Unit of equilibrium constant for the reversible reaction `H_(2) + I_(2)hArr 2HI` is

The value of equilibrium constant for the reaction H_(2)(g)+ I_(2) (g) hArr 2HI (g) is 48 at 720 K What is the value of the equilibrium constant for the reaction 2HI (g) hArr H_(2) (g) + I_(2) (g)

Equilibrium concentration of HI, I_(2) and H_(2) is 0.7, 0.1 and 0.1 M respectively. The equilibrium constant for the reaction, I_(2)+H_(2)hArr 2HI is :

Equilibrium concentration of HI, I_(2) and H_(2) is 0.7, 0.1 and 0.1 M respectively. The equilibrium constant for the reaction, I_(2)+H_(2)hArr 2HI is :

On a given condition, the equilibrium concentration of HI , H_(2) and I_(2) are 0.80 , 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction H_(2) + I_(2) hArr 2HI will be

On a given condition, the equilibrium concentration of H, H_(2) and I_(2) are 0.80 , 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction H_(2) + I_(2) hArr 2HI will be

Suppose the value of the equilibrium constant for the given reaction H_2(g) + I_(2)(g) hArr 2HI(g) at 717 K is 64, then calculate the same (eq. const.) for the reaction HI hArr 1/2 H_2 + 1/2 I_2 .

Suppose the value of the equilibrium constant for the given reaction H_2(g) + I_(2)(g) hArr 2HI(g) at 717 K is 64, then calculate the same (eq. const.) for the reaction HI hArr 1/2 H_2 + 1/2 I_2 .

If the equilibrium constant of the reversible reaction HI(g)hArr1//2H_2(g)+1//2I_2(g) is 7.4 , the equilibrium constant for the reversible reaction 2HI(g)hArrH_2(g)+I_2(g) will be