In the reaction, `H_(2)(g)+I_(2)(g)hArr2HI(g)` The concentration of `H_(2), I_(2)`, and `HI` at equilibrium are `8.0, 3.0` and `28.0` mol per `L` respectively. Determine the equilibrium constant.

In the reaction, `H_(2)(g)+I_(2)(g)hArr2HI(g)` The concentration of `H

1.	In the reaction, `H_(2)(g)+I_(2)(g)hArr2HI(g)` The concentration of `H_(2), I_(2)`, and `HI` at equilibrium are `8.0, 3.0` and `28.0` mol per `L` respectively. Determine the equilibrium constant.
Answer» `H_(2)(g)+I_(2)(g)hArr2HI(g)` Applying the law of mass action `K_(c )=([HI]^(2))/([H_(2)][I_(2)])` Given `[H_(2)]=8.0 "mol" L^(-1)` `[I_(2)]=3.0 "mol" L^(-1)` `[HI]=2.8 "mol" L^(-1)` So, `K_(c )=((28.0)^(2))/((8.0)xx(3.0))=32.66`

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In the reaction, `H_(2)(g)+I_(2)(g)hArr2HI(g)` The concentration of `H_(2), I_(2)`, and `HI` at equilibrium are `8.0, 3.0` and `28.0` mol per `L` respectively. Determine the equilibrium constant.

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