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

Answer 1

You gave the combustion reaction (for 2 mol C2H2):
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)

Using standard enthalpies of formation (ΔHf°):
- C2H2(g) = +226.7 kJ·mol⁻¹
- CO2(g) = −393.5 kJ·mol⁻¹
- H2O(l) = −285.8 kJ·mol⁻¹
- O2(g) = 0 kJ·mol⁻¹

ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants)
= [4(−393.5) + 2(−285.8)] − [2(226.7) + 5(0)]
= (−1574.0 − 571.6) − 453.4
= −2145.6 − 453.4
= −2599.0 kJ (for 2 mol C2H2)

Per mole of C2H2: −2599.0 / 2 = −1299.5 kJ·mol⁻¹

Answer: ΔHcombustion ≈ −1.30 × 10^3 kJ·mol⁻¹ (exothermic).

Note: if H2O is taken as gas (ΔHf° = −241.8 kJ·mol⁻¹) the value becomes ≈ −1256 kJ·mol⁻¹.