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

Explanation

Let the given equilibrium be established when the pressure of HBr decreases by p bar.
So, at equilibrium the partial pressures of the gases involved in the equilibrium will be as follows:
`{:(,H_(2)(g),+,Br_(2)(g),hArr,2HBr),("Initial pressure (bar)",0,,0,,10),("Equilibrium pressure (bar)",p,,p,,10-2p):}`
`therefore K_(p)=(p_(HBr)^(2))/(p_(H_(2))xxp_(Br_(2)))=((10-2p)^(2))/(pxxp)=1.6xx10^(5)`
or, `(10-2p)/(p)=4xx10^(2)=400`
`therefore p=0.025` bar
`therefore`At equilibrium, `p_(H_(2))=p_(Br_(2))=0.025` bar and `p_(HBr)=(10-0.025)"bar"=9.975` bar