For reaction HI `hArr1//2 H_(2) + 1//2 I_(2)` value of `K_(c)` is `1//8`, then value of `K_(c)` for `H_(2) + I_(2) hArr2HI`.

For the reaction , 2HI hArr H_(2) + I_(2)

For the reaction 2HI hArr H_(2)+I_(2)

For reaction H_(2)(g) +I_(2)(g) hArr 2HI (g) The value of K_(p) changes with

the value of equilibrium constant for the reaction HI(g) hArr 1//2 H_(2) (g) +1//2 I_(2) (g) " is " 8.0 The equilibrium constant for the reaction H_(2) (g) +I_(2) (g) hArr 2HI(g) will be

For the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) . The value of K_(c) is 4. If 2 moles of H_(2)," 2 moles of "I_(2) and 2 moles of HI are present in one litre container then moles of I_(2) present at equilibrium is :

For the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) . The value of K_(c) is 4. If 2 moles of H_(2)," 2 moles of "I_(2) and 2 moles of HI are present in one litre container then moles of I_(2) present at equilibrium is :

The value of equilibrium constant of reaction HI_((g))hArr1/2H_(2(g))+1/2I_(2(g)) is 8.0. The equilibrium constant of the reaction , H_(2(g))+I_(2(g))hArr2HI_((g)) will be

In the reversible reaction, 2HI(g) hArr H_(2)(g)+I_(2)(g), K_(p) is

In the reversible reaction, 2HI(g) hArr H_(2)(g)+I_(2)(g), K_(p) is