The enthalpy change for a balanced chemical reaction as written may be calculated by subtracting the sum of the enthalpies of formation for the products from the sum of the enthalpies of formation for the reactants. Because the enthalpies of formation are given per mole, they each need to be multiplied by the stoichiometric coefficients from the balanced chemical equation!ΔH°rxn = Σ [ΔHf°(products)] - Σ [ΔHf°(reactants)]Table of Enthalpies of formation (kJ/mol) Formula Name ΔH°f (kJ/mol) CaCO3(s) calcium carbonate -1207.6 CaO(s) calcium oxide -635.1 CO2(g) carbon dioxide -393.51 C3H5(NO3)3(l) nitroglycerin -364.0 N2(g) nitrogen 0 O2(g) oxygen 0 H2O(l) water -285.83 H2O(g) water -241.82 C3H8(g) propane -103.8 C6H6(l) benzene 49 C6H12O6(s) D-galactose -1286.3 Mg(OH)2(s) magnesium hydroxide -924.66 HF(aq) hydrogen fluoride -332.6 Mg2+(aq) magnseium ion -466.9 F-(aq) fluoride ion -332.6 Using the enthalpies of formation tabulated above, calculate the enthalpy change for the given reactions. CaCO3(s) → CaO(s) + CO2(g) 178.99 kJ/reaction C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) kJ/reaction 4 C3H5(NO3)3(l) → 6 N2(g) + O2(g) + 12 CO2(g) + 10 H2O(g) kJ/reaction Mg(OH2)(s) + 2 HF(aq) → Mg2+(aq) + 2 F-(aq) + 2 H2O(l) kJ/reaction C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) kJ/reaction Based on your calculations above, how much heat energy would the combustion of 1 mole of solid galactose produce? kJ Based on your calculations above, how much heat energy would the combustion of 1 mole of gaseous propane produce? kJ Based on your calculations above, how much heat energy would the decomposition of 1 mole of liquid nitroglycerin produce? kJ
The enthalpy change for a balanced
| Formula | Name | ΔH°f (kJ/mol) |
| CaCO3(s) | calcium carbonate | -1207.6 |
| CaO(s) | calcium oxide | -635.1 |
| CO2(g) | carbon dioxide | -393.51 |
| C3H5(NO3)3(l) | nitroglycerin | -364.0 |
| N2(g) | nitrogen | 0 |
| O2(g) | oxygen | 0 |
| H2O(l) | water | -285.83 |
| H2O(g) | water | -241.82 |
| C3H8(g) | propane | -103.8 |
| C6H6(l) | benzene | 49 |
| C6H12O6(s) | D-galactose | -1286.3 |
| Mg(OH)2(s) | magnesium hydroxide | -924.66 |
| HF(aq) | hydrogen fluoride | -332.6 |
| Mg2+(aq) | magnseium ion | -466.9 |
| F-(aq) | fluoride ion | -332.6 |
Using the enthalpies of formation tabulated above, calculate the enthalpy change for the given reactions.
CaCO3(s) → CaO(s) + CO2(g) 178.99 kJ/reaction
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) kJ/reaction
4 C3H5(NO3)3(l) → 6 N2(g) + O2(g) + 12 CO2(g) + 10 H2O(g) kJ/reaction
Mg(OH2)(s) + 2 HF(aq) → Mg2+(aq) + 2 F-(aq) + 2 H2O(l) kJ/reaction
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) kJ/reaction
Based on your calculations above, how much heat energy would the combustion of 1 mole of solid galactose produce? kJ
Based on your calculations above, how much heat energy would the combustion of 1 mole of gaseous propane produce? kJ
Based on your calculations above, how much heat energy would the decomposition of 1 mole of liquid nitroglycerin produce? kJ
A city of 100,000 people uses approximately 1.0 x 1011 kJ of energy per day. Suppose all of that energy comes from the combustion of liquid pentane (C5H12, density = 0.626 g/mL, molar mass = 72.151 g/mol) to form gaseous carbon dioxide and gaseous water. Use the standard enthalpies of formation (in the tables above and below) to determine how much energy is provided from the combustion reaction, and then determine how many gallons of pentane each person would require for their daily energy consumption. Much of this problem can be solved quite readily using dimensional analysis.
Table of Enthalpies of formation (kJ/mol)
| Formula | Name | ΔH°f (kJ/mol) |
| C5H12(l) | pentane | -173.5 |
| C6H14(l) | hexane | -198.8 |
| C7H16(l) | heptane | -224.2 |
| C8H18(l) | octane | -250.1 |
| C2H6O (l) | ethanol | -277.6 |
gallons of pentane per person per day.
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