Question

What is the enthalpy of combustion when 1 mol C6H6(g) completely reacts with oxygen?
2C6H6(g) + 15O2(g) mc001-1.jpg 12CO2(g) + 6H2O(g)

–6339 kJ/mol

–3169 kJ/mol

1268 kJ/mol

6339 KJ/mol
A 2-column table with 6 rows. Column 1 is labeled Compound with entries upper C a upper O solid, Upper C a upper C upper O subscript 3 solid, Upper C subscript 6 upper H subscript 6 gas, Upper C upper O subscript 2 gas, Upper H subscript 2 upper O gas, Upper N upper O subscript 2 gas. Column 2 is labeled Delta H subscript f in kilojoules per molecule with entries negative 157.3, negative 1,207.1, 82.90, negative 393.5, negative 241.82, 33.84.

Answer 1

Use ΔH°rxn = ΣνΔH°f(products) − ΣνΔH°f(reactants).

For 2 C6H6(g) + 15 O2(g) → 12 CO2(g) + 6 H2O(g):

ΔH°rxn = 12(−393.5) + 6(−241.82) − 2(82.90)
= −4722.0 − 1450.92 − 165.80
= −6338.72 kJ (for 2 mol C6H6).

Per 1 mol C6H6: ΔH° = −6338.72 / 2 ≈ −3169 kJ/mol.

Answer: −3169 kJ/mol.