Standard Enthalpies of Formation

Standard Enthalpy of Formation the quantity of energy associated with the formation of one mole of a substance from its elements in their standard states (at SATP) Symbol: ΔH°f Units: kJ/mol For example: C(s) + ½ O2 (g) + 2 H2 (g)  CH3OH (l) Δ H°f = - 239.1 kJ/mol 2 Al (s) + 3/2 O2 (g)  Al2O3 (s) Δ H°f = -1675.7 kJ/mol All of the elements on the left sides of the equations are in their standard states (SATP)

Writing Formation Equations Step 1: Write one mole of the product in the state that has been specified. Example: FeSO4 (s) Step 2: Write the reactant elements in their standard states (reference periodic table. Pg 838). Fe (s) + S (s) + O2 (g)  FeSO4 (s) Step 3: Write coefficients for the reactants to give a balanced equation yielding one mole of product. Fe (s) + S (s) + 2 O2 (g)  FeSO4 (s)

H2O (g) + C (s)  H2 (g) + CO (g) ΔH = ? Calculating ΔH A third way of calculating ΔH is possible using standard enthalpies of formation. Table C.6 pg 799 gives standard enthalpies of formation for common molecules. According to Hess’s Law, the enthalpies of known equations may be used to calculate the enthalpy of an unknown reaction. For example: The formation of hydrogen and carbon monoxide gas from water and graphite: H2O (g) + C (s)  H2 (g) + CO (g) ΔH = ? Use the formation equations for each of the products and reactants to create the target equation.

Calculating ΔH0, continued H2O (g) + C (s)  H2 (g) + CO (g) Reactants H2(g) + 1/2 O2(g) --> H2O(g) H0f H2O = - 241.8 kJ/mol C (s) – graphite in standard state H0f = 0 kJ/mol Products C(s) + 1/2 O2(g) --> CO(g) H0f CO = - 110.5 kJ/mol H2 (g) – hydrogen in standard state H0f = 0 kJ/mol H2O(g) --> H2(g) + 1/2 O2(g) H0 = 241.8 kJ/mol C(s) + 1/2 O2(g) --> CO(g) H0 = - 110.5 kJ/mol H2O (g) + C (s)  H2 (g) + CO (g) H0 = 131.3 kJ/mol

In general, when all enthalpies of formation are known, the enthalpy of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. where n = the amount in moles of each product or reactant

Sample Problem Calculate the heat of combustion of methanol. CH3OH (l) + 3/2 O2 (g) --> CO2 (g) + 2 H2O (g) H0comb = ? Horxn = nHof (prod) - nHof (react) Horxn = Hof (CO2) + 2 Hof (H2O) - {3/2 Hof (O2) + Hof (CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-239.1 kJ)} Horxn = -638 kJ per mol of methanol

Multistep Energy Calculations Several energy calculations may be required to solve multistep problems Heat flow q = mcΔT Enthalpy changes ΔH = n ΔHx Hess’s Law ΔHtarget = ∑ΔHknown Enthalpies of Formation

Sample Problem, pg 336

Sample problem cont’d

Sample problem cont’d