What is #K_2#...for #2CO(g) + O_2(g) rightleftharpoons 2CO_2(g)#?

Question

Given....#2CO(g) + O_2(g) rightleftharpoons 2CO_2(g)#
....for which #K_1=6#...what is #K_2# for the reaction...
#2CO(g) + O_2(g) rightleftharpoons 2CO_2(g)#?:
#A.# #K_2=36#; #B.# #K_2=6#; #C.# #"unknown"#; #D.# #2.50#?

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Answer 1 · 2017-05-09T17:25:39

#"Option D"#

For #2CO(g) + O_2(g) rarr 2CO_2(g)#

We have #K_(c1)=6=([CO_2]^2)/([O_2][CO]^2)#

But for...........

#CO(g) + 1/2O_2(g) rarr CO_2(g)#

We have #K_(c2)=([CO_2(g)])/([CO][O_2]^(1/2))#

But #K_(c2)=sqrt(([CO_2]^2)/([O_2][CO]^2))=sqrt(K_"c1")=sqrt(6)=2.45#, close enuff to #D.#