Thermochemistry

Thermochemistry Standard Enthalpies of Formation

Standard Enthalpies of Formation • Tabulated enthalpy changes called standard enthalpy of formation (∆Hºf) can also be used to determine enthalpies of reaction (Nelson textbook pp. 799-800) • Standard enthalpy of formation is the quantity of energy associated with the formation of one mole of a substance from its elements in their standard states C(s) + O2(g) CO2(g) ∆Hºf = -393.5 kJ/mol

How to write formation equations • Write one mole of product in the state that has been specified. • Write the reactant elements in their standard states. Remember, • Most metals are monoatomic solids (Mg(s), Ca(s), Fe(s), Au(s), Na(s)) • Some nonmetals are diatomic gases (N2(g), O2(g), H2(g)) • Halogen family show a variety of states (F2(g), Cl2(g), Br2(l), I2(s))

Choose equation coefficients for the reactants to give a balanced equation yielding one mole of product. Example 1 Write the equation for the formation of liquid water directly from its elements. H2(g) + ½ O2(g)H2O(l) ∆Hºf = -285.8 kJ

Example 2 Write the equation for the formation of solid calcium carbonate directly from its elements. Ca(s) + C(s) + 3/2 O2(g) CaCO3(s) ∆Hºf = -1206.9 kJ

Using Standard Enthalpies of Formation • Generalization to all elements in their standard states  ∆Hºf for Elements, the standard enthalpy of formation of an element already in its standard state is zero • Thus, the ∆Hºf for Fe(s), O2(g), and Br2(l) are all zero

We can apply Hess’s Law to predicting the energy changes for many reactions • The enthalpy change for any given equation equals the sum of the enthalpies of formation of the products MINUS the sum of the enthalpies of formation of the reactants ∆H°r = ∑ n∆Hºf (products) - ∑ n∆Hºf (reactants)

Using Standard Enthalpies of Formation to Calculate Enthalpy Changes ∆H°r = ∑ n∆Hºf (products) - ∑ n∆Hºf (reactants) Determine the enthalpy change for the complete combustion of methane, CH4(g) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H = [1mol (∆Hºf of CO2(g)) + 2mol (∆Hºf of H2O(g))] - [1mol (∆Hºf of CH4(g)) + 2mol (∆Hºf of O2(g))]

Substitute the standard enthalpies of formation from Appendix C to get the following calculation. ∆H°r = [1mol (-393.5 kJ/mol) + 2mol (-241.8 kJ/mol)] – [1mol (-74.8 kJ/mol) + 2mol (0 kJ/mol)] = -802.3 kJ/mol of CH4(g)

Using Standard Enthalpies of Formation to Calculate Enthalpy Change ∆H°r = ∑ n∆Hºf (products) - ∑ n∆Hºf (reactants) Practice Use standard enthalpies of formation to calculate the standard enthalpy change for the oxidation of ammonia represented by the following balanced equation: 4 NH3(g) + 5 O2(g)  6 H2O(g) + 4 NO(g)

How does this method of adding heats of formation relate to Hess’s Law?

Target Equation: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Consider the equations for the formation of each compound that is involved in the reaction of methane with oxygen. (1) H2(g) + ½ O2(g)  H2O(g) ∆Hºf = -241.8 kJ (2) C(s) + O2(g)  CO2(g) ∆Hºf = -393.5 kJ (3) C(s) + 2H2(g)  CH4(g) ∆Hºf = -74.6 kJ There is no equation for the formation of oxygen, because oxygen is an element in its standard state.

Learning Checkpoint p. 332 Practice UC # 1 p. 335 Practice UC # 2, 3, 4 p. 335 Correct Answers 4(a) 205.7 kJ 4(b) -41.2 kJ

Thermochemistry