Home
Class 12
CHEMISTRY
[" Pre-Medical: Chemistry "],[" alPMT/NE...

[" Pre-Medical: Chemistry "],[" alPMT/NEET & AIIMS "(2006-2018)],[" The value of equilibrium constant of the reaction "],[qquad [" HI (g) "rightleftharpoons(1)/(2)H_(2)(g)+(1)/(2)I_(2)" is "8.0],[" The equilibrium constant of the reaction "],[H_(2)(g)+I_(2)(g)rightleftharpoons2HI" (g) will be: "],[[" (1) "(1)/(64)," (2) "16," (3) "(1)/(8)," (4) "(1)/(16)]]]

Promotional Banner

Similar Questions

Explore conceptually related problems

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

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

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

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