Respuesta :
This is an incomplete question, here is a complete question.
The reduction of iron(III) oxide to iron during steel-making can be summarized by this sequence of reactions:
[tex]2C(s)+O_2(g)\rightleftharpoons 2CO(g)[/tex]; [tex]K_1[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightleftharpoons 2Fe(l)+3CO_2(g)[/tex]; [tex]K_2[/tex]
The net reaction is:
[tex]2Fe_2O_3(s)+6C(s)+3O_2(g)\rightleftharpoons 4Fe(l)+6CO_(g)[/tex]; [tex]K[/tex]
Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K be sure you use their standard symbols.
Answer : The equilibrium expression will be:
[tex]K=(K_1)^3\times (K_2)^2[/tex]
Explanation :
The intermediate reaction are:
(1) [tex]2C(s)+O_2(g)\rightleftharpoons 2CO(g)[/tex]; [tex]K_1[/tex]
(2) [tex]Fe_2O_3(s)+3CO(g)\rightleftharpoons 2Fe(l)+3CO_2(g)[/tex]; [tex]K_2[/tex]
The net reaction is:
[tex]2Fe_2O_3(s)+6C(s)+3O_2(g)\rightleftharpoons 4Fe(l)+6CO_(g)[/tex]; [tex]K[/tex]
Now we are multiplying reaction 1 by 3 and reaction 2 by 2 and then adding both the reaction we get the net reaction.
(1) [tex]6C(s)+3O_2(g)\rightleftharpoons 6CO(g)[/tex]; [tex](K_1)^3[/tex]
(2) [tex]2Fe_2O_3(s)+6CO(g)\rightleftharpoons 4Fe(l)+6CO_2(g)[/tex]; [tex](K_2)^2[/tex]
If the equation is multiplied by a factor of '2' or '3', the equilibrium constant will be the power of 2 or 3 of the equilibrium constant of initial reaction.
If the equations are added then the equilibrium constant of the reactions will be multiplied.
Thus the equilibrium expression will be:
[tex]K=(K_1)^3\times (K_2)^2[/tex]
