`0.5 mol` of `H_(2)` and `0.5` mol of `I_(2)` react in `10 L` flask at `448^(@)C`. The equilibrium constant `(K_(c ))` is `50` for `H_(2)+I_(2) hArr 2HI` a. What is the value of `K_(p)`? b. Calculate the moles of `I_(2)` at equilibrium.

`0.5 mol` of `H_(2)` and `0.5` mol of `I_(2)` react in `10 L` flask at

1.	`0.5 mol` of `H_(2)` and `0.5` mol of `I_(2)` react in `10 L` flask at `448^(@)C`. The equilibrium constant `(K_(c ))` is `50` for `H_(2)+I_(2) hArr 2HI` a. What is the value of `K_(p)`? b. Calculate the moles of `I_(2)` at equilibrium.
Answer» `{:(,H_(2),+,I_(2),hArr,2HI),("moles at t=0",0.5,,0.5,,0),("moles at equilibrium",(0.5-x),,(0.5-x),,2x):}` `:. K_(p)=K_(c )=(4x^(2))/((0.5-x)^(2))` Note: Volume term is eliminated, if `Deltan=0`. a. `K_(p)=K_(c ) ( :. Deltan=0)` `:. K_(p)=50` b. `50=(4x^(2))/((0.5-x)^(2))` or `(2x)/((0.5-x))=sqrt(50)` `:. x=0.39` `:.` "moles" of `I_(2)` at equilibrium `=0.50-0.39=0.11 "mol"`

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`0.5 mol` of `H_(2)` and `0.5` mol of `I_(2)` react in `10 L` flask at `448^(@)C`. The equilibrium constant `(K_(c ))` is `50` for `H_(2)+I_(2) hArr 2HI` a. What is the value of `K_(p)`? b. Calculate the moles of `I_(2)` at equilibrium.

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