Đáp án:

`H_2+I_2`$\rightleftharpoons$`2HI`

Theo đề:

`[H_2]_{"bđ"}=1,6(M)`

`[I_2]_{"bđ"}=1(M)`

Gọi:

`[H_2]_{"pứ"}=[I_2]_{"pứ"}=x(M)`

`=>[HI]_{"pứ"}=[HI]_{"cb"}=2x(M)`

`=>[H_2]_{"cb"}=1,6-x\ (M)`

`=>[I_2]_{"cb"}=1-x\ (M)`

Ta có:

`K_C={[HI]^2}/{[H_2].[I_2]}=51,52`

`<=>{(2x)^2}/{(1,6-x).(1-x)}=51,52`

`=>x~~0,908(M)`

`=>[H_2]_{"cb"}=1,6-x=1,6-0,908=0,692(M)`