In the dissociation of HI, `20%` of HI is dissociated at equilibrium. Calculate `K_(p)` for
`HI(g) hArr 1//2 H_(2)(g)+1//2 I_(2)(g)`

256 g of HI were heated in a sealed bulb at 444°C till the equilibrium was attained. The acid was found to be 22% dissociated at equilibrium. Calculate the equilibrium constant for the reaction 2HI(g) hArr H_2(g) +I_2 (g)

2 mol of HI were heated in a sealed tube at 440^(@)C until the equilibrium was reched. HI was found to be 22% dissociated. Calculate the equilibrium constant for the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) .

If the equilibrium constant for the reaction, H_(2)(g) +I_(2)(g) hArr 2HI(g) is K What is the equilibrium constant of HI(g) rarr 1/2 H_(2)(g) + 1/2I_(2)(g) ?

At a certain temperature , only 50% HI is dissociated at equilibrium in the following reaction: 2HI(g)hArrH_(2)(g)+I_(2)(g) the equilibrium constant for this reaction is:

At a certain temperature , only 50% HI is dissociated at equilibrium in the following reaction: 2HI(g)hArrH_(2)(g)+I_(2)(g) the equilibrium constant for this reaction is:

At a certain temperature , only 50% HI is dissociated at equilibrium in the following reaction: 2HI(g)hArrH_(2)(g)+I_(2)(g) the equilibrium constant for this reaction is:

For the equilibrium reaction H_(2(g))+I_(2(g))hArr 2HI_((g))

For the equilibrium reaction H_(2(g))+I_(2(g))hArr 2HI_((g))