The equilibrium constant for the following reactions is `1.6xx10^(5)` at `1024 K`,
`H_(2(g))+Br_(2(g))hArr2HBr_((g))`
Find the equilibrium pressure of all gases, if `10.0 bar` of `HBr` is introduced into a sealed container at `1024 K`.

Explanation

`H_(2(g))+Br_(2(g))hArr2HBr`,
`K_(c)=([HBr]^(2))/([H_(2)][Br_(2)])=1.6xx10^(5)`
Given: `P_(HBr)=10.0` bar initially
`{:(,2HBr_((g)),hArr,H_(2(g)),+,Br_(2(g))),("Initial pressure",10,,0,,0),("Equilibrium pressure",(10-P),,(P//2),,(P//2)):}`
`:. K_(p)=(P'_(H_(2))xxP'_(Br_(2)))/([P'_(HBr]^(2))`
`=((P//2)xx(P//2))/(10-P)^(2))=(1)/(1.6xx10^(+5))`
or `(P)/(2(10-P))=sqrt((1)/(1.6xx10^(+5)))`
`:. P=0.050 `bar
`P_(H_(2))=(0.50)/(2)=`0.025 bar=`P_(Br_(2))`,
`P_(HBr)=10-0.05=9.95` bar