Bond enthalpy is an important chemical concept that defines the amount of energy needed to break the covalent bond between two gases. This type of energy does not apply to ionic bonds. When two atoms join together to form a new molecule, it is possible to calculate the strength of their bond by measuring the amount of energy it takes to separate them. Remember that only one atom does not have this energy, which exists only in the presence of two atoms. To find the bond enthalpy of a reaction, simply determine how many bonds have been broken and subtract the total number of those that have formed.
Steps
Part 1 of 2: Determine the Broken and Formed Bonds
Step 1. Define the equation to calculate the bond enthalpy
This energy is the difference between the sum of the broken bonds and that of the bonds formed: ΔH = ∑H(broken) - ∑H(formats). ΔH expresses the change in enthalpy and ∑H is the sum of the energies in each side of the equation.
- This equation is an expression of Hess's law.
- The unit of measurement for bond enthalpy is the kilojoule per mole (kJ / mol).
Step 2. Draw the chemical equation showing all the bonds between molecules
When an equation written simply with numbers and chemical symbols is provided, it is worth drawing it so that all the bonds that form between the various elements and molecules are visible. The graphical representation allows you to calculate all the bonds that break and form on the reactant side and on the product side.
- Don't forget that the left side of the equation contains all reactants and the right side all products.
- Single, double or triple bonds have different enthalpies, so remember to draw the diagram with the correct bonds between the elements.
- For example, draw the following chemical equation: H.2(g) + Br2(g) - 2 HBr (g).
- H-H + Br-Br - 2 H-Br.
Step 3. Learn the rules for counting the bonds that break and form
In most cases, the enthalpy values you use for these calculations are averages. The same bond can have different enthalpies based on the molecule that is formed, therefore average data are generally used.
- A single, double or triple bond that breaks is always treated as if it were one; the bonds have different enthalpies, but "they are worth" as a single one that dissolves.
- The same rule also applies in their training process.
- In the example described above, the reaction involves only single bonds.
Step 4. Find the broken links on the left side of the equation
This section describes the reactants and the bonds that dissolve during the reaction. It is an endothermic process that requires the absorption of energy to break the bonds.
In the example above, the left side shows an H-H and a Br-Br bond
Step 5. Count the bonds that have formed on the right side of the chemical equation
In this side there are all the products of the reaction and therefore the bonds that have formed. It is an exothermic process that releases energy, usually in the form of heat.
In the above example there are two H-Br bonds
Part 2 of 2: Calculate the Bond Enthalpy
Step 1. Look for the energies of the bonds in question
There are several tables that report the average enthalpy value of specific bonds and you can find them online or in chemistry textbooks. It is important to note that these data always refer to molecules in the gaseous state.
- Consider the example given in the first part of the article and find the enthalpy for the H-H, Br-Br and H-Br bond.
- H-H = 436 kJ / mol; Br-Br = 193 kJ / mol; H-Br = 366 kJ / mol.
-
To calculate the energy for liquid molecules, you also need to consider the change in enthalpy of vaporization. This is the amount of energy needed to transform a liquid into a gas; this number must be added to the total bond enthalpy.
For example: if you are given information about water in the liquid state, you have to add up the change in enthalpy of vaporization of this substance (+41 kJ / mol)
Step 2. Multiply the bond enthalpies by the number of broken unions
In some equations the same bond is dissolved several times; for example, if you have 4 hydrogen atoms in a molecule, the enthalpy of hydrogen must be taken into account by 4 times, i.e. multiplied by 4.
- Always consider the previous example where there is only one bond for each molecule; in this case, the enthalpy of each bond must be multiplied by 1.
- H-H = 436 x 1 = 436 kJ / mol.
- Br-Br = 193 x 1 = 193 kJ / mol.
Step 3. Add up all the values for the broken bonds
Once you have multiplied the values by the number of individual bonds, you need to find the sum of the energies present on the reactant side.
In the case of the example: H-H + Br-Br = 436 + 193 = 629 kJ / mol
Step 4. Multiply the enthalpies by the number of bonds that have formed
Just as you did for the reactant side, multiply the number of bonds that have been created by the respective energies and that are present on the products side; if 4 hydrogen bonds have developed, multiply the amount of enthalpy by 4.
In the example you can see that there are two 2 H-Br bonds, so you have to multiply their enthalpy (366kJ / mol) by 2: 366 x 2 = 732 kJ / mol
Step 5. Add up all the enthalpies of the new bonds
Repeat the same procedure on the product side as you did on the reagent side. Sometimes, you only have one product and can then skip this step.
In the example considered so far there is only one product, therefore the bond enthalpy that has formed concerns only the two H-Br, therefore 732 kJ / mol
Step 6. Subtract the enthalpy of the bonds formed from that of the broken bonds
Once you have found the total energies on both sides of the chemical equation, simply proceed to the subtraction by remembering the formula: ΔH = ∑H(broken) - ∑H(formats); replace the variables with the known values and subtract.
For example: ΔH = ∑H(broken) - ∑H(formats) = 629 kJ / mol - 732 kJ / mol = -103 kJ / mol.
Step 7. Determine if the entire reaction is endothermic or exothermic
The final step in calculating the bond enthalpy is to evaluate whether the reaction releases or absorbs energies. An endothermic (energy-consuming) reaction has a positive total enthalpy, while an exothermic (energy-releasing) reaction has a negative enthalpy.
