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 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

The value of equilibrium constant of the reaction, H(g) hArr(1)/(2) H_(2)(g) is 0.8 . The equilibrium constant of the reaction H_(2)(g) + I_(2)(g)hArr2HI(g) will be

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

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

If the equilibrium constant of the reaction 2HIhArr H_(2)+I_(2) is 0.25, then the equilibrium constant for the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) would be

If the equilibrium constant of the reaction 2HIhArr H_(2)+I_(2) is 0.25, then the equilibrium constant for the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) would 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)