Answer :
iii) < ΔU0
According to the first law of thermodynamics,
∆H°=∆U°+∆ngRT
we need to find ∆ng by writing the chemical equation of combustion of methane.
where, ∆ng=total mole of gaseous products - total mole of gaseous reactants.
Combustion of methane:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)
∆ng = 1 - (1 + 2) = -2
Now,
∆H°=∆U°+∆ngRT
=-X°-2RT
=-(X°+RT) < -X°
hence, ∆H°< ∆U°
According to the first law of thermodynamics,
∆H°=∆U°+∆ngRT
we need to find ∆ng by writing the chemical equation of combustion of methane.
where, ∆ng=total mole of gaseous products - total mole of gaseous reactants.
Combustion of methane:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)
∆ng = 1 - (1 + 2) = -2
Now,
∆H°=∆U°+∆ngRT
=-X°-2RT
=-(X°+RT) < -X°
hence, ∆H°< ∆U°
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