Given the reaction:

\[ \text{C}_2\text{H}_2(g) + \frac{5}{2} \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + \text{H}_2\text{O}(g) \]

ΔH°rxn = −1255.8 kJ

Given:
- ΔH°f of CO₂(g) = −393.5 kJ/mol
- ΔH°f of H₂O(g) = −241.8 kJ/mol

Find ΔH°f of C₂H₂(g).

A) −52.2 kJ/mol
B) −97.8 kJ/mol
C) −78.9 kJ/mol
D) −64.6 kJ/mol

The calculation for the standard enthalpy of formation for acetylene (C2H2) using the given enthalpy change of the reaction and the known enthalpies of formation for CO2 and H2O yielded a value of 227 kJ/mol, which does not match any of the provided multiple choice options.

Explanation:

To find the standard enthalpy of formation (ΔH°f) for acetylene (C2H2), we can use the reaction's standard enthalpy change (ΔH°rxn) and the standard enthalpies of formation for the products provided in the reaction equation C2H2(g) + 5/2 O2(g) → 2 CO2(g) + H2O(g). The enthalpy change for this reaction is given as −1255.8 kJ. We know the ΔH°f of CO2(g) is −393.5 kJ/mol and the ΔH°f of H2O(g) is −241.8 kJ/mol. The reaction produces 2 moles of CO2 and 1 mole of H2O.

To calculate the ΔH°f of C2H2(g), we can use the following equation derived from Hess's law:

ΔH°rxn = [2 * ΔH°f(CO2(g)) + ΔH°f(H2O(g))] − ΔH°f(C2H2(g))

Plugging in the values:

−1255.8 kJ = [2 * (−393.5 kJ/mol) + (−241.8 kJ/mol)] − ΔH°f(C2H2(g))

−1255.8 kJ = −1028.8 kJ − ΔH°f(C2H2(g))

ΔH°f(C2H2(g)) = −1028.8 kJ + 1255.8 kJ

ΔH°f(C2H2(g)) = 227 kJ or 227,000 J

However, since we need the enthalpy per mole and the balanced equation shows 1 mole of C2H2 reacting, the calculation should already reflect the amount per mole. Therefore, the correct option is not provided in the multiple choices A) -52.2 kJ/mol, B) -97.8 kJ/mol, C) -78.9 kJ/mol, and D) -64.6 kJ/mol as we have calculated a positive value of 227 kJ/mol.