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Answer :
Final answer:
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.
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