Final answer:
To calculate the standard enthalpy of formation of acetylene, we use Hess's Law to rearrange the given reactions. We correct the enthalpy values for the one mole of C2H2 and sum the enthalpies of formation for CO2 and H2O, resulting in the standard enthalpy of formation for acetylene being 226.7 kJ/mol.
Step-by-step explanation:
To calculate the standard enthalpy of formation of acetylene (C2H2) from its elements, we can use Hess's Law which states that the total enthalpy change is the same, no matter how the chemical change occurs, as long as the initial and final conditions are the same.
We can rearrange the given reactions to develop a route to form one mole of C2H2 from its elements, C (graphite) and H2 (g).
The combustion reaction provided for acetylene is for two moles, so we need to divide all its related values by two to correct for one mole of C2H2. This makes the enthalpy change for combustion of one mole of acetylene half of -2598.8 kJ/mol, which is -1299.4 kJ/mol.
Now, we use the provided enthalpies of formation for CO2 and H2O to write the thermochemical equations for one mole of product:
2 C (graphite) + 2 O2 (g) → 2 CO2 (g) ΔH° = 2(-393 kJ/mol)
H2 (g) + ½ O2 (g) → H2O (l) ΔH° = -285.8 kJ/mol
Adding the enthalpies of the above reactions should be equal to the heat of combustion of acetylene corrected for one mole, which includes the formation of 2 moles of CO2 and 2 moles of H2O.
Therefore, the enthalpy for forming one mole of acetylene is calculated as follows:
-2(393 kJ/mol) - 285.8 kJ/mol = -2(393 kJ/mol) - 285.8 kJ/mol + 1299.4 kJ/mol
= ΔH°f[C2H2(g)]
= 226.7 kJ/mol.