A mixture of 0.09186 mol of C2H4, 0.05291 mol of N2, 0.03896 mol of NH3, and 0.1119 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 1013 K. The following equilibrium is established: 3 C2H4(g) + 1 N2(g) <--> 2 NH3(g) + 1 C6H6(g) At equilibrium 0.03239 mol of NH3 is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of C2H4, N2, NH3, and C6H6. Peq(C2H4) = . Peq(N2) = . Peq(NH3) = . Peq(C6H6) = . (b) Calculate KP for this reaction. KP = .