Respuesta :
Explanation:
[tex]\Delta G^o=-RT\ln K_1[/tex]
where,
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]600^oC=[273.15+600]K=873.15 K[/tex]
[tex]K_p[/tex] = equilibrium constant at 600°C = 0.900
Putting values in above equation, we get:
[tex]\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )[/tex]
[tex]\Delta G^o=764.85 J/mol[/tex]
The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.
Equilibrium constant at 600°C = [tex]K_1=0.900[/tex]
Equilibrium constant at 1000°C = [tex]K_2=0.396[/tex]
[tex]T_1=[273.15+600]K=873.15 K[/tex]
[tex]T_2=[273.15+1000]K=1273.15 K[/tex]
[tex]\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}][/tex]
[tex]\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}][/tex]
[tex]\Delta H^o=-18,969.30 J/mol[/tex]
The ΔH° of the reaction at 600 C is -18,969.30 J/mol.
ΔG° = ΔH° - TΔS°
764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°
ΔS° = -22.60 J/K mol
The ΔS° of the reaction at 600 C is -22.60 J/K mol.
[tex]FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)[/tex]
Partial pressure of carbon dioxide = [tex]p_1=P\times \chi_1[/tex]
Partial pressure of carbon monoxide = [tex]p_2=P\times \chi_2[/tex]
Where [tex]\chi_1\& \chi_2[/tex] mole fraction of carbon dioxide and carbon monoxide gas.
The expression of [tex]K_p[/tex] is given by:
[tex]K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}[/tex]
[tex]0.900=\frac{\chi_1}{\chi_2}[/tex]
[tex]\chi_1=0.900\times \chi_2[/tex]
[tex]\chi_1+\chi_2=1[/tex]
[tex]0.9\chi_2+\chi_2=1[/tex]
[tex]1.9\chi_2=1[/tex]
[tex]\chi_2=\frac{1}{1.9}=0.526[/tex]
[tex]\chi_1=1-\chi_2=1-0.526=0.474[/tex]
Mole fraction of carbon dioxide at 600°C is 0.474.
