Question

Using the information in the table to the right, calculate the enthalpy of combustion of 1 mole of acetylene for the reaction:

2Ces001-1.jpgHes001-2.jpg + 5Oes001-3.jpg es001-4.jpg 4COes001-5.jpg + 2Hes001-6.jpgO

kJ/mol
A 2-column table with 5 rows. Column 1 is labeled Compound with entries methane (C H 4), Acetylene (C 2 H 2), Ethanol (C 2 H 6 O), Carbon dioxide (C O 2), Water (H 2 O). Column 2 is labeled Delta H subscript f in kilojoules per molecule with entries negative 74.8, 226.77, negative 235.1, negative 393.5, negative 241.82.

Answer 1

Write the reaction enthalpy from formation enthalpies:

ΔH°rxn = Σ n ΔHf(products) − Σ n ΔHf(reactants)

Using ΔHf(O2) = 0:
ΔH°rxn = [4(−393.5) + 2(−241.82)] − [2(226.77) + 5(0)]
= (−1574.0 − 483.64) − 453.54
= −2057.64 − 453.54
= −2511.18 kJ

This is for 2 mol C2H2, so per 1 mol C2H2:
ΔH°combustion = −2511.18 / 2 = −1255.59 kJ ≈ −1.256 × 10^3 kJ mol−1

(Exothermic, so negative.)