Question

For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

Answer 1

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy  = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy  = 95636.8 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

$k=1.70\times 10^{15}$

Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$