Having a library of values for standard enthalpies of formation allows calculating the enthalpy of reaction without doing calorimetric analysis.

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:

where *m *and *n *are the stoichiometric coefficients for the reactants and products, and Σ (sigma) means “the sum of.”

So, the enthalpy of a reaction is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants.

**2) The Indirect Approach**

If the enthalpies of formation of all the components in chemical reactions are not known, then the Hess’s law is used.

The Hess’s law is covered in a separate post, and our focus today is going to be on the direct method for determining the enthalpy of a reaction.

**For example**,

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

**Solution:**

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.

Let’s **another example**, where the **heat of formation needs to be calculated** from the enthalpy of the reaction.

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

**Solution:**

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.

**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
- Standard Enthalpies of Formation
**Thermochemistry Practice Problems**