CH3CH2OH(l) –277.7 × = +499.8 H2O(g) –2 –249.9 –241.8 × 0 = 0 H2O(l) –25 –285.8 kJ –10940.6 -1 pt for 10156.5 kJ; used H2O(g) instead of (l) (2 pts) The enthalpy change that you determined in the previous problem is (circle one): exothermic b) endothermic c) neither (heat neutral) Must be consistent with answer above (8 pts) Determine the enthalpy of reaction of 2H2S(g) + SO2(g) 3S(s) + 2H2O(l) using the following standard enthalpies of formation: H2S(g) = -20.63 kJ/mol H°f SO2(g) = -296.83 kJ/mol H°f H2O(g) = -241.82 kJ/mol H°f H2O(l) = -285.83 kJ/mol H°f H° = 2 mol × (-285.83 kJ/mol) 2 mol × (-20.63 kJ/mol) 1 mol × (-296.83 kJ/mol) – – = -233.57 kJ -2 pts for -145 kJ; used H2O(g) instead of H2O(l)
–2 × –249.9 = +499.8 –25 × 0 = 0 –10940.6 kJ -1 pt for 10156.5 kJ; used H2O(g) instead of (l) CH3CH2OH(l) –277.7 H2O(g) –241.8 H2O(l) –285.8 (2 pts) The enthalpy change that you determined in the previous problem is (circle one): exothermic b) endothermic c) neither (heat neutral) Must be consistent with answer above (8 pts) Determine the enthalpy of reaction of 2H2S(g) + SO2(g) 3S(s) + 2H2O(l) using the following standard enthalpies of formation: H2S(g) H°f = -20.63 kJ/mol SO2(g) H°f = -296.83 kJ/mol H2O(g) H°f = -241.82 kJ/mol H2O(l) H°f = -285.83 kJ/mol H° = 2 mol × (-285.83 kJ/mol) – 2 mol × (-20.63 kJ/mol) – 1 mol × (-296.83 kJ/mol) = -233.57 kJ -2 pts for -145 kJ; used H2O(g) instead of H2O(l)
General%20Chemistry%20I%20Test%20Bank.docx