The combustion reaction of hexane is: Type numbers in the boxes_ 10 points 19 02(9) 4 6 CO2(g) C6H14(l) 7 HzO() The heat of combustion of hexane is -4163 kJ/mol. Given the following information, calculate the heat of formation of hexane_ AH? ; (CO_(g)) ~393.5 kJ/mol AH? (H,O()) -285.8 kJ/mol kJ/mol
The Correct Answer and Explanation is:
To calculate the heat of formation (ΔHf∘\Delta H_f^\circΔHf∘) of hexane (C6H14\text{C}_6\text{H}_{14}C6H14), we can use Hess’s Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps, regardless of the pathway taken.
Step 1: Write the combustion reaction for hexane.
The combustion reaction for hexane is:C6H14(l)+O2(g)→6CO2(g)+7H2O(l)\text{C}_6\text{H}_{14(l)} + \text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 7\text{H}_2\text{O(l)}C6H14(l)+O2(g)→6CO2(g)+7H2O(l)
Step 2: Use the given data for the heat of combustion.
The heat of combustion of hexane (ΔHcomb\Delta H_\text{comb}ΔHcomb) is −4163 kJ/mol-4163 \, \text{kJ/mol}−4163kJ/mol. This represents the enthalpy change when one mole of hexane undergoes complete combustion.
Step 3: Use the heats of formation of the products and reactants.
The standard heats of formation are:
- ΔHf∘(CO2(g))=−393.5 kJ/mol\Delta H_f^\circ (\text{CO}_2(g)) = -393.5 \, \text{kJ/mol}ΔHf∘(CO2(g))=−393.5kJ/mol
- ΔHf∘(H2O(l))=−285.8 kJ/mol\Delta H_f^\circ (\text{H}_2\text{O(l)}) = -285.8 \, \text{kJ/mol}ΔHf∘(H2O(l))=−285.8kJ/mol
- The heat of formation for O2(g)\text{O}_2(g)O2(g) is 0 kJ/mol0 \, \text{kJ/mol}0kJ/mol because it is in its elemental form.
- We are solving for ΔHf∘(C6H14(l))\Delta H_f^\circ (\text{C}_6\text{H}_{14(l)})ΔHf∘(C6H14(l)), which is the unknown.
Step 4: Apply Hess’s Law.
Hess’s Law states that:ΔHcomb=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_\text{comb} = \sum \Delta H_f^\circ (\text{products}) – \sum \Delta H_f^\circ (\text{reactants})ΔHcomb=∑ΔHf∘(products)−∑ΔHf∘(reactants)
Substitute the known values:−4163=[6×(−393.5)+7×(−285.8)]−[ΔHf∘(C6H14(l))+0]-4163 = [6 \times (-393.5) + 7 \times (-285.8)] – [\Delta H_f^\circ (\text{C}_6\text{H}_{14(l)}) + 0]−4163=[6×(−393.5)+7×(−285.8)]−[ΔHf∘(C6H14(l))+0]−4163=[−2361−2000.6]−ΔHf∘(C6H14(l))-4163 = [-2361 – 2000.6] – \Delta H_f^\circ (\text{C}_6\text{H}_{14(l)})−4163=[−2361−2000.6]−ΔHf∘(C6H14(l))−4163=−4361.6−ΔHf∘(C6H14(l))-4163 = -4361.6 – \Delta H_f^\circ (\text{C}_6\text{H}_{14(l)})−4163=−4361.6−ΔHf∘(C6H14(l))
Step 5: Solve for ΔHf∘(C6H14(l))\Delta H_f^\circ (\text{C}_6\text{H}_{14(l)})ΔHf∘(C6H14(l)).
ΔHf∘(C6H14(l))=−4361.6+4163\Delta H_f^\circ (\text{C}_6\text{H}_{14(l)}) = -4361.6 + 4163ΔHf∘(C6H14(l))=−4361.6+4163ΔHf∘(C6H14(l))=−198.6 kJ/mol\Delta H_f^\circ (\text{C}_6\text{H}_{14(l)}) = -198.6 \, \text{kJ/mol}ΔHf∘(C6H14(l))=−198.6kJ/mol
Final Answer:
The heat of formation of hexane is ΔHf∘(C6H14(l))=−198.6 kJ/mol\Delta H_f^\circ (\text{C}_6\text{H}_{14(l)}) = -198.6 \, \text{kJ/mol}ΔHf∘(C6H14(l))=−198.6kJ/mol.
Explanation:
This calculation uses Hess’s Law to break down the combustion reaction into its individual steps, comparing the heat of formation of the products and reactants. The negative sign indicates that the formation of hexane from its elements releases energy. The reaction is exothermic, as expected for combustion processes.
