At a given temperature, the equilibrium concentrations in a closed vessel for the reaction 2 HI _ ( g ) H _ 2( ~g ) + I _ 2( ~g ) are [ H ₂] = 1.5 10⁻³ ~M , [ I ₂] = 6.0 10⁻³ ~M , and [ HI ] = 2.0 10⁻³ ~M . If 0.5 ~mol L ⁻¹ of pure HI _ ( g ) is taken in an empty closed vessel at the same temperature, what will be the degree of dissociation ( ) of HI _ ( g ) at equilibrium?

Step-by-step solution

The equilibrium constant K_c for the reaction 2 HI _ ( g ) H _ 2( ~g ) + I _ 2( ~g ) is given by: K_c = [ H ₂][ I ₂] [ HI ]^2 Substituting the given equilibrium concentrations: K_c = (1.5 10⁻³)(6.0 10⁻³) (2.0 10⁻³)^2 = 9.0 10⁻⁶ 4.0 10⁻⁶ = 9 4 Let the initial concentration of HI be C = 0.5 ~M and its degree of dissociation be . At equilibrium, the concentrations are: [ HI ] = C(1 - ) [ H ₂] = C 2 [ I ₂] = C 2 Substituting these into the equilibrium constant expression: K_c = ( C 2 ) ( C 2 ) (C(1 - ))^2 = ^2 4(1 - )^