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 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 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 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 unit of equilibrium constant for the reaction , H_2+I_2 hArr 2HI IS :

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: