Final answer:
The ΔH∘rxn for the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) is -1369 kJ/mol.
Step-by-step explanation:
We can use the standard enthalpies of formation to calculate the standard enthalpy change (ΔH∘rxn) for the given reaction. The standard enthalpy of formation (ΔH∘f) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. To calculate ΔH∘rxn, we can use the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants. In this case, the ΔH∘rxn of the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) is equal to the sum of the ΔH∘f values of the products minus the sum of the ΔH∘f values of the reactants.
The ΔH∘f of C2H5OH(l) is given as -278 kJ/mol, the ΔH∘f of CO2(g) is -393.5 kJ/mol, and the ΔH∘f of H2O(g) is -286 kJ/mol. To calculate ΔH∘rxn, we multiply the ΔH∘f values of the products by their respective coefficients and subtract the sum from the sum of the ΔH∘f values of the reactants. Plugging in the values, we get:
2(-393.5 kJ/mol) + 3(-286 kJ/mol) - (-278 kJ/mol) = -1369 kJ/mol
Therefore, the ΔH∘rxn for the given reaction is -1369 kJ/mol.