δ*H*^{o}_{f} (ZnS) + 3 x Δ*H*^{o}_{f} (O_{2})]

Δ*H*°_{rxn} = [2 x (-350.5 kJ) + 2 x (-296.8 kJ)] – [2 x (-206.0 kJ) + 3 x 0]

Δ*H*°_{rxn} = -701 kJ – 593.6 kJ + 412 kJ = -883 kJ

Δ*H*°_{rxn }is negative which means it is an exothermic reaction releasing 883 kJ of heat.

Using the standard heats of formation given below, calculate the heat of reaction between barium carbonate (BaCO_{3}) and sulfuric acid (H_{2}SO_{4}).

BaCO_{3}(*s*) + H_{2}SO_{4}(*aq*) → BaSO_{4}(*s*) + CO_{2}(*g*) + H_{2}O(*l*)

Δ*H*_{f}° for BaSO_{4}(*s*) = -1473.2 kJ/mol

Δ*H*_{f}° for CO_{2}(*g*) = –393.5 kJ/mol

Δ*H*_{f}° for H_{2}O(*l*) = –285.8 kJ/mol

Δ*H*_{f}° for BaCO_{3}(*s*) = -1213.0 kJ/mol

Δ*H*_{f}° for 3H_{2}SO_{4}(*aq*) = -814.0kJ/mol

-126 kJ

To calculate 𝚫*H*°_{rxn}, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

We can now enter the numbers to determine the heat of our reaction:

Δ*H*°_{rxn} = [1 x Δ*H*^{o}_{f} (BaSO_{4}(*s*)) + 1 x Δ*H*^{o}_{f} (CO_{2}(*g*)) + 1 x Δ*H*^{o}_{f} (H_{2}O(*l*))] – [1 x Δ*H*^{o}_{f} (BaCO_{3}(*s*)) + 1 x Δ*H*^{o}_{f} (H_{2}SO_{4}(*aq*))]

Δ*H*°_{rxn} = [-1473.2 kJ –393.5 kJ –285.8 kJ] – [-1213.0 kJ -814.0kJ]

Δ*H*°_{rxn} = -2,152.5 kJ + 2,027 kJ = -125.5 kj

Δ*H*°_{rxn }is negative which means it is an exothermic reaction releasing 126 kJ of heat.