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`.

Answer

`[P_(H_(2))]_(eq) = [P_(Br_(2))]_(eq.) = 2.5 xx 10^(-2) "bar", [P_(HBr)] = 10.0` bar

Explanation

Given,
`K_(P)` for the reaction i.e, `H_(2(g)) + Br_(2(g)) harr 2HBr_((g))` is `1.6 xx 10^(5)`
Therefore, for the reaction `2HBr_((g)) harr H_(2(g)) + Br_(2(g))`, the equilibrium constant will be,
`K'_(P) = 1/(K_(P))`
`= (1)/(1.6 xx 10^(5))`
`= 6.25 xx 10^(-6)`
Now let p be the pressure of both `H_(2)` and `Br_(2)` at equilibrium
`{:(,2HBr_((g)),harr,H_(2(g)),+,Br_(2(g))),("Initial conc.",10,,0,,0),("At equilibrium",10-2p,,p,,p):}`
Now, we can write,
`(p_(HBr)xxp_(2))/(p_(HBr)^(2)) = K'_(2)`
`(pxxp)/((10-2p)^(2)) =6.25xx10^(-6)`
`(p)/(10-2p)=2.5xx10^(-3)`
`p=2.5xx10^(-2)-(5.0xx10^(-3))p`
`p+(5.0xx10^(-3))p=2.5xx10^(-2)`
`(1005xx10^(-3))p=2.5xx10^(-2)`
`p=2.49xx10^(-2)"bar"=2.5xx10^(-2)`bar (approximately)
Therefore `[H_(2)] = [Br_(2)] = 2.49 xx 10^(-2)` bar
`[HBr] = 10 - 2 xx (2.49 xx 10^(-2))` bar
`= 9.95 "bar" = 10 "bar"` (approximately)