PRACTICE PROBLEMS Sample Problem

1. PRACTICE PROBLEMS Sample Problem Calculate the enthalpy change for the reaction in which hydrogen gas, H2(g), is combined with fluorine gas, F2(g), to produce 2 moles of hydrogen fluoride gas, HF(g). This reaction is represented by the balanced chemical equation H2(g)+F2(g)→ 2HF(g) Given: for H2(g): nH–H = 1 mol; DH–H = 432 kJ/mol; for F2(g): nF–F = 1 mol; DF–F = 154 kJ/mol; Required:ΔH Analysis: ∆H = Σn x D bonds broken –Σn x D bonds formed ΔH = (nH-HDH-H + nF-FDF-F) – (nH-FDH-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= (nH-HDH-H + nF-FDF-F) – nH-FDH-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. Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f DH0 (O3) = 142 kJ/mol DH0 (C, diamond) = 1.90 kJ/mol f f DH0 (C, graphite) = 0 f 5.5: Standard enthalpy of formation The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f 5.5

4. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn mDH0 (reactants) dDH0 (D) nDH0 (products) cDH0 (C) aDH0 (A) bDH0 (B) f f f f f f 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.) 5.5

5. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn. 5.5

6. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3267.44 kJ/mol C6H6 6DH0 (H2O) - 6534.9 kJ f 2 mol mDH0 (reactants) nDH0 (products) f f Benzene (C6H6) 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. 5.5

7. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3267.44 kJ/mol C6H6 6DH0 (H2O) - 6534.9 kJ f 2 mol mDH0 (reactants) nDH0 (products) f f Benzene (C6H6) 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. 5.5

8. The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? NaCl, KCl, NH4Cl, NH4NO3 Which substance(s) could be used for a cold pack? LiCl & CaCl2 6.6