A mixture of 0007878 mol of c2h6 0008025 mol of n2 001096, Ask an Expert

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A mixture of 0.007878 mol of C2H6, 0.008025 mol of N2, 0.01096 mol of NH3, and 0.02646 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 3324 K. The following equilibrium is established: 3 C2H6(g) + 1 N2(g) 2 NH3(g) + 3 C2H4(g) At equilibrium 0.003983 mol of C2H6 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 =.

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