For the gas-phase decomposition reaction A(g) 2 B(g) + C(g) the degree of dissociation is very small ( 1 ). If the reaction is carried out at an equilibrium pressure P , the expression for in terms of the equilibrium constant K_p and P is approximately:

Step-by-step solution

Let the initial moles of A be 1 . At equilibrium: Moles of A = 1 - Moles of B = 2 Moles of C = Total moles at equilibrium = 1 - + 2 + = 1 + 2 Since 1 , total moles 1 . The partial pressures are: P_A = 1- 1+2 P P P_B = 2 1+2 P 2 P P_C = 1+2 P P The equilibrium constant K_p is given by: K_p = P_B^2 P_C P_A Substituting the partial pressures: K_p = (2 P)^2 ( P) P = 4 ^2 P^2 P P = 4 ^3 P^2 Rearranging for : ^3 = K_p 4P^2 = ( K_p 4P^2 )^ 1 3 Answer: ( K_p 4P^2 )^ 1 3