Answer :
The value of standard Gibbs free energy is, 60.8 kJ
This reaction is reactant favored under standard conditions at 267 K.
Explanation :
As we know that,
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = ?
[tex]\Delta H^o[/tex] = standard enthalpy = 98.8 kJ = 98800 J
[tex]\Delta S^o[/tex] = standard entropy = 142.5 J/K
T = temperature of reaction = 267 K
Now put all the given values in the above formula, we get:
[tex]\Delta G^o=(98800J)-(267K\times 142.5J/K)[/tex]
[tex]\Delta G^o=60752.5J=60.8kJ[/tex]
As we know that:
As, the value of [tex]\Delta G[/tex] is more than zero that means the reaction is non-spontaneous and reaction will be favored in the backward direction that means favored in reactants.