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

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)

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)

The value of equilibrium constant of the reaction. HI(g)hArr(1)/(2)H_2(g)+(1)/(2)I_2(g) is 8.0 The equilibrium constant of the reaction. H_2(g)+I_2(g)hArr2HI(g) will be

The value of equilibrium constant of the reaction. HI(g)hArr(1)/(2)H_2(g)+(1)/(2)I_2(g)is 8.0 The equilibrium constant of the reaction. H_2(g)+I_2(g)hArr2HI(g) will be

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

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

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