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.

practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.

ugh fiel iente of insistí is did kiwis jaj