When the system `2HI(g) hArr H_(2)(g)+I_(2)(g)` is at equilibrium, inert gas is introduced. Dissociation of `HI` is ………….

In an experiment , 2 moles of HI are taken into an evacuated 10.0 litre container at 720 K. The equilibrium constant equals to 0.0156for the gaseous reaction, 2 HI (g) hArr H_(2) (g) + I_(2)(g). find equilibrium concentration of HI (g) , H_(2) (g) , I_(2)(g).

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

In an experiment 5 moles of HI were enclosed in a 5 litre container . At 717 K equilibrium constant for the gaseous reaction , 2HI (g) hArr H_2(g) +I_2(g) is 0.025 . Calculate the equilibrium concentration of HI H_2 and I_2 . What is the fraction of HI that decomposes ?

For 2HI(g) hArr H_2(g)+I_2(g) write the expression of k_c .

To the system H_(2)(g)+I_(2)(g) hArr 2HI(g) in equilibrium, some N_(2) gas was added at constant volume. Then,

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 H_(2)(g)+I_(2)(g)hArr2HI(g) , the equilibrium constant will change with

If the equilibrium constant of the reaction 2HI(g)hArrH_2(g)+I_2(g) is 0.25 , the equilibrium constant of the reaction 1/2H_2(g)+1/2I_2(g)hArr HI(g) will be

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:

If the equilibrium constant of the reaction 2HI(g)hArrH_(2)(g)+I_(2)(g) is 0.25 , find the equilibrium constant of the reaction. (1)/(2)H_(2)+(1)/(2)I_(2)hArrHI(g)