Respuesta :
Answer : The value of equilibrium constant for this reaction at 328.0 K is [tex]1.70\times 10^{15}[/tex]
Explanation :
As we know that,
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = ?
[tex]\Delta H^o[/tex] = standard enthalpy = 151.2 kJ = 151200 J
[tex]\Delta S^o[/tex] = 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:
[tex]\Delta G^o=(151200J)-(328.0K\times 169.4J/K)[/tex]
[tex]\Delta G^o=95636.8J=95.6kJ[/tex]
The relation between the equilibrium constant and standard Gibbs free energy is:
[tex]\Delta G^o=-RT\times \ln k[/tex]
where,
[tex]\Delta G^o[/tex] = 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:
[tex]95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k[/tex]
[tex]k=1.70\times 10^{15}[/tex]
Therefore, the value of equilibrium constant for this reaction at 328.0 K is [tex]1.70\times 10^{15}[/tex]
