The equilbrium constant of the reaction `H_(2)(g)+I_(2)(g)hArr 2HI(g)` is 50. If the volume of the container is reduced to half of its original value, the value of equilibrium constant will be

The equilibrium constant of the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) is 64. If the volume of the container is reduced to half of its original volume, the value of equilibrium constant will be :

The equilibrium constant of the reaction, H_(2)+I_(2)hArr 2HI is 50. If the volume of the container is reduced to half of its original value, the value of equilibrium constant will be

The K_(c) for H_(2(g)) + I_(2(g))hArr2HI_(g) is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be

The K_(c) for H_(2(g)) + I_(2(g))hArr2HI_(g) is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant 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 equilibrium constant (K_(p)) for the reaction, PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)) is 16 . If the volume of the container is reduced to half of its original volume, the value of K_(p) for the reaction at the same temperature will be:

The equilibrium constant (K_(p)) for the reaction, PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)) is 16 . If the volume of the container is reduced to half of its original volume, the value of K_(p) for the reaction at the same temperature will be: