A mixture of 0.06061 mol of C2H4, 0.04095 mol of N2, 0.05684 mol of NH3, and 0.0

Question

A mixture of 0.06061 mol of C2H4, 0.04095 mol of N2, 0.05684 mol of NH3, and 0.01542 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 1035 K. The following equilibrium is established:

3 C2H4(g) + 1 N2(g) --> 2 NH3(g) + 1 C6H6(g)

At equilibrium 0.01082 mol of C6H6 is found in the reaction mixture.


(a) Calculate the equilibrium partial pressures of C2H4, N2, NH3, and C6H6.


(b) Calculate KP for this reaction.


Please show work!! or explain.

Explanation / Answer

3 C2H4(g) + 1 N2(g) --> 2 NH3(g) + 1 C6H6(g) 0.06061-----0.04095----0.05684-----0.01542 at eqm 0.07441------0.04555-----0.04764------0.01082 total moles = 0.17842 PV = nRT => P = 0.17842 * 8.314 * 1035 / 10^-3 = 1535302 Pa = 15.15 atm a) Eqm Partial Pressure C2H4 = (0.07441/0.17842) * 15.15 = 6.32 atm N2 = 3.87 atm NH3 = 4.83 atm C6H6 = 0.92 atm b) Kp = 0.92 * 4.83^2 / 3.87 * 6.32^3 = 0.02197

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