Calculating the Enthalpy of Formation

This is part of Year 11 HSC Chemistry course under the topic of Enthalpy and Hess's Law

HSC Chemistry Syllabus

  • Explain the enthalpy changes in a reaction in terms of breaking and reforming bonds, and relate this to:

– The Law of Conservation of Energy


        Calculating the Enthalpy of Formation

        This video will discuss ideas relating to the Law of Conservation of Energy, and also demonstrate how the enthalpy of formation can be determined given the bond energy values for various types of bonds which exist within a compound. 


        Enthalpy in Chemical Reactions

        Contrary to what might seem intuitive, the breaking of chemical bonds is an energy absorbing (endothermic) process, while the formation of bonds is an energy-releasing (exothermic) process. This fundamental concept is key to understanding how chemical reactions store or release energy. 


        The Law of conservation of energy:

        The law of conservation of energy tells us that energy cannot be created nor destroyed, only transformed from one form to another. This principle is crucial in understanding why not all energy absorbed in endothermic reactions leads to an increase in temperature. Instead this energy can be converted into various other forms, either kinetic or potential energy. 


        Bond Energy and Enthalpy of Formation

        Bond energy is s term which refers to the amount of energy required to break a chemical bond. Take methane (`CH_4`) as an example. It contains four C-H bonds. With a. bond energy of 415.5 `kJ` `mol^{–1}` per C-H bond, breaking all bonds in one mole of `CH_4` would require 1662 `kJ` (`4 \times 415.5 kJ` `mol^{–1}`) 

        The enthalpy of formation is the amount of energy that is required to reverse the breaking of bonds. Because it is bond formation, typically the formation of bonds are an inherently exothermic process. If we reused the example of methane `CH_4`, we know that forming all the C-H bonds is the opposite of the energy required to break all the C-H bonds. This release is 1662 `kJ` which is the same amount needed to break them. This release is represented by a negative enthalpy of formation, –1662  `kJ` `mol^{–1}`/ 


        The general formula for calculating the standard enthalpy of formation is:


        $$\Delta H_{rxn} = \Sigma \Delta H_f(products) – \Sigma \Delta H_f(reactants)$$


        Elemental Substances and Their Enthalpy Values

        Elemental forms of substances such as hydrogen, oxygen, nitrogen, chlorine, carbon, copper, and sodium exhibit an enthalpy change of zero. This is an important consideration when calculating the enthalpy changes in reactions involving these events. 


        Consider the formation of water (`H_2O`) from hydrogen gas (`H_2`) and oxygen gas (`O_2`). The reaction can be represented as:


        $$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$


        Here the Δ`H_f` of `H_2` and `O_2` is considered zero because they are in their elemental forms. The standard enthalpy of formation of liquid water is approximately -285.8 `kJ` `mol^{–1}` 

        Since the above reaction produces 2 moles of `H_2O`, the enthalpy change for the reaction can be calculated using the standard enthalpies of formation. If we use the formula for calculating the enthalpy of formation, we get:


        $$ \Delta H_{\text{rxn}} = \left[ (2 \times -285.8 \text{ kJ/mol}) \right] - \left[ 2 \times (0 \text{ kJ/mol}) + 0 \text{ kJ/mol} \right] $$


        $$ = -571.6 \text{ kJ/mol} $$


        Calculating Reaction Enthalpy Example (answered in video)

         Consider a reaction involving the formation of carbon monoxide (CO) and carbon dioxide (`CO_2`) from elemental forms. If 222 kJ is released in forming 2 moles of CO, then each mole releases 111 kJ. Subtracting 222 kJ from –564 kJ and dividing by 2 (the stoichiometric ratio), we can find the enthalpy change of `CO_2` formation as –393 kJ/mol. 


        Previous Section: Enthalpy Change (ΔH) in Ionic Compound Dissolution

        Next Section: Hess's Law