Respuesta :
Answer:
The value of [tex]K_p[/tex] for this reaction at 1200 K is 4.066.
Explanation:
Partial pressure of water vapor at equilibrium = [tex]p^o_{H_2O}=15.0 Torr[/tex]
Partial pressure of hydrogen gas at equilibrium = [tex]p^o_{H_2}=?[/tex]
Total pressure of the system at equilibrium P = 36.3 Torr
Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:
[tex]P=p^o_{H_2O}+p^o_{H_2}[/tex]
[tex]p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr[/tex]
[tex]3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)[/tex]
The expression of [tex]K_p[/tex] is given by:
[tex]K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}[/tex]
[tex]K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066[/tex]
The value of [tex]K_p[/tex] for this reaction at 1200 K is 4.066.
