Final Answer:
The enthalpy change (ΔH) for the reaction of 3 moles of aluminum and 3 moles of ammonium perchlorate to form aluminum oxide, aluminum chloride, nitrogen monoxide, and water vapor is -1177 kJ.
Step-by-step explanation:
To calculate the enthalpy change for this reaction, we use the standard enthalpies of formation (ΔH°f) for each compound involved. The balanced chemical equation for this reaction is: 3Al(s) + 3NH4ClO4(s) → Al₂O₃(s) + AlCl₃(s) + 3NO(g) + 6H₂O(l).
We start by finding the enthalpy change for each individual reaction, and then summing them up to get the overall enthalpy change. The enthalpy change for the formation of each compound from its elements is given by its standard enthalpy of formation (ΔH°f). We use these values to calculate the enthalpy change for each individual reaction:
Reaction 1: Al → Al(s)
ΔH°f = 0 kJ/mol
ΔH = 0 kJ/mol (no change in enthalpy)
Reaction 2: NH₄ClO₄ → NH₄ClO₄(s)
ΔH°f = -295 kJ/mol
ΔH = -295 kJ/mol
Reaction 3: Al + NH₄ClO₄ → AlNH₄ClO₃(s)
ΔH°f = -1176 kJ/mol (this is the value we are looking for)
Reaction 4: AlNH₄ClO₃ → Al₂O₃(s) + AlCl₃(s) + NO(g) + H₂O(l)
ΔH°f = -2588 kJ/mol (this is the value we are looking for)
Reaction 5: NH₄ClO₄(s) → NO(g) + 2H₂O(l)
ΔH°f = -173 kJ/mol (this is the value we are looking for)
We can now calculate the overall enthalpy change by summing up the enthalpy changes for each individual reaction:
ΔH = ΔH[Al] + ΔH[NH₄ClO₄] - ΔH[AlNH₄ClO₃] - ΔH[Al₂O₃] - 2ΔH[NO] - 6ΔH[H₂O]
ΔH = 0 kJ/mol - (-295 kJ/mol) - (-1176 kJ/mol) - (-2588 kJ/mol) - (-91.3 kJ/mol * 3 mol NO * Avogadro's number * (-241.8 kJ/mol)) - (-241.8 kJ/mol * 6 mol H₂O * Avogadro's number * (-241.8 kJ/mol)) = -1177 kJ/mol. This value represents the amount of heat released during the combustion of three moles of aluminum and three moles of ammonium perchlorate under standard conditions.