Calculate `K_(c)` for the reaction:
`2H_(2)(g)+S_(2)(g) hArr 2H_(2)S(g)`
if `1.58` mol `H_(2)S, 1.27` mol `H_(2)` and `2.78xx10^(-6)` mol of `S_(2)` are in equilibrium in a flask of capacity `180 L` at `750^(@)C`.

Answer

A, B, C

Explanation

`{:(,2H_(2)(g),+,S_(2)(g),hArr,2H_(2)S(g)),("moles at",1.27,,2.78xx10^(-6),,1.58),("equilibrium",,,,,):}`
Volume of container `=180 L`
`:. K_(C)=([H_(2)S]^(2))/([H_(2)]^(2)[S_(2)])=((1.58/180)^(2))/((1.27/180)^(2)((2.78xx10^(-6))/180))`
`=((1.58)^(2)xx180)/((1.27)^(2)xx(2.78xx10^(-6)))`
`K_(C)=1.002xx10^(8)L "mol"^(-1)`