A gaseous substance A dissociates as A ( g ) 2 B ( g ) at a constant temperature. The theoretical (undissociated) vapour density of A is D₀ and the observed vapour density of the equilibrium mixture is D . If the total pressure at equilibrium is P , which of the following is the correct expression for the equilibrium constant K_p ?

Step-by-step solution

For the dissociation reaction A ( g ) n B ( g ) , the degree of dissociation is related to the theoretical vapour density D₀ and observed vapour density D by the formula: = D₀ - D (n-1)D Here, n = 2 , so: = D₀ - D D For the reaction A ( g ) 2 B ( g ) , the equilibrium constant K_p in terms of and total pressure P is: K_p = 4 ^2 P 1 - ^2 Substituting the expression for into the K_p equation: K_p = 4 ( D₀ - D D )^2 P 1 - ( D₀ - D D )^2 Simplify the denominator: 1 - ( D₀ - D D )^2 = D^2 - (D₀ - D)^2 D^2 = D^2 - (D₀^2