The `K_(p)` values for the reaction, `H_(2)+I_(2)hArr2HI`, at `460^(@)C` is `49`. If the initial pressure of `H_(2)` and `I_(2)` is `0.5atm` respectively, determine the partial pressure of each gas at equilibrium.

Explanation

`{:(,H_2+,I_2 hArr , 2HI),("Initial pressure", 0.5, 0.5 ,0),("Equi. Pressure",(0.5-x),(0.5-x),2x):}`
`K_p=(pHI)^2/((pH_2)(pI_2))`
`49=(2x)^2/((0.5-x)(0.5-x))`
or `(2x)/(0.5-x)=7`
`x=3.5/9`=0.389
`therefore pH_2` =0.5-0.389=0.111 atm
`pI_2` =0.5-0.389 =0.111 atm
pHI=2 x 0.389 = 0.788 atm