Use the given standard enthalpies of formation to determine the heat of reaction of the reaction:C3H6O(l) + 4O2(g) → 3 CO2(g) + 3 H2O(l)ΔH°f CO2(g) = -394 kJ/moleΔH°f H2O(l) = -286 kJ/moleΔH°f C3H6O(l) = -250 kJ/mole
Correct answer: D. -1790 kJ
- A. -1566 kJ
- B. -1285 kJ
- C. -1856 kJ
- D. -1790 kJ
Explanation
To determine the heat of reaction, use the formula: ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants). First, calculate the enthalpy of the products: (3 moles of CO2 × -394 kJ/mole) + (3 moles of H2O × -286 kJ/mole) = -1182 kJ + -858 kJ = -2040 kJ. Next, calculate the enthalpy of the reactants: 1 mole of C3H6O × -250 kJ/mole = -250 kJ. Then, subtract the enthalpy of the reactants from the enthalpy of the products: -2040 kJ - (-250 kJ) = -1790 kJ. Therefore, the correct answer is -1790 kJ. Other options result from calculation errors in summing or subtracting the enthalpies.
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About Thermochemistry and Energetics of Chemical Reactions
Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
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