Question

The equilibrium constants of the reactions

FeO(s) + CO(g) = Fe(s) + CO2; K1

FeO (s) + H2(g) = Fe(s) + H2O (g) ; K2

temp (oC) 600 700 800 900 1000

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K1 0.900 0.673 0.535 0.453 0.396

K2 0.332 0.422 0.499 0.594 0.668

Calculate the equilibrium constant of the reaction

CO(g) + H2O(g) = CO2 (g) + H2(g)

at the above temperature Also calculate the heat of the reaction.

Answer 1

The equilibrium constants of the reactions

FeO(s) + CO(g) <==> Fe(s) + CO₂ (g) ; K1

K1 = [Fe] [CO₂]/ [FeO] [CO] 

FeO (s) + H₂(g) = Fe(s) + H₂O (g) ;    K2

K2 = [Fe] [ H₂O]/ [FeO] [ H₂] 

CO(g) + H₂O(g)   <==> CO₂ (g) + H₂(g) ; K3

K3 = [CO₂] [ H₂]/ [CO] [ H₂O]

we can write ,   K3 = K1 /K2 = [CO₂] [ H₂]/ [CO] [ H₂O]

So, from given data :

Temp. (C)	600	700	800	900	1000
K1	0.900	0.673	0.535	0.453	0.396
K2	0.332	0.422	0.499	0.594	0.668
K3 = K1 /K2	2.79	1.59	1.07	0.763	0.593

we have,    G (T) = -RT ln K

From integrated form of Gibbs-helmholtz equation :

we can write :    we assume enthalpy remains constant over temperature range)

G (T2)/ T2 -   G (T1)/ T1 = H (1/T2 - 1/T1)

we choose, T2 = 800 C    and T1 = 600 C

G (T1) = -8.314 J/K-mol *873 K ln 2.79 = - 7.45 kJ/mol

G (T2) = -8.314 J/K-mol *1073 K ln 1.07 = - 0.604 kJ/mol

putting value in equation :

(- 0.604 kJ/mol/ 1073 K ) - ( - 7.45 kJ/mol/ 873 K) = H (1/1073 - 1/873)

H (reaction) = -37.33 kJ/mol.