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0.5 mol of H(2) and 0.5 mol of I(2) reac...

`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.

Text Solution

Verified by Experts

`{:(,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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