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Calculating Enthalpy Change Using 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 Enthalpy Change using Enthalpy of Formation 

        This video will explore the Law of Conservation of Energy and demonstrate how enthalpy change values can be calculated using enthalpy of formation values. It will also cover practical examples to illustrate how the enthalpy of formation values for reactants and products can be used to determine the overall energy change in a chemical reaction. 

         

        VIDEO COMING SOON

         

        What is Enthalpy Change and Enthalpy of Formation? (Review) 

        Enthalpy is a measurement of energy in a chemical system that changes as chemical reactions proceed. Specifically, the enthalpy change (ΔH) is the total change in potential energy when reactants turn into products. 

        The enthalpy of formation (ΔHf) of a compound represents the energy released or absorbed when a compound forms from constituent elements in elemental form. For example, the following equation demonstrates the formation of methane from carbon and hydrogen

        $$C(s) + 2H_2(g) \rightarrow CH_4(g) \hspace{2cm} \Delta H°_f \text{  }–74.8 \text{ } kJ/mol$$

        The specific reaction is exothermic, meaning energy is released.

         

        Relationship between Enthalpy Change and Enthalpy of Formation

        The following formula can be used to calculate the enthalpy change (ΔH) in a reaction using enthalpy of formation values

        $$ \Delta H \text{ } = \text{ } \Sigma \Delta H°_f \text{ } (products) \text{ } – \text{ } \Sigma \Delta H°_f \text{ }(reactants)$$

          

        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. The formation of methane's bonds thus releases 1662 `kJ` which is the same amount that would be required to break them. To represent that this enthalpy change is an exothermic release of energy, we add a negative sign to give a value of –1662  `kJ` `mol^{–1}` for the formation of methane's bonds.

         

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

         

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

          

        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 elements.  

         

        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 we previously looked at for calculating the enthalpy of formation.

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

        we get:

         

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

         

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

         

        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 forming the bonds of one mole of CO 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 of formation and Bond Energy (COMING)

        Next Section: Calculating Enthalpy Change using Enthalpy of Formation (COMING)

         

         

        BACK TO MODULE 4: DRIVERS OF REACTION