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Answer :
The question is incomplete , the complete question is ;
In a study of the following reaction :
[tex]3Fe(s)+4H_2O(g)\rightleftharpoons Fe_2O_4(s)+4H_2(g)[/tex]
at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the [tex]K_p[/tex] foe this reaction at 1200 K.
Answer:
The [tex]K_p[/tex] for this reaction at 1200 K is 4.066[/tex].
Explanation:
Partial pressure of the water vapor at equilibrium = [tex]p_1=15.0 Torr[/tex]
Partial pressure of the hydrogen gas at equilibrium = [tex]p_2[/tex]
Total pressure at equilibrium = [tex]P=36.3 Torr[/tex]
[tex]P_1+P_2[/tex]
[tex]p_2=P-P_1=36.3 Torr-15.0 Torr=21.3 Torr[/tex]
[tex]3Fe(s)+4H_2O(g)\rightleftharpoons Fe_2O_4(s)+4H_2(g)[/tex]
The expression if [tex]K_p[/tex] is given as;
[tex]K_p=\frac{(p_2)^4}{(p_1)^4}[/tex]
(the partial pressure of the gas will be taken along with which partial pressure of the solids are taken as unity)
[tex]K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}[/tex]
[tex]K_p=4.066[/tex]
The [tex]K_p[/tex] for this reaction at 1200 K is 4.066[/tex].
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