Question

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

T 600°C 1000°C
Kp 0.900 0.396

a. Calculate ?G°reaction, ?S°reaction, and ?H°reactionnfor this reaction at 600°C. Assume that ?Hreaction°is independent of temperature.

b. Calculate the mole fraction of CO2(g) present in the gas phase at 600°C.

Answer 1

To calculate ΔG° reaction, ΔS° reaction, and ΔH° reaction for the given reaction at 600°C, we can use the relationship ΔG° = -RTlnKp, where Kp is the equilibrium constant at constant pressure.

Step-by-step explanation:

To calculate ΔG° reaction, ΔS° reaction, and ΔH° reaction for the given reaction at 600°C, we can use the relationship ΔG° = -RTlnKp, where Kp is the equilibrium constant at constant pressure. The values of Kp at 600°C are given as 0.900.

ΔG° reaction = -RTlnKp = -(8.314 J/mol·K) * (600 + 273.15) K * ln(0.900) = -9.02 kJ/mol.

Since ΔH reaction° is assumed to be independent of temperature, ΔH° reaction = ΔG° reaction + TΔS° reaction = -9.02 kJ/mol + TΔS° reaction.

Answer 2

Step-by-step 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)* 873.15 K* \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 (K_2)/(K_1)=(\Delta H^o)/(R)* [(1)/(T_1)-(1)/(T_2)]`

`\ln (0.396)/(0.900)=(\Delta H^o)/(8.314 J/mol K)* [(1)/(873.15 K)-(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* \chi_1`

Partial pressure of carbon monoxide =
`p_2=P* \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=(p_1)/(p_2)=(P* \chi_1)/(P* \chi_2)`

`0.900=(\chi_1)/(\chi_2)`

`\chi_1=0.900* \chi_2`

`\chi_1+\chi_2=1`

`0.9\chi_2+\chi_2=1`

`1.9\chi_2=1`

`\chi_2=(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.