The equilibrium constant for the following reaction is `1.6 xx 10^(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 1024K.

Explanation

`2HBr(g) hArr H_(2)(g)+Br_(2)(g), K=(1)/((1.6xx10^(5)))`
`{:(Initial,10 "bar",,,),(At Eq,10-p,,p//2,p//2):}`
`K_(p)=((p//2)(p//2))/((10-p)^(2))=1/(1.6xx10^(5))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))`
or `(4xx10^(2))p=2(10-p) or 402p=20`
or `p=20/402=4.98xx10^(-2)` bar
Hence, at equilibrium
`p_(H_(2))=p_(Br_(2))=p//2=2.5xx10^(-2)` bar,
`p_(HBr)=10-p~~10` bar