1. PRACTICE PROBLEMS Sample Problem Calculate the enthalpy change for the reaction in which hydrogen gas, H 2 (g), is combined with fluorine gas, F 2(g), to produce 2 moles of hydrogen fluoride gas, HF (g). This reaction is represented by the balanced chemical equation H 2(g) +F 2(g) → 2HF (g) Given: for H 2(g) : nH–H = 1 mol; DH–H = 432 kJ/mol; for F 2(g) : nF–F = 1 mol; DF–F = 154 kJ/mol; Required:ΔH Analysis: ∆H = Σn x D bonds broken –Σn x D bonds formed Δ H = (n H-H D H-H + n F-F D F-F ) – (n H-F D H-F 5.3

2. Solution: 1 mol each of H–H and F–F bonds are broken The bonds formed are 2 mol of H–F bonds ∆H= (n H-H D H-H + n F-F D F-F ) – n H-F D H-F (1 mol x 432KJ) + (1 mol x 154 KJ) - (2 mol x 565 KJ mol mol mol ∆H = -544 KJ The enthalpy change for the reaction of 1 mol hydrogen gas and 1 mol fluorine gas to ptoduce 2 mol. Hydrogen fluoride is -544 KJ 5.3

3. 5.5: Standard enthalpy of formation f Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 5.5

4. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - 5.5 Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

5. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 5.5

6. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ - 6534.9 kJ 2 mol = - 3267.44 kJ/mol C 6 H 6 5.5

7. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ - 6534.9 kJ 2 mol = - 3267.44 kJ/mol C 6 H 6 5.5

8. The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.  H soln = H soln - H components 6.6 Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? LiCl & CaCl 2 NaCl, KCl, NH 4 Cl, NH 4 NO 3