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

(I).Calculation of `K_(p)`
`H_(2)(g) +Br_(2)(g) hArr 2HBr(g)`
`K_(p) =K_(c) (RT)^(Deltan) =K_(c)(RT)^(@)`
`K_(p) =K_(c) =1.6 xx 10^(5)`
(II). Calculation of partial pressure of gases

`K_(p) =(pH_(2)xxpBr_(2))/(P^(2)HBr) " or " (1)/(1.6 xx 10^(5)) =(P//2 xxP//2)/((10-P)^(2)) =(P^(2))/(4(10-P)^(2))`
On taking square root , `(P)/(2(10-P)) =((1)/(1.6 xx 10^(5)))^(1//2) " or " (2(10 -P))/(P) =(1.6xx 10^(5))^(1//2) =4 xx 10^(2)`
`20- 2P =4 xx 10^(2)P " or " P(4 xx 10^(2)+ 2) =20`
`" or " " " P= (20)/((400 +2)) =(20)/(402) =0.050 " bar "`
`pH_(2) =0.025 " bar " , pBr_(2) =0.025 " bar " : pHBr =10- 0.05 =9.95 " bar "~~ 10.0 " bar ".`