The equilibrium constant for the following reaction is `1.6xx10^(5)` at `1024 K`
`H_(2)(g)+Br_(2)(g) hArr 2HBr(g)`
find the equilibrium pressure of all gases if `10.0` bar of `HBr` is introduced into a sealed container at `1024 K`.

Explanation

`2HBr(g) hArr H_2(g)+Br_2(g)`
`K=1/(1.6xx10^5)`
Let at equilibrium , the pressure of HBr is decreased by p bar so that
`{:(,2HBr(g)hArr , H_2(g)+, Br_2(g)),("Initial conc.",10,0,0),("At. equi.",10-p,"p/2","p/2"):}`
`therefore K_p=((pH_2)(pBr_2))/(pHBr)^2`
`=((p/2)xx(p/2))/(10-p)^2 =1/(1.6xx10^5)`
or `p^2/(4(10-p)^2)=1/(1.6xx10^5)`
Taking square root of both sides , we get
`p/(2(10-p))=1/(4xx10^2)`
`4xx10^2p =20-2p`
or 402p=20
`therefore p=20/402=4.98xx10^(-2)` bar
Hence at equilibrium ,
`pH_2=(4.98xx10^(-2))/2=2.49xx10^(-2)` bar
`pBr_2=(4.98xx10^(-2))/2=2.49xx10^(-2)` bar
pHBr=`10-4.98xx10^(-2)` = 10 bar .