The first one is for (CH 3) 3C-Cl where the Δ H°=331 kJ/mol and the second one is the H-Cl bond (Δ H°=431 kJ/mol).

Next, identify the bonds that are formed. The other bond that is broken is the Cl-Cl which has Δ H° of 243 kJ/mol. However, you need specifically look for the bond connecting the hydrogen to a tertiary carbon.Ī quick reminder about primary, secondary and tertiary carbon atoms:įrom the tale we find that the Δ H° for (CH 3) 3C-H bond is 381 kJ/mol. In this case, they are highlighted in red and blue respectively. The broken bonds are going to be in the starting materials and the forming bonds are in the products. The first thing you need to do when working with the bond dissociation energies is to identify the bonds that are broken, and the ones formed during the reaction. Let’s look in the table below to see if this conclusion matches with the numeric data of the bond dissociation energies.

Looking at this from the bond strength’s prospective, we can say that the bonds formed in the products are stronger than the bonds broken in the starting materials. This means more energy is released in forming bonds than is consumed to break the ones in the starting materials. And because these energies are not equal, there is a difference when they are summarized for the net transformation.įor example, the following reaction between chlorine and 2-methylpropane is an exothermic reaction Δ H° = −138 kJ/mol. We mentioned, the total change in enthalpy (Δ H°) for the reaction, called the heat of reaction, occurs because of the difference between the bond strengths of the starting materials and products.Įach of the bonds broken during the chemical reaction requires energy while forming a new bond releases energy. In the previous post, we talked about the main principle of energy, heat and enthalpy associated with chemical reactions.

Formula and bond chemistry calculator