The `K_(p)` value for the reaction.
`H_(2) +I_(2) hArr 2Hi`
at `460^(@)C` is 49. If the initial pressure of `H_(2) " and " I_(2)` is 0.5 atm respectively , what will be the partial pressure of `H_(2)` at equilibrium ?

Options

A. `0.111` atm
B. `0.123` atm
C. `0.133` atm
D. `0.222` atm

Answer

A

Explanation

`{:(,H_(2) (g) ,+, I_(2) ,hArr, 2HI(g)),("Initial (atm)",0.5,,0.5,,0),("pressure",,,,,),("Final (atm)",0.5-x,,0.5-x,,2X),("pressure",,,,,):}`
`K_(p) =((pHI)^(2))/((pH_(2))(pI_(2)))`
`49 = ((2X)^(2))/((0.5-X)^(2)) " or " 7 =(2X)/((0.5 -X))`
`3.5 xx -7X =2X " or " X =(3.5)/(9) =0.389`
`pH_(2)` at eqm. =(0.5 - 0.389) = 0.111