Answer:
2.864 times (Almost 3 times)
Explanation:
The equation for the combustion is
C₆H₆ (l) + 15/2 O₂ (g) -> 6CO₂ (g) + 3H₂O (l)
Where l represents liquid and g represents gas
Maximum Nonexpansion work = ∆G°R
∆G°R = 3∆G°f (H₂O,l) + 6G°f (CO₂,g) -15/2G°f(O₂,g) - ∆G°f(C₆H₆ ,l)
Converting the equation above to amount of energy per mole of hydrogen
= 3 * (-237.1kJ/mol) + 6 * (-394.4kJ/mol) - 15/2 * 0 - 124.5kJ/mol
= -711.3kJ/mol - 2366.4kJ/mol - 0 - 124.5kJ/mol
= -3202.2 kJ/mol
Under a standard condition,
1 mol per g = 1/78
-3202.2kJ/mol =
-3202.2 * 1/78 (KJ/mol * mol/g)
= -41.054kJ/g
Similarly,
H₂(g) + ½O(g) -> H₂O (l)
Maximum Nonexpansion Work = = ∆G°R
∆G°R = ∆G°f (H₂O,l) - ½G°f (O₂,g) - G°f (H₂,g)
= -237.1kJ/mol - ½ * 0 - 0
= -237.1kJ/mol
Under standard condition,
-237.1kJ/mol = -237.1kJ/mol * 1mol/2.016g
= -117.609kJ/g
On a per gram basis, there are -117.609/-41.054
= 2.864
On a per gram basis, there are 2.864 (almost 3) times as much work can be extracted from the oxidation of hydrogen than benzene.
