Calculating ΔH using molar heats of formation Chem 12

If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hof . Standard conditions (standard state): 1 atm and 25 oC (298 K). Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

Examples – write formation reactions for each: *Remember the compounds are formed directly from their elements. H2SO4 NH4Cl

You want only one mole of the product being formed. Look up the ΔHf on the table H2 + S + 2O2  H2SO4 ΔHf = -814 kJ/ mol 2. ½ N2 + 2H2 + ½ Cl2  NH4Cl ΔHf = -314.4 kJ/ mol

Enthalpies of Formation

Using Enthalpies of Formation for Calculating Enthalpies of Reaction For a reaction

By definition, the enthalpy of formation of an element in its standard state is zero. Example, oxygen (O2) and chlorine (Cl2) both have Hof of zero.

Sample Problem 1 Calculate ΔH for the following reaction using standard molar heats of formation, ΔH°f . 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g) ΔH = ?

ΔH°f for NH3(g) = -45.9 kJ/mol ΔH°f for HCl(g) = -92.3 kJ/mol ΔH°f for Cl2(g) and N2(g) is 0 ΔHrxn = Σ nΔH°f(product) - Σ nΔH°f(reactant) ΔHrxn= (0 + 6(-92.3 kJ)) - (2(-45.9 kJ) + 0) = (-553 kJ) - (-91.8 kJ) = -461 kJ

Examples Use the tables in the back of your book to calculate ΔH for the following reactions: 4 CuO (s)  2 Cu2O (s) + O2 (g) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) NH3 (g) + HCl (g)  NH4Cl (s) 1. +292 kJ 2. -2220. kJ 3. -176.2 kJ

Practice: Page 685 #15 Page 687 #19-21 Page 691 # 5