Answer:
[tex]\Delta G_{rxn}^{0}=-117.4kJ/mol[/tex]
Explanation:
Gibbs free energy is an additive property.
[tex]2ClF(g)\rightarrow Cl_{2}(g)+F_{2}(g)[/tex] ; [tex]\Delta G_{1}^{0}=-115.4kJ/mol[/tex]
[tex]Cl_{2}(g)+Br_{2}(g)\rightarrow 2ClBr(g)[/tex] ; [tex]\Delta G_{2}^{0}=-2.0kJ/mol[/tex]
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[tex]2ClF(g)+Br_{2}(g)\rightarrow 2ClBr(g)+F_{2}(g)[/tex] ; [tex]\Delta G_{rxn}^{0}=\Delta G_{1}^{0}+\Delta G_{2}^{0}=(-115.4-2.0)kJ/mol=-117.4kJ/mol[/tex]
So, standard gibbs free enrgy change for the given reaction is -117.4 kJ/mol
The change in free energy is called G (∆G). The [tex]\rm \Delta G^\circ rxn= -117.4\; kj/mol[/tex]
Gibb free energy is the measure of the maximum amount of reversible work done in a thermodynamic system.
It means only when the temperature change, but the pressure is constant.
[tex]\rm Cl_2(g) + F_2(g) \longrightarrow 2ClF(g) \;delta\; G\;degree_rxn = 115.4 kJ/mol[/tex]
[tex]\rm Cl_2(g) + Br_2(g) \longrightarrow 2ClBr(g) \;delta\; G \;degree_r_x_n = -2.0 kJ/mol[/tex]
[tex]\rm \Delta G^\circ rxn= \Delta G^\circ1 +G^\circ2[/tex]
[tex]2ClF(g) + Br_2(g) \longrightarrow 2ClBr(g) + F_2(g)[/tex]
[tex]\rm \Delta G^\circ rxn= -115.4\;kj/mol - 2.0 \;kj/mol = -117.4\;kj/mol[/tex]
Thus, the [tex]\rm \Delta G^\circ rxn= -117.4\; kj/mol[/tex].
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