For a reaction `2HI hArr H_(2)+I_(2)`, at equilibrium `7.8 g, 203.2 g`, and `1638.4 g` of `H_(2), I_(2)`, and HI, respectively were found. Calculate `K_(c)`.

Answer

A, C

Explanation

`{:(,2HI,hArr,H_(2),+,I_(2)),("moles at equilibrium",1638.4/128,,7.8/2,,203.2/254),(,=12.8,,3.9,,0.8):}`
Let volume of container be VL
`[H_(2)]=3.9/V,[HI]=12.8/V,[I_(2)]=0.8/V`
`:. K_(c)=([H_(2)][I_(2)])/([HI]^(2))=(3.9xx0.8)/(VxxVxx(12.8/V))=0.019`
`:. K_(c)=0.019`