Understanding Standard Molar Enthalpies of Formation
Formation reactions involve substances being created from elements in their standard states, with the enthalpy change known as the standard molar enthalpy of formation (Hf). This enthalpy represents the energy released or absorbed when one mole of a compound is formed from its elements in their standard states. Formation equations are written to show the formation of exactly one mole of the compound of interest, and enthalpy changes can be calculated by summing the heats of formation of products and subtracting the heats of formation of reactants, in alignment with Hess's Law.
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Standard Molar Enthalpies of Formation Hfo
Focus Questions 1) What are formation reactions? 2) What is standard molar enthalpy of formation? 3) How do we write a formation equation? 4) How do we calculate the H using the standard molar enthalpy of formations? 5) How does this method relate to Hess s Law?
Formation Reactions In a formation reaction, a substance is formed from elements in their standard states. From what elements is water formed? H2(g)+ O2 (g) H2O (l) Hfo= -285.8 kJ The enthalpy change of a formation reaction is called the standard molar enthalpy of formation, H f.
Definition of Hf The standard molar enthalpy of formation is the quantity of energy that is absorbed or released when one mole of a compound is formed directly from its elements in their standard states.
Note The standard enthalpies of formation of most compounds are negative. By definition, the enthalpy of formation of an element in its standard state is zero
Writing a formation equation Always write the elements in their standard state (l, g, or s). A formation equation should also show the formation of exactly one mole of the compound of interest.
Calculating Enthalpy Changes You can calculate the enthalpy change of a chemical reaction by adding the heats of formation of the products and subtracting the heats of formation of the reactants. H = (n H f products) - (n H f reactants)
Note As usual, you need to begin with a balanced chemical equation. If a given reactant or product has a molar coefficient that is not 1, you need to multiply its H f by the same molar coefficient.
Try Solving using Standard enthalpy of formation equation CH4(g) + 2O2(g) C02(g) + 2H2O(g) H = [(n H f of C02(g) ) + 2(n H f of H2O(g) )] [(n H f of CH4(g)) + 2(n H f of O2(g) )] H = [(-393.5 kJ/mol) + 2(-241.8 kJ/mol )] [(-74.4 kJ/mol) + 2(0 kJ/mol)] = -802.7 kJ/mol of CH4
How does this method of adding heats of formation relate to Hess s law? Solve previous question using Hess s Law (1) H2(g) + O2(g) H2O2(g) H f= -241.8 kJ (2) C(s) + O2(g) CO2(g) H f = -393.5 kJ (3) C(s) + 2H2(g) CH4(s) H f =-74.4 kJ 2 x (1) 2H2(g) + O2(g) 2H2O2(g) H f= 2(-241.8) kJ (2) C(s) + O2(g) CO2(g) H f = -393.5 kJ -1 x (3) CH4(s) C(s) + 2H2(g) H f =-1(-74.4) kJ CH4(g) + 2O2(g) 2H2O(g) + CO2(g) H f = -802.7 kJ
Note It is important to realize that, in most reactions, the reactants do not actually break down into their elements and then react to form products. Since there is extensive data about enthalpies of formation, however, it is useful to calculate the overall enthalpy change this way.
Challenging Question: When octane burns in a car engine, heat is released to the air and to the metal of the engine, but a significant portion is absorb by the liquid in the cooling system-an aqueous solution of ethylene glycol. What mass of octane is completely burned to cause the heating of 20kg of ethylene glycol from 10oC to 70oC? Assume water is produced as a gas and that all the heat flows into the coolant. ( ethylene glycol 3.5J/goC)
