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`.

For the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) The equilibrium constant K_(p) changes with

For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) the equilibrium constant K_(p) changes with

k_(P) & k_(C) are ….. for reaction at equilibrium of type H_(2)(g) + Br_(2)(g) hArr 2HBr(g) .

The equilibrium constant for the reaction H_(2)+Br_(2)hArr2HBr is 67.8 at 300(@)K . The equilibrium constant for the dissociation of HBr is:

At 500 K, equlibrium constant, K_(c) for the following reaction is 5. 1//2 H_(2)(g)+ 1//2(g)hArr HI (g) What would be the equilibrium constant K_(c) for the reaction 2hi(g)hArrH_(2)(g)+l_(2)(g)

At 700 K , hydrogen and bromine react to form hydrogen bromine. The value of equilibrium constant for this reaction is 5xx10^(8) . Calculate the amount of the H_(2), Br_(2) and HBr at equilibrium if a mixture of 0.6 mol of H_(2) and 0.2 mol of Br_(2) is heated to 700K .