Answer:
Explanation:
From the given reaction:
[tex]6Cl_{2(g)}+2Fe_2O_{3(s)} \to 4FeCl_{3(s)}+3O_2[/tex]
From the Gibbs Free Energy table at standard conditions, the value of each compound is as follows:
[tex]G_f^0 \ of \ Cl_2 = 0 \ KJ/mol[/tex] [tex]G_f^0 \ of \ Fe_2O_3 = -742.24 \ KJ/mol[/tex]
[tex]G_f^0 \ of \ Fe_2Cl_3 = -334.05 \ KJ/mol[/tex] [tex]G_f^0 \ of \ O_2 = 0 \ KJ/mol[/tex]
Now, the standard Gibb's Free energy for the given reaction can be estimated as follows:
[tex]\mathtt{\Delta G^0 = (4 *G_f^0(FeCl_3) +3*G_f^0(O_2)) - (6*G_f^0 (Cl_2) +2*G_f^0(Fe_2O_3))}[/tex]
[tex]\mathtt{\Delta G^0 = (4 *(-334.05) +3*(0)) - (6(0) +2(-742.24))}[/tex]
[tex]\mathtt{\Delta G^0 = 148.28 \ kJ/mol}[/tex]
using the following formula:
[tex]\mathtt{\Delta G^0 =-RTIn K_{eq}}[/tex]
the equilibrium constant can be determined as:
[tex]\mathtt{ In K_{eq} =\dfrac{\Delta G^0 }{-RT}}[/tex]
[tex]\mathtt{ In K_{eq} =\dfrac{148.28*10^3 J/mol }{-(8.314 \ J/k mol )*298 \ K}}[/tex]
[tex]\mathtt{ In K_{eq} =-59.85}[/tex]
[tex]\mathtt{ K_{eq} =e^{-59.85}}[/tex]
[tex]\mathtt{ K_{eq} =1.0*10^{-26}}[/tex] to 2 significant figures.
