(Eqn 3) H2(g) + ½ O2(g) H2O(l) H = - 286 kJ Calculate the enthalpy of reaction for the combustion to propane: C3H6(g) + 9/2O2(g) 3CO2(g) + 3 H2O(l) H = ? Reverse Eqn 1: H = - 53.3 kJ Multiply Eqn 2 by 3 = - 1182 kJ H Multiply Eqn 3 by 3 H = - 858 kJ Hrxn = - 2093 kJ 9. (8 pts) The enthalpy change for the oxidation of styrene, C8H8, is measured by calorimetry: C8H8 + 10 O2(g) 8 CO2(g) + 4 H2O(l) = -4395.0 kJ Hrxn Given the following standard enthalpies of formation, calculate the molar enthalpy of formation (Hof) of styrene: CO2(g) = -394 kJ Hof H2O(l) = -286 kJ Hof [8 + 4 [HfC8H8] HfCO2 HfH2O] – -4395.0 kJ = [8(-394) + 4(-286)] - [HfC8H8] -4395.0 = -3152 1144 - [HfC8H8] – -99 kJ = - [HfC8H8] = + 99 kJ HfC8H8
(Eqn 3) H2(g) + ½ O2(g) H2O(l) H = - 286 kJ Calculate the enthalpy of reaction for the combustion to propane: C3H6(g) + 9/2O2(g) 3CO2(g) + 3 H2O(l) H = ? Reverse Eqn 1: H = - 53.3 kJ Multiply Eqn 2 by 3 H = - 1182 kJ Multiply Eqn 3 by 3 H = - 858 kJ Hrxn = - 2093 kJ 9. (8 pts) The enthalpy change for the oxidation of styrene, C8H8, is measured by calorimetry: C8H8 + 10 O2(g) 8 CO2(g) + 4 H2O(l) Hrxn = -4395.0 kJ Given the following standard enthalpies of formation, calculate the molar enthalpy of formation (Hof) of styrene: Hof CO2(g) = -394 kJ Hof H2O(l) = -286 kJ [8 HfCO2 + 4 HfH2O] – [HfC8H8] -4395.0 kJ = [8(-394) + 4(-286)] - [HfC8H8] -4395.0 = -3152 – 1144 - [HfC8H8] -99 kJ = - [HfC8H8] HfC8H8 = + 99 kJ
C103Exam2Spring2009Solutions.docx