Đáp án:

$CO(k)+H_2O(k)\rightleftharpoons CO_2(k)+H_2(k)\quad K_c=1$ 

Ban đầu:

$C_{CO}=0,1M$ 

$C_{H_2O}=0,4M$ 

Đặt $[CO_2]=[H_2]=xM$ 

$\Rightarrow [CO]=0,1-x M, [H_2O]=0,4-x M$ 

$\Rightarrow \dfrac{x^2}{(0,1-x)(0,4-x)}=1$ 

$\Rightarrow x^2=x^2-0,5x+0,04$ 

$\Rightarrow x=0,08$

Vậy:

$[CO_2]=[H_2]=0,08M$ 

$[CO]=0,02M$ 

$[H_2O]=0,32M$