Calculate the maximum nonexpansion work that can be gained from the combustion of benzene(l) and of H2(g) on a per gram and a per mole basis under standard conditions. Is it apparent from this calculation why fuel cells based on H2 oxidation are under development for mobile applications

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.

points

explanation: