Respuesta :
Answer:
[tex]2CO(g)+2Br_2(g)\rightleftharpoons 2COBr_2(g)[/tex]
27.70 is [tex]K_c'[/tex] for the reaction.
Explanation:
[tex]COBr_2(g)\rightleftharpoons CO(g) + Br_2(g)[/tex]
The equilibrium constant of the reaction = [tex]K_c=0.190[/tex]
The expression of equilibrium constant is given by :
[tex]K_c=\frac{[CO][Br_2]}{[COBr_2]}[/tex]..[1]
[tex]2CO(g)+2Br_2(g)\rightleftharpoons 2COBr_2(g)[/tex]
The equilibrium constant expression for above reaction can be written as:
[tex]K_c'=\frac{[COBr_2]^2}{[CO]^2[Br]^2}[/tex]
[tex]K_c'=\frac{1}{(K_c)^2}[/tex] ( from [1])
[tex]K_c'=\frac{1}{(0.190)^2}=27.70[/tex]
27.70 is [tex]K_c'[/tex] for the reaction.
The kc for the given reaction is 27.70.
Equilibrium constant:
Since
[tex]COBr_2(g) \rightleftharpoons CO(g) + Br_2(g)[/tex]
The equilibrium constant of the reaction should be 0.190
Now the expression for the same should be
[tex]K_c = \frac{[CO][Br_2]}{[COBr_2]} .......(1)[/tex]
[tex]2CO(g) + 2Br_2(g) \rightleftharpoons 2COBr_2(g)[/tex]
Now
The equilibrium constant expression for above reaction could be expressed
[tex]K_c' = \frac{[COBr_2]^2}{[CO]^2[Br]^2} \\\\K_c' = \frac{1}{(K_e)^2} \\\\K_c' = \frac{1}{(0.190)^2}[/tex]
= 27.70
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