Respuesta :
Answer: 5.85kJ/Kmol.
Explanation:
The balanced equilibrium reaction is
[tex]A(g)+2B(g)\rightleftharpoons 4C(g)+D(g)[/tex]
The expression for equilibrium reaction will be,
[tex]K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}[/tex]
Now put all the given values in this expression, we get the concentration of methane.
[tex]K_p=\frac{(4.85)\times [(4.11)^4}{(5.04)^2\times (5.16)}[/tex]
[tex]K_p=10.6[/tex]
Relation of standard change in Gibbs free energy and equilibrium constant is given by:
[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]
where,
R = universal gas constant = 8.314 J/K/mole
T = temperature = [tex]25^0C=(25+273)K=298 K[/tex]
[tex]K_c[/tex] = equilibrium constant = 10.6
[tex]\Delta G^o=-2.303\times 8.314\times 298\times \log (10.6)[/tex]
[tex]\Delta G^o=5850.23J/Kmol[/tex]
[tex]\Delta G^o=5.85kJ/Kmol[/tex]
Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.
