This equation is a form of Hess’s Law. The unit for bond energy is kilojoules per mol or kJ/mol. [3] X Research source

Remember, the left side of the equation is all of the reactants and the right side is all of the products. Single, double, and triple bonds have different bond energies, so be sure to draw your diagram with the correct bonds between elements. [4] X Research source For example, if you were to draw out the following equation for a reaction between 2 hydrogen and 2 bromine: H2(g) + Br2(g) —> 2 HBr(g), you would get: H-H + Br-Br —> 2 H-Br. The hyphens represent single bonds between the elements in the reactants and the products.

A single, double, and triple bond are all treated as 1 break. They all have different bond energies, but count as only a single break. The same is true for the formation of a single, double, or triple bond. It will be counted as single formation. For our example, all of the bonds are single bonds.

For our example, the left side has 1 H-H bond and 1 Br-Br bond.

For our example, the right side has 2 H-Br bonds.

For our example, you need to find the bond energy for an H-H bond, a Br-Br bond, and an H-Br bond. H-H = 436 kJ/mol; Br-Br = 193 kJ/mol; H-Br = 366 kJ/mol. To calculate bond energy for molecules in a liquid state, you need to also look up the enthalpy change of vaporization for the liquid molecule. This is the amount of energy needed to convert the liquid into a gas. [10] X Research source This number is added to the total bond energy. For example: If you were given liquid water, you would need to add the enthalpy change of vaporization of water (+41 kJ) to the equation. [11] X Research source

In our example, there is only 1 bond of each molecule, so the bond energies are simply multiplied by 1. H-H = 436 x 1 = 436 kJ/mol Br-Br = 193 x 1 = 193 kJ/mol

For our example, the sum of the bonds broken is H-H + Br-Br = 436 + 193 = 629 kJ/mol.

For our example we have 2 H-Br bonds formed, so the bond energy of H-Br (366 kJ/mol) will be multiplied by 2: 366 x 2 = 732 kJ/mol.

In our example, there is only 1 product formed, so the energy of the bonds formed is simply the energy of the 2 H-Br bonds or 732 kJ/mol.

For our example: ΔH = ∑H(bonds broken) - ∑H(bonds formed) = 629 kJ/mol - 732 kJ/mol = -103 kJ/mol.

In our example, the final bond energy is negative, therefore, the reaction is exothermic.