Question

Consider the reaction FeO (S) + CO(g) <-----> Fe(s) + CO2(g) for which KP is found to have the following values:

Answer 1

Explanation:

$\Delta G^o=-RT\ln K_1$

where,

R = Gas constant = $8.314J/K mol$

T = temperature = $600^oC=[273.15+600]K=873.15 K$

$K_p$ = equilibrium constant at 600°C =  0.900

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )$

$\Delta G^o=764.85 J/mol$

The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.

Equilibrium constant at 600°C = $K_1=0.900$

Equilibrium constant at 1000°C = $K_2=0.396$

$T_1=[273.15+600]K=873.15 K$

$T_2=[273.15+1000]K=1273.15 K$

$\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

$\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]$

$\Delta H^o=-18,969.30 J/mol$

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.

$FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)$

Partial pressure of carbon dioxide = $p_1=P\times \chi_1$

Partial pressure of carbon monoxide = $p_2=P\times \chi_2$

Where $\chi_1\& \chi_2$ mole fraction of carbon dioxide and carbon monoxide gas.

The expression of $K_p$ is given by:

$K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}$

$0.900=\frac{\chi_1}{\chi_2}$

$\chi_1=0.900\times \chi_2$

$\chi_1+\chi_2=1$

$0.9\chi_2+\chi_2=1$

$1.9\chi_2=1$

$\chi_2=\frac{1}{1.9}=0.526$

$\chi_1=1-\chi_2=1-0.526=0.474$

Mole fraction of carbon dioxide at 600°C is 0.474.