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

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 for the reaction, H_(2)(g) +I_(2)(g) hArr 2HI(g) is K What is the equilibrium constant of HI(g) rarr 1/2 H_(2)(g) + 1/2I_(2)(g) ?

The value of equilibrium constant for the reaction H_(2)(g)+ I_(2) (g) hArr 2HI (g) is 48 at 720 K What is the value of the equilibrium constant for the reaction 2HI (g) hArr H_(2) (g) + I_(2) (g)

The equilibrium constant at 717 K for the reaction H_(2)(g)+I_(2)(g)hArr 2HI(g) is 50. The equilibrium constant for the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) is :

The value of the equilibrium constant for the reaction : H_(2) (g) +I_(2) (g) hArr 2HI (g) at 720 K is 48. What is the value of the equilibrium constant for the reaction : 1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)

The value of the equilibrium constant for the reaction : H_(2) (g) +I_(2) (g) hArr 2HI (g) at 720 K is 48. What is the value of the equilibrium constant for the reaction : 1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)

Define equilibrium constant. Write the equilibrium constant expression for the reaction of H_(2)(g)+I_(2)(g)hArr2HI(g) and its reverse reaction. How are the two equilibrium constants related?

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