It is true that the complete combustion of a hydrocarbon can do more work than incomplete combustion. Compare the work ΔG done by the combustion of one mole of ethane undeer these conditions.
Incomplete combustion: 2 C2H6 (g) + 5 O2 (g) → 4 CO(g) + 6 H2O (g)
Complete combustion: 2 C2H6 (g) + 7 O2 (g) → 4 CO(g) + 6 H2O (g)
can someone please help me solve this problem? and show how it is done.
thank you
2007-12-09 13:14:17 · 1 answers · asked by abc 1 in Science & Mathematics ➔ Chemistry
Your second equation should give CO2, not CO
Use change = (sum of products) - (sum of reactants) and plug in values of DeltaG0(formation) . Since Q specifies one mole ethane, divide through by 2
Also notice that 2nd eqn is first eqn followed by
4 CO + 2O2 = 4 CO2
which is energy-releasing.
2007-12-09 15:42:29 · answer #1 · answered by Facts Matter 7 · 0⤊ 0⤋