$\rm ~~~~~~~~~~~~~~~~~~~~CO~~~~ +~~~~~~H_2O \rightleftharpoons ~~~~~CO_2 +~~~~H_2\\Ban~đầu~~~~~~0,2~~~~~~~~~~~~~~~~0,8~~~~~~~~~~~~~~0~~~~~~~~~~~~0\\Phản~ứng~~~~~~~a~~~~~~~~~~~~~~~~~~a~~~~~~~~~~~~~~~~a~~~~~~~~~~~~~a\\Cân~bằng~(0,2-a)~~~~~(0,8-a)~~~~~~~~~a~~~~~~~~~~~a$

Ta có $\rm K_c=1$ 

`<=>` $\rm \dfrac{a^2}{(0,2-a)(0,8-a)}=1$ 

`->` $\rm a=0,16$ 

Vậy $\rm [H_2]=a=0,16M$ 

Đặt $\rm [CO]$ phản ứng $\rm = a(M)$

$\rm \kern1,2cmCO \kern0,2cm+ \kern0,2cmH_2O\kern0,3cm\overset{}\leftrightarrows \ \ CO_2 + H_2\\BD:\kern,15cm 0,2\kern1,1cm 0,8\kern1,9cm \\PU:\kern,4cma\kern1,6cma\kern1,85cma\kern1,1cma\\CB:0,2-a\kern0,25cm0,8-a\kern1,05cm a\kern1,1cma$

Hằng số cân bằng:

$\rm K_c = \dfrac{[CO] \times [H_2]}{[CO]\times[H_2O]}$

$\rm ⇔1 = \dfrac{a \times a}{(0,2-a)(0,8-a)}$

$\rm \to a = 0,16$

$\rm \to [H_2] = a = 0,16M$