(18.4 ~g ~N ₂ O ₄ ) was placed in (1 ~L ) vessel at (400 ~K ) and allowed to attain the following equilibrium ( N ₂ O ₄(g) 2 NO ₂(g) ). If the total pressure at equilibrium was 10.64 bar, approximate (K_p ) is ((R=0.083 ~L ) bar ( K ⁻¹ ~mol ⁻¹ ) ) (Assume ( N ₂ O ₄, NO ₂ ) as ideal gases)

Step-by-step solution

From ideal gas equation, ( aligned p V & =n R T p_ N ₂ O ₄ l & = 18.4 92 0.083 400 p_ N ₂ O ₄ & =6.64 bar aligned ) For the equilibrium reaction, ( aligned p_T & =p-p_i+2 p_i p_T & =p+p_i p_i=p_T-p p_i & =10.64-6.64 & =4.00 bar aligned ) At equilibrium, ( aligned p_ N ₂ O ₄ & =p-p_i=6.64-4 & =2.64 bar p_ NO ₂ & =2 p_i=2 4 & =8 bar K_p & = [p_ NO ₂ ]^2 [p_ N ₂ O ₄ ] & = [8]^2 2.64 =24.29 aligned )