A gas-phase reaction 2 A(g) B(g) + 3 C(g) is carried out in a closed vessel starting with pure A. At equilibrium, the partial pressure of A is found to be equal to the partial pressure of C. If the total equilibrium pressure is P , the equilibrium constant K_p for the reaction is:

Step-by-step solution

Let the initial moles of A be 2 . At equilibrium: Moles of A = 2 - 2 Moles of B = Moles of C = 3 Given that P_A = P_C , their mole fractions and hence their number of moles must be equal. 2 - 2 = 3 5 = 2 = 0.4 Substituting back to find the equilibrium moles: n_A = 2 - 2(0.4) = 1.2 n_B = 0.4 n_C = 3(0.4) = 1.2 Total moles at equilibrium = 1.2 + 0.4 + 1.2 = 2.8 The partial pressures are: P_A = 1.2 2.8 P = 3 7 P P_B = 0.4 2.8 P = 1 7 P P_C = 1.2 2.8 P = 3 7 P The equilibrium constant K_p is: K_p = P_B P_C^3 P_A^2 Sinc