A mixture of 0.07207 mol of C2H6, 0.04213 mol of N2, 0.04392 mol of NH3, and 0.03782 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 1298 K. The following equilibrium is established: 3 C2H6(g) + 1 N2(g) 2 NH3(g) + 3 C2H4(g) At equilibrium 0.004044 mol of C2H4 is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of C2H6, N2, NH3, and C2H4. Peq(C2H6) = . Peq(N2) = . Peq(NH3) = . Peq(C2H4) = . (b) Calculate KP for this reaction. KP =